Download Advanced Quantum Mechanics - TU Graz

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Advanced Quantum Mechanics
Last updated February 2, 2017
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Partially taken from the lecture notes of Prof. Wolfgang von der Linden.
Thanks to Katharina Rath for help with the translation
Contents
1
Rotations and the angular momentum operator
1.1 Main goals in this chapter . . . . . . . . . . . . . . . . . . . .
1.2 Algebra of angular momentum operators . . . . . . . . . . .
1.2.1 Rotation matrices in R3 . . . . . . . . . . . . . . . . .
1.2.2 Commutation rules of angular momentum operators
1.3 Scalar and Vector Operators . . . . . . . . . . . . . . . . . . .
1.4 Eigenvalue Problem for the Angular Momentum Operator .
1.5 The Orbital Angular Momentum . . . . . . . . . . . . . . . .
1.5.1 Ortsraumeigenfunktionen des Bahndrehimpulses . .
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Schrödinger equation in a central potential
2.1 Main results in this chapter (until Sec. 2.5) .
2.2 Radial- und Drehimpulsanteil . . . . . . . . .
2.3 Produktansatz für die Schrödingergleichung
2.4 Entartung bei unterschiedlichen m . . . . . .
2.5 Wasserstoff und H-ähnliche Probleme . . . .
2.5.1 Summary . . . . . . . . . . . . . . . .
2.5.2 Center of mass coordinates . . . . . .
2.5.3 Eigenvalue equation . . . . . . . . . .
2.5.4 Entartung . . . . . . . . . . . . . . . .
2.5.5 Energieschema des H-Atoms (Z=1) .
2.5.6 Lichtemission . . . . . . . . . . . . . .
2.5.7 Wasserstoff-Wellenfunktion . . . . .
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Erweiterungen und Anwendungen
51
3.1 Main results/goals in this chapter (until Sec. 3.4) . . . . . . . 51
3.2 Kovalente Bindung . . . . . . . . . . . . . . . . . . . . . . . . 52
52
3.2.1 Das H+
2 Molekül. . . . . . . . . . . . . . . . . . . . . .
3.3 Optimierung der (Variations-)Wellenfunktion in einem Teilraum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
57
3.4 Back to the variational treatment of H+
2 . . . . . . . . . . . .
3
4
CONTENTS
3.5
4
5
6
7
3.4.1 Muonisch katalysierte Fusion . . . . . . . . . . . . . . 61
Van-der-Waals-Wechselwirkung . . . . . . . . . . . . . . . . 62
Several degrees of freedon and the product space
4.1 Das Tensor-Produkt . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Vollständige Basis im Produkt-Raum . . . . . . . . . . . . . .
4.3 Orthonormierung im Produkt-Raum . . . . . . . . . . . . . .
4.4 Operatoren im direkten Produktraum . . . . . . . . . . . . .
4.5 Systeme mit zwei Spin 12 Teilchen . . . . . . . . . . . . . . . .
4.6 Addition of angular momenta . . . . . . . . . . . . . . . . . .
4.6.1 Determining the allowed values of j . . . . . . . . . .
4.6.2 Construction of the eigenstates . . . . . . . . . . . . .
4.6.3 Application to the case of two spin 21 . . . . . . . . . .
4.6.4 Scalar product . . . . . . . . . . . . . . . . . . . . . . .
4.6.5 How to use a table of Clebsch-Gordan coefficients . .
4.7 Matrix elements of vector operators (Wigner-Eckart’s theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7.1 Applications . . . . . . . . . . . . . . . . . . . . . . . .
Time dependent perturbation theory
5.1 Zeitabhängige (Diracsche) Störungstheorie . .
5.1.1 Das Wechselwirkungsbild . . . . . . . .
5.1.2 Harmonische oder konstante Störung .
5.1.3 Adiabatisches Einschalten der Störung
Identical particles
6.1 Pauli exclusion principle . . . . . .
6.2 Anyonen (Optional) . . . . . . . . .
6.3 Electron and spin . . . . . . . . . .
6.4 The Helium atom . . . . . . . . . .
6.5 Excited states of helium . . . . . .
6.6 Occupation number representation
6.6.1 Fock Space . . . . . . . . . .
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85
. . . 85
. . . 86
. . . 90
. . . 95
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Charged particle in an electromagnetic field
115
7.1 Classical Hamilton function of charged particles in an electromagnetic field . . . . . . . . . . . . . . . . . . . . . . . . . 115
7.2 Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . 116
7.3 Landau Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
CONTENTS
8
9
5
Field quantisation
8.1 Continuum systems: classical treatment . .
8.2 Quantisation . . . . . . . . . . . . . . . . . .
8.2.1 Hamiltonian in diagonal form . . .
8.2.2 Creation and destruction operators .
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Second quantisation
9.1 Quantisation of the Schrödinger field . . . . . . . . . . . . . .
9.1.1 Lagrangian of the Schrödinger field . . . . . . . . . .
9.1.2 Quantisation . . . . . . . . . . . . . . . . . . . . . . .
9.2 Transformation of operators to second quantisation . . . . .
9.2.1 Single particle operators . . . . . . . . . . . . . . . . .
9.2.2 Two-particle operators . . . . . . . . . . . . . . . . . .
9.3 Second quantisation for fermions . . . . . . . . . . . . . . . .
9.3.1 Useful rules for (anti) commutators . . . . . . . . . .
9.4 Summary: Fock space . . . . . . . . . . . . . . . . . . . . . . .
9.5 Unitary transformations . . . . . . . . . . . . . . . . . . . . .
9.5.1 Application: tight-binding hamiltonian . . . . . . . .
9.5.2 Field operators . . . . . . . . . . . . . . . . . . . . . .
9.6 Heisenberg time dependence for operators . . . . . . . . . .
9.6.1 Time dependence of field operators . . . . . . . . . .
9.6.2 Time dependence of creation and annihilation operators . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7 Momentum space . . . . . . . . . . . . . . . . . . . . . . . . .
9.8 Free fermions . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.8.1 Single-particle correlation function . . . . . . . . . . .
9.8.2 Pair distribution function . . . . . . . . . . . . . . . .
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154
10 Quantisation of the free electromagnetic field
157
10.1 Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . 157
10.2 Normal modes . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
10.3 Quantisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
11 Interaction of radiation field with matter
11.1 Free radiation field . . . . . . . . . .
11.2 Electron and interaction term . . . .
11.3 Transition rate . . . . . . . . . . . . .
11.3.1 Photon emission . . . . . . .
11.3.2 Photon absorption . . . . . .
11.3.3 Electric dipole transition . . .
11.3.4 Lifetime of an excited state .
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6
CONTENTS
11.4
11.5
11.6
11.7
11.8
11.9
Nonrelativistic Bremsstrahlung . . . . . . . . .
Kinetic energy of the electron . . . . . . . . . .
Interaction between charge and radiation field
Diamagnetic contribution(Addendum) . . . . .
Potential term . . . . . . . . . . . . . . . . . . .
Transistion amplitude . . . . . . . . . . . . . .
11.9.1 First order . . . . . . . . . . . . . . . . .
11.9.2 Second order . . . . . . . . . . . . . . .
11.10Cross-section . . . . . . . . . . . . . . . . . . . .
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170
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182
12 Eine kurze Einführung in die Feynman’schen Pfadintegrale
187
12.1 Aharonov-Bohm-Effekt . . . . . . . . . . . . . . . . . . . . . . 190
12.2 Quanten-Interferenz aufgrund von Gravitation . . . . . . . . 191
A Details
A.1 Proof of eq. 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.2 Evaluation of (1.9) . . . . . . . . . . . . . . . . . . . . . . . . .
A.3 Discussion about the phase ωz in (1.14) . . . . . . . . . . . . .
A.4 Transformation of the components of a vector . . . . . . . . .
A.5 Commutation rules of the orbital angular momentum . . . .
A.6 Further Commutation rules of J . . . . . . . . . . . . . . . .
A.7 Uncertainty relation for j = 0 . . . . . . . . . . . . . . . . . .
A.8 J × J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.9 Explicit derivation of angular momentum operators and their
eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.10 Relation betwen L2 and p2 . . . . . . . . . . . . . . . . . . . .
A.11 Proof that σ ≤ 0 solutions in (2.17) must be discarded . . . .
A.12 Details of the evaluation of the radial wave function for hydrogen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.13 Potential of an electron in the ground state of a H-like atom
A.14 Löwdin orthonormalisation . . . . . . . . . . . . . . . . . . .
A.15 Some details for the total S . . . . . . . . . . . . . . . . . . . .
A.16 Proof of Wigner Eckart’s theorem for vectors . . . . . . . . .
A.17 Proof of the projection theorem . . . . . . . . . . . . . . . . .
A.18 A representation of the delta distribution . . . . . . . . . . .
A.19 Integral evaluation of (6.7) . . . . . . . . . . . . . . . . . . . .
A.20 Example of evaluation of matrix elements in product states .
A.21 Some proofs for Chap.(8) . . . . . . . . . . . . . . . . . . . . .
A.22 Proof of (9.3) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.23 Gauge transformation for the wave function . . . . . . . . .
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215
215
CONTENTS
A.24 Commutators and relation for the Schrödinger field quantisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.25 Normalisation of two-particle state . . . . . . . . . . . . . .
A.26 Commutations rules . . . . . . . . . . . . . . . . . . . . . .
A.27 Lagrangian for electromagnetic field . . . . . . . . . . . . .
A.28 Integrals vs sums . . . . . . . . . . . . . . . . . . . . . . . .
A.29 Commutators and transversality condition . . . . . . . . .
A.30 Commutators of fields . . . . . . . . . . . . . . . . . . . . .
A.31 Fourier transform of Coulomb potential . . . . . . . . . . .
A.32 (No) energy conservation for the first-order process (11.34)
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221
8
CONTENTS
List of Figures
1.1
A plot of the first few spherical harmonics ((50)). The radius
is proportional to |Ylm |2 , colors gives arg(Ylm ), with green=
0, red= π. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
2.1
2.2
Coulomb-Potential . . . . . . . . . . . . . . . . . . . . . . . .
Energieniveaus des H-Atoms . . . . . . . . . . . . . . . . . .
40
45
3.1
3.2
3.3
3.4
Skizze des H2+ Moleküls . . . . . . . . . . . . . . . . . . . . . 53
H2+ -Wellenfunktionen mit gerader und ungerader Parität . . 58
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Geometrie zur Berechnung der van-der-Waals-Wechselwirkung
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.1
j1 = 2, j1 = 1 Clebsch-Gordan coefficients from http://pdg.lbl.gov/2002/clebrpp.pdf.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
6.1
6.2
6.3
6.4
Particle exchange in R2 . . . . . . . . . . . .
Particle exchange in R2 . . . . . . . . . . . .
Helium atom . . . . . . . . . . . . . . . . . .
Energy splitting of excited states for helium
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111
9.1
9.2
9.3
9.4
9.5
9.6
Schematic representation of an interacting process
Scattering by a potential . . . . . . . . . . . . . . .
Fermi sphere . . . . . . . . . . . . . . . . . . . . .
Particle-hole excitation. . . . . . . . . . . . . . . .
Equal time correlation function (9.86) . . . . . . .
Equal spin pair distribution function gσ,σ (9.90) . .
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11.1 Feynman diagramms for the two Scattering events of the
interaction term of the Hamiltonian (11.29). . . . . . . . . . . 173
9
10
LIST OF FIGURES
11.2 The contributions to H 00 ((11.30)). The lower two terms contribute to Compton scattering. The upper terms describe
emission and absorption of two photons. . . . . . . . . . . . 175
11.3 Second order processes for bremsstrahlung (cf. (11.29), (11.31)).
On the left, with intermediate state n1 and on the right with
intermediate state n2 . . . . . . . . . . . . . . . . . . . . . . . . 178
12.1 Aharonov-Bohm-Effekt . . . . . . . . . . . . . . . . . . . . . . 193
A.1 Plot der Funktion ∆t (ω) . . . . . . . . . . . . . . . . . . . . . 211
List of Tables
1.1
1.2
Quantum numbers of angular momentum operators . . . . . 23
Die ersten Kugelflächenfunktionen . . . . . . . . . . . . . . . 30
2.1
Quantenzahlen des H-Atoms mit Wertebereichen . . . . . .
43
4.1
Schematic structure of the product space of two systems
with angular momentum quantum numbers j1 and j2 . . . .
76
11
12
LIST OF TABLES
Chapter 1
Rotations and the angular
momentum operator
1.1
Main goals in this chapter
• Motivation: in Chap. 2 we shall deal with central potentials for which
angular momentum is conserved. We thus need to study the properties of angular momentum in quantum mechanics
• Determine algebra i.e. commutation rules of a generic angular momentum operator Ĵ .
This is achieved by looking at the algebra of rotation matrices.
• From the commutation rules only find eigenvalues and properties
of eigenvectors.
• Find expression for orbital angular momentum operator when acting on ψ(r).
• Find its eigenvalues and properties of eigenfunctions, i.e. spherical
harmonics.
1.2
Algebra of angular momentum operators
We have already seen that transformations, for example translations and
time evolution, in quantum mechanics are represented by unitary transformations. These continuous transformations form a so-called Lie groups
(see QM1) and have the general form
Û (ϕ) = e−iϕÂ .
13
(1.1)
14CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
with an hermitian operator  (generator) and a contiunuous real parameter ϕ. The transformation (1.1) can also be seen as a combination of infinitesimal transformations:
ϕ N
1
ϕ
N →∞
Û (ϕ) = e−i N Â
= (1 − i Â)N = (1 − iϕÂ − ϕ2 Â2 + · · · ) . (1.2)
N
2
A rotation in R3 can be specified by a corresponding rotation matrix Rn (ϕ)
Definition: Rn (ϕ): rotation matrix by an angle ϕ around the axis n.
The corresponding action on a quantum mechanical state is expressed by
an unitary operator Û (Rn (ϕ)), which according to (1.1), can be expressed
as
R OTATION OPERATOR
ˆ
Û (Rn (ϕ)) = e−iϕJn /~ .
(1.3)
where Jˆn is the generator of rotations around the axis n, i.e. the AngularMomentum operator (Drehimpulsoperator). The form of these operators
depend on the vector space one is considering. For example, in QM1
we have already met the corresponding operators ~Ŝx , ~Ŝy , ~Ŝz acting on
Spin- 12 states. For rotation in coordinate space, we will see that Ĵ corresponds to the orbital angular momentum L̂
In the next sections we will study the properties of these operators.
1.2.1
Rotation matrices in R3
Let us start with what we know, namely ordinary rotation matrices in real
space. These can also be expressed in a form similar to (1.3), namely
Rn (ϕ) = e−iϕAn .
(1.4)
To illustrate this, we consider a rotation by an infinitesimal angle ϕ about
the z axis:


1 −ϕ 0
1 0  + O(ϕ2 )
Rz (ϕ) =  ϕ
(1.5)
0
0 1
1.2. ALGEBRA OF ANGULAR MOMENTUM OPERATORS
using (1.2) to order O(ϕ) with (1.4) identifies in this case


0 −1 0
0 0 
Az = i  1
0
0 0
It is straightforward to show (proof): Sec. A.1
recovers the well known expression

cos ϕ − sin ϕ
−iϕAz

sin ϕ
cos ϕ
Rz (ϕ) = e
=
0
0
15
(1.6)
that for a finite ϕ one

0
0  .
1
(1.7)
Similarly, by cyclic permutation, 1 one obtains the two remaining generators:




0 0
0
0 0 1
Ax = i  0 0 −1 
Ay = i  0 0 0 
(1.8)
0 1
0
−1 0 0
This tells us the action of the rotation operators and of its generators on
a three-dimensional vector space. We also know, from QM1, its action on a
two-dimensional Spin- 21 system. The question to address is now, what is
their action on the infinite-dimensional Hilbert space of wavefunctions, in
other words, what are the properties of its generators Ĵ .
To learn more about this issue, we first consider the fact that the combination of two rotations around different axes does not commute. For example, the combination
Ry (−ϕ)Rx (−ϕ)Ry (ϕ)Rx (ϕ)
(1.9)
is not equal to the identity. We evaluate (1.9) up to order ϕ2 , obtaining for
(1.9) (details): Sec. A.2
1 + [Ax , Ay ]ϕ2 + · · · .
(1.10)
The commutator is readily evaluated from (1.8)
[Ax , Ay ] = iAz .
(1.11)
This gives for (1.9)
Ry (−ϕ)Rx (−ϕ)Ry (ϕ)Rx (ϕ) ≈ 1 + iϕ2 Az + · · · ≈ Rz (−ϕ2 ) + · · · . (1.12)
1
Here and below we will obtain results for a fixed cartesian component, and then generalize it to arbitrary components by means of a cyclic permutation, i.e. by exchanging
everywhere x → y → z → x → · · ·
16CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
The meaning of this is the following: Assume that one first rotates around
the x-axis and then around the y-axis by a (small) angle ϕ, respectively,
and then repeats the sequence of rotations in the same order by the angle
−ϕ. This does not correspond to an identity transformation but results
into a rotation around the z-axis by an angle −ϕ2 .
In general, we can write
Rβ (−ϕ)Rα (−ϕ)Rβ (ϕ)Rα (ϕ) = εαβγ Rγ (−ϕ2 ) + O(ϕ3 )
1.2.2
(1.13)
Commutation rules of angular momentum operators
If we now carry out these rotations on a quantum mechanical system, then
both sides of Eq. (1.13) mut produce the same physical result. I. e.
!
Û (Ry (−ϕ))Û (Rx (−ϕ))Û (Ry (ϕ))Û (Rx (ϕ)) = eiωz Û (Rz (−ϕ2 )) + O(ϕ3 )
(1.14)
where ωz is a yet unknown ϕ-dependent phase. We exploit this property in
ˆ
order to learn more about J.
Again, we expand (1.14) up to second order in ϕ. Comparing (1.3) with
(1.4), it is clear that the result of the left hand side of (1.14) is given by
(1.10) whith Aα → Jˆα /~, i.e.
2
ϕ
1 + [Jˆx , Jˆy ] 2
~
(1.15)
Similarly, the r.h.s. of (1.14) becomes up to the same order, and neglecting
the phase ωz (See for a discussion): Sec. A.3
1+i
ϕ2 ˆ
~Jz
~2
(1.16)
comparing (1.15) with (1.16) gives [Jˆx , Jˆy ] = i~Jˆz . In general one gets the
C OMMUTATION RULES OF THE ANGULAR MOMENTUM OPERATORS
[Jˆα , Jˆβ ] = i~εαβγ Jˆγ
.
(1.17)
1.3. SCALAR AND VECTOR OPERATORS
1.3
17
Scalar and Vector Operators
Scalar and vector quantities (observables) have well defined transformation properties under rotation. Here we show that these imply well defined commutation rules of the corresponding operators with the angular
momentum operators.
Let us consider a scalar or vector operator Â, as well as one of its eigenstates |ψi:
 |ψi = a |ψi
.
Let us now consider a rotation R of the coordinate system. The operator
is transformed into Â0 and the state vector |ψi into |ψ 0 i = Û (R) |ψi. The
eigenvalue equation in the rotated system now reads
Â0 |ψ 0 i = a |ψ 0 i
Â0 Û (R) |ψi = a Û (R) |ψi
Û † (R)Â0 Û (R) |ψi = a |ψi
.
This observation holds for all eigenvectors of Â. Since these constitute a
complete basis set, one has the identity Û † (R)Â0 Û (R) = Â, or
Â0 = Û (R)ÂÛ † (R)
.
We assume again an infinitesimal rotation angle ϕ around the axis α and
thus replace the unitary rotation operator
ϕ
Û (R) = e−i ~
Jˆα
with a series expansion up to linear order in ϕ
i
i
Â0 = (11 − ϕJˆα )Â(11 + ϕJˆα ) + O(ϕ2 )
~
~
i ˆ
0
 =  − ϕ[Jα , Â] + O(ϕ2 )
~
(1.18)
If  is a scalar operator, the it is invariant under rotations, i.e. Â0 = Â. From
(1.18) it follows
S CALAR O PERATORS Â
[Jˆα , Â] = 0 .
(1.19)
18CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
An operator  with vector-character consists in a set of three operators (for
example in cartesian coordinates Âx , Ây , Âz ) which transform appropriately under rotations. A rotation R of the coordinate system corresponds to an
inverse rotation of the components of Â, i.e. by R−1 = RT (details): Sec. A.4
. We now consider specifically rotations around the z-axis by an angle ϕ.
The corresponding rotation matrix is (1.5). The rotation of the cartesian
0
components  = RT  is given explicitly by
Â0x = Âx + ϕ Ây
Â0y = Ây − ϕ Âx
Â0z = Âz
.
We compare this with (1.18) and obtain
[Jˆz , Âx ] = i~Ây
[Jˆz , Ây ] = −i~Âx
[Jˆz , Âz ] = 0
.
This can be summarized into
[Jˆz , Âβ ] = i~εzβγ Âγ
.
A cyclical permutation of the above results leads to generalized transformation properties of
V ECTOR - OPERATORS
[Jˆα , Âβ ] = i~ εαβγ Âγ
.
Notice that (1.17) is a special case of (1.20), since Ĵ is a vector.
(1.20)
1.4. EIGENVALUE PROBLEM FOR THE ANGULAR MOMENTUM OPERATOR19
1.4
Eigenvalue Problem for the Angular Momentum Operator
Systems with a central force are invariant under rotation. This means that
their Hamiltonian Ĥ is a scalar, i.e. it commutes with each component Jˆα
(see (1.19)). As discussed in QM1, this means that it is possible to find a
common set of eigenfunctions (eigenvectors) of Ĥ and one of the Jˆα . As
we will see, this allows to simplify the problem considerably and provides
the opportunity to classify the eigenstates of the Hamiltonian.
Of course, one should try to take the maximum number of mutually commuting operators. Unfortunately, since the different Jˆα do not commute
with each other, we can take only one of them. However, there is one
additional operator, namely Jˆ2 , that commutes with all Jˆα (and with a
rotation-invariant Hamiltonian):
[Jˆ2 , Jˆα ] = 0 .
(1.21)
To prove this it is sufficient to observe that Jˆ2 , the square length of a vector,
is a scalar, and thus (1.19) holds. (For a mathematical proof ): Sec. A.6 .
As a consequence, Jˆ2 and, e.g., Jˆz (as well as the rotation invariant Ĥ) admit a
complete set of common eigenvectors. The remaining angular momentum operators cannot, however, be simultaneously diagonalized. Alternatively,
one could also have chosen Jˆ2 and Jˆx or Jˆ2 and Jˆy . One can, however,
have just one of the three cartesian component. It has become established
to use the z-component. The eigenvalue problem to be solved is, thus,
Jˆ2 |j, mi = ~2 aj |j, mi
Jˆz |j, mi = ~ m|j, mi .
The eigenstates |j, mi are characterized by two quantum numbers j and m,
with corresponding eigenvalues ~2 aj und ~m. We collected factors ~ from
the eigenvalues, so that aj and m are dimensionless (Ĵ has the same dimensions as ~). Notice that we haven’t yet specified what j, m (or even aj )
are, for the moment they are just dimensionless (in principle real) numbers
used to specify the states |j, mi.
Ladder operators
In many treatments of the angular momentum operator it is convenient to
use a different representation of its components. We define
20CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
L ADDER OPERATORS
Jˆ± = Jˆx ± iJˆy
.
(1.22)
Clearly
Jˆ±† = Jˆ∓ .
(1.23)
From this we can obtain back the cartesian components of the angular
momentum operator
(Jˆ+ + Jˆ− )
Jˆx =
2
ˆ
(J+ − Jˆ− )
Jˆy =
2i
The commutation rules for the new operators read Proof: Sec. A.6.
[Jˆz , Jˆ± ] = ±~Jˆ±
[Jˆ+ , Jˆ− ] = 2~Jˆz
(1.24)
[Jˆ2 , Jˆ± ] = 0
The solution of the eigenvalue problem of the angular momentum operator from the results (1.24) follows a similar procedure as for the harmonic
oscillator. The Jˆ± are termed ladder operators, since they modify the quantum number m by ±1:
Jˆz (Ĵ± |j , mi) = Jˆ± Jˆz |j, mi + [Jˆz , Jˆ± ] |j, mi
| {z } | {z }
m ~·|j,mi
(1.25)
±~Jˆ±
= ~(m ± 1)(Ĵ± |j , mi)
(1.26)
I.e. Jˆ± |j, mi is an eigenstate of Jˆz with eigenvalue ~(m ± 1). This is the property of a ladder operator. Jˆ± does not change aj , the eigenvalue of Jˆ2 , as
1.4. EIGENVALUE PROBLEM FOR THE ANGULAR MOMENTUM OPERATOR21
[Jˆ2 , Jˆ± ] = 0, i.e.
Jˆ2 (Jˆ± |j, mi) = Jˆ± Jˆ2 |j, mi
= ~2 aj (Jˆ± |j, mi)
(1.27)
(1.28)
and therefore Jˆ± |j, mi is eigenvector of Jˆ2 with eigenvalue ~2 aj . From this
it follows
Jˆ± |j, mi = Cm± |j, m ± 1i
(1.29)
The proportionality constants Cm± will be determined later via normalisation.
Can Jˆ± be applied arbitrarily often, as in an harmonic oscillator? The answer is no, as can be shown by the following arguments. First of all, notice
that
Jˆ2 − Jˆz2 = Jˆx2 + Jˆy2 ≥ 0 ,
and therefore must be a nonnegative operator. Therefore,
hj, m| Jˆ2 − Jˆz2 |j, mi = ~2 (aj − m2 ) ≥ 0 ⇒ aj ≥ m2
(1.30)
We therefor have following conditions:
1. aj ≥ 0, since the eigenvalues of Jˆz are real
2. |m| ≤
√
aj
From the requirement that Jˆ+ |j, mi must not lead to to large unallowed
values of m, it follows that it exists a maximum (j-dependent) mmax for
which
Jˆ+ |j, mmax i = 0
.
(1.31)
In order to determine mmax , we first transform the Operator Jˆ− Jˆ+ as follows:
Ĵ− Ĵ+ = (Jˆx − iJˆy )(Jˆx + iJˆy ) = Jˆx2 + Jˆy2 + i [Jˆx , Jˆy ] = Ĵ 2 − Ĵz2 − ~Ĵz (1.32)
| {z }
i~Jˆz
One then has
0 = Jˆ− Jˆ+ |j, mmax i = ~2 (aj − m2max − mmax ) |j, mmax i
and
aj = mmax (mmax + 1 )
(1.33)
22CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
Due to (1.33) mmax determines uniquely the eigenvalue aj . Therefore, we
can identify the former with the quantum number j.
mmax = j
(1.34)
Similarly, a mmin must exist with
Jˆ− |j, mmin i = 0 .
Similar arguments as above lead to
Jˆ+ Jˆ− = Jˆ2 − Jˆz2 + ~Jˆz
(1.35)
aj = mmin (mmin − 1)
(1.36)
mmin = −j
(1.37)
and finally to
Combining (1.33) with (1.36) one gets. 2
Starting with |j, mmin i and repeatedly applying Jˆ+ one obtains (see (1.29))
n Mal
z }| {
Jˆ+ · · · Jˆ+ |j, mmin i ∝ |j, mmin + ni .
(1.38)
This works until |j, ji is reached. The next Jˆ+ destroys the state, according
to (1.31). In order to exactly reach |j, ji, j − mmin := n ∈ N0 must be integer.
Together with (1.37) one obtains
n
n ∈ N0
(1.39)
j=
2
j is therefore half-integer or integer. (1.39) and (1.33) determine the possible
eigenvalues of Jˆ2 und Jˆz as well as their relation. Accordingly, eigenstates
are classified with following convention which represents the Quantisation of Angular Momentum.
Eigenvalues and Eigenvectors of Angular Momentum Operators
Jˆ2 |j, mi = ~2 j(j + 1)|j, mi
Jˆz |j, mi = ~ m|j, mi
2
A second solution would be j = mmin − 1, which is obviously inconsistent
(1.40a)
(1.40b)
1.4. EIGENVALUE PROBLEM FOR THE ANGULAR MOMENTUM OPERATOR23
The notation and allowed values for these Quantum numbers are summarized in table 30. This result is completely general and follows simply
Symbol
Name
range of values
j
Angular momentum quantum n.
m
magnetic quantum number
j=
n
2
n ∈ N0
m ∈ {−j, −j + 1, . . . , j − 1, j}
Table 1.1: Quantum numbers of angular momentum operators
from the commutation rules (1.17).
We make the following interesting observation:
Jzmax = ~ · j; ⇒
Jˆ2 = ~2 · j(j + 1)
(Jzmax )2 =
~2 j 2
(1.41)
=
~2 j 2 + ~2 j
(1.42)
I.e., structureJˆ2 is strictly larger than Jˆz2 : the angular momentum operator
cannot be completely aligned. This should be like that, since suppose one
had Jˆ2 = Jˆz2 , then it would follow Jˆx = 0 and Jˆy = 0. This would contradict
Heisenberg’s uncertainty principle. An exception is provided for j = 0 See
why: Sec. A.7 .
Normalisation
Next we determine the proportionality constants Cm± in (1.29) from normalisation (we use (1.32) and (1.35))
!
hj, m|Jˆ±† Jˆ± |j, mi = |Cm± |2 hj, m ± 1|j, m ± 1i
|
{z
}
=1
|Cm± |2 = hj m|Jˆ∓ Jˆ± |j mi
= hj m|(Jˆ2 − Jˆ2 ∓ ~Jˆz )|j mi
z
2
= ~ (j(j + 1) − m2 ∓ m)
= ~2 (j(j + 1) − m(m ± 1))
The fact that Cm± = 0 for m = ±j is consistent with (1.34) and (1.37).
The phase of Cm± is in principle arbitrary. In the usual convention this is
chosen real and positive. We thus obtain the following result
24CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
1
Jˆ±
|j, mi
|j, m ± 1i = p
j(j + 1) − m(m ± 1) ~
.
(1.43)
With the help of (1.43) one can, starting from |j, −ji (or from an arbitrary
|j, m0 i) explicitly construct all other |j, mi
We have derived all possible eigenvalues of Jˆ2 from the commutator algebra. This does not yet mean that all possible values of j are realized in
nature. In particular j = 21 does not correspond to a orbital angular momentum. On the other hand, we have identified j = 12 with the spin of an
electron in a Stern-Gerlach Experiment (QM1). The general rule is that orbital angular momenta have integer values of j (see Sec. 1.5). Half-integer
values originate from the summation of spin with spin and/or with orbital
angular momentum.
The vectorial angular momentum operator have a peculiar property which
originates from the fact that its cartesian components do not commute. For
normal vectors v the cross product v × v vanishes. This does not hold in
the case of the operator Ĵ . Instead one has Proof: Sec. A.8 :
Ĵ × Ĵ = i~Ĵ
1.5
The Orbital Angular Momentum
Orbital angular momentum operator in real space
We have not jet specified the expression of the angular momentum operator Ĵ . Indeed, this depends on the space it acts on. When acting on wave
functions without spin this indeed acquires the (quantized) form of the
(orbital) angular momentum in classical physics.
To show this let us consider a wave function of the coordinates ψ(r). This
is a scalar, so that under a rotation its value on corresponding points in space
remains unchanged:
ψ(r) = ψ 0 (r 0 ) = Û (R)ψ(Rr) ⇒ Û (R)† ψ(r) = ψ(Rr) .
1.5. THE ORBITAL ANGULAR MOMENTUM
25
For an infinitesimal rotation ϕ, say around the z-axis (see (1.5),(1.3)) this
gives (r ≡ (x, y, z))
ϕ ˆ
1 + i Jz ψ(x, y, z) = ψ(x − ϕy, y + ϕx, z) + O(ϕ2 ) =
~
(1.44)
= (1 + ϕ(−y∂x + x∂y )) ψ(r) + O(ϕ2 ) .
This identifies
Ĵz = −i~(−y∂x + x∂y ) = x̂ p̂y − ŷp̂x .
(1.45)
the latter is indeed the known expression for the z component of the orbital
angular momentum operator L̂z . Generalisation to the other components is
obtained by cyclic permutations.
Thus, in the case of a scalar wave function, the generator of rotations Ĵ is the
orbital angular momentum
Ĵ ⇒ L̂ = r̂ × p̂ .
(1.46)
Notice that the order of the operators does not matter here. When the
wave function has components, like in the case of a spinor considered in
QM1, Ĵ is the sum of L̂ plus other terms (e.g., the spin Ŝ).
It is straightforward to show (and should be expected from the discussion
up to now) that L̂, being an operator generating rotations, has to obey
the correct commutation relations (1.17). In addition, the commutation
relations (1.20) hold for the known vector operators r̂ and p̂ (proofs here):
Sec. A.5 .
Eigenwerte
Von den Vertauschungsrelationen alleine wissen wir bereits, daß die Eigenwerte von L̂z ganzzahlige oder halbzahlige Vielfache von ~ sind. Wir werden noch zeigen, daß die halbzahligen Drehimpulse aufgrund der inneren
Struktur des Bahndrehimpulses ausscheiden.
As discussed, the (orbital) angular momentum L̂ is useful for problems
with rotation symmetry, i.e. when the potential V (r) only depends on r ≡
|r|. In this case, it is also useful to work in spherical coordinates (r, θ, ϕ). It
is, therefore, useful to express the operators L̂ in these coordinates. First,
one can readily see that all components of L̂ only act on the angular part
of the wave function, i.e. on Ω ≡ (θ, ϕ). This can be deduced from the fact
that (cf. (1.3)) e−iϕL̂α /~ produces a rotation of the coordinates, so it only
acts on Ω, and it cannot act on r. As a consequence, also the L̂α only act
26CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
on Ω. A second, more simple, argument comes from the fact that L̂α /~ is
dimensionless and, therefore, can only act on dimensionless quantities.
The consequence is that if we write a wave function ψ(r, Ω) in the product
form r ≡ |r|
ψ(r, Ω) = f (r)Y (Ω)
then applying an arbitrary component of L̂ (or L̂2 )
L̂ f (r)Y (Ω) = f (r) L̂ Y (Ω) .
(1.47)
As a consequence, the search for the common eigenfunctions of L̂2 , L̂z corresponding to the vectors |l, mi, can be restricted to functions Y (Ω) of the
angles Ω only. But (1.47) also means that given a common eigenfunction
Y (Ω) of L̂2 , L̂z , then one can multiply it by any function of r, and this will
remain an eigenfunction of L̂2 , L̂z with the same eigenvalues. There is a
large degeneracy. This also means that one chosen set of common eigenfunctions of L̂2 , L̂z do not represent a complete basis in the space of the
ψ(r, Ω). It is, however, complete in the space of the Y (Ω), as we shall see
below.
These eigenfunctions, classified according to their quantum numbers, are
termed Ylm (Ω), and are the well-known spherical harmonics (Kugelflächenfunktionen). Here, m has the same meaning as in (1.40b), while it is convention to use l instead of j ((1.40a)) for the orbital angular momentum.
In agreement with (1.40a) and (1.40b) one thus has
E IGENVALUE EQUATION FOR SPHERICAL HARMONICS
L̂2 Ylm (Ω) = ~2 l(l + 1) Ylm (Ω)
m
m
L̂z Yl (Ω) = ~m Yl (Ω) ,
(1.48)
(1.49)
Below, we will show this result and derive the explicit expression of the
operators L̂z , L̂± as well as of their eigenfunctions Ylm in spherical coordinates.
1.5. THE ORBITAL ANGULAR MOMENTUM
1.5.1
27
Ortsraumeigenfunktionen des Bahndrehimpulses
Goals of this section are (a) to express the angular momentum operators
in spherical coordinates and (b) to obtain the expression for the Ylm (θ, ϕ).
The derivation is rather lengthy and tedious, so we will here only point out
the steps one has to carry out, following the logics of the previous sessions,
and show only the simpler calculations. The more tedious details can be
found in the appendix: Sec. A.9 , In order to understand the procedure,
we will explicitly derive the lowest spherical harmonics, i.e. the ones with
l = 0 and l = 1.
The steps are the following
• From the expression of the orbital angular momentum operator in
cartesian coordinates (cf. (1.45)),
L̂ = r̂ × p̂ = −i~ r̂ × ∇
(1.50)
write down the expression of the relevant operators (L̂z , L̂± ) in spherical coordinates. They will have the form of linear differential operators
in θ, ϕ.
• L̂z produces rotations around the z axis, and, thus, only acts on the
angle ϕ. It has a very simple form,
L̂z =
~ ∂
i ∂ϕ
so we start by solving for the ϕ-part of the Ylm by using (1.49). The
eigenvalue equation
∂ m
Yl (θ, ϕ) = im Ylm (θ, ϕ)
∂ϕ
has a simple solution
Ylm (θ, ϕ) ∝ eimϕ .
Since the wave function must be single-valued, m must be integer.
As a consequence, l must be integer as well. This fact excludes halfinteger values for the orbital angular momentum quantum numbers l and
m. In contrast , as seen in QM1, the spin angular momentum can
be half-integer.
(1.51)
28CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
• Once we have the ϕ part, we write a differential equation (just in θ)
for the wavefunction with maximum m (cf. (1.31)):
L̂+ Yll (θ, ϕ) = 0
We solve it, the solution is relatively simple, and normalize the Yll .
• Once we have the Yll , we obtain the Yll−1 , Yll−2 , · · · by applying L̂−
and by normalizing, i.e. with (1.43) where Ĵ → L̂ and j → l.
The final result, i.e. the general form for the spherical harmonics, is given
in (A.14).
Being a complete orthonormal basis in the space of functions f (θ, ϕ), the
Ylm are
Orthonormal:
Z Z
∗
0
d ϕ d cos θ Ylm (θ, ϕ)Ylm
(1.52)
0 (θ, ϕ) = δl,l0 δm,m0 .
Notice that the integration measure d cos θ = sin θdθ is the one appropriate
for spherical coordinates: dΩ ≡ dϕ sinθdθ.
Complete
l
∞ X
X
l=0 m=−l
∗
Ylm (θ, ϕ)Ylm (θ0 , ϕ0 ) = δ(cos θ − cos θ0 )δ(ϕ − ϕ0 )
(1.53)
Equivalently, this means that any function f (θ, ϕ) on the unit sphere can
be expanded in terms of the Ylm :
f (θ, ϕ) =
l
∞ X
X
cm
Ylm (θ, ϕ)
l
(1.54)
l=0 m=−l
The coefficient, as usual, are evaluated by scalar multiplication from left
with hl0 , m0 |:
Z Z
0
cm
l0 =
0
∗
dϕ d cos θ Ylm
0 (θ, ϕ) f (θ, ϕ)
Explicit calculation for l = 0 and l = 1
For simplicity, we do this without normalisation
l = 0:
For l = 0, we have m = 0, therefore
L̂z Y00 = 0 L̂± Y00 = 0 ⇒ L̂x|y Y00 = 0 .
(1.55)
1.5. THE ORBITAL ANGULAR MOMENTUM
29
Thus since all generators of rotation (cf. (1.3)) give zero, this means that
Y00 (θ, ϕ) is invariant under rotations, i.e. it is a constant.
l = 1:
Wir suchen zunächst nach Y1m=+1 , was die Gleichung
L̂z m=1
∂
!
Y1
= −i Y1m=1 = 1 · Y1m=1
~
∂ϕ
erfüllt, also
Y11 = eiϕ f (θ)
Die Gleichung (mit (A.12))
L̂+ m=1
Y
= eiϕ
~ 1
∂
cos θ
f (θ) − 1 ·
f (θ)
∂θ
sin θ
!
=0
hat die Lösung
f (θ) ∝ sin θ
Also
Y11 ∝ eiϕ sin θ
Anwenden des Leiteroperators (A.12)
cos θ
∂
0
1
−iϕ iϕ
sin θ ∝ cos θ
Y1 ∝ L̂− Y1 = e e
− −
∂θ
sin θ
ein zweites Mal
Y1−1 ∝ L̂− Y10 = e−iϕ (−
∂
) cos θ ∝ e−iϕ sin θ
∂θ
Die niedrigsten Kugelflächenfunktionen sind in Tab. 50 angegeben
30CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
l
m
0
0
Y00 =
0
Y10
±1
Y1±1
= ∓
0
Y20
q
±1
Y2±1
= ∓
±2
Y2±2
q
1
2
Kugelflächenfunktion
=
=
=
√1
4π
q
3
4π
q
cos θ
3
8π
sin θe±iϕ
5
(3 cos2
16π
q
15
8π
15
32π
θ − 1)
sin θ cos θe±iϕ
sin2 θe±2iϕ
Table 1.2: Die ersten Kugelflächenfunktionen.
1.5. THE ORBITAL ANGULAR MOMENTUM
31
Figure 1.1: A plot of the first few spherical harmonics ((50)). The radius is
proportional to |Ylm |2 , colors gives arg(Ylm ), with green= 0, red= π.
32CHAPTER 1. ROTATIONS AND THE ANGULAR MOMENTUM OPERATOR
Chapter 2
Schrödinger equation in a central
potential
2.1
Main results in this chapter (until Sec. 2.5)
• The Schrödinger equation for a particle in a central potential (i.e.
rotation invariant) is written in polar coordinates r, θ, ϕ.
• Hereby one notices that the kinetic part can be splitted in a part acting
only on r and one acting only on θ, ϕ.
2
• The part acting only on θ, ϕ is proportional to L̂ .
• Therefore, one can look for solutions of the form R(r)Ylm (θ, ϕ), since
2
the Ylm are eigenfunctions of L̂ .
• After doing that one ends up with a one-dimensional eigenvalue
equation for R(r).
2
• Since the eigenvalue of L̂ does not depend on m, one has the general
result that for a central potential eigenfunctions with a given l have a
2l + 1 degeneracy.
2.2
Radial- und Drehimpulsanteil
Der Hamiltonian für ein quantenmechanisches Teilchen im kugelsymmetrischen
Potential (Zentralfeld) lautet
H=
p̂2
+ V (r)
2m
33
.
(2.1)
34CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Hierbei ist r = |r| die Norm des Ortsvektors. In der klassischen Mechanik
ist der Drehimpuls l eine Erhaltungsgröße. Er ist durch die Anfangsbedingungen gegeben.
2
2
2
p̂r
p̂
L̂
wird in einem zentrifugal-Beitrag 2mr
Der Term 2m
2 und einem radial-Teil 2m ,
aufgespaltet. Die Bewegungsgleichung des Teilchens reduziert sich somit
auf die Radialgleichung in einem effektiven Potential
2
L̂
Vef f (r) = V (r) +
.
(2.2)
2mr2
Wir werden versuchen, auch die Schrödingergleichung auf ein Radialproblem zu reduzieren. Dazu brauchen wir den Zusammenhang zwischen
2
p̂2 und L̂ . Durch explizite Berechnung (details): Sec. A.10 erhalten wir
2
L̂ = r̂ 2 p̂ 2 + i~ r̂ · p̂ − (r̂ · p̂)2
also
2
L̂ + (r̂ · p̂)2 − i~ (r̂ · p̂)
(2.3)
p̂ =
r̂ 2
Man erkennt den Zusammenhang aus der klassischen Mechanik für ~ →
0. Der Hamiltonoperator nimmt hiermit die folgende Gestalt an
2
2
(r̂ · p̂)2 − i~ (r̂ · p̂)
L̂
Ĥ =
+V (r) +
2
2m
2m r̂2
|
{zr̂
}
(2.4)
T̃ˆ
Man erkennt bereits, dass T̃ˆ nur die Komponente von p entlang r enthält.
Der Übergang in die Ortsdarstellung, mit der Zuordnung r̂ → r und p̂ → −i~∇,
liefert aus (2.4)
2
(r · ∇) + (r · ∇)
2m ˆ
− 2 T̃ →
~
r2
In Kugelkoordinaten r = r er und daher benötigen wir von ∇ nur die
Komponente entlang r:
∂
r·∇=r
∂r
also
2m
1 ∂ ∂
1 ∂
1 ∂ ∂
∂
− 2 T̃ˆ → 2 (r r ) + 2 r
= ( r + )
~
r ∂r ∂r
r ∂r
r ∂r ∂r ∂r
1 ∂2
=
(
r)
(2.5)
r ∂ r2
2.3. PRODUKTANSATZ FÜR DIE SCHRÖDINGERGLEICHUNG
35
ˆ2
As expected, (2.5) is the radial part of the Laplace operator,1 while − r2L ~2
(cf. (2.4)) its angular part.
Der Hamiltonoperator ist somit
2
~2 1 ∂ 2
1 L̂
Ĥ = −
r + V (r) +
2
2
| 2m r ∂r{z
} 2m r
.
Â(r)
Man kann auch hier den letzten Term als Zentrifugalbeitrag erkennen (cf.
(2.2)).
2.3
Produktansatz für die Schrödingergleichung
Die Schrödingergleichung
2
1 L̂
Â(r) +
2m r2
!
ψ(r, θ, ϕ) = Eψ(r, θ, ϕ)
kann durch einen Produktansatz ψ(r, θ, ϕ) = R(r) · Y (θ, ϕ) weiter vereinfacht werden. Nach Multiplikation mit r2 wird diese zu
2
L̂
Y (θ, ϕ) = Y (θ, ϕ)r2 (E − Â(r))R(r)
R(r)
2m
da L2 nur auf θ, ϕ und Â(r) nur auf r wirkt. Nach Multiplikation mit
(R(r) · Y (θ, ϕ))−1 von links erhält man
2
1
L̂
1
2
Y (θ, ϕ) =
r (E − Â(r))R(r) = κ
Y (θ, ϕ) 2m
R(r)
.
Da die linke Seite nur θ und ϕ, die rechte Seite hingegen nur r enthält, muß
κ eine Konstante sein und wir erhalten zwei in Winkel- und Radialanteil
getrennte Differentialgleichungen
2
L̂ m
Y (θ, ϕ) = κYlm (θ, ϕ)
2m l
κ
(Â(r) + 2 )R(r) = ER(r)
r
1
(2.6)
(2.7)
This result is only valid provided the wave function is non singular for r → 0. Consider for example, the case ψ ∝ 1/r. In that case T̂ 2 ψ ∝ ∇2 ψ ∝ δ(r).
36CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
(2.6) ist die bereits gelöste Eigenwertgleichung des Drehimpulsoperators,
~2
deren Eigenwerte den Parameter κ = 2m
l(l +1) mit der Drehimpulsquantenzahl in Verbindung bringen.
Einsetzen in die Radialgleichung (2.7) ergibt
~2 1 d2
~2 l(l + 1)
R(r) = ER(r)
−
(r R(r)) + V (r) +
2m r dr2
2m r2
Multiplikation von links mit r und Verwendung der Abkürzung χ(r) := r R(r)
liefert schließlich die
S CHRÖDINGERGLEICHUNG FÜR DEN
R ADIALANTEIL DER W ELLENFUNKTION
~2 d2
~2 l(l + 1)
−
χ(r) + V (r) +
χ(r) = E χ(r)
2m dr2
2m r2
|
{z
}
.
(2.8)
Vef f (r)
Die Lösung dieser Differentialgleichung hängt vom jeweiligen Potential
V (r) ab und muß je nach Problem neu gelöst werden. Die gesamte Wellenfunktion ψ ist dann
1
ψlm (r, θ, ϕ) = χ(r) · Ylm (θ, ϕ)
r
,
(2.9)
wobei nur die Quantenzahlen des Drehimpulses explizit angegeben wurden. Die Quantenzahlen, die sich aus dem Radialanteil ergeben, werden
später eingeführt. Allgemein können wir bereits erkennen, daß die Energien, wegen der Rotationsinvarianz in m, (2l + 1)-fach entartet sind .
2.4
Entartung bei unterschiedlichen m
Die Entartung der Zuständen mit unterschiedlichen m bei festem l ist eine
direkte Folgerung der Rotationsinvarianz. Diese gilt also allgemein für Sys-
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
37
teme, in der der Hamiltonian H mit allen Komponenten des Drehimpulsoperators Ĵ (Erinnerung, hier Ĵ = L̂) vertauscht. Das kann folgendermaßen gezeigt werden:
2
Wir haben bereits gezeigt, dass Ĵ und Jˆz zusammen mit dem Hamiltonian diagonalisiert werden können. Die gemeinsame Eigenzustände von
2
H, Ĵ , Jˆz können also dargestellt werden als
|n, j, mi
wo n eine zusätzliche Quantenzahl ist. Die Eigenwertgleichung lautet
Ĥ |n, j, mi = En,j,m |n, j, mi .
(2.10)
wendet man auf beiden Seiten den Leiteroperator Jˆ− an, und betrachtet
man die Tatsache, dass [Ĥ, Jˆ− ] = 0, erhält man
Ĥ Jˆ− |n, j, mi = En,j,m Jˆ− |n, j, mi .
mit (1.43)
Jˆ− |n, j, mi = const. |n, j, m − 1i
erhält man
Ĥ |n, j, m − 1i = En,j,m |n, j, m − 1i .
also, verglichen mit (2.10) haben wir
En,j,m−1 = En,j,m ≡ En,j
unabhängig von m, wie gesagt.
2.5
2.5.1
Wasserstoff und H-ähnliche Probleme
Summary
• For the Hydrogen Atom, one has a Coulomb potential V ∝ −1/r.
• We look for bound states, i.e. states with E < 0. The solution of the
corresponding equation for R(r) is obtained in two steps.
• First one looks for the asymptotic solution at r → ∞. Here the requirement is that R(r) vanishes exponentially.
• For the short-r part one makes a polynomial Ansatz. Inserting into the
Schrödinger equation, one obtains a recursive equation for the coefficient of the polynomial.
38CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
• The requirement that the recursive equation stops at some point (i.e. the
polynomial has finite order), leads to the eigenvalue condition for
the energy.
• For the case that the recursive equation does not stop the sequence leads
to an exponential behavior of the wave function, which is not allowed.
Das Wasserstoffatom H und seine Isotopen 2 H = D und 3 H = T, sowie
die Ionen He+ ,Li2+ ,Be3+ sind die einfachsten atomaren Systeme.
Ihre Kernladung ist ein ganzzahliges Vielfaches der Elementarladung und
die Elektronenhülle besteht ausschließlich aus einem Elektron.
Ohne äußere Kräfte wirkt nur die Coulombwechselwirkung zwischen Kern
und Elektron.
2.5.2
Center of mass coordinates
This is a two-body-problem. However, as in classical mechanics one can
introduce center of mass coordinates R ≡ m1 r 1 +m2 r 2 , P = p1 +p2 (1, 2 inp1
−
dicate the two particles), and relative coordinates r ≡ r 1 − r 2 , p = mred ( m
1
1 −1
1
, where mred ≡ ( m1 + m2 ) is the reduced mass. This is a canonical transformation, so it is easy to show that the correct commutation rules for the
new variables hold. The advantage is that now the Hamiltonian separates
H=
p2
P2
+
+ V (r)
2(m1 + m2 ) 2mred
and one can solve separately for the center of mas motion, which is free,
and the relative motion, which describes a single body with mass mred in
the potential V (r).
In the case of Hydrogen, since the nucleus is much heavier than the electron, mred is essentially given by the electron mass. In the following , for
simplicity, we will use the electron mass m instead of mred . We will also
leave out the center of mass part of the Hamiltonian, which is trivial. Results for two particles with similar masses, such as for example positronium, are easily recovered by replacing m with mred .
2.5.3
Eigenvalue equation
Wir verwenden folgende Abkürzungen
p2
)
m2
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
39
m = Masse des Elektrons
+Ze = Ladung des Kerns
-e = Ladung des Elektrons
Die potentielle Energie (Coulombenergie) des Systems ist in Gaußschen
Einheiten2 gegeben durch
V (r) = −
Ze2
r
Die Schrödingergleichung wird dann zu
~2 2
∇ + V (r))ψ(r) = Eψ(r)
Ĥψ(r) = (−
2m
Aus (2.9) wissen wir bereits, daß
ψ(r) = ψ(r, θ, ϕ) =
χ(r)
· Ylm (θ, ϕ)
r
,
wobei χ(r) der Radialanteil der Schrödingergleichung ist, dessen Quantenzahlen noch nicht spezifiziert sind.
−
Ze2
~2 l(l + 1)
~2 d2
χ(r)
+
(−
+
)χ(r) = Eχ(r)
2m dr2
r
2m r2
Ze2 2m 1 l(l + 1) 2mE
χ (r) + (
−
+ 2 )χ(r) = 0
2
r2
~
| ~
{z } r
00
.
(2.11)
=: 2Z
a
0
In dieser Gleichung tritt eine charakteristische Länge atomarer Systeme
auf, der sogenannte Bohrsche Radius.
a0 =
◦
~2
=
0.529
A
me2
.
(2.12)
Für r → ∞ vereinfacht sich (2.11) zu
00
χ (r) = −
2
2mE
χ(r)
~2
.
Für mikroskopische Phänomene sind Gaußsche Einheiten bequemer als die SIEinheiten, da viele Vorfaktoren einfacher werden.
40CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
l=1
r/a0
5
10
15
20
V
-0.5
l=0
-1.0
-1.5
Figure 2.1: Effective Coulomb-Potential Vef f (r) in atomic units for l = 0
and l = 1
Wir untersuchen hier gebundene Zustände, das sind solche mit negativer
Energie. Wenn die Energie negativ ist, wird E − V (r) für r → ∞ negativ und die Schrödingergleichung führt zu einem exponentiellen Abfall
der Wellenfunktion. Neben den gebundenen Zuständen gibt es noch
Streuzustände mit positiver Energie. Für r → ∞ lautet die Differentialgleichung (2.11) somit
00
χ (r) = −
2mE
χ(r) = γ 2 χ(r)
~2
.
Die Lösung dieser Differentialgleichung ist bekanntlich
χ(r) = ae−γr + be+γr
.
Der zweite Summand ist nicht normierbar. Er beschreibt also keinen gebundenen Zustand und wir setzen deshalb aus physikalischen Gründen b =
0.
Für beliebige r machen wir den Ansatz
χ(r) = F (r)e−γr
und setzen ihn in (2.11) ein. Zusammen mit
d2
00
0
χ(r) = e−γr (F (r) − 2γF (r) + γ 2 F (r))
2
dr
(2.13)
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
41
erhalten wir
2Z 1 l(l + 1)
00
0
2
−γr
2
F (r) − 2γ F (r) + γ F (r) + (
−
e
− γ )F (r) = 0
a0 r
r2
1
2Z
00
0
F (r) − 2γ F (r) + ( r − l(l + 1)) 2 F (r) = 0
a0
r
(2.14).
Die Lösung der Differentialgleichung läßt sich als eine Potenzreihe ansetzen
F (r) = r
σ
∞
X
cµ r µ
(2.15)
µ=0
=
∞
X
cµ rµ+σ
µ=0
Einsetzen von (2.15) in die Differentialgleichung (2.14) ergibt
∞
X
µ=0
cµ (µ + σ)(µ + σ − 1)rµ+σ−2 − 2γ
+
2Z
a0
∞
X
µ=0
∞
X
cµ (µ + σ)rµ+σ−1
µ=0
cµ rµ+σ−1 − l(l + 1)
∞
X
cµ rµ+σ−2 = 0
.
µ=0
Wir fassen Terme gleicher Potenz in r zusammen
c0 (σ(σ − 1) − l(l + 1))rσ−2 +
∞ X
+
cµ+1 (µ + 1 + σ)(µ + σ) − 2γcµ (µ + σ)+
µ=0
2Z
cµ − l(l + 1)cµ+1 rµ+σ−1 = 0
+
a0
(2.16)
Die Koeffizienten der Potenzen rl müssen individuell verschwinden, da
die Gleichung für beliebige Werte von r gelten muß und die {rl } ein
vollständiges, linear unabhängiges Basissystem bilden. Zunächst folgt
aus der Bedingung c0 6= 0 für σ die Bedingung
σ(σ − 1) = l(l + 1)
.
(2.17)
Das ergibt (the second solution must be discarded, see: Sec. A.11 )
σ =l+1.
(2.18)
42CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Einsetzen in (2.16) liefert die Bestimmungsgleichungen der Koeffizienten
cµ . Für alle µ ≥ 0 gilt
2γ(µ + l + 1) − 2 aZ0
cµ+1
=
cµ
(µ + l + 2)(µ + l + 1) − l(l + 1)
.
(2.19)
Das Verhalten für µ 1 ist
cµ+1
2γ
−→
cµ µ1 µ + 1
.
Das heisst, der Beitrag aus großen µ liefert
∞
X
(2γ)µ
µ!
µ
rµ ∼ e2γr
(2.20)
Da die hohen Potenzen µ 1 das Verhalten der Funktion für große r
bestimmen, verhält sich χ für große r wie
χ(r) = F (r) e−γr = rl+1 e2γr · e−γr = rl+1 eγ·r
.
Wenn die Potenzreihe nicht abbricht, divergiert χ(r) für r → ∞ und beschreibt
wieder keinen normierbaren, gebundenen Zustand. Wir müssen also erreichen, daß die Reihe abbricht. Es kann in der Tat erreicht werden, daß ein
Koeffizient cµ∗ +1 in (2.19) , und somit alle nachfolgenden, verschwinden.
Das ist genau dann der Fall, wenn
γ=
2Z
a0
2(µ∗
+ l + 1)
=
a0
(µ∗
Z
+ l + 1)
;
µ∗ ∈ N0
.
(2.21)
Für die Energie bedeutet das
Z2
~2
~2 m2 e4
Z2
~2 2
a20
=
−
)
E = −
γ = −
(
2m
2m (µ∗ + l + 1)2
2m ~4 (µ∗ + l + 1)2
.
Die Energie ist also quantisiert. Es ist üblich eine modifizierte Quantenzahl
n = µ∗ + l + 1
(2.22)
anstelle von µ∗ einzuführen. Aus µ∗ ≥ 0 und l ≥ 0 folgt n ≥ 1 und die
erlaubten Energien der gebundenen Zustände sind
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
me4 Z 2
2~2 n2
Z2
= −Ry 2
n
En = −
En
n = 1, 2, 3, . . .
43
(2.23)
.
Die natürliche Einheit der Energie ist das
RYDBERG
1Ry =
me4
2~2
.
(2.24)
Wegen µ∗ + l + 1 = n sind bei gegebener Hauptquantenzahl n nur Drehimpulsquantenzahlen l < n erlaubt.
Symbol
Name
erlaubte Werte
n Hauptquantenzahl
n = 1, 2, 3, . . .
l
l = 0, 1, 2, . . . , n − 1
m
Drehimpulsquantenzahl
magnetische Quantenzahl m = {−l, −l + 1, . . . , +l − 1, l}
2l + 1 mögliche Werte
Table 2.1: Quantenzahlen des H-Atoms mit Wertebereichen.
2.5.4
Entartung
Wir haben schon gesehen (Sec. 2.4), dass für ein rotationsinvariantes Hamiltonian, die Energie nicht von der Quantenzahl m abhängig ist.
44CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Im Fall vom Wasserstoffatom hängt aber die Energie En nur von der Hauptquantenzahl n, also auch nicht von l ab. Zu festem n, also für eine gegebene
Energie, kann die Drehimpulsquantenzahl die Werte l = 0, 1, . . . , n − 1
annehmen. Zu jedem l wiederum sind 2l + 1 Werte für die magnetische
Quantenzahl möglich. Die Anzahl der entarteten Zustände ist somit
Entartung =
n−1
X
(2l + 1)
l=0
n−1
X
= 2
l+n
l=0
= 2
n(n − 1)
+ n = n2
2
Die Entartung ist also n2 .
• Die Drehimpulserhaltung erklärt nur die (2l+1)-fache Entartung der
magnetischen Q.Z.
• Die höhere Entartung bedeutet, daß es hier eine weitere Erhaltungsgröße, nämlich den Runge-Lenz Vektor gibt. Er ist klassisch definiert
als
N = p × L − e2 Zm er
Quantenmechanisch
N̂ =
1
(p̂ × L̂ + L̂ × p̂) − e2 Zmer
2
Dieser Vektor ist nur im 1r -Potential eine Erhaltungsgröße. Man spricht
daher auch von zufälliger Entartung.
2.5.5
Energieschema des H-Atoms (Z=1)
Das H-Atom definiert charakteristische Werte für Energie und Länge
me4
= 13.6 eV
2~2
◦
~2
=
=
0.529
A
me2
1 Ry =
a0
.
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
45
Figure 2.2: Energieniveaus des H-Atoms in Ry und Coulombpotential (durchgezogene Kurve).
2.5.6
Lichtemission
Nach den Gesetzen der klassischen Elektrodynamik strahlt beschleunigte
Ladung Energie ab. Das hieße, daß das Elektron, das klassisch auf einer
Ellipsenbahn um den Kern kreist, permanent Energie abstrahlen würde.
Es müßte dadurch spiralförmig in den Kern stürzen. Das steht natürlich
im Widerspruch zur Beobachtung stabiler Atome.
Zudem erwartet man klassisch ein kontinuierliches Emissionsspektrum.
Man findet aber experimentell isolierte Spektrallinien. Quantenmechanisch
sind im Atom nur die Energien En erlaubt. Wenn ein Elektron einen Übergang Eni → Enf (initial → final) macht, wird die freiwerdende Energie
Eni − Enf in Form eines Photons mit der Energie
~ ω = −Ry(
1
1
− 2) =
2
ni
nf
n2i − n2f Ry
n2i n2f
emittiert. Experimentell wurden anfänglich drei Typen von Übergängen
beobachtet, die Lyman, Balmer und Paschen Serien
Zur Erinnerung
E = h·ν =
⇒
h·c
λ
h·c
E
h · c = 1.2 · 10−6
λ =
eV · m
46CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Lyman Serie (nf = 1) :
h · c n2
(
)
n = 2, 3, . . .
Ry n2 − 1
λ = (9 . . . 12) · 10−8 m
UV : (1 . . . 40) · 10−8 m
λ =
Balmer Serie (nf = 2) :
h · c n2 · 4
(
)
n = 3, 4, . . .
Ry n2 − 4
λ = (3.6 . . . 6.6) · 10−7 m
Sichtbar : (4 . . . 8) · 10−7 m
λ =
Paschen Serie (nf = 3) :
h · c n2 · 9
(
)
n = 4, 5, . . .
Ry n2 − 9
λ = (0.8 . . . 1.9) · 10−6 m
Infrarot = (1 . . . 100) · 10−6 m
λ =
Isotopeneffekt
Für Wasserstoff bzw Deuterium gilt
H : M =Mp
wo M die Masses des Kernes ist. Da das MassenD : M ≈ 2Mp
,
verhältnis
1
me
=
Mp
1836
sehr klein ist, kann man die effektive Masse mred schreiben als
mred ≈ m(1 −
m
)
M
Die Berücksichtigung der Kernbewegung führt also nur zu geringen
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
47
Modifikationen der Energie (2.23)
me · e2
me
)
(1 −
2
2~ · n
M
Ry
me
)
=
(1 −
2
n
M
E =
⇒
∆En ≡ EnD − EnH =
=
Ry me
me
−
)
(
2
n Mp 2Mp
Ry
me
·
2
n
2Mp
| {z }
2,7·10−4
◦
∆λ = 2, 7 · 10−4 ⇒ |∆λ| = O(1 A)
⇒ λ .
Weitere Korrekturen zum Wasserstoffspektrum rühren von relativistischen
◦
Effekten her. Diese Korrekturen sind von der Ordnung O(0.1 A). Weitere
Details sind Inhalt der Atom- und Molekülphysik.
2.5.7
Wasserstoff-Wellenfunktion
Aus (2.22) hat der letzte nicht verschwindende Term in der Reihe ((2.15))
den Index µ∗ = n − l − 1. Die Reihe lautet somit
F (r) = r
l+1
n−l−1
X
cµ r µ
.
(2.25)
µ=0
Die Funktion F (r) ist somit ein Polynom n-ten Grades.
Die Wasserstoff-Wellenfunktion lautet
Ψnlm (r) =
χnl (r) m
Yl (θ, ϕ)
r
.
Wir fassen nun die Ergebnisse für den Radialteil der Wellenfunktion zusammen, wobei wir nun alle Quantenzahlen explizit berücksichtigen (cf. (2.15),(2.18))
n−l−1
X
χnl (r)
−γr l
=e
r
cµ r µ
Rnl (r) =
r
µ=0
(2.26)
Z
na0
(2.27)
γ=
.
48CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Eine detailiertere Rechnung (details in: Sec. A.12 ) führt zum
Ortsanteil der Wasserstoff-Wellenfunktionen
χnl (r)
= (2γ)3/2
r
(n − l − 1)!
2n(n + l)!
1/2
(2γr)l e−γr L2l+1
n−l−1 (2γr)
1/2
(n + l)!
χnl (r)
1
3/2
= (2γ)
(2γr)l e−γr
r
2n(n − l − 1)!
(2l + 1)!
∗
.
(2.28)
1 F1 (l + 1 − n, 2l + 2; 2γr)
Normierung
Die Normierung der Wellenfunktion ist so gewählt, daß
2
Z ∞
Z ∞
χnl (r)
2 2
r2 dr = 1
Rnl (r) r dr =
r
0
0
.
(2.29)
Der Faktor r2 stammt vom Volumenelement d3 r = r2 drdΩ und stellt sicher,
daß die Wellenfunktionen zusammen mit dem Winkelanteil (cf. (1.52)) orthonormal sind:
Z
|Ψnlm (r, θ, ϕ)|2 r2 dr d cos θ dϕ = 1
Spezialfälle
Interessant im Vergleich mit der Bohr-Sommerfeld-Theorie ist der Fall
mit maximalem Drehimpuls l = n − 1 (siehe (2.26)). In diesem Fall ist
1 F1 (0, 2n; r) = 1 und der Radialanteil der Wellenfunktion vereinfacht sich
zu
1/2
χn,n−1 (r)
1
3/2
= (2γ)
(2γr)n−1 e−γr
.
r
2n(2n − 1)!
Die radiale Wahrscheinlichkeitsdichte ist
p(r) = |χn,n−1 (r)|2 ∝ (2γr)2n e−2γr = e−2γr+(2n) ln(2γr)
.
Sie hat das Maximum bei r = nγ = n2 a0 . Diese Werte stimmen mit denen
der Bohr-Sommerfeld-Theorie überein. Diese Übereinstimmung sollte aber
2.5. WASSERSTOFF UND H-ÄHNLICHE PROBLEME
49
nicht überbewertet werden, da die Elektronen nicht auf klassischen stationären Bahnen umlaufen.
Mit der Wellenfunktion Ψnlm können folgende, oft benötigten Erwartungswerte
berechnet werden
a0
2
3n − l(l + 1)
(2.30a)
hri =
2Z
a20 n2
2
2
hr i =
5n + 1 − 3l(l + 1)
(2.30b)
2Z 2
1
Z
h i= 2
.
(2.30c)
r
n a0
Für spätere Rechnungen benötigen wir die Grundzustandswellenfunktion, i.e. n = 1, l = m = 0.
G RUNDZUSTANDSWELLENFUNKTION DES H- ÄHNLICHEN ATOMS
ψ100 (r, θ, ϕ) =
Z3
a30 π
1/2
e−Zr/a0
.
(2.31)
The first few radial wave functions Rnl (normalized according to (2.29))
are
32
Z
(2.32)
R10 (r)
=2
e−Zr/a0
a0
3 Z 2
Zr
R20 (r)
=2
1−
e−Zr/(2a0 )
2a0
2a0
3
1
Z 2 Zr −Zr/(2a0 )
R21 (r)
=√
e
a0
3 2a0
50CHAPTER 2. SCHRÖDINGER EQUATION IN A CENTRAL POTENTIAL
Chapter 3
Erweiterungen und
Anwendungen
3.1
Main results/goals in this chapter (until Sec. 3.4)
• We will consider here two simple models for bondings which often
occur in chemistry: the covalent and the van-der-Waals bonding.
• Althoug the system are simple, they are complex enough so that they
cannot be treated analytically, therefore we will have to resort to approximations.
• In the first case of a covalent bonding we will consider a system
of two protons sharing one electron. The first approximation BornOppenheimer consists in initially neglecting the dynamics (i.e. the
momentum) of the two much heavier protons so that one is left
with a single-particle problem.
• The problem left is still complex since it is not central symmetric.
Again one carries out a variational Ansatz by choosing a physically
motivated form of the wave function (see (3.2) with (3.3)).
• The distance between the protons is obtained by minimizing the
energy
• The final results is a very good approximation (in comparison with
experiment) for the binding energy of the molecule and for the equilibrium distance between the protons
51
52
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
3.2
Kovalente Bindung
1
In Molekülen und Festkörpern gibt es verschiedene Arten von Bindungsmechanismen. Einer hiervon ist die kovalente Bindung. Hierbei teilen sich benachbarte Atome ein Elektron. Durch diesen Teilchenaustausch kommt es
zu einer attraktiven Wechselwirkung. Im Gegensatz hierzu gibt es auch
die ionische Bindung, bei der sich zwei anfangs neutrale Atome so beeinflussen, daß ein energetisch günstigerer Zustand entsteht, wenn ein Atom
ein Elektron an das andere abgibt. Hierbei entstehen unterschiedlich geladene
Ionen, die sich nun elektrostatisch anziehen. Bei der kovalenten Bindung
wird kein Elektron abgegeben, sondern die beiden Atome teilen sich ein Elektron. Am einfachsten ist dieser Effekt am ionisierten H+
2 zu verstehen.
3.2.1
Das H+
2 Molekül.
Das H+
2 Molekül besteht aus zwei einfach positiv geladenen Atomkernen
der Masse M und einem Elektron der Masse m. Selbst dieses zweiatomige
Molekül ist ein relativ komplexes System, das exakt nur mit Mühe zu lösen
ist. Man kann dieses Problem allerdings sehr gut mit Näherungsverfahren
behandeln. Hier werden wir die Variationsrechnung anwenden.
me
< 10−3 , bewegen sich die
Wegen des großen Massenunterschiedes, m
p
Elektronen sehr viel schneller als die Atomkerne, und man kann die beiden Bewegungen entkoppeln. D.h., wir halten zunächst die beiden Atomkerne an den Positionen R1 und R2 fest und lösen die Schrödingergleichung in dem daraus resultierenden Potential. Daraus erhalten wir die
möglichen Energien des elektronischen Systems En (R) als Funktion des
Abstandes R = R1 − R2 der beiden Kerne. Diese Energie stellt für die
Dynamik der Kerne den elektronischen Beitrag zum Potential dar, in dem
sie sich bewegen.
V eff (R) = V Kern−Kern (|R2 − R1 |) + En (R)
.
Diese Born-Oppenheimer-Näherung ist insbesondere in der Festkörperphysik
weitverbreitet und extrem zuverlässig. Da die Kerne eine sehr viel größere
Masse als die Elektronen besitzen, kann die Bewegung in guter Näherung
bereits klassisch behandelt werden
M R̈ = −∇R V eff (R)
1
.
Wir werden von nun an weniger streng sein mit der Benutzung vonˆfür Operatoren.
Es wird nur dort verwendet, wo es sonst zu Verwechslung führen könnte.
3.2. KOVALENTE BINDUNG
53
Diese Näherung wird in der Quantenchemie und in der Festkörperphysik
verwendet, um Gleichgewichtskonfigurationen von Molekülen und deren
Schwingungen, sowie Oberflächengeometrien und Phononmoden zu bestimmen. Diese Vorgehensweise geht auf R. Car und M. Parinello zurück.
Hier werden wir für H+
2 den Grundzustand und den elektronischen Beitrag
zum effektiven Kernpotential bestimmen. Der Hamiltonoperator hat die
Figure 3.1: Positionierung der Atome und des Elektrons im H2+ Molekül.
Gestalt
H = −
d2
d2
e2 e2 e2
~2 d2
( 2 + 2 + 2) − − +
2m dx
dy
dz
r1 r2
R
,
wobei rα und R für die Längen der Vektoren r α und R stehen. Im weiteren geben wir alle Längen in Einheiten von a0 an, d.h. rα = r̃α · a0 und
R = R̃ · a0 . In den dimensionslosen Größen x̃, R̃ lautet der Hamiltonoperator dann
H
a0
1)
2)
e2
a0
~2
2m a20
e2
1
1
1
~2 ˜ 2
∇
+
(− − + )
2
2m a0
a0 r̃1 r̃2 R̃
2
~
=
⇒
me2
=−
= 2Ry
=
. . . Rydberg
~2 m2 e4
me4
=
= 1Ry
2m~4
2~2
54
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
Daraus folgt
˜2
∇
1
1
1
H = 2Ry(−
− − + )
2
r̃1 r̃2 R̃
In der weiteren Rechnung werden wir Energien in Einheiten von 2Ry (
Rydberg) angeben, H = H̃ 2Ry , womit sich der Hamilton-Operator noch
weiter vereinfacht Diese Einheiten nennt man atomare Einheiten.
H AMILTON -O PERATOR IN ATOMAREN E INHEITEN
H̃ = −
˜2
∇
1
1
1
− − +
2
r̃1 r̃2 R̃
.
(3.1)
In atomaren Einheiten sind die numerischen Werte von e, ~ und me Eins, d.h.
diese Naturkonstanten können überall weggelassen werden. Wir werden
im weiteren die Tilden weglassen und davon ausgehen, daß alle Größen
in atomaren Einheiten vorliegen. Das ist eine in der theoretischen Physik
weitverbreitete, sinnvolle Vorgehensweise, da in diesen Einheiten alle Größen
von der Ordnung 1 sind.
Um eine Vorstellung zu bekommen, wie die Variationsfunktion aussehen
könnte, betrachten wir zunächst R >> a0 . Dann gibt es zwei Möglichkeiten:
Entweder ist das Elektron beim Kern 1 oder beim Kern 2. Die Wellenfunktionen sind jeweils die Grundzustandsfunktionen des Wasserstoffatoms
(2.31)
Da beide Möglichkeiten vorliegen können, setzen wir an
ψ(r) = c1 ψ1 (r) + c2 ψ2 (r)
.
(3.2)
Dieser Ansatz ist mit leichten Modifikationen auch für kleine Abstände
sinnvoll. Da für R → 0 die exakte Grundzustandswellenfunktion die des
H-ähnlichen Atoms mit Z = 2 ist, verwenden wir den Ansatz (3.2) mit 2
ψα (r) =
Z3
π
12
e−Z rα
(3.3)
und den Variationsparametern c1 , c2 und Z, deren Werte aus der Minimierung
2
Es sei daran erinnert, daß in atomaren Einheiten a0 = 1 gilt.
3.3. OPTIMIERUNG DER (VARIATIONS-)WELLENFUNKTION IN EINEM TEILRAUM55
der Energie folgen.
Z3 1
Z3 1
ψ(r) = c1 · ( ) 2 e−Z r1 +c2 · ( ) 2 e−Z r2
| π {z
| π {z
}
}
ψ1 (r)
ψ2 (r)
Wir betrachten zunächst das Teilproblem
min E(c1 , c2 , Z) = min
c1 ,c2
c1 ,c2
hψ|Ĥ|ψi
hψ|ψi
zu festem Z. Das hieraus resultierende Eigenwertproblem trifft man in
der Q.M. sehr häufig an und es soll daher etwas allgemeiner behandelt
werden.
3.3
Optimierung der (Variations-)Wellenfunktion
in einem Teilraum
Wir entwickeln die gesuchte Wellenfunktion ψ nach einer beliebigen “Basis”. Im Gegensatz zu den bisher verwendeten Basissätzen, muß die Basis hier weder vollständig, noch normiert noch orthogonal sein 3 . Allerdings
müssen wir linear abhängige Vektoren zuvor eliminieren.
|ψi =
N
X
i=1
ci |ψi i
Die bestmögliche Linearkombination für den Grundzustand finden wir,
indem wir die Energie bzgl. der Entwicklungskoeffizienten minimieren
Hij
P ∗ z }| {
hψ|Ĥ|ψi
i,j ci hψi |Ĥ|ψj i cj
= P ∗
E(c) =
hψ|ψi
i,j ci hψi |ψj i cj
| {z }
Sij
∂
E=
∂c∗i
3
P
Hij cj
hψ|Ĥ|ψi X
!
·
Sij cj = 0
−
2
hψ|ψi
hψ|ψi
j
j
Wir hatten bisher unter einer Basis immer eine vollständige Basis verstanden. Das ist
auch die gängige Definition. Man findet aber in der Literatur, insbesondere in der Quantenchemie, auch den Begriff Basis für einen Satz von Funktionen, nach dem entwickelt
wird, der keine vollständige Basis dartellen muß.
56
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
Multiplikation mit hψ|ψi liefert
X
j
Hij cj −
hψ|Ĥ|ψi X
!
·
Sij cj = 0
hψ|ψi
| {z } j
=E
V ERALLGEMEINERTES E IGENWERTPROBLEM
Hc=ESc
(3.4)
Die stationären Punkte der Funktion E(c) erhält man aus dem verallgemeinerten Eigenwertproblem (3.4). Ein möglicher Lösungsweg besteht in
der Löwdin-Orthogonalisierung
1
x := S 2 c
1
1
HS − 2 x = ES 2 x
|S
− 21
(3.5)
− 12
HS } x = Ex
{z
H̃
Die Matrix H̃ ist auch hermitesch und wir haben ein normales Eigenwertproblem vorliegen. Wie hier: Sec. A.14 gezeigt, entspricht die LöwdinOrthogonalisierung der orthonormierung einer beliebigen, nicht orthonormalen Basis. Genau das wurde in (3.5) ausgenutzt. Wir hätten also genausogut die Basis vorab orthonormieren können. In dem Fall wäre S = 11
gewesen. In diesem Fall ist am einfachsten zu erkennen, daß das Ergebnis
(3.4) besagt, daß die niedrigste Energie in einem Teilraum des vollen Hilbertraums dadurch erreicht wird, daß in dem Teilraum das Eigenwertproblem gelöst
wird.
3.4. BACK TO THE VARIATIONAL TREATMENT OF H+
2
3.4
57
Back to the variational treatment of H+
2
Für das H+
2 Problem bedeutet obige Überlegung, daß wir
hψ1 |H|ψ1 i = hψ2 |H|ψ2 i
hψ1 |H|ψ2 i = hψ2 |H|ψ1 i
S11 = S22 = hψ2 |ψ2 i
S12 = S21 = hψ1 |ψ2 i
=: H1
=: H2
=1
=: S
benötigen, wobei die Symmetrie des Problems ausgenutzt wurde. Es
bleibt zu lösen (cf. (3.4))
c1
1 S
c1
H1 H2
·
=E
(3.6)
c2
c2
S 1
H2 H1
Clearly the Hamilton operator commutes with the parity operator
P |ψ1 i = |ψ2 i
P |ψ2 i = |ψ1 i
So that the eigenvectors of H (3.6) are also eigenvectors of P , i.e. even
(gerade):
ψ1 (r) + ψ2 (r)
H1 + H2
ψb (r) = p
Eb =
(3.7)
1+S
2(1 + S)
or odd (ungerade).
ψa (r) =
ψ1 (r) − ψ2 (r)
p
2(1 − S)
Ea =
H1 − H2
1−S
(3.8)
Wir überprüfen leicht, daß diese Zustände ortho-normal sind. Aus Gründen, die gleich klar werden, nennt man (3.8) den antibindenden und (3.7) den
bindenden Zustand. Die Zustände sind in Abbildung (3.2) dargestellt. Man
nennt die so gewonnenen Wellenfunktionen auch Molekular-Orbitale.
Matrixelemente von (3.1)
Wir wissen, daß ψ1 (r) die exakte Grundzustandsenergie (cf. (2.23)) eines
H-ähnlichen Atoms mit Kernladungszahl Z ist
∇2
Z
Z2
−
−
ψ1 (r) = − ψ1 (r)
2
r1
2
.
58
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
Figure 3.2: Wellenfunktionen mit gerader und ungerader Parität.
2
− Z2 ist die Energie des H-ähnlichen Atoms in atomaren Einheiten. Die
Berechnung der Matrixelemente läßt sich damit vereinfachen.
∇2
Z
Z −1
1
1
H=−
−
+
− +
2
r1
r1
r2 R
Z −1
Z2
1
1
+
Hψ1 (r) = −
− +
ψ1 (r)
2
r1
r2 R
H1 = hψ1 |H|ψ1 i
1
Z2
1
1
= (−
+ ) hψ1 |ψ1 i + (Z − 1)hψ1 | |ψ1 i − hψ1 | |ψ1 i
2
R | {z }
r1
r2
{z
}
|
=1
(3.9)
(3.10)
.
(3.11)
:=K
Der Erwartungswert
1
hψ1 | |ψ1 i =
r2
Z
|ψ1 (r)|2
1 3
dr
r2
mit (3.3) entspricht (minus) der klassischen Coulomb-Wechselwirkungsenergie
des zweiten Kernes im Coulombfeld des ersten Elektrons. Das Ergebnis
kann man mit Hilfe des Gausschen Satzes herleiten here: Sec. A.13 :
1
1
1 − (1 + ZR)e−2ZR
hψ1 | |ψ1 i =
r2
R
Dieser Beitrag fällt wie
1
R
ab.
Der zweite Erwartungswert hψ1 | r11 |ψ1 i ist die Coulomb-Wechselwirkungsenergie
des ersten Kernes im Coulombfeld des ersten Elektrons, also das obere
Ergebnis mit R → 0:
1
hψ1 | |ψ1 i = Z .
(3.12)
r1
3.4. BACK TO THE VARIATIONAL TREATMENT OF H+
2
59
Der Beitrag aus diesen beiden Termen ergibt (cf. (3.10))
K = Z(Z − 1) −
H1 = −
1
(1 − (1 + ZR)e−2ZR )
R
Z2
1
+ +K
2
R
(3.13)
Nun wenden wir uns der Berechnung von H2 zu (cf. (3.9))
H2 = hψ2 |H|ψ1 i = (−
Z2
1
1
1
+ ) · S + (Z − 1)hψ2 | |ψ1 i − hψ2 | |ψ1 i (3.14)
2
R
r1
r2
|
{z
}
:=A
1
1
hψ2 | |ψ1 i = hψ2 | |ψ1 i =
r1
r2
Z
ψ1 (r)ψ2 (r) 3
dr=
r1
Z p
ρ1 (r)ρ2 (r) 3
dr
r1
.
hψ2 | r12 |ψ1 i ist die sogenannte Austauschwechselwirkung, zu der es klassisch
kein Pendant gibt. Die Austauschwechselwirkung ist ein Interferenzeffekt
und ist proportional zum Überlapp der Wellenfunktionen und daher fällt
wie e−ZR ab. Die Beiträge der Austauschwechselwirkung sind
A = (Z − 1)hψ2 |
1
1
|ψ1 i − hψ2 | |ψ1 i = (Z − 2) · Z(1 + RZ)e−RZ
r1
r2
.
Wir benötigen noch das Überlappintegral S
Z
1
S = ψ1 (r)ψ2 (r) d3 r = (1 + ZR + (ZR)2 )e−ZR
3
Insgesamt mit (3.13) und (3.14) lauten die bonding und antibonding Energien (3.7) und (3.8) zu festem Variationsparameter Z und festem Kernabstand R:
Eb/a (R, Z) =
H1 ± H2
Z2
1
K ±A
= (−
+ )+
1±S
2
R
1±S
Im oberen Teil von Fig. (3.3) ist Z opt (R) als Funktion von R für den bindenden (durchgezogene Kurve) und den anti-bindenden (gestrichelte Kurve) Zustand gezeigt.Der untere Teil enthält E opt (R, Z opt (R)) als Funktion von
R für den bindenden (durchgezogene Kurve) und den anti-bindenden
(gestrichelte Kurve) Zustand. Zusätzlich ist noch der Austauschbeitrag A(r)
als dünn gestrichelte Kurve gezeigt.
60
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
Figure 3.3:
• Zu festem R muß Z opt (R) für beide Lösungen a/b aus min Eb/a sepaZ
opt
opt
rat ermittelt werden. ⇒ Zb/a
(R). In Abbildung (3.3) ist Zb/a
(R) aufgetragen. Man erkennt, daß nur der bindende Zustand gegen den richtigen Wert für R → 0 strebt.
• Im unteren Teil von Abbildung (3.3) ist Eα (R, Zαopt (R)) für α = a und
α = b aufgetragen. Nur die bindende Kombination ψb = √ψ1 +ψ2
2(1+S)
liefert eine Gesamtenergie (Emin = −0.586a.E.), die unter der des
dissoziierten Moleküls E dis = E(H + H+ ) = −1Ry = − 21 a.E. liegt.
Die Differenz ist die Bindungsenergie
+
∆E = E(H+
2 ) − E(H + H ) = 0.086a.u. = 0.086 · 27.2 eV = 2.35 eV
Experimentell findet man eine Bindungsenergie von 2.8 eV . Man beachte
aber, daß Eges = −16.95 eV , d.h. der relativer Fehler der Gesamtenergie beträgt nur ≈ 3%. Das ist für diese grobe Näherung der Wellenfunktion
sehr gut.
3.4. BACK TO THE VARIATIONAL TREATMENT OF H+
2
61
◦
◦
• Der Gleichgewichtsabstand Ropt = 2.00a.E. = 2.00 ∗ 0.529 A= 1.06 A
stimmt mit dem experimentell gemessenen Wert überein.
• Hält man die Kernladungszahl bei Z = 1 fest, so erhält man
– Ebind = 1.8 eV
◦
– Ropt = 1.3 A
.
Die Freigabe des Kernladungszahl als Variationsparameter führt zu
einem deutlich besseren Ergebnis.
• Die Ursache der Bindung im Fall von ψb ist eine Anhäufung von
Elektronendichte zwischen den Kernen (siehe Abbildung(3.2)). Hier
profitiert die Elektronenladung am meisten von der anziehenden Energie beider Kerne.
• Das Verfahren, das wir verwendet haben um eine molekulare Wellenfunktion aus atomaren Wellenfunktionen aufzubauen heißt Linear
Combination of Atomic Orbitals (LCAO) und wird vielseitig in der¯ Quan¯
¯
tenchemie
und¯ der Festkörperphysik
eingesetzt.
3.4.1
Muonisch katalysierte Fusion
Man beachte, daß der Gleichgewichtsabstand der Protonen gegeben ist durch
R∗ = 2.0a.E. = 2.0
~
me2
Hätten wir anstelle der Elektronen die 210-mal so schweren Muonen verwendet, (ebenfalls Spin 21 , Ladung -e) so hätten wir für den Gleichgewichtsabstand
Ropt/Elektron
Ropt/Muon =
210
gefunden.
Es ist folgender Effekt denkbar
Deuterium+ + Deuterium+ + µ
Fusion!
−→ Tritium + p + 4.0M eV
Es wurden solche Reaktionen schon experimentell beobachtet. Deuterium
ist im Meer zur Genüge vorhanden.
62
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
Im Gegensatz zur konventionellen Kernfusion, bei der die DeuteriumAtome durch große kinetische Energie (Hitze) dazu gebracht werden, die
Coulomb-Barriere zu überwinden, handelt es sich bei der muonisch katalysierten
Fusion um kalte Fusion.
Muonen sind allerdings instabil4 und kommen natürlich nicht vor (außer
in kosmischen Strahlen). Sie müßten erst in Beschleunigern erzeugt werden. Bislang ist diese Art der Fusion nicht machbar.
3.5
Van-der-Waals-Wechselwirkung
• In the second part we consider a simple (yet unrealistic) model to understand the van der Waals interaction. This consists of two Hidrogen atoms at large distances.
• Here the first order energy correction is zero due to the fact that the
ground state of the two atoms doesn’t have an electric dipole.
• To find a nonzero result we need to go to second order, which consists in a first order change of the wave function leading to the formation of a dipole.
Eine weitere interessante Anwendung der zeitunabhängigen Störungstheorie stellt die langreichweitige van-der-Waals-Wechselwirkung zwischen
neutralen Atomen dar. Diese aus der Quantenmechanik resultierende Wechselwirkung ist zwar bei geladenen Atomen ebenfalls vorhandenen, wird
aber dort von der Coulomb-Wechselwirkung dominiert.
Wir betrachten hier den Fall zweier Wasserstoffatome im Grundzustand.
Wie beim H+
2 Molekül und mit derselben Begründung verwenden wir
die Born-Oppenheimer-Näherung. Die Atomkerne ruhen im festen großen
Abstand r auf der z-Achse (siehe Abbildung (3.4)). r 1 ist der Vektor vom
ersten Proton zu „seinem Elektron”, r 2 ist der Vektor vom zweiten Proton
zu „seinem Elektron”. Der Hamilton-Operator in der Ortsdarstellung für
4
Halbwertszeit = 2.2 µsec
3.5. VAN-DER-WAALS-WECHSELWIRKUNG
63
Elektron
2
000
111
111
000
000
111
000
111
Elektron 1
111
000
000
111
000
111
000
111
r2
11
00
r1
+
Kern 1
+
11
00
r
Kern 2
z
Figure 3.4: Geometrische Anordnung zur Berechnung der van-der-WaalsWechselwirkung zweier H-Atome.
die beiden Kerne mit Elektronen lautet in atomaren Einheiten
∇21
1
∇22
1
H = (−
− ) + (−
− )
2
r1
2
r2
1
1
1
1
+
+
−
−
r
|r + r 2 − r 1 | |r + r 2 | |r − r 1 |
|{z}
|
{z
} | {z } | {z }
K1 −K2
E1 −E2
K1 −E2
=:H0
=:H1 (r, r 1 , r 2 )
.
K2 −E1
(3.15)
Hierbei steht Kα für Kern α und Eα für Elektron α. Da uns nur große
Kernabstände interessieren, ist H1 klein und kann störungstheoretisch behandelt werden.
Die Schrödingergleichung von H0 kann leicht gelöst werden, da H0 =
H01 + H02 aus identischen Hamilton-Operatoren für die beiden getrennten
H-Atome aufgebaut ist. Daraus folgt die Grundzustandswellenfunktion
des durch H0 beschriebenen Systems
ψ0 = ψ 1 (r 1 )ψ 2 (r 2 )
H0α ψ α (r α ) = Eψ α (r α )
1
1
ψ0 = ( π1 ) 2 e−r1 ( π1 ) 2 e−r2
Die hochgestellten Indizes an den Wellenfunktionen geben an, bei welchem
Atomkern die Funktion lokalisiert ist. Die Energiekorrektur erster Ord-
64
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
nung E 1 (r) ≡ hψ0 |H1 |ψ0 i ergibt also
Z
1
E (r) = ρ(r 2 )ρ(r 1 )H1 (r, r 1 , r 2 )d3 r 1 d3 r 2
Wir benötigen folgendes Integral was nicht anderes ist als das CoulombPotential des Elektrons im Grundzustand des H-Atoms: Die Herleitung ,
mittels des Gausschen Satzes, haben wir bereits hier: Sec. A.13 gemacht.
Z
1
1
1
ρ(r1 ) d3 r1 = − e−2r 1 +
.
(3.16)
Ṽ (r) =
|r − r1 |
r
r
(3.16) übergeht für große Abstände in das Coulombpotential einer Punktladung. Wie eingangs erwähnt interessiert man sich bei der van-der-WaalsWechselwirkung nur für große Abstände, d.h. Abstände sehr groß im Vergleich zur Ausdehnung der Atome (r 1). Wir können also für das
Coulomb-Potential der Ladungsverteilung
Ṽ (r) ≈
1
r
verwenden, bis auf Korrekturen, die exponentiell klein in r sind.
Die Beiträge zur Energiekorrektur erster Ordnung , die von der Kern-Kern
und Kern-Elektron-Wechselwirkung herrühren (3.15), laufen alle auf Integrale vom Typ Ṽ heraus und liefern bis auf Vorzeichen einen Beitrag 1/r
(1)
1
r
1
=−
r
1
=−
r
EK1 ,K2 =
(1)
EK1 ,E2
(1)
EK2 ,E1
Die Elektron-Elektron-Wechselwirkung aus (3.15) enthält zunächst ein ZweizentrenIntegral, das unter Ausnutzen von Ṽ in zwei Schritten berechnet werden
kann
Z
Z
1
(1)
3
3
EE1 E2 :=
d r2
d r1
ρ(r1 ) ρ(r2 )
|r + r 2 − r 1 |
|
{z
}
Ṽ (|r+r 2 |)
≈
Z
d3 r2
1
1
ρ(r2 ) = Ṽ (r2 ) ≈
|r + r 2 |
r
3.5. VAN-DER-WAALS-WECHSELWIRKUNG
65
Die berechneten Beiträge zu E (1) (r) addieren sich zu Null, bis auf Terme,
die exponentiell (e−2r ) mit dem Abstand r abfallen. Das geschiet aufgrund
der Kugelsymmetrie der Grundzustands-Wellenfunktion.
Nichtverschwindende Beitrage, genauer gesagt Beitrage, die nicht exponentiell abfallen, kommen aus den führenden Beiträge der zweiten Ordnung
Störungsrechnung. Dazu benötigen wir H1 (r, r 1 , r 2 ) ((3.15)) zur führenden Ordnung in ri /r. Notice that H1 describes the interaction of the electron
and nucleus “N.2” in the electric potential ϕ1 of the “N.1” (or vice-versa). To
leading order r1 /r (see Elektromagnetische Felder, Chap. 1) this is the
corresponding dipole field, with the dipole d1 = −r 1 . On the other hand,
the potential energy of the system N.2 in the potential ϕ1 to leading order
in r2 /r is given by (see Elektromagnetische Felder, Chap. 2)
H1 = −E 1 · d2
where E 1 is the electric field produced by the charges N.1, and d2 = −r 2
is the dipole of N.2. Using the well known expression for the electric field
of a dipole, one gets (in cgs)
x1 x2 + y1 y2 − 2z1 z2
1
1
1
2
)
=
+
O(
),
3(r
·
r)(r
·
r)/r
−
r
·
r
+
O(
1
2
1
2
r3
r4
r3
r4
where in the last term we have taken for definiteness r = rez , i.e. along
the z-axis.
Die zweite Ordnung Störungstheorie liefert
0
X
| ψn1 1 ψn2 2 H1 |ψ01 ψ02 i |2
∆E = −
=
E
+
E
−
2E
n
n
1
1
2
n1 ,n2
0
1 X | ψn1 1 ψn2 2 x1 x2 + y1 y2 − 2z1 z2 |ψ01 ψ02 i |2
= 6
r n ,n
En1 + En2 − 2E1
H1 = −
1
2
Der Term n1 = n2 = 1 ist von der Summe ausgenommen, und die Energiekorrektur ist, wie bereits allgemein gezeigt, negativ.
VAN - DER -WAALS -W ECHSELWIRKUNG
EvdW = −c
1
r6
.
Die wesentlichen Merkmale der van-der-Waals-Wechselwirkung sind
66
CHAPTER 3. ERWEITERUNGEN UND ANWENDUNGEN
• Die Wechselwirkung ist attraktiv F (r) = −∇E =
−6c rr8
• Das Potential fällt wie 1/r6 ab.
– Es stammt aus zwei Dipolen, die je wie 1/r3 abfallen.
– Da ψ0 kugelsymmetrisch ist, sind es induzierte Dipole, die aus
virtuellen Anregungen zu den angeregten Zuständen entstehen.
• Die vdW-Wechselwirkung wird bei neutralen Atomen beobachtet.
Sie existiert zwar auch für geladene Atome, wird hier aber vom MonopolBeitrag der Coulomb-Wechselwirkung dominiert.
Chapter 4
Several degrees of freedon and
the product space
Viele physikalischen Probleme bestehen aus Teilsystemen bzw. mehreren
Freiheitsgraden. Z.B. bestehen Atome aus Atomkern und Elektronen. Elektronen wiederum besitzen eine Ortskoordinate und einen Spin-Freiheitsgrad.
Es ist zweckmäßig, von den Hilberträumen H1 , H2 , · · · der einzelnen Teilsysteme/Freiheitsgrade auszugehen und daraus den Hilbertraum des gesamten
Systems aufzubauen.
Beispiel:
Elektron: HR für den Ort (gebundener Zustand): R
Menge aller Funktionen f (x) mit |f (x)| d3x < ∞
HS für Spin:
Vektorraum, der durch die Eigenzustände von S z
der Spin-Operatoren aufgespannt wird.
4.1
Das Tensor-Produkt
Wir greifen aus den Vektoren zweier verschiedener Hilberträume 1 und
2 je einen Vektor |Φi1 und |Ψi2 heraus und bilden aus ihnen formal ein
Produkt (kein Skalarprodukt und auch kein äußeres Produkt). Für das
Produkt schreibt man (es gibt unterschiedliche Notationen):
|Φ, Ψi := |Φi1 ⊗ |Ψi2 := |Φi ⊗ |Ψi := |Φi|Ψi
.
(4.1)
Das soll einfach bedeuten: Das Teilsystem 1 befindet sich im Zustand |Φi
das Teilsystem 2 befindet sich im Zustand |Ψi. Die Indizes 1 und 2 werden
67
68CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
letztendlich eingespart und durch die Reihenfolge verdeutlicht. Der “bra”
Vektor schreibt man als
hΦ, Ψ| := hΦ| ⊗ hΨ|
.
(4.2)
Dieses Produkt, das direktes oder Tensor- Produkt genannt wird, soll wieder
ein Vektor aus einem linearen Vektorraum sein, der die Physik zusammengesetzter Systeme beschreiben soll. Offensichtlich liegt der Vektor
weder in H1 noch in H2 . Er liegt in einem sogenannten Produktraum
H = H1 ⊗ H2
.
H wird von allen Vektoren (4.1) und deren Linearkombinationen aufgespannt.
Aus der Distributivität von |Φi soll folgen:
Wenn: |Φi = a|Φ1 i + b|Φ2 i
|Φi ⊗ |Ψi = a|Φ1 i ⊗ |Ψi + b|Φ2 i ⊗ |Ψi
Dasselbe gilt für |Ψi.
Def. 4.1 (Skalarprodukt im Produktraum). Das Skalarprodukt zweier Produktvektoren ist definiert als das Produkt der Skalarprodukte in den beiden Teilräumen:
hΦA , ΨA |ΦB , ΨB i = hΦA |ΦB i hΨA |ΨB i
(4.3)
Aus der Linearität der Quantenmechanik sind auch Linearkombinationen
von Produktzustände (4.1) erlaubt.
a |ΦA ΨA i + b |ΦB ΨB i
(4.4)
Dieser beschreibt in der Regel einen Verschränkten Zustand.
verschränkter Zustand: kann nicht als direktes Produkt der Gestalt |Φi ⊗
|Ψi geschrieben werden.
4.2
Vollständige Basis im Produkt-Raum
Wenn {|Φk i, k = 1, 2, · · · } eine vollständige Basis in {H1 } und {|Ψl i, l =
1, 2, · · · } einen vollständige Basis in H2 ist, dann ist
{|Φk i ⊗ |Ψl i, l = 1, 2, · · · , k = 1, 2, · · · } vollständige Basis des Produktraumes
H1 ⊗ H2 .
Im Produktraum numerieren die Indextupel (k,l) die Basisvektoren durch.
Wenn die Dimension von Hα durch Nα gegeben ist, dann ist die Dimension des Produktraumes H1 ⊗ H2 gleich N1 · N2 . Dieses Konzept gilt auch
auf ∞-dimensionale Teilräume, und kann leicht auf kontinuierlichen Räumen verallgemeinert werden.
4.3. ORTHONORMIERUNG IM PRODUKT-RAUM
4.3
69
Orthonormierung im Produkt-Raum
Aus den orthonormierten Vektoren der Teilräume folgt mit (4.3) sofort
hΦk , Ψl |Φk0 , Ψl0 i = hΦk |Φk0 ihΨl |Ψl0 i = δk,k0 δl,l0
Wenn |Φk i, |Ψl i vollständige Orthonormalbasen in H1 , H2 sind, dann gilt
Z X
Z
X
|Φk , Ψl ihΦk , Ψl | dk dl = 11
k
Z
X
l
soll entweder eine Summe oder ein Integral über die Variable k darstellen,
k
je nachdem, ob es sich um eine diskrete oder kontinuierliche Größe handelt.
Ein beliebiger Vektor |Ωi im Produktraum läßt sich entwickeln als
Z X
Z
X
hΦk , Ψl |Ωi |Φk , Ψl i dk dl
|Ωi =
k
l
Beispiele:
• Elektron mit Spin:
|Ψi ist die räumliche Wellenfunktion: hx|Ψi = Ψ(x)
|χi beschreibt den Spin (Basis: S z -Eigenzustände |σi) 1
|Ψ, χi = |Ψi ⊗ |χi beschreibt den Gesamtzustand eines Elektrons mit
räumliche Wellenfunktion Ψ(x) und spin χ.
Die Gesamt-Wellenfunktion ist in diesem Produkt-Zustand
ψ(x, σ) := hx, σ|Ψ, χi = hx|Ψi hσ|χi = Ψ(x) · χ(σ)
Eine mögliche Wellenfunktion ist ein verschränkter Zustand bestehend aus einer lineare Kombination von Produktzustände:
ψ(x, σ) = ΨA (x)χA (σ) + ΨB (x)χB (σ)
.
• Zwei Elektronen ohne Spin:
|Ψ1 i beschreibt den Ort des ersten Elektrons.
1
Für dieDarstellung
von Spinzuständen gibt es verschiedene äquivalente Notationen:
|±1i, oder ± 21 , oder |±zi, oder |↑i , |↓i. Diese bezeichen die Eigenzustände von S z mit
Eigenwert ±~/2.
70CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
|Ψ2 i beschreibt den Ort des zweiten Elektrons.
|Ψ1 , Ψ2 i beschreibt den Gesamtzustand. (z.B. Basis: |x1 , x2 i)
Die Wellenfunktion ist
ψ(x1 , x2 ) = hx1 , x2 |Ψ1 , Ψ2 i = Ψ1 (x1 ) · Ψ2 (x2 )
Auch hier ist eine verschränkte Wellenfunktion möglich
ψ(x1 , x2 ) = Ψ1 (x1 )Ψ2 (x2 ) + Ψ̃1 (x1 )Ψ̃2 (x2 )
Produktzustände vom |Ψi ⊗ |Φi-Typ beschreiben unkorrelierte Teilchen.
Man nennt diese Vektoren auch elementare Tensoren. Zustände, die nicht so
dargestellt werden können, nennt man “verschränkte” (entangled) Zustände.
Diese entstehen aus der Wechselwirkung zwischen den Teilchen.
Zum Beispiel können sich zwei Spins im Zustand
|Ψi = |+zi ⊗ |−zi + |−zi ⊗ |+zi
befinden. Es ist keines der Teilchen im Eigenzustand der zugehörigen
Operatoren 2 S1z , S2z :
z
z
S1 |Ψi = S1 |+zi ⊗ |−zi + |−zi ⊗ |+zi
−~
~
|+zi ⊗ |−zi +
|−zi ⊗ |+zi
2
2
~
~
=
|+zi ⊗ |−zi−|−zi ⊗ |+zi 6= |Ψi
2
2
=
Unitäre Transformationen lassen sich direkt auf Produkträume übertragen.
Das Konzept des direkten Produktes läßt sich sofort auf eine beliebige
Zahl von Teilräumen erweitern.
|Ψ1 , Ψ2 , · · · , Ψn i = |Ψ1 i ⊗ |Ψ2 i ⊗ · · · ⊗ |Ψn i =
n
N
i=1
|Ψi i
H = H1 ⊗ H2 ⊗ · · · ⊗ Hn
2
Um nicht den Indizes der Teilräume in die Quere zu kommen, werden wir nun die
Bezeichnung der Komponenten von Spin und Drehimpuls als “superscript” angeben,
z.B. S z , J x , L+ , J − , etc.
4.4. OPERATOREN IM DIREKTEN PRODUKTRAUM
4.4
71
Operatoren im direkten Produktraum
Alle bisherigen Überlegungen über lineare Operatoren gelten auch für
Operatoren in Produkträumen. Betrachtet man einen Operator L in den
Basiszuständen
|Φk , Ψl i := |Φk i ⊗ |Ψl i
L(kl; k 0 l0 ) := hΦk , Ψl |L|Φk0 , Ψl0 i
Wenn L nur in H1 wirkt, wird er in H1 ⊗ H2 eingebettet als:
L1 := L ⊗ 11 ⇒ L ⊗ 11 |Φk , Ψl i = L|Φk i ⊗ |Ψl i
.
Analog, wenn der Operator M nur auf Vektoren im Raum H2 wirkt, wird
er im Produktraum wie folgt eingebettet
M2 := 11 ⊗ M ⇒ 11 ⊗ M |Φk , Ψl i = |Φk i ⊗ M|Ψl i
.
Die Matrixelemente von L1 bzw. M2 sind in der Produktraumbasis
L1 (kl; k 0 l0 ) ≡ hΦk , Ψl |L1 |Φk0 , Ψl0 i = hΦk |L|Φk0 i · δll0
M2 (kl; k 0 l0 ) ≡ hΦk , Ψl |M2 |Φk0 , Ψl0 i = δkk0 · hΨl |M|Ψl0 i
Operatoren verschiedener Teilsysteme kommutieren!
[L1 , M2 ] = 0
(4.5)
Für die Matrixelemente des Produktoperators L1 M2 = L ⊗ M gilt:
hΦk , Ψl |L1 M2 |Φk0 , Ψl0 i = hΦk |L|Φk0 i · hΨl |M|Ψl0 i
Als Beispiel betrachten wir die Spin-Bahnkopplung, die bei der relativistischen
Behandlung der Atom-Wellenfunktionen ein wichtige Rolle spielt. Der
Operator, der die Kopplung beschreibt, ist proportional zu S · L, dem
Produkt aus Spin- und Drehimpuls-Vektoroperator. Als Basiszustände für
den Spin können die Eigenvektoren |σi von S z verwendet werden. Eine
mögliche Basis für den Bahnanteil bilden die gemeinsamen Eigenvektoren
|l, mi von L2 und Lz .
72CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
Die elementaren Tensorprodukte sind |l, m, σi = |l, mi ⊗ |σi. In dieser
Basis sind die Matrixelemente des z-Anteils der Spin-Bahnkoppluing
hl, m, σ|Lz S z |l0 , m0 , σ 0 i = hl, m|Lz |l0 , m0 i · hσ|S z |σ 0 i
~
σ δσ,σ0
= ~ m δm,m0 δl,l0 ·
2
4.5
.
Systeme mit zwei Spin 21 Teilchen
Um einen Zustand mit zwei Spin 12 Teilchen zu beschreiben, verwenden
wir die elementare Produktbasis |σ1 i |σ2 i. Zur Erinnerung
Sβz |σiβ =
~
σ|σiβ
2
1 1
3
Sβ2 |σiβ = ~2 ( + 1)|σiβ = ~2 |σiβ
2 2
4
,
Hierbei gibt der Index β an, zu welchem Teilchen der Zustandsvektor
bzw. der Spin-Operator gehört. Die Produktbasis ist vollständig und die
gesamte Physik der beiden Spin-1/2-Teilchen kann darin beschrieben werden.
Solange kein äußeres B-Feld vorliegt ist der Gesamtspin erhalten, d.h.
seine Komponenten
S = S1 + S2
vertauschen mit dem Hamiltonoperator. Dieses gilt aber nicht für die
einzelnen Spinkomponenten S 1 und S 2 : Wechselwirkungen zwischen den
beiden Teilchen können bewirken, daß diese nicht mehr mit dem Hamiltonoperator vertauschen, und deshalb werden die Eigenwerte von den einzelnen z-Komponenten Sαz keine guten Quantenzahlen (Erhaltungsgrößen)
mehr sein. Die z-Komponente des Gesamtspins S z ≡ S1z + S2z hingegen
bleibt eine gute Quantenzahl. Es ist deshalb sinnvoll und interessant, das
Eigenwertproblem des Gesamtspin-Operators zu lösen.
Da S 1 /~ (S 2 /~) Generatoren der Drehungen für den ersten (zweiten) Spin
sind S/~ are die Generatoren für die Drehungen des gesamten System
(product Raum) details: Sec. A.15 . Daher müssen Sie die üblichen Vertauschungsregeln folgen
[S α , S β ] = αβγ S γ
.
(4.6)
Es muß also eine gemeinsame Basis für S z und S 2 geben, da diese beiden
Drehimpuls-Operatoren generell miteinander vertauschen. Wir wollen
4.5. SYSTEME MIT ZWEI SPIN
1
2
TEILCHEN
73
nun den Zusammenhang (Transformation) zwischen diesen Eigenvektoren
und der Produktbasis |σ1 i |σ2 i bestimmen.
Zunächst stellen wir fest, daß
[S1α , S2β ] = 0
da die beiden Spinoperatoren auf die unterschiedliche Teilsysteme wirken
(cf.(4.5)). Daraus folgt unmittelbar
[S 21 , S 22 ] = 0
wie auch
[S 2α , S 2 ] = 0
Schließlich vertauschen S 21 , S 22 auch noch mit S z = S1z + S2z ,
[S z , S 2α ] = [S1z + S2z , S 2α ] = 0
α = 1, 2 .
Allerdings ist zu beachten, dass
[Sαz , S 2 ]6= 0
Wir können also eine Basis gemeinsamer Eigenvektoren von den Operatoren
und S z , aber nicht von den Sαz , zusammenstellen.
Wir werden zeigen, dass diese gemeinsame Eigenbasis den gesamten Hilbertraum spannt. Die Vektoren dieser Basis bezeichen wir als |s1 , s2 , j , mi, mit
folgenden Eigenwertbedingungen:
S 21 , S 22 , S 2
a) S 21 |s1 , s2 , j , mi = ~2 s1 (s1 + 1) |s1 , s2 , j , mi
b) S 22 |s1 , s2 , j , mi = ~2 s2 (s2 + 1) |s1 , s2 , j , mi
2
2
c) S |s1 , s2 , j , mi =
~ j(j + 1)
|s1 , s2 , j , mi
d) S z |s1 , s2 , j , mi =
~m
|s1 , s2 , j , mi
(4.7)
,
mit s1 = s2 = 1/2. Die Transformation zwischen der Produktbasis und
die neue Basis wird durch die, noch zu bestimmenden, Entwicklungsko(jm)
effizienten Cσ1 ;σ2 gegeben
|s1 , s2 , j , mi =
X
σ1 ,σ2 ∈{±1}
Cσ(jm)
|σ1 i |σ2 i
1 ,σ2
74CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
4.6
Addition of angular momenta
The above issue can be made more general. Specifically, one can consider,
instead of two spin 12 systems, two systems with angular momenta J 1 and
J 2.
For example, J 1 can describe the orbital angular momentum of an electron
and J 2 its spin. In that case the corresponding quantum numbers will
have the possible values j1 = 0, 1, · · · , ∞, and j2 = 21 . The two angular
momenta interact via spin-orbit coupling. This term does not commute with
J 1 or J 2 separately, but it does commute with J ≡ J 1 + J 2 . Another
example is represented by two interacting atoms, each one with its angular
momentum J i .
The discussion of Sec. 4.5 can be readily taken over, specifically, we have
the following set of mutually commuting operators:
J 21 , J 22 , J 2 , J z
Accordingly, one can derive a common eigenbasis specified by the vectors |j1 , j2 , j , mi with the following eigenvalue properties:
a) J 21 |j1 , j2 , j , mi
= ~2 j1 (j1 + 1) |j1 , j2 , j , mi
b) J 22 |j1 , j2 , j , mi
= ~2 j2 (j2 + 1) |j1 , j2 , j , mi
2
2
c) J |j1 , j2 , j , mi
=
~ j(j + 1)
|j1 , j2 , j , mi
d) J z |j1 , j2 , j , mi
=
~m
|j1 , j2 , j , mi
(4.8)
,
The basis transformation is given by
|j1 , j2 , j , mi =
ji
X
mi =−ji
C(j, m|j1 , m1 ; j2 , m2 ) |j1 , m1 i |j2 , m2 i
(4.9a)
where the expansion coefficients
C(j, m|j1 , m1 ; j2 , m2 ) = hj1 , m1 | hj2 , m2 | |j1 , j2 , j , mi
(4.9b)
are termed Clebsch-Gordan coefficients.
4.6.1
Determining the allowed values of j
For the follwing discussion we work at fixed j1 , j2 . Therefore we omit, for
simplicity of notation, the j1 , j2 in the |j1 , j2 , j , mi basis, i.e. |j1 , j2 , j , mi → |j , mi.
This is also the standard notation.
4.6. ADDITION OF ANGULAR MOMENTA
75
There are some restrictions in the sum (4.9a). First, the z components are
additive: J z = J1z + J2z so that m = m1 + m2 . This does not hold for
j: j 6= j1 + j2 . However, for given j1 and j2 , not all values of the total
angular momentum quantum number j are allowed: The maximum value
mmax of m is given by the sum of the maximum values of m1 and m2 , i.e.
mmax = j1 + j2 . Since there are no larger values of m, this state must have
j = j1 + j2 , and there are no larger values of j. So the first constraint is
j ≤ j1 + j2 .
This discussion clearly identifies (see (4.13))
|j = j1 + j2 , m = j1 + j2 i = |j1 , j1 i |j2 , j2 i
(4.10)
Smaller values of j
Now going to the next smaller value m = j1 + j2 − 1, there are two states
with this m, namely
|j1 , j1 − 1i |j2 , j2 i
and
|j1 , j1 i |j2 , j2 − 1i
(4.11)
in this 2 dimensional subspace one vector ( a particular linear combination
) must have j = j1 + j2 , since if one vector of the multiplet is present,
(here |j = j1 + j2 , m = j1 + j2 i) then all other must be present. The other
vector must have j = j1 + j2 − 1. The expressions of these two vectors |j = j1 + j2 , m = j1 + j2 − 1 i and |j = j1 + j2 − 1 , m = j1 + j2 − 1 i are
nontrivial, since in general, they will be linear combinations of (4.11). This
discussion is shown schematically in Tab. (4.13). One can proceed in the
same way for smaller m, noticing that there are three independent vectors
with m = j1 + j2 − 2, and that the subspace will contain the vectors with
that m and j = j1 + j2 , j1 + j2 − 1, and j1 + j2 − 2 (see (4.13)). The story
continues for all m = j1 + j2 − p with p ≤ 2j2 (for definiteness we assume
j2 ≤ j1 ). The subspace is p + 1-dimensional, containing the vectors with
j = j1 + j2 , j1 + j2 − 1, · · · , j1 + j2 − p.
However, the situation changes when p = 2j2 + 1. In that case, there are
not enough m2 to have a total of p + 1 states. This means that there will be
no additional values of j. This shows that the
76CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
POSSIBLE VALUES OF
j
ARE
j = j1 + j2 , j1 + j2 − 1, j1 + j2 − 2, · · · |j1 − j2 | .
m
j1 + j2
product states with that m
|j
1 , j1 i |j2 , j2 i
|j1 , j1 − 1i |j2 , j2 i
j1 + j2 − 1
 |j1 , j1 i |j2 , j2 − 1i
 |j1 , j1 − 2i |j2 , j2 i
|j1 , j1 − 1i |j2 , j2 − 1i
j1 + j2 − 2

|j1 , j1 i |j2 , j2 − 2i
···
·· ·
 |j1 , j1 − 2 ∗ j2 , i |j2 , j2 i
···
j1 − j2

|j1 , j1 i |j2 , −j2 i
···
···
combine to j values
j1 + j2
j1 + j2
 j1 + j2 − 1
 j1 + j2
j1 + j2 − 1

j1 + j2 − 2
·· ·
 j1 + j2
···

j1 − j2
···
(4.12)
(4.13)
Table 4.1: Schematic structure of the product space of two systems with
angular momentum quantum numbers j1 and j2
4.6.2
Construction of the eigenstates
We illustrate here how to explicitly construct the eigenstates, i. e., in the
end, the coefficients in (4.9a). For practical purposes, these coefficients are
available in tables.
As shown above, the first vector is given by
|j = j1 + j2 , m = j1 + j2 i = |j1 , j1 i |j2 , j2 i
We can now find the next vector with the same j by applying the ladder
operator (1.43):
|j = j1 + j2 , m = j1 + j2 − 1 i ∝ J − |j1 , j1 i |j2 , j2 i
since
J − = J1− + J2−
4.6. ADDITION OF ANGULAR MOMENTA
77
we have (cf (1.43))
p
j1 (j1 + 1) − j1 (j1 − 1) |j1 , j1 − 1i |j2 , j2 i +
p
+ j2 (j2 + 1) − j2 (j2 − 1) |j1 , j1 i |j2 , j2 − 1i ,
(4.14)
|j = j1 + j2 , m = j1 + j2 − 1 i ∝
where the constant is obtained by normalisation. Since the subspace
with m = j1 + j2 − 1 is spanned by the two vectors (see (4.13)), the vector
with j = j1 + j2 − 1 is simply the one orthogonal to (4.14) in this subspace.
The procedure goes on recursively like this: one applies the ladder operators to the two vectors
|j = j1 + j2 , m = j1 + j2 − 1 i and |j = j1 + j2 − 1 , m = j1 + j2 − 1 i to obtain the two corresponding vectors with m = j1 + j2 − 2. The vector
orthogonal to both of them in this three dimensional subspace is the one
with j = j1 + j2 − 2. An so on.
4.6.3
Application to the case of two spin
1
2
The procedure in Sec. 4.6.2 is quite complicated, but relatively easy for two
spin 21 systems. Here j1 = j2 = 21 do not need to be specified. We have (cf.
(4.10)):
1 1
+
|j = 1 , m = +1 i = +
(4.15)
2 2
By applying the ladder operator (cf. (4.14))
1 1
1 1
−
−
|j = 1 , m = 0 i ∝ +
+
2 + 2 ,
2 2
and the proportionality (normalization) constant is clearly √12 .
|j = 0 , m = 0 i is obtained by orthogonalizing to (4.16):
1 1
1 1
−
−
|j = 0 , m = 0 i ∝ +
−
2 + 2 ,
2 2
(4.16)
(4.17)
again with proportionality constant √12 . The sign (or the phase) is of course
arbitrary. Finally, we could obtain |j = 1 , m = −1 i by applying J − to
(4.16). However, it is easier to observe that |j = 1 , m = −1 i is the only
state having m = −1, and thus it is given by
1 1
−
|j = 1 , m = −1 i = −
.
(4.18)
2 2
78CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
In problems in which the two spins interact with each other, but the Hamiltonian still commutes with the total spin, the four possible states split into the
singlet |j = 0 , m = 0 i and the three degenerate triplet
|j = 1 , m = +1 i, |j = 1 , m = 0 i, |j = 1 , m = −1 i states.
S INGLET AND TRIPLET STATES OF TWO SPIN -1/2
j=0 m=0
√1
2
j=1 m=0
m = −1
4.6.4
√1
2
|↑i |↓i − |↓i |↑i
m = +1
PARTICLES
|↑i |↑i
|↑i |↓i + |↓i |↑i
singlet
(4.19)
triplet
|↓i |↓ii
Scalar product
An object which often occurs in these kinds of problems is the scalar product
J1 · J2 .
For example, the spin-orbit coupling reads L · S.
The term is straightforward to evaluate, since
J 2 = J 21 + J 22 + 2J 1 · J 2 .
Therefore, one readily obtains
J1 · J2 =
4.6.5
~2
[j(j + 1) − j1 (j1 + 1) − j2 (j2 + 1)]
2
(4.20)
How to use a table of Clebsch-Gordan coefficients
Let us recall the generic expansion (4.9a) of a state with given total angular momentum quantum numbers j, m in terms of product states with
quantum numbers j1 , m1 , j2 , m2 .
4.6. ADDITION OF ANGULAR MOMENTA
79
A table of the coefficient of the expansion (Clebsch-Gordan coefficients)
(4.9b)
C(j, m|j1 , m1 ; j2 , m2 ) = hj1 , m1 | hj2 , m2 | |j1 , j2 , j , mi
can be found in http://pdg.lbl.gov/2002/clebrpp.pdf (just for me here)
A javascript tool to produce them:
http://www.correlati.de/jsim/zhang/JSim/chapter5.html
j1 × j2
Info for coefficient: add square root, with sign outside!
ä
j
m
m1
m2
Figure 4.1:
j1 = 2, j1 = 1 Clebsch-Gordan coefficients from
http://pdg.lbl.gov/2002/clebrpp.pdf.
To understand how to use this table (161) let us look at an example for
j1 = 2, j1 = 1 This table gives information to get the coefficients of the
expansion (4.9a)
hj1 = 2, m1 | hj2 = 1, m2 | |j1 = 2 , j2 = 1 , j , mi .
The inner rectangle contains, in principle, the coefficients, whereby one
has to put a square root in front and pull the sign in front of it. In addition,
when exchanging j1 , m1 with j2 , m2 one can use the relation
hj1 , m1 | hj2 , m2 | |j1 , j2 , j , mi = (−1)j−j1 −j2 hj2 , m2 | hj1 , m1 | |j2 , j1 , j , mi
(4.21)
For example, the green mark shows a −2/5. This means that
p
hj1 = 2, m1 = 0| hj2 = 1, m2 = 0| |j1 = 2 , j2 = 1 , j = 1 , m = 0 i = − 2/5 .
80CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
As another example, let us write a j = 2, m = −1 state (see violet arrow):
|j1 = 2 , j2 = 1 , j = 2 , m = −1 i
1
1
1
= √ | 2 , 0 > | 1 , −1 > − √ |2, −1i |1, 0i − √ |2, −2i |1, +1i
↑
↑
2 ↑ m↑1
6
3
m2
j1
4.7
j2
Matrix elements of vector operators (WignerEckart’s theorem)
Here, we want to look at the application of the theory of the addition of
angular momenta to other problems in atomic and molecular physics (and
beyond), whenever we need to calculate matrix elements of vector operators between atomic states. For example, in optics the transition rate between two states contains the dipole matrix element
d = e hn, l, m| r̂ |n0 , l0 , m0 i .
(4.22)
This is also needed to evaluate the effects of an electric field (Stark effect),
and the corresponding lifting of degeneracy of atomic states.
Another application is the Zeeman effect, in which one considers the perturbation due to a magnetic field
Matrix elements of the form (4.22) can involve the calculation of a large
number of integrals. For example, for a transition 2p → 3d orbitals one has
in (4.22) l0 = 1, l = 2, i.e. 3 × 5 combinations of m, m0 . In addition, there
are 3 components of r̂, for a total of 45 integrals. In addition, many of
these integrals are zero due to symmetry, which corresponds to forbidden
transitions, i.e. selection rules. It would be a useful information to figure
out which one are zero without doing calculations.
Wigner Eckart’s Theorem allows to obtain all 45 integrals by just evaluating one
of them. The idea is that all those states and integrals can be obtained by
symmetry transformations. The generalisation of (4.22) consists of a matrix element of a vector operator V̂ between angular momentum eigenstates
ha, j, m| V̂ |b, j1 , m1 i
Here, a, b are additional quantum numbers, e.g. n, n0 for hydrogen atom. j
could be just l, or the total angular momentum j, or spin s depending on
the problem at hand.
It is convenient to rewrite the vector operator in analogy to the ladder
operators (1.22) (there are slightly different factors):
√
V̂0 = V̂z
(4.23)
V̂±1 = ∓(V̂x ± iV̂y )/ 2
4.7. MATRIX ELEMENTS OF VECTOR OPERATORS (WIGNER-ECKART’S THEOREM)81
Now the theorem states that Proof here: Sec. A.16
W IGNER -E CKART ’ S THEOREM FOR VECTORS
ha, j, m|V̂M |b, j1 , m1 i
(4.24)
= hj1 , m1 | hj2 = 1, m2 = M | |j1 , j2 = 1 , j , mi ha, j| |V̂ | |b, j1 i ,
where hj1 , m1 | hj2 = 1, m2 = M | |j1 , j2 = 1 , j , mi are the Clebsch-Gordan coefficients (4.9b). The last term ha, j| |V̂ | |b, j1 i is simply a proportionality
constant, and can be determined by evaluating just one of the (nonvanishing) matrix elements on the left h.s., i.e. for one set of m, M, m1 . Then
one automatically has all of them. (4.24), compared with (4.9b) suggests
that V̂M behaves similarly to a state with angular momentum quantum numbers
j2 = 1, m2 = M .
The more general version of Wigner-Eckart’s theorem applies to higherdimensional tensors such as quadrupoles. A special case is a 0-dimensional
tensor, i.e. a scalar operator Ŝ , we already knew that
W IGNER -E CKART ’ S THEOREM FOR SCALARS
ha, j, m| Ŝ |b, j1 , m1 i = δj,j1 δm,m1 ha, j| |Ŝ| |b, ji
(4.25)
Selection rules and parity
(4.24) provides an easy rule to determine which matrix elements are allowed. From the rule j = j1 + j2 , · · · , |j1 − j2 |, and m = m1 + m2 , we get for
the present case |j − j1 | ≤ 1 and m = m1 + M . The latter observation leads
to the selection rules for optical dipole transitions between states of a spherical
atom: lf inal = linitial ± 1.
In principles (4.24) only states |lf inal − linitial | ≤ 1 so also lf inal = linitial
would be allowed. This, however, is excluded by the fact that lf inal and
linitial must have different parity, i.e. lf inal − linitial = odd. To see this,
82CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
remember that the parity operator P̂ transforms r → −r:
P̂ ψ(r) = ψ(−r)
Now, by looking at the spherical harmonics it is easy to see that states
with a given l have a fixed parity (−1)l , i.e.
P̂ Ylm (θ, ϕ) = (−1)l Ylm (θ, ϕ) .
On the other hand, for the position operator, we have
P̂ r̂ P̂ = −r̂
which indicates the transformation of r̂ under P̂ . We thus have for (4.22)
hn, l, m| r̂ |n0 , l0 , m0 i = − hn, l, m| P̂ r̂ P̂ |n0 , l0 , m0 i
(
0
for l − l0 even
0
= (−1)l−l +1 hn, l, m| r̂ |n0 , l0 , m0 i =
6= 0 for l − l0 odd
(4.26)
this gives the mentiones condition lf inal − linitial = odd.
4.7.1
Applications
Stark effect in Hydrogen
As an application we consider the perturbation of the degenerate n = 2
Hydrogen states by an homogeneous electric field:
V̂ = −eE ẑ .
(4.27)
According to degenerate perturbation theory, we ought to calculate all 16
matrix elements
h2, l, m| ẑ |2, l0 , m0 i
First of all, due to (4.26), only matrix elements between l = 0 and l = 1 are
nonzero. Second, since (cf. (4.23)) ẑ = r̂0 , we have from (4.24) m = m0 . So
in the end, the only nonvanishing matrix elements are
d ≡ h2, 1, 0| ẑ |2, 0, 0i = h2, 0, 0| ẑ |2, 1, 0i ,
(4.28)
We don’t calculate d here. From the 4 × 4 matrix we are left with a 2 × 2
between these states, so the eigenstates/eigenenergy corrections are (cf.
(4.27), (4.28))
eigenstate
energy correction
|2, 0, 0i + |2, 1, 0i
−eEd
(4.29)
|2, 1, ±1i
0
|2, 0, 0i − |2, 1, 0i
eEd
4.7. MATRIX ELEMENTS OF VECTOR OPERATORS (WIGNER-ECKART’S THEOREM)83
Land’e g-factor
This deals with the problem of an atom in which electrons have a total orbital angular momentum L̂ and total spin Ŝ, characterized by corresponding quantum numbers l and s (in the case of one electron, s = 21 ).
They first couple via spin-orbit coupling
HSO = λ L̂ · Ŝ
(4.30)
so that the total angular momentum
(4.31)
Ĵ = L̂ + Ŝ
commutes with the hamiltonian, and eigenenergies depend on the corresponding quantum number j only, with degeneracy 2j + 1 (fine-structure
states).
One then studies the influence of an external magnetic field B which couples as (Zeeman effect)
HB = µ̂ · B = µB (L̂ + gs Ŝ) · B
µB =
q
2m
(4.32)
due to the electron’s gyromagnetic ratio gs = 2 the term in brackets is not
in the same direction as Ĵ .
Here, the perturbation due to B is considered to be smaller than the effect
of spin-orbit coupling, so that it is sufficient to consider first-oder perturbation theory, which is, however, degenerate, due to the degeneracy of the
j-multiplets. Therefore, we ought to calculate all matrix elements of HB ,
i.e. of Lz and Sz between the degenerate states |n, j, mj i for all mj .
But again, Wigner-Eckart’s theorem gives us an easier solution. From the
proportionality (4.24), it is clear that within a fixed set of a, j, b, j1 all matrix
elements of vectors are proportional to each other, so in our case
(4.33)
n, j, mj V̂ |n, j, mj i = γ(n, j) n, j, mj Ĵ |n, j, mj i .
Here we prove : Sec. A.17 that the constant γ(n, j) is given by
n, j, mj V̂ · Ĵ |n, j, mj i
γ(n, j) =
~2 j(j + 1)
and is independent of mj (projection theorem).
(4.32) with (4.33) gives
n, j, mj µ̂ n, j, m0j = gnj µB n, j, mj Ĵ n, j, m0j
(4.34)
(4.35)
84CHAPTER 4. SEVERAL DEGREES OF FREEDON AND THE PRODUCT SPACE
with the Landé factor (cf. (4.34)
n, j, mj (L̂ + gs Ŝ) · Ĵ |n, j, mj i
gnj ≡
~2 j(j + 1)
(4.36)
To determine this we consider that
2
2
2
1 2
L̂ · Ĵ = L̂ · (L̂ + Ŝ) = L̂ + (Ĵ − L̂ − Ŝ )
2
and similarly
2
2
2
1 2
Ŝ · Ĵ = Ŝ + (Ĵ − L̂ − Ŝ )
2
So that (4.36)
gj =
j(j + 1)(1 + gs ) + l(l + 1)(2 − 1 − gs ) + s(s + 1)(−1 + gs )
2j(j + 1)
(4.37)
Chapter 5
Time dependent perturbation
theory
5.1
Zeitabhängige (Diracsche) Störungstheorie
Häufig interessiert man sich für zeitabhängige Hamilton-Operatoren. Z.B.
könnte man daran interessiert sein, was mit einem Atom passiert, wenn
man elektromagnetische Wellen einstrahlt. Man kann versuchen, die zeitabhängige Schrödingergleichung analytisch exakt zu lösen. Das gelingt allerdings nur in den seltensten Fällen. Der Ausweg sind entweder numerische
Verfahren, die in den letzten Jahren rasant an Leistungsfähigkeit und Bedeutung zugenommen haben. Alternativ hat man die Möglichkeit, das
Problem perturbativ zu lösen. Wir gehen davon aus, daß der HamiltonOperator H = H0 + H1 (t) aus einem zeitunabhängigen Teil H0 und einer
zeitabhängigen Störung H1 (t) besteht. Für das Folgende gehen wir davon
aus, daß das Eigenwertproblem von H0
H0 |Φn i = n |Φn i
gelöst ist. In der Praxis benutzt man die Lösung des zeitabhängigen Problems, um experimentell Rückschlüsse auf das Eigenwertspektrum von H0 zu gewinnen. Z.B. kann man einen klassischen Oszillator von außen mit einer periodischen Kraft mit einer Frequenz ω anregen. Wenn wir ω kontinuierlich variieren, wird die Amplitude der erzwungenen Schwingung bei der
Eigenfrequenz des ungestörten Oszillators maximal sein. Wir können also
auf diese Weise auf die Eigenfrequenz (bzw. Federkraft) und die Reibungskräfte
rückschließen. In Quantensystemen ist diese Vorgehensweise die einzig
mögliche, das mikroskopische System zu untersuchen. In diesem Zusammenhang hat man allerdings die Stärke des Störterms H1 unter Kontrolle
85
86
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
und kann erreichen, daß „ H1 H0 ” Die Aufteilung der Dynamik in Anteile, die von H0 und solche, die von H1 herrühren, führt zur Einführung
des Wechselwirkungsbildes.
5.1.1
Das Wechselwirkungsbild
Die Zeitentwicklung eines Zustandes nach dem Gesamthamiltonian H =
H0 +H1 (t) wird durch die zeitabhängige Schrödingergleichung bestimmt:
i~
d S
|Ψ (t)i = H0 + H1 (t) |ΨS (t)i .
dt
(5.1)
der superscript “S” steht für “Schrödinger Bild”.
Die Zeitentwicklung eines beliebigen Anfangszustands |Φi im zeitunabhängigen ungestörten System lautet bekanntlich
i
|Φ(t)i = e− ~ H0 t |Φ(t = 0)i
.
Es ist sinnvoll, den “einfachen” Teil der Dynamik, beschrieben durch H0 ,
explizit im wechselwirkenden Zustandsvektor |Ψi zu eliminieren durch Einführung des Wechselwirkungsbildes (Interaction representation). Der
Zustand im Wechselwirkungsbild wird mit dem index “I” (interaction)
bezeichnet.
i
|ΨI (t)i =: e ~ H0 t |ΨS (t)i .
(5.2)
Die Zeitableitung von (5.2) liefert
i~
i
i
d I
d |Ψ (t)i = −H0 e ~ H0 t ΨS (t) + e ~ H0 t i~ ΨS (t)
dt
dt
(5.3)
wir benutzen (5.1) und erhalten
i~
i
i
d I
|Ψ (t)i = −H0 e ~ H0 t Ψ S (t) + e ~ H0 t H0 Ψ S (t)
dt
i
i
i
+ e ~ H0 t H1 (t)e− ~ H0 t e ~ H0 t ΨS (t)
|
{z
} |
{z
}
:=HI1 (t)
=
|ΨI (t)i
H1I (t) ΨI (t)
(5.4)
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
87
W ECHSELWIRKUNGSBILD
i
i
H1I (t) = e+ ~ H0 t H1 (t)e− ~ H0 t
d
i~ |ΨI (t)i = H1I (t)|ΨI (t)i
dt
(5.5)
Let us now derive an expansion of |ΨI (t)i in powers of H1I . This means
that we formally replace H1I → εH1I in (5.5). 1 Since |ΨI (t)i now depends
on ε, we can carry out a Taylor expansion.
I
|Ψ (t)i =
∞
X
`=0
ε` |ΨI(`) (t)i .
(5.6)
Inserting this into (5.5), and collecting same powers of ε` on both sides
yields
d I(`+1 )
|Ψ
(t)i = ε H1I (t) ε` |ΨI(`) (t)i
`≥0.
(5.7)
dt
Assuming that H1I (t) vanishes before a certain time t0 (when the perturbation is switched on), then ΨI (t0 ) does not depend on H1 (t), and thus on
ε, i.e. only the ` = 0 is nonzero. We therefore have the initial conditions
ε(`+1) i~
|ΨI(`) (t0 )i = δ`,0 |Φi ,
(5.8)
where |Φi = |Φ(t = 0)i = |ΨI(`=0) (t)i is the unperturbed state evaluated in
the interaction representation, which is, by definition, time independent.
(5.7) can thus be solved by iteration:
I(`+1)
|Ψ
i
(t)i = −
~
Z
t
H1I (τ )|ΨI(`) (τ )i dτ
(5.9)
t0
|ΨI(0) (t)i = |Φi .
1
This is just a formal way to introduce an expansion for “small” H1I . In the end, we
will be interested in the case ε = 1
88
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
For example, the first-order term reads
I(1)
|Ψ
i
(t)i = −
~
t
Z
t0
H1I (τ )dτ |Φi
(5.10)
and the second-order term
I(2)
|Ψ
Z
i t
(t)i = −
dt1 H1I (t1 )|ΨI(1) (t1 )i =
~ t0
2 Z t
Z t1
i
I
−
dt2 H1I (t2 )|Φi .
dt1 H1 (t1 )
~
t0
t0
(5.11)
Wir entwickeln nun die |ΨI(`) (t)i nach den Eigenzuständen |Φn i von H0 :
|ΨI(`) (t)i =
X
n
c(`)
n (t)|Φn i
Entsprechend (5.6) sind
cn (t) ≡
∞
X
c(`)
n (t)
(5.12)
`=0
die Entwicklungskoeffizienten von |ΨI (t)i.
Einsetzen in (5.9) und multiplikation von links mit hΦm |: liefert:
c(`+1)
(t)
m
iX
=−
~ n
Z
t
t0
i
hΦm |H1I (τ )|Φn i
{z
}
|
c(`)
n (τ ) dτ
(5.13)
i
e+ ~ m τ hΦm |H1S (τ )|Φn ie− ~ n τ
Mit der Definition
m − n
~
lautet die recursive Gleichung (5.13) für die Entwicklungskoeffizienten:
ωmn :=
(t)
c(`+1)
m
iX
= −
~ n
Z
t
dτ eiωmn τ Hmn (τ ) c(`)
n (τ )
t0
Hmn (τ ) = hΦm |H1S (τ )|Φn i
.
(5.14)
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
89
Die ` = 0 Koeffiziente sind die Zeitunabhängige Entwicklungkoeffiziente
der Zustandes (in der Wechselwirkungsdarstellung) vor einschalten der
Störung
c(`=0)
(t0 ) = cm (t0 ) ≡ cm
m
Wir nehmen von nun an zur Vereinfachung t0 = 0. Aus (5.14) lauten die
Koeffizienten erster Ordnung
c(1)
m (t)
iX
= −
cn
~ n
Z
t
Hmn (τ )eiωmn τ dτ
0
Der nächste Iterationsschritt liefert:
c(2)
m (t)
Z
iX t
dτ Hmn (τ )eiωmn τ c(1)
(5.15)
=−
n (τ )
~ n 0
2 X Z t
X Z τ
i
0
iωmn τ
dτ 0 Hnn0 (τ 0 )eiωnn0 τ
dτ Hmn (τ )e
cn 0
= −
~
0
0
n
n0
Übergangswahrscheinlichkeit
Es interessiert uns folgende Frage: Wenn das System zur Zeit t = 0 im
Zustand |Φi i ist, wie groß ist die Wahrscheinlichkeit, daß es dann zur Zeit
t > 0 in dem Zustand |Φf i ist? (|Φi i, |Φf i sind Eigenzustände von H0 )
Pi→f
2
2 X
X
2
= |hΦf |Ψ(t)i| = hΦf |
cn (t)Φn i = cn (t) hΦf |Φn i
n
| {z }
n
δf n
= |cf (t)|2
.
(5.16)
Die Anfangsbedingung lautet cn (0) = δn,i . Daraus folgt in zweiter Ordnung in H1 (siehe(5.15))
cf (t) = δf,i
Z
i t
Hf i (τ )eiωf i τ dτ
−
~ 0
Z
Z τ
1 X t
0
iωf n τ
dτ 0 Hni (τ 0 )eiωni τ
− 2
dτ Hf n (τ )e
~ n 0
0
+ O(H13 )
(5.17)
90
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
5.1.2
Harmonische oder konstante Störung
Eine sehr wichtige Anwendung der zeitabhängigen Störungstheorie sind
Probleme, bei denen zur Zeit t = 0 eine konstante oder harmonische Störung
eingeschaltet wird
H1 = Θ(t)2V cos(ωt)
• Θ(t) schaltet die Störung zur Zeit t=0 ein
• für ω = 0 beschreibt diese Gleichung eine konstante Störung
Es gilt 2
H1 (t) = Θ(t) V eiωt + V † e−iωt
∗ −iωt
Hmn (t) = Vmn eiωt + Vmn
e
=
X
s
Vmn
eisωt
s=±1
s
Vmn
mit
(
Vmn
=
∗
Vmn
für
für
s = +1
.
s = −1
Θ(t) wird in diesem Zusammenhang nicht mehr benötigt, da die Integrale
in (5.17) ohnehin erst bei t = 0 beginnen!
Nullte Ordnung
In Nullter Ordnung ist nicht viel zu rechnen. Der Beitrag lautet
(0)
cf = δif
und er kommt nur zum Tragen, wenn Anfangs- und Endzustand gleich
sind.
Erste Ordnung
Das erste Integral in (5.17) lautet
Z t
Z t
X
iωmn τ
s
dτ Hmn (τ )e
=
Vmn
dτ ei(ωmn +sω)τ
0
=
s
2
0
s
X
s
Vmn
i(ωmn +sω)t
e
−1
i(ωmn + sω)
(5.18)
Für nicht-selbstadjungiertes V ist diese Form allgemeiner und enthält eine Phasenverschiebung
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
91
D.h. der Term erster Ordnung liefert:
i X s i ωf i +sω t sin
(1)
cf (t) = −
V e 2
~ s fi
ωf i +sω
t
2
ωf i +sω
2
(5.19)
Zweite Ordnung
Wir wollen uns zunächst mit der ersten Ordnung begnügen. Den Beitrag
zweiter Ordnung werden wir später mit einer leicht modifizierten Methode berechnen, mit der das Ergebnis leichter hergeleitet werden kann.
Konstante Störung
Wir betrachten zunächst eine konstante Störung (ω = 0). Damit vereinfachen sich die Ausdrücke (5.19) bzw. (5.17)
ω
i X s i ωf i t sin 2f i t
(1)
V e 2
(5.20)
cf (t) = −
ωf i
~ s=±1 f i
2
| {z }
Hf i
(2)
cf (t)
0
1 X X Vfsn Vnis eiωf i t − 1 eiωf n t − 1
= 2
−
~ s,s0 =±1 n
ωni
ωf i
ωf n
1 X Hf n Hni eiωf i t − 1 eiωf n t − 1
= 2
−
~ n
ωni
ωf i
ωf n
(5.21)
Die Wahrscheinlichkeit, dass das System in der Zeit t in den Zustand f
übergegangen ist, lautet auf niedrigster Ordnung aus (5.20)
Pi→f
2 ωf i
t
t2
2 sin
2
= |cf (t)| = 2 |Hf i |
ωf i 2
~
t
2
.
(5.22)
2
Falls das Spektrum diskret und i nicht entartet ist, gibt es zwischen der
Energie i und der Energie der Endzustände f eine Energielücke ∆f . Die
Übergangswahrscheinlichkeit lautet
Pi→f
4 |Hf i |2
=
sin2
(∆f )2
∆f t
2~
.
92
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
. Das ist
Diese Wahrscheinlichkeit oszilliert also mit der Frequenz ω = ∆
~
3
ein charakteristisches Phänomen diskreter Systeme. Die Periode der Os|H |2
zillationen ist T = 2π~
und die Amplitude nimmt proportional zu (∆ffi )2
∆
ab.
Wenn das diskrete Spektrum in i entartet ist, d.h. ∃f 6= i : f = i , dann
liefert (5.22) stattdessen
t2 |Hf i |2
Pi→f =
.
~2
Die Wahrscheinlichkeit wächst proportional zu t2 an. Ab einer bestimmten
Zeit wird die Wahrscheinlichkeit größer Eins. Das ist natürlich unsinnig
und zeigt an, daß in Fall von Entartung
t
~
|Hf i |
(5.23)
erfüllt sein muß, damit die Störungstheorie erster Ordnung anwendbar
ist.
Kontinuierliches Spektrum
Wir wenden uns nun dem Fall zu, daß das Spektrum bei i kontinuierlich
ist. Es ist hierbei zweckmäßig, die Übergangsrate (Übergangswahrscheinlichkeit pro Zeiteinheit)
P
(5.24)
W :=
t
einzuführen, die aus (5.22) folgt
Wi→f
2
sin(ωfi t/2 )
2π
2 t
|Hf i |
=
~2
2π
ωfi t/2
In diesem Ausdruck erscheint die Funktion
2
)
t sin( ωt
2
∆t (ω) :=
,
ωt
2π
2
.
(5.25)
die sich im Limes t → ∞ wie die Delta-Funktion verhält Diskussion hier:
Sec. A.18 (siehe Fig.(A.1)):
lim ∆t (ω) = δ(ω)
t→∞
3
.
(5.26)
Erst im Fall kontinuierlicher Spektren verschwindet die Periodizität, wie wir gleich
sehen werden.
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
93
Hiermit wird die Übergangsrate (5.25) für große t zu
F ERMI ’ S G OLDENE R EGEL
Wi→f =
2π
|Hf i |2 δ(f − i )
~
.
(5.27)
Diese Darstellung mit der δ-Funktion ist nur dann sinnvoll, wenn in der
Nachbarschaft von i ein Kontinuum von Endzuständen vorhanden ist.
Das ist in physikalischen Anwendungen oft der Fall. Im Limes t → ∞
beschreibt (5.27) die Energieerhaltung. Für kurze Zeiten ist die Energieerhaltung aufgeweicht, d.h. es hat eine Unschärfe (siehe Fig.(A.1)) ∆E/~ ∝
2π/t also die Heisenberg Unschärfe zwischen Zeit und Energie
t ∆E ∝ h
Wir finden im Fall der Entartung eine zeitlich konstante Übergangsrate,
d.h. die Übergangswahrscheinlichkeit wächst linear mit der Zeit und führt
zu unphysikalischen Ergebnissen, wenn die Störungstheorie nicht mehr
anwendbar ist, d.h. wenn die Ungleichung (5.23) verletzt wird.
Bei einem kontinuierlichen Spektrum kann man nicht mehr die Übergangswahrscheinlichkeit in individuelle Zustände angeben. P = |hΦf |Ψ(t)i|2
in (5.16) hat dann vielmehr die Bedeutung einer Wahrscheinlichkeitsdichte.
In diesem Fall interessieren wir uns für die Übergangsrate in einem Intervall von Endzustandsenergien (n ∈ ∆If := [f , f + ∆E]).
Wir führen nun die Zustandsdichte ρ(E) ein . Für ein infinitesimales Energieintervall ∆E ist ρ(E) die Anzahl der Zustände dividiert durch ∆E.
Für ein endliches Intervall
Z
X
ρ(E)dE =
1
(5.28)
∆E
n
n ∈∆E
Man kann auch schreiben
ρ(E) =
X
n
δ(n − E)
(5.27) wird dann zu:
Wi→∆If =
X 2π
|Hni |2 δ(n − i )
~
n
n ∈∆If
(5.29)
94
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
In der Annahme, dass |Hni |2 If nicht wesentlich variiert, kann man diesen
durch einem Mittelwert |Hf i |2 ersetzen. Wir erhalten
Wi→∆If = Θ(i ∈ ∆If ) |Hf i |2
2π
ρ(i )
~
(5.30)
Harmonische Störung mit kontinuierlichem Spektrum
Der Fall einer harmonischen Störung läßt sich nun leicht diskutieren. Hier
ist ω 6= 0 und es gibt in (5.19) zwei Beiträge: af (ωf i + ω) + bf (ωf i − ω). Für
große t oszillieren diese Terme stark und verschwinden im Durschnitt, es
sei denn, der Argument ωf i ± ω ist klein, was nur bei einer der beiden
geschehen kann. Die Übergangswahrscheinlichkeit enthält den Term
|af (ωf i + ω) + bf (ωf i − ω)|2 = |a|2 |f (ωf i + ω)|2 + |b|2 |f (ωf i − ω)|2
+(a∗ b + ab∗ )f (ωf i + ω) · f (ωf i − ω)
Der Interferenzterm hat (für ω 6= 0) immer eine stark oszillierende Komponente und kann daher vernachlässigt werden. Die anderen Beiden Termen ergeben eine Übergangsrate in der Form (5.25) mit ωf i → ωf i ± ω, was
letzendlich zur δ-Form (5.27) führen.
Für endliche Frequenzen kann immer einer der beiden Peaks vernachlässigt werden. Voraussetzung hierfür, und auch für die Vernachlässigung
des Interferenzterms ist, daß der Abstand 2ω der beiden δ-artigen Peaks
groß ist gegen die Peakbreite 2π/t (siehe Abbildung (A.1)). Das heißt,
t
~
π
≈
ω
|f − i |
.
Gleichzeitig muß immer noch (5.23) erfüllt sein, damit die erste Ordnung
Störungstheorie gültig ist. Für die Zeit t erhalten wir also die Bedingung
~
~
t
|f − i |
|Hf i |
.
Voraussetzung dafür, daß diese Bedingung für t überhaupt erfüllt werden
kann, ist
|Hf i |
1
.
|f − i |
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
95
Das ist genau die gleiche Voraussetzung, die für die Gültigkeit der zeitunabhängigen Störungstheorie notwendig ist.
Analog zu (5.27) erhalten wir


Wi→f =

2π 
|Wf i |2 δ(f − i + ~ω) + Wf∗i 2 δ(f − i − ~ω) 
~ 
|
|
{z
}
{z
} 
f =i −~ω⇒Emission
f =i +~ω⇒Absorption
Diese Formel ist der Ausgangspunkt zur Beschreibung vieler physikalischer Effekte, wie z.B.: (inverse) Photoemission, Augerspektroskopie, CoulombStreuung und Compton-Streuung.
5.1.3
Adiabatisches Einschalten der Störung
Wir multiplizieren den Störterm H1 mit einem adiabatischen Einschaltfak+
tor eO t . Die Störung wird nun bereits für t → −∞ eingeschaltet. Jedoch
sehr langsam, da O+ eine infinitesimal kleine positive Größe ist. Die Änderung im Vergleich zum abrupten Einschalten bei t = 0 ist, dass die Integrale sich alle bis t → −∞ erstrecken. Der Einschaltfaktor stellt sicher,
dass die Integrale konvergieren. Der einzige Unterschied ist folgender
Z t
Z t
+
iωτ
ei(ω−iO )τ dτ
e dτ →
0
iωt
e
−∞
i(ω−iO+ )t
e
−1
→
.
iω
i(ω − iO+ )
Ganz am Schluss, lassen wir dann O+ → 0 gehen. Zur Vereinfachung der
Schreibarbeit verwenden wir die Abkürzung ω + := ω − iO+
Für den Term erster Ordnung erhalten wir dann anstelle von (5.18)
Z t
+
X
ei(ωmn +sω)t
s
iωmn τ
Vmn
dτ Hmn (τ )e
=
(5.31)
+ + sω)
i(ωmn
−∞
s
Der erste Ordnungsterm in (5.17) liefert somit bei konstanter Störung
(ω = 0)
+
(1)
cf
i X s ei(ωf i )t
=−
V
~ s f i iωf+i
+
1
eiωf i t
= − Hf i + .
~
ωf i
(5.32)
96
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
Damit gilt für die Wahrscheinlichkeit
+
Pi→f
1
e2O t
2
= |cf | = 2 |Hni | 2
~
ωf i + O+ 2
2
Daraus folgt
+
Wi→f
dPi→f
2
O+ e2O t
=
= 2 |Hni |2 2
.
dt
~
ωf i + O+ 2
Nun können wir den Grenzübergang O+ → 0 durchführen und erhalten
für die Übergangsrate
O+
2 = πδ(ω)
→0 ω 2 + O +
2π
|Hni |2 δ(Ef − Ei ) .
Wi→f =
~
mit lim
+
O
Das ist wieder das bekannte Ergebnis (5.27). Bestätigt durch dieses Ergebnis berechnen wir nun hiermit die Korrektur zweiter Ordnung für eine
ω = 0 Störung.
Z τ
Z
+ 0
1 X t
(2)
iωf+n τ
dτ 0 Hni eiωni τ
dτ Hf n e
cf (t) = − 2
~ n −∞
−∞
Z
+ t
iωni
τ
1 X
iωf+n τ e
=− 2
Hf n Hni
dτ e
+
~ n
iωni
−∞
+
1 X Hf n Hni eiωf i t
= 2
.
+
~ n
ωni
ωf+i
Wir addieren nun die Beiträge erster (5.32) und zweiter Ordnung
+
+
eiωf i t X Hf n Hni eiωf i t
cf = −Hf i + +
+
~ωf i
~ωni
~ωf+i
n
X Hf n Hni eiωf+i t
= − Hf i +
.
+
−~ωni
~ωf+i
n
+
Der erste Nenner lässt sich auch schreiben als −~ωni
= Ei − En + iO+ . Der
Unterschied zum Ausdruck erster Ordnung (5.32) besteht darin, dass Hf i
durch
X
Hf n Hni
Hf i → Mf i = Hf i +
(5.33)
+
E
−
E
+
iO
i
n
n
5.1. ZEITABHÄNGIGE (DIRACSCHE) STÖRUNGSTHEORIE
zu ersetzen ist. Die Übergangsrate ist demnach
wi→f =
2π
|Mf i |2 δ(Ef − Ei )
~
97
98
CHAPTER 5. TIME DEPENDENT PERTURBATION THEORY
Chapter 6
Identical particles
In many-body systems (e.g. atoms with more than one electron, molecules)
we deal with a system of identical particles. In classical mechanics, identical particles are always distinguishable in the sense that their trajectories can be traced. However, in quantum mechanics, identical particles are
truly indistinguishable, because due to the uncertainty relation, we cannot
follow their trajectory. A paradigmatic example is here the double-slit experiment The fact that particles are indistinguishable has a number of important consequences. We consider a system of N identical particles. We
initially assume that the particles do not interact with each other.
H = H1 + H2 + · · · + HN
.
Particle α must satisfy the eigenvalue equation
Hα |ϕν iα = ν |ϕν iα
.
An eigenstate of H is therefore the product of the eigenstates of the subsystems
|Ψi =
N
O
α=1
|ϕνα iα = |ϕν1 , ϕν2 , · · · , ϕνN i
H |Ψi = E |Ψi mit E =
N
X
να
α=1
[Note: We have to distinguish between many-particle states, such as |Ψi, and
single-particle states such as the |ϕi i. In many-body physics, the former are
99
100
CHAPTER 6. IDENTICAL PARTICLES
often simply refereed to as “states” and the latter as “orbitals” or “levels”,
although they must not necessarily refer to atomic orbitals.]
Example
N = 2
|Ψi = |ϕν , ϕµ i
E = ν + µ
.
The point is that also the state vector |ϕµ , ϕν i with interchanged particles is
an eigenstate with energy E = µ +ν . Hence, the two states are degenerate.
Therefore, a state consisting of N particles with different energies has a
very high degeneracy of N !. This is termed exchange degeneracy.
We now consider, independently of the Hamilton-operator, the consequences of the indistinguishability of particles. If the interchange of two particles should not lead to measurable consequence, initial and final states can
differ only by a phase.
Pij |Ψi = eiα |Ψi
def.Exchange operator
Pij is the exchange operator, which interchanges particles i and j.
Pij |ϕ1 , ϕ2 , · · · , ϕi , · · · , ϕj , · · · , ϕn i = |ϕ1 , ϕ2 , · · · , ϕj , · · · , ϕi , · · · , ϕn i (6.1)
Applying Pij twice to an N-particle state obviously leads back to the initial
state. As a consequence
Pij2
⇒
Pij2 |Ψi
2iα
= 11
= e2iα |Ψi = |Ψi
= 1 ⇔ α = 0, π
⇒ Pij |Ψi = ±|Ψi
e
This holds for arbitrary indices i, j. Thus, |Ψi has to by symmetric / antisymmetric with respect to all pairs i, j. Only two of the N ! possibilities are
physically distinguishable.
101
PARTICLE STATISTICS
When Pij |Ψi = +|Ψi one refers to as bosons.
When Pij |Ψi = −|Ψi one refers to as fermions.
(6.2)
From relativistic quantum theory, particles with integer spin are bosons and particles with half integer spin are fermions. Two combined fermions form a
boson (e.g. superconductivity).
We define
Def. 6.1 (Permutation operator). P is the permutation operator, that has the
following effect on an arbitrary N-particle wave function.
P |ϕ1 , ϕ2 , · · · , ϕn i = |ϕP1 , ϕP2 , · · · , ϕPn i
Def. 6.2 (Symmetrisation operator).
1 X
P
S := √
N! P
Def. 6.3 (Antisymmetrisation operator).
1 X
A := √
sign(P )P
N! P
Starting from an arbitrary N -particle state vector |Ψi, bosonic and fermionic
states can be obtained by applying the above defined operators :
Bosons:
|ΨB i = S|Ψi
Fermions: |ΨF i = A|Ψi
Example for N = 2
1
|ΨB i = √ (|ϕ1 , ϕ2 i + |ϕ2 , ϕ1 i)
2
1
|ΨF i = √ (|ϕ1 , ϕ2 i − |ϕ2 , ϕ1 i)
2
102
CHAPTER 6. IDENTICAL PARTICLES
If |ϕ1 , ϕ2 , · · ·i is a tensor product state |ϕ1 i1 ⊗ |ϕ2 i2 · · · , then |ΨF i is a socalled Slater determinant
|ϕ1 i1 |ϕ2 i1 · · · |ϕN i1 1 |ϕ1 i2 |ϕ2 i2 · · · |ϕN i2 |ΨF i = √ ···
···
· · · N! · · ·
|ϕ1 iN |ϕ2 iN · · · |ϕN iN in coordinate representation
hx1 , · · · , xN |ΨF i = ΨF (x1 , · · · , xN )
ϕ1 (x1 ) ϕ2 (x1 )
1 ϕ1 (x2 ) ϕ2 (x2 )
= √ ···
N! · · ·
ϕ1 (xN ) ϕ2 (xN )
· · · ϕN (x1 )
· · · ϕN (x2 )
···
···
· · · ϕN (xN )
Example for N = 2
1 |ϕ1 i1 |ϕ2 i1 |ΨF i = √ 2 |ϕ1 i2 |ϕ2 i2 1
= √ (|ϕ1 i1 ⊗ |ϕ2 i2 − |ϕ1 i2 ⊗ |ϕ2 i1 )
2
1
= √ (|ϕ1 , ϕ2 i − |ϕ2 , ϕ1 i)
2
in coordinate representation
ΨF (x1 , x2 ) = ϕ1 (x1 )ϕ2 (x2 ) − ϕ1 (x2 )ϕ2 (x1 )
Example for N = 3
|ϕ1 i1 |ϕ2 i1 |ϕ3 i1 1 |ΨF i = √ |ϕ1 i2 |ϕ2 i2 |ϕ3 i2 3! |ϕ i |ϕ i |ϕ i 1 3
2 3
3 3
1
= √ |ϕ1 i1 ⊗ |ϕ2 i2 ⊗ |ϕ3 i3 + |ϕ2 i1 ⊗ |ϕ3 i2 ⊗ |ϕ1 i3
6
+|ϕ3 i1 ⊗ |ϕ1 i2 ⊗ |ϕ2 i3 − |ϕ3 i1 ⊗ |ϕ2 i2 ⊗ |ϕ1 i3
−|ϕ2 i1 ⊗ |ϕ1 i2 ⊗ |ϕ3 i3 − |ϕ1 i1 ⊗ |ϕ3 i2 ⊗ |ϕ2 i3 .
(6.3)
6.1. PAULI EXCLUSION PRINCIPLE
6.1
103
Pauli exclusion principle
An immediate consequence of antisymmetry is that the electron must satisfy the Pauli exclusion principle:: Two identical fermions cannot occupy the
same state. If |ϕi i = |ϕj i where i 6= j, the exchange (6.1) results in the same
state, but because of (6.2) with a negative sign, so it gives the zero state.
Analogously, in this case two columns in the Slater determinant are identical, thus the determinant vanishes.
Example: He atom, two electrons
Quantum numbers: n; l; ml ; ms
Two electrons may not have the same set of quantum numbers.
The ground state has the quantum numbers: n = 1; l = 0; ml = 0; ms = ± 21
Two electrons with opposite spin can be in the ground state. (Therefore,
all states are pairwise occupied.)
Adding more electrons, they have to occupy the next shell:
n = 2:
l = 0; ml = 0=> 2 electrons
l = 1; ml = −1, 0, +1 => 6 electrons
In this way, the Aufbau of the periodic table is explained.
6.2
Anyonen (Optional)
Pij2 = 11 is not valid in two dimensions because of the special topology.
Exchanging two particles in R2 twice, the way in which the particles are
exchanged matters. The two ways to swap the particles back cannot be distorted
into each other. Either particle I goes once around particle II or it goes down and
up again on the same side. (Abb. 6.1).
A very demonstrative example is given in fig. 6.2. Two laces are fixed at
one end, the loose end is ’exchanged’. In three dimensions, the two ways
(fig. 6.1) are identical, because they can be distorted into each other without touching the second particle. In two dimensions, Pij2 = 11 is not valid.
Thus they can have any phase upon particle exchange rather than just ±1.
Particles which get an arbitrary phase upon exhange are called Anyons.
These play an important role in the theory of fractional quantum hall effect (FQH).
104
CHAPTER 6. IDENTICAL PARTICLES
1
t
Ia
v
iP
P
II
)
t
Ib
Figure 6.1: Double exchange of two particles (I and II) in R2 .
t
v
v
t
t
v
Figure 6.2: double exchange of two particles in R2 .
6.3
Electron and spin
Electrons have an intrinsic degree of freedom (spin). The complete state of
an electron consists of a orbital part ϕi and a spin part |χα i. The general
form of the state vector of an electron can be written as
X
|Ψi =
ci,α |ϕi χα i
i,α
Ψ(x, σ) := hx, σ|Ψi =
X
ci,α ϕi (x)χα (σ)
.
i,α
What states can two electrons occupy? For both electrons, we use the same
complete basis for space ({|ϕi i}) and spin ({|χα i}). An elementary tensor
product, thus, reads
|Ψi = |ϕi χα , ϕj χβ i = |ϕi i1 ⊗ |ϕj i2 ⊗ |χα i1 ⊗ |χβ i2
Two situations can be distinguished:
,
6.3. ELECTRON AND SPIN
105
1. Identical orbital parts (i = j): Due to the Pauli exclusion principle,
we must have α 6= β. The antisymmetrisation of the state yields
(without normalization):
|ϕi χα , ϕi χβ i − |ϕi χβ , ϕi χα i
1 = |ϕi i1 ⊗ |ϕi i2 ⊗ √ |χα i1 ⊗ |χβ i2 − |χβ i1 ⊗ |χα i2
2
|
{z
}
Singlet
.
The singlet term (4.19) has the same form in any quantisation direction n̂ for the spin (base states |±n̂i). This state is equivalent to
a total angular momentum of zero and is therefore invariant under
rotations and independent of the quantisation direction.1 .
S INGLET STATE OF TWO SPINS
1 |Singleti = √ |+n̂, −n̂i − |−n̂, +n̂i
2
1 = √ |+z, −zi − |−z, +zi
2
1. Different orbital parts (i 6= j): There are four possibilities for spins
I) α = β = +z
II) α = β = −z
III) α = −β = +z
IV) α = −β = −z
The resulting antisymmetrized states can be combined in antisymmetric in spin and symmetric in space; or symmetric in spin and antisymmetric in space. (cf. (4.19)):
1
Proof: Exercise sessions.
(6.4)
106
CHAPTER 6. IDENTICAL PARTICLES
S INGLET AND T RIPLET STATES OF TWO ELECTRONS
6.4
√1
2
|ϕi , ϕj i + |ϕj , ϕi i ⊗ |S = 0, ms = 0i
√1
2
spin-singlet
|ϕi , ϕj i − |ϕj , ϕi i ⊗ |S = 1, ms ∈ {0, ±1}i spin-triplet
The Helium atom
As an application, let us consider the example of two electrons in the Helium atom. As shown in figure (6.3), Helium consists of a nucleus (Z = 2)
and two electrons. The Hamiltonian for He is given by (atomic units)
Figure 6.3: Schematic sketch of an Helium atom
(6.5)
6.4. THE HELIUM ATOM
107
H0,1
H0,2
z }| {
z }|
p21
p2
Z
H=
−
+ 2−
2
r1
2
{z
|
H0
{
1
Z
+
r2
r12
} |{z}
(6.6)
H1
We first focus on H0 . The two terms H0,1 and H0,2 are identical, except that
they act on different particles. Each separate Hamilton operator describes
a H-like atom with Z = 2. The lowest energy is obtained when both electrons are in ground state. Due to (6.5) the spin part has to be in a singlet
state.
|ψorb i = |n = 1, l = 0, ml = 0i1 ⊗ |n = 1, l = 0, ml = 0i2
= |1, 0, 0i1 ⊗ |1, 0, 0i2
|ψspin i = |s = 0, ms = 0i
The corresponding energy is
2
Z
∧
= −4 = −4 · 27, 2eV = −108.8eV
E = −2
2
Now, we take the interaction term H1 = H1orb ⊗ 11spin into account within
first-order perturbation theory:
∆E (1) = hΨorb |H1orb |Ψorb ihΨspin |11spin |Ψspin i
= hΨorb |H1orb |Ψorb i
ZZ
1
=
|hr 1 |1, 0, 0i1 |2 |hr 2 |1, 0, 0i2 |2
d3 r1 d3 r2
|r 1 − r 2 |
ZZ
1
ρ(r 2 ) d3 r1 d3 r2
=
ρ(r 1 )
|r 1 − r 2 |
It describes the electrostatic interaction of two identical charge distributions ρ(r). The electron density for the ground state is
1 3/2 −Zr 2
1
ρ(r) = √ Z e = Z 3 e−2Zr
π
π
The energy correction reads
∆E
(1)
=
Z 3 2
π
ZZ
e−2Zr1
1
e−2Zr2 d3 r1 d3 r2
|r 1 − r 2 |
We substitue 2Zr α = xα and get
ZZ
Z 3 2
1
(1)
−5
e−x1
∆E =
(2Z)
e−x2 d3 x1 d3 x2
|x1 − x2 |
| π {z
}
=
Z
25 π 2
(6.7)
108
CHAPTER 6. IDENTICAL PARTICLES
This integral is not dependent on Z. The calculation (details : Sec. A.19 )
gives
5 ∧ 5
∆E (1) = Z = · 27, 2eV = 34.0eV
.
(6.8)
8
4
Thus, the ground state energy is E ∼
= −74.8eV. The experimental value is
EExp = −78.8eV. The calculated value does not yet agree very well with
the experimental results due to the crude approximation we made. An
improvement consists in adopting the variational method. As for the H2+
molecule, we consider the atomic number of the wave function ϕn=1,l=0,ml =0 of
the unperturbed problem as a variational parameter Z. As the energy in first
order perturbation theory
E (1) =
hΨ|H0 + H1 |Ψi
hΨ|Ψi
is equal to the energy term for Z = 2, that is minimized in the variational method, the result can only improve. This choice also has a physical interpretation. The effective charge Z seen by each electrons is smaller
than Z, because the positive charge is partially screened by the negatively
charged cloud of the other electron. By using Z → Z we take this effect
into account. We have already evaluated the required contributions to the
energy. As in the case of the H2+ molecule we rewrite the Hamiltonian as
H0,2 (Z)
H0,1 (Z)
z }| {
z }| {
Z −Z
Z −Z
1
p21 Z
p2 Z
+
+ 2−
+
+
−
H=
2
r1
r1
2
r2
r2
r12
The contributions from H0,1 (Z) and H0,2 (Z) result in
(1)
(1)
(2)
(2)
hϕ0 (Z)|H0,1 (Z)|ϕ0 (Z)i + hϕ0 (Z)|H0,2 (Z)|ϕ0 (Z)i = −2
(i)
Z2
,
2
where |ϕ0 i is the wave function of electron i. For this calculation, we used
the fact that the hamiltonian as well as the wave function describe an atom
with atomic number Z similar to hydrogen.
The contribution from the modified potential was already calculated in
(3.12).
(1)
hϕ0 (Z)|
1 (1)
1 (2)
(2)
|ϕ0 (Z)i + hϕ0 (Z)| |ϕ0 (Z)i = 2 Z
r1
r2
The remaining term was already evaluated in (6.8).
(1)
hϕ0 (Z)|
1 (1)
5
|ϕ0 (Z)i = Z
r12
8
.
.
6.5. EXCITED STATES OF HELIUM
109
The total variational energy is
5
5
E(Z) = −Z 2 + 2(Z − Z)Z + Z = Z 2 − (2Z − )Z
8
8
.
Minimisation of the energy yields
∂E(Z)
5 !
= 2(Z − Z + ) = 0
⇒
∂Z
16
5
= 1.69
Z∗ = Z −
16
5
5
∧
E(Z ∗ ) = −Z 2 + Z −( )2 = −2.848 · 27.21 eV = −77.5 eV
8
16
|
{z
}
.
E (1)
The first two terms coincide with the results of perturbation theory. Additionally, the energy is lowered due to correlation effects mentioned above.
The result is already very close to the experimental value (−78.8 eV).
6.5
Excited states of helium
Finally, we consider the lowest excited states of helium, to show interesting physical quantum phenomena.
For an excited state, one particle stays in the ground state n = 1, l = 0, ml = 0,
whereas the second particle is in n = 2, l, ml . According to (6.5), there are four
combinations
1
√ (|100i1 ⊗ |2lmi2 + |2lmi1 ⊗ |100i2 ) |s = 0, ms = 0i
2
|
{z
}
Singlet
Ψ+
B
1
√ (|100i1 ⊗ |2lmi2 − |2lmi1 ⊗ |100i2 ) |s = 1, ms ∈ {0, ±1}i Triplet
2
|
{z
}
Ψ−
B
(6.9)
These states are degenerate, if we do not take electron electron interaction
into account, and have the following energy (in zeroth order)
Z2
Z2
Z2
1
5 ∧
E0 =
−
−
=−
1+
= − = −68.0eV
2
2
2·1
2·2
2
4
2
The energy does not dependend neither on the spin state (singlet, triplet)
nor on the angular momentum quantum numbers (l, m). There are four
110
CHAPTER 6. IDENTICAL PARTICLES
spin states and for each state, there are for n = 2 four angular momentum
quantum numbers (l = 0, m = 0) and (l = 1, m = 0, ±1). Thus, there
is a sixteenth-fold degeneracy. To estimate the energy correction in first
order, we have to diagonalize H1 in this subspace. As the perturbation
hamiltonian is already diagonal in the states (6.9), this step can be omitted.
H1 is independent of spin, therefore matrix elements of H1
0
0
hψB | ⊗ hψS | H1 |ψB i ⊗ |ψS i = hψB |H1 |ψB0 ih ψS |ψS0 i
are proportional to the overlap of spin states. The spin states are orthonormal, thus the perturbation term of the spin quantum numbers is diagonal.
Similar considerations apply to the orbital part: Clearly, H1 is invariant
under rotations of the coordinate system (see (6.6)). Both position vectors,
r 1 and r 2 , are rotated simultaneous and kr 1 − r 2 k remains constant. The
generator of the rotations is the total angular momentum
L = L1 + L2
As the permutation operator is invariant under rotations, H1 commutes
with L and Lz commutes with L2 . The four orbital parts |ΨB i in (6.9) are
eigenstates of L2 and Lz .
As the orbital angular momentum is zero for the particle in state |n = 1, l =
0, m = 0i, the whole orbital angular momentum is determined by the particle in
state |n = 1, l, mi. The same holds for interchanged particle indices in the
tensor product. Meaning, the states of (6.9) are eigenstates of L2 and Lz with
eigenvalues ~2 · l(l + 1) und ~m. The orbital states have different quantum
numbers (l, m). Therefore, H1 is diagonal for the states of (6.9) and their
expectation values are only dependent on l. In other words, the states are
already adapted to the symmetry of the perturbation and we only need
for the energy correction (cf. (6.9)):
S/T
S/T
±
±
±
E (1) = hΨ±
B |H1 |ΨB ihΨS |ΨS i = hΨB |H1 |ΨB i
1
h100|1 ⊗ h2lm|2 H1 |100i1 ⊗ |2lmi2
=
2
1
h2lm|1 ⊗ h100|2 H1 |2lmi1 ⊗ |100i2
+
2
1
h2lm|1 ⊗ h100|2 H1 |100i1 ⊗ |2lmi2
± 2·
2
(6.10)
The first two terms are equal, the factor 2 occuring in the third term is due
6.5. EXCITED STATES OF HELIUM
111
to the equality of the mixed terms (for details see here : Sec. A.20 ).
Z
1
(1)
|ϕ2lm (r 2 )|2 dr 1 dr 2
E = |ϕ100 (r 1 )|2
|r 1 − r 2 |
|
{z
}
classical
Coulomb
interaction
Z
1
(6.11)
ϕ∗ (r 2 )ϕ2lm (r 2 )dr 1 dr 2
± ϕ∗2lm (r 1 )ϕ100 (r 1 )
|r 1 − r 2 | 100
|
{z
}
Exchange interaction
=: C ± A
C originates from the electrostatic Coulomb interaction and corresponds
to the term causing the energy shift in the calculation for the ground state.
The exchange term A describes another quantum phenomenon. It is caused
by the interference of the different involved orbital states. The integral in (6.11)
gives A > 0. Therefore, the spin singlet state (symmetric orbital part) is raised
energetically, whereas the spin triplet state (antisymmetric orbital part) is lowered
by A. This can be qualitatively understood: For an antisymmetric orbital
state, the probability to find both electrons very close to each other is strongly reduced. Thus, the Coulomb repulsion is small. For a symmetric orbital state,
the opposite is true. The energy levels are outlined in figure 6.4. Contrary
6
+A
6
C
E100 + E2lm
@
@
−A
?
?
Figure 6.4: Energy splitting of excited states for helium.
to the ground state, the lowest excited state is a spin triplet, since it is energetically favorable to have an antisymmetric orbital part. Even though there
is no spin interaction in H, the spin-dependent splitting comes into play due to
fermi statistics. This exchange interaction is the reason for collective magnetism. A similar mechanism leads to Hund’s rules. The singlet-triplet
112
CHAPTER 6. IDENTICAL PARTICLES
splitting is an order of magnitude higher than the electrostatic interaction in a
solid. By calculating the integrals in (6.11), we get
(1)
E1s,2s = C1s,2s ± A1s,2s
(1)
E1s,2p
= C1s,2p ± A1s,2p
= 11.4eV ± 1.2eV
= 13.2eV ± 0.9eV
(6.12a)
.
(6.12b)
By adding the zero-order energy, we get the total energy
E1s,2s = −56.6eV ± 1.2eV
E1s,2p = −54.8eV ± 0.9eV
Exp: − 58.8eV ± 0.4eV
Exp: − 57.9eV ± 0.1eV
.
(6.13a)
(6.13b)
The degeneracy is lifted for the spin s (singlet-triplet) as well as for the
angular momentum quantum number l (s-,p-Orbitals). Only the degeneracy for ms and m survives due to the rotation invariance in spin and
position space. The quantitative agreement between experiments and firstorder perturbation theory is not perfect. Nevertheless, two important effects can be qualitatively explained: (i) singlet-triplet splitting of the energy levels (ii) The splitting for the 2s state is larger than for the 2p state.
6.6
Occupation number representation
As discussed above, a basis vector for a Hilbert space HN describing N
identical particles consists of a symmetrized (Bosons) or antisymmetrized
(Fermions) tensor product of single-particle basis states (“orbitals”) |ϕm i:
S± |ϕm1 i ⊗ |ϕm2 i · · · ⊗ |ϕmN i
where we have denoted S+ = S (symmetrisation) for bosons, and S− = A
(antisymmetrisation) for fermions. The wave function is
Ψ(x1 , · · · , xN ) = S± ϕm1 (x1 )ϕm2 (x2 ) · · · ϕmN (xN ) .
For simplicity, we have consider here a discrete basis for a single particle
system, and taken, mi = 1, 2, · · · , ∞. In general, m can describe a set of
quantum number (e.g. m ≡ (n, l, m)), or be continuous.
Each basis element is, therefore, specified by the set of positive integers {m1 , · · · , mN }.
This provides the information in which orbital each particle is. However, due
to the indistinguishability of the particles, this is a redundant information,
since any permutation describes the same many body-state, up to a constant.
6.6. OCCUPATION NUMBER REPRESENTATION
113
A more appropriate specification of a many-body basis state is the socalled “occupation number representation” (Besetzungszahldarstellung).
Here one specifies for each orbital |ϕm i, the number of particles Nm which are
present in (occupy) that orbital.:
|N1 , N2 , N3 , · · · i
(6.14)
Obviously, for bosons each Nm can be any integer from 0 to ∞, for fermions
just 0 or 1 due to the Pauli principle. Notice that, especially for fermions,
this specifies the many-body state only up to ± sign which depends on
the order in which the orbitals are occupied. However, one can fix a convention according to which, for example, single-particle states are sorted
in order of increasing m from left to right.
6.6.1
Fock Space
For a Hilbert space
HN with a given particle number N , the Nm are conP
strained to be m Nm = N
However, one can consider the space obtained as the direct sum of all the
HN for N = 0, · · · , ∞. This space with variable particle number is called
Fock space. A possible basis for the Fock space is given by the set of states
(6.14) with Nm = 0, 1 for fermions or Nm = 0, · · · , ∞ for bosons but otherwise no restriction on the sum of the Nm .
The state |0i with no particles is called the vacuum state. This state is normalized h0|0i = 1 and should not be confused with the 0 “vector” which
has norm 0.
The scalar product in the Fock space is naturally defined as
hN10 , N20 , N30 , · · · |N1 , N2 , N3 , · · ·i = δN10 ,N1 δN20 ,N2 δN30 ,N3 · · ·
(6.15)
A particular useful formalism to describe a Fock space of bosons or fermions
is given by second quantisation which will be introduced in Sec. 9.
114
CHAPTER 6. IDENTICAL PARTICLES
Chapter 7
Charged particle in an
electromagnetic field
7.1
Classical Hamilton function of charged particles in an electromagnetic field
Remember from classical mechanics
H AMILTONIAN FOR A CHARGE IN A EM
H=
FIELD
2
1 q
p − A(r, t) + qφ(r, t)
2m
c
(7.1)
We check that it gives the correct equation of motion. First consider that
the velocity
d
∂
1 q v= r=
H=
p− A
(7.2)
dt
∂p
m
c
is not the usual p/m. Therefore, the Lorentz force ought to be
F =
d
d
q d
mv = p −
A
dt
dt
c dt
(7.3)
From the Hamilton equations we have
1
q
q
d
pi = −∇i H = (p − A) · ∇i A − q∇i φ
dt
|m {z c } c
v
115
(7.4)
116CHAPTER 7. CHARGED PARTICLE IN AN ELECTROMAGNETIC FIELD
Consider that
d
∂
A(r(t), t) = vj ∇j A + A
dt
∂t
Putting everything ((7.3),(7.4),(7.5)) together, we obtain
q
1∂
Fi = (∇i Aj vj − vj ∇j Ai ) − q
Ai + ∇i φ
c
c ∂t


q
= v × (∇ × A) + qEi
| {z }
c
B
(7.5)
i
The correct expression for the Lorentz Force. So indeed (7.1) is the correct
Hamilton function.
As usual, the correspondence principle of quantum mechanics tells us that
(7.1) is also the quantum mechanical Hamilton operator for a particle in an electromagnetic field.
The Schrödinger equation, thus, becomes
2
1 q
∂
p − A(r, t) + qφ(r, t) Ψ
i~ Ψ =
(7.6)
∂t
2m
c
The operator for the velocity
d
1
q
r = i[H, r] = (p − A)
dt
m
c
(7.7)
has the same expression as the classical one (7.2)
For a particle with spin S there is an additonal interaction between spin
and magnetic field
∆H = −µB · S
7.2
µ=g
q
2m
(7.8)
Gauge invariance
Due to the gauge invariance in classical electrodynamics, physical properties should not change under the gauge transformations
1∂
χ
(7.9)
c ∂t
The point here is that A and φ are not observable. (7.6) seems in contradiction with that, since A and φ enter explicitly in the equation for the
A0 = A + ∇χ
φ0 = φ −
7.3. LANDAU LEVELS
117
wavefunction, in contrast to classical mechanics in which only fields enter
the expression for the Lorentz force.
The answer to this issue is that in fact also the wave function is not observable,
and it will turn out that any observable quantity does not change under a gauge
transformation (7.9).
To show this let us consider a modified wave function Ψ0
q
Ψ0 (r, t) = ei ~c χ(r,t) Ψ(r, t)
(7.10)
Now we show that, if Ψ is a solution of (7.6), then Ψ0 ((7.10)) is a solution of
the Schrödinger equation for the modified potentials (7.9):
1 q 2
∂
p − A0 Ψ0 + qφ0 Ψ0
i~ Ψ0 =
(7.11)
∂t
2m
c
We prove it here : Sec. A.23 . Obviously (7.10) has the same probability
density as Ψ. It can be easily shown that all expectation values of observables are equal in the two gauges, as it should be.
For example consider the velocity (7.2)
qc
qc
q
q
q
m v 0 Ψ 0 = (p − A0 )Ψ0 = (p − A)Ψ0 = ei ~ χ (p − A)Ψ = e i ~ χ mvΨ
c
c
c
(7.12)
where we have used (A.38), (A.39), and (A.40).
This means that matrix elements of v are gauge invariant:
hΨ01 | v 0 |Ψ02 i = hΨ1 | v |Ψ2 i
Notice that, in contrast to v, the canonical momentum p is not a gauge invariant operator. It is, therefore, not an observable.
7.3
Landau Levels
The situation of a constant, homogeneous magnetic field is relevant for
the quantum Hall effect. Reminder: In classical physics, charged particles
in an homogeneous magnetic field move along circles.
Let us consider the case of a constant field in the z direction:
B = Bez
We have several choices for the vector potential, which are connected by
gauge transformations. One possibility is
A = −B y ex
118CHAPTER 7. CHARGED PARTICLE IN AN ELECTROMAGNETIC FIELD
(Other possibilities are A = B x ey , or the symmetric case A = 12 r × B =
1
(xey − yex )).
2
The Hamiltonian thus becomes
1 q
H=
(px + By)2 + p2y + p2z
2m
c
The operators px and pz (but not py ) commute with the hamiltonian. Therefore we can find common eigenvalues of H, px , pz . The corresponding
eigenfunctions will have the form:
Ψ(r) = ei(kx x+kz z) f (y)
Therefore, the time independent Schrödinger equation becomes
qB 2
i(kx x+kz z)
i(kx x+kz z) 1
2
2 2
He
f (y) = e
ŷ) + p̂y + ~ kz f (y)
(~kx +
2m
c
= ei(kx x+kz z) Ef (y)
which is the Schrödinger equation for a shifted harmonic oscillator.
We can introduce the shifted y coordinate:
qB
qB
ȳ =
y + ~kx
c
c
which has the same commutation rules with py :
[ȳ, py ] = i~
The Schrödinger equation becomes
p2y
q2B 2 2
+
ȳ f (y) = (E − ~2 kz2 )f (y)
2m 2mc2
(7.13)
This describes an harmonic oscillator with frequency
m 2
q2B 2
ωc =
2
2mc2
ωc =
|qB|
mc
(7.14)
which is the well-known cyclotron frequency (in c.g.s. units). The energy
eigenvalues are, thus , given by
1
~2 kz2
En,kz = ~ωc (n + ) +
2
2m
(7.15)
7.3. LANDAU LEVELS
119
There is no dependence of the energies on kx so there is a large degeneracy. Typically, the quantum Hall effect is interesting in two-dimensional
systems, where the z coordinate is confined. 1 In that case, there are discrete, highly degenerate energy levels separated by the energy ~ωc , the
Landau levels.
The energy separation between the Landau levels depends on the magnetic field and can be estimated as
~ωc ≈ 10−4 eV × (B/T esla)
Notice that 10−4 eV corresponds to approximatively 1 Kelvin temperature.
The wave function of the ground state of the harmonic oscillator in (7.13)
has the behavior
ȳ 2
f (ȳ) ∼ exp − 2
2a
where
r
a=
1
s
~
=
mω0
◦
1
~c
≈ 100A (B/T esla)− 2
qB
One could, for example, consider a “box” potential in the z directions
120CHAPTER 7. CHARGED PARTICLE IN AN ELECTROMAGNETIC FIELD
Chapter 8
Field quantisation
8.1
Continuum systems: classical treatment
On the way to the concept of photons, we have to figure out how to quantize fields, i.e. continuum variables of position x and time t. As a first example, we consider a vibrating string. We start by summarizing some of the
classical results. See e.g. the lecture “Analytical Mechanik” by W. von der
Linden with lecture notes here. The notation here may be slightly different.
The vertical displacement u(x, t) of an infinitesimal string segment at horizontal position x along the string (0 < x < L) obeys the equations of
motion EOM
∂ 2u
∂ 2u
=
.
(8.1)
c2
∂x2
∂t2
This equation can be obtained from the corresponding Lagrange functional
Z L
L[u̇, u, t] =
`(u(x, t), u0 (x, t), u̇(x, t), x, t) dx
(8.2)
0
with the Lagrange density
`(u(x, t), u0 (x, t), u̇(x, t), x, t) =
ρ 2
u̇ − c2 u02 .
2
(8.3)
Here, ρ is the mass density of the string, and
u̇ ≡
∂u
∂t
u0 ≡
∂u
.
∂x
The Hamilton (minimal action) principle leads to the continuum Euler equa121
122
CHAPTER 8. FIELD QUANTISATION
tions of motion1
d ∂`
d ∂`
∂`
+
(8.4)
=
dt ∂ u̇ dx ∂u0
∂u
which can be easily shown to give the EOM (8.1).
Equation (8.4) can be obtained by discretising the values of x to xi , with
spacing ∆x. Then L can be seen as a function of the ui ≡ u(xi ) and u̇i . In
the continuum (∆x → 0) limit, the usual (discrete) Lagrange II equation
d ∂L
∂L
=
dt ∂ u̇i
∂ui
(8.5)
can be shown to go over to (8.4). One way to prove this ((proof here): Sec.
A.21 ) is to start from a discrete version of (8.2):
X
L = ∆x
`(ui , u0i , u̇i , xi , t)
(8.6)
i
8.2
Quantisation
The starting point for quantising a set of independent discrete variables ui
is to identify their canonical momenta pi , and postulating commutation rules
between the variables and their momenta
[ûi , p̂j ] = i~ δij .
(8.7)
We thus need the canonical momenta. Once we have the Lagrange function, we know from classical mechanics that the pi are given by
pi =
∂L
.
∂ u̇i
Thus, for the present case (8.6)
pi = ∆x
∂`
∂ u̇i
Introducing the momentum field
π(xi ) =
∂`
pi
=
,
∆x
∂ u̇(xi )
d
Here the “total derivatives” dt
and
account the dependence of u on x and t.
1
d
dx
(8.8)
are meant in the sense that one takes into
8.2. QUANTISATION
123
which in this case (8.3) equals ρu̇, the commutation rules (8.7) become
[ûi , π̂j ] = i~
δij
⇒
∆x ∆x→0
[û(x ), π̂(x 0 )] = i ~ δ(x − x 0 ) .
(8.9)
The commutation rules between the u and the π are zero:
[û(x), û(x0 )] = [π̂(x), π̂(x0 )] = 0 .
(8.10)
(8.8) with (8.9) and (8.10) determine the quantisation of the field u.
The Hamilton function is obtained as usual from the Legendre transformation. Replacing classical fields with their operators yields the Hamilton
operator
Z
Z
Ĥ = u̇(x)π̂(x) dx − L ≡ ĥ (π̂(x), û(x)) dx
2
2 #
Z " 1 L 1
dû(x)
=
π̂(x) + c2 ρ
dx
(8.11)
2 0 ρ
dx
whith the Hamiltonian density ĥ.
This example summarizes the generic procedure (which we will use later
for other cases) to quantize a field
• Determine the Lagrange density (here (8.3)) which gives the correct
classical wave equation
• With (8.8) determine the canonical momenta π(x)
• Postulate the commutation rules (8.9) (which classically were the
Poisson brackets)
• Write the Hamilonian in terms of the π̂ and û. This will determine
the quantum mechanical time evolution in the usual way.
After quantising the field, we have to solve the problem i.e. first put the
8.2.1
Hamiltonian in diagonal form
In order to carry out calculations it is convenient to bring the Hamiltonian
(8.11) in diagonal form. This is achieved by starting from the set of classical
solutions of the homogeneous wave equation
u(x, t) = Un (x) cos(ωn t)
124
CHAPTER 8. FIELD QUANTISATION
q
2
In the present case, Un (x) =
sin(kn x) and ωn = ckn with kn = πn/L.
L
The important point is the Un (x) build a complete orthonormal basis with the
usual properties
Z L
Un (x) Um (x) dx = δn,m
(8.12)
0
∞
X
n=1
Un (x)Un (x0 ) = δ(x − x0 ) .
One can use these relation to introduce new operators
Z L
Um (x) û(x) dx
q̂m =
0
Z L
Um (x) π̂(x) dx .
p̂m =
(8.13)
(8.14)
0
The inverse relations read
û(x) =
π̂(x) =
∞
X
n=1
∞
X
Un (x)q̂n
Un (x) p̂n
(8.15)
n=1
It is straightforward to show (see here): Sec. A.21 that the q̂m and p̂m obey
canonical commutation rules
[q̂m , p̂m0 ] = i~δm,m0
[q̂m , q̂m0 ] = [p̂m , p̂m0 ] = 0
(8.16)
The Hamiltonian in terms of these operators has a particularly simple form
(see here): Sec. A.21:
1X 1 2
2 2
Ĥ =
p̂ + ρ ωn q̂n .
(8.17)
2 n
ρ n
with ωn = ckn .
This is the Hamilton operator of a set of uncoupled harmonic oscillators
with mass ρ and frequencies ωn . The solution goes via the usual ladder
operators:
r
1
ρωn
1
†
q̂n − i √
p̂n
(8.18)
b̂n = √
~
~ρωn
2
r
1
ρωn
1
b̂n = √
q̂n + i √
p̂n .
~
~ρωn
2
8.2. QUANTISATION
125
with the known commutation relations
[b̂n , b̂m ] = 0
[b̂†n , b̂†m ]
[b̂n , b̂†m ]
(8.19)
=0
= δnm .
The hamiltonian finally becomes
∞
X
1
†
Ĥ =
~ωn b̂n b̂n +
.
2
n=1
Its eigenstates can be written in terms of the set of quantum numbers
{Nn , n = 1, 2, · · · } of each one of the harmonic oscillators as
|N1 , N2 , · · · i .
(8.20)
where each Nn can be 0, 1, · · · , ∞.
These states have the same structure as the states of the many-particle Fock
space (9.43). The scalar product is the same as (6.15).
The interesting aspect of this result is that we started from a field, whose
excitations are waves, and we come out with a description in which the
wave modes are similar to particles. This is an important characteristic of
quantum mechanics in which, on the one hand, particles are described by
waves but also the opposite waves are described by particles.
8.2.2
Creation and destruction operators
We already know from Quantum Mechanics I the properties of the ladder
operators b̂n and b̂†n . Acting on the basis states they have the properties of
increasing or decreasing the excitations:
p
b̂†n |N1 , N2 , · · · Nn , · · · i = Nn + 1 |N1 , N2 , · · · (Nn + 1), · · · i ,
p
(8.21)
b̂n |N1 , N2 , · · · Nn , · · · i = Nn |N1 , N2 , · · · (Nn − 1), · · · i .
Obviously the same operators with the same properties can be formally
introduced for a bosonic Fock space. They are completely defined by their
commutation rules (8.19) and, due to their properties (8.21) to create and
destroy particles are termed creation and destruction (or annihilation)
operators.
One important composite operator is the number operator b̂†n b̂n which
counts the number of particles in level n:
b̂†n b̂n |N1 , N2 , · · · Nn , · · · i = Nn |N1 , N2 , · · · Nn , · · · i
(8.22)
126
CHAPTER 8. FIELD QUANTISATION
Chapter 9
Second quantisation
9.1
Quantisation of the Schrödinger field
We could now consider the wave function in the Schrödinger equation,
also known as S CHRÖDINGER FIELD, as a classical field, and adopt the
procedure described in Chap. (8)) to quantise it. Since the system is somewhat quantised twice, this is referred to as Second Quantisation. A more
correct term is Field Quantisation.
9.1.1
Lagrangian of the Schrödinger field
Starting from the Schrödinger equation of a single particle in a potential
V (x),
−
~2 2
∇ Ψ + V (x)Ψ = i~Ψ̇ .
2m
we can consider the real and the imaginary part as independent fields
since Ψ and Ψ∗ are complex functions of x and t. Another possibility is
to take Ψ and Ψ∗ as indepenent fields.
We are looking for a Lagrangian, which produces the Schrödinger equation via the Lagrange II equations. A possible form satisfying this is easily
seen to be
Z
~2 2
∗
L = Ψ i~Ψ̇ +
∇ Ψ − V (x)Ψ d3 x .
(9.1)
2m
The Lagrange II equations are now evaluated with respect to Ψ∗ . The canonical momentum for Ψ∗ vanishes, due to
π̃ :=
δL
=0.
δ Ψ̇∗
127
(9.2)
128
CHAPTER 9. SECOND QUANTISATION
Therefore we have
δL
~2 2
∇ Ψ − V (x)Ψ
=
i~
Ψ̇
+
δΨ∗
2m
~2 2
∇ Ψ + V (x)Ψ .
i~Ψ̇ = −
2m
0=
i.e.
This is indeed the Schrödinger equation. We now derive the Lagrangian II
equation with respect to the second independent variable Ψ. The canonical
momentum is (see (9.1))
π :=
δL
= i~Ψ∗ (x, t)
δ Ψ̇
(9.3)
and the Lagrangian II equation gives the complex conjugate Schrödinger
equation.((proof here): Sec. A.22 ). As usual, we obtain the Hamilton
function by a Legendre transformation.
(9.3)
(9.2)
= i~Ψ∗
=0
HSQ
z}|{
Z z}|{
δL
δL ∗
Ψ̇ +
Ψ̇ − L d3 x
=
δ Ψ̇∗
δ Ψ̇
2
Z ~
2
∗
∗
∗
=
i~Ψ Ψ̇ − i~Ψ Ψ̇ − Ψ
∇ − V (x) Ψ d3 x .
2m
The first two terms cancel out and therefore we get the Hamilton function
Z
~2 2
∗
∇ + V (x) Ψ d3 x .
(9.4)
HSQ = Ψ −
2m
Despite of the fact that the Lagrangian is antisymmetric between Ψ and
Ψ∗ , we obtained a symmetric form of the Hamilton function. We recognize
this as the energy expectation value.
HSQ = hΨ|Ĥ|Ψi
(9.5)
Up to now, Ψ and Ψ∗ were classical fields. As next step, we are going to
quantise them.
9.1.2
Quantisation
We use the relation (9.3) π = i~Ψ∗ to find the canonical momentum to Ψ.
We replace both fields with field operators and we require as before
[Ψ̂(x), π̂(x0 )] = i~δ(x − x0 ) .
(9.6)
9.1. QUANTISATION OF THE SCHRÖDINGER FIELD
129
This results in
[Ψ̂(x), Ψ̂† (x0 )] = δ(x − x0 ) .
(9.7)
[Ψ̂(x), Ψ̂(x0 )] = 0
(9.8a)
[Ψ̂† (x), Ψ̂† (x0 )] = 0 .
(9.8b)
†
Additionally we require that Ψ̂(x) and Ψ̂ (x) commute with each other.
As shown in Chapter (8) we expand the field operators in terms of the
stationary solutions of the field equation, i.e. the eigenfunctions of the
Schrödinger equation ϕn (x) with operator coefficients b̂n .
X
Ψ̂(x) =
b̂n ϕn (x)
(9.9)
n
†
Ψ̂ (x) =
X
b̂†n ϕ∗n (x) .
n
The canonical commutation rules for the fields (9.6) and (9.7)), become for
the operators bn (we will omit theˆfor bn ) proof here: Sec. A.24 :
[bn , b†m ] = δn,m
[bn , bm ] = 0
(9.10a)
(9.10b)
[b†n , b†m ] = 0 .
(9.10c)
Since we quantised the wave functions, the Hamilton function becomes
the Hamiltonian operator proof here: Sec. A.24
H AMILTON OPERATOR IN SECOND QUANTISATION
IN ITS EIGENBASIS
~2 2
∇ + V (x) Ψ̂(x) d3 x
= Ψ̂ (x) −
2m
X
=
En b†n bn .
Z
ĤSQ
†
(9.11a)
(9.11b)
n
Again these kinds of “ harmonic oscillator” operators b†n describe the creation of a particle in the eigenstate of the Hamiltonian (“level”) with singleparticle wave function ϕn (x). The advantage of (9.11a) is, that one can create
130
CHAPTER 9. SECOND QUANTISATION
many (identical) particles with different wave functions. In first quantisation, this is only possible with a symmetrized function with more than one
variable Ψ(x1 , · · · , xn ).
It is thus convenient to use the one-to-one correspondence with a quantummechanical system consisting of a set of harmonic oscillators. Its (unnormalized) eigenstates are
|ψi =
Y
(b†n )Nn |0i ,
(9.12)
n
where Nn is the occupation of the level n.
Here, |0i is vacuum state with the following properties:
bn |0i = 0
for all n
(9.13)
and, due to (9.9)
Ψ̂(x)|0i = 0
for all x
(9.14)
Another possibility is to expand the field operators in terms of an alternative set of orthonormal basis functions χn .
Ψ̂(x) =
X
ĉn χn (x)
Ψ̂† (x) =
X
ĉ†n χ∗n (x) .
(9.15)
n
n
One can transform directly from the operators b to c:
bn =
X
m
Unm ĉm
b†n =
X
∗
Unm
ĉ†m ,
m
R
using the unitary matrix Unm = ϕ∗n (x)χm (x) d3 x. This follows from (9.9)
by multiplying from the left with ϕn and integrating (Exercise).
The creation and destruction (or annihilation) operators c†n (cn ) (as for the
operators b we will omit theˆfrom now on) obey the same commutation
rules as the operators b due to the orthonormality of the basis. proof here:
Sec. A.24 The Hamiltonian is no longer diagonal in the c operators Instead, we get for the
9.2. TRANSFORMATION OF OPERATORS TO SECOND QUANTISATION131
H AMILTONIAN IN SECOND QUANTISATION
FOR AN ARBITRARY ORTHONORMAL BASIS
ĤSQ =
X
hn,m c†n cm
(9.16)
n,m
with matrix elements
Z
hn,m =
χ∗n (x)
~2 2
∇ + V (x) χm (x) d3 x :
−
2m
(9.17)
We could expand the field operators in terms of a arbitrary linearly independent basis functions, e.g. 1, x, x2 , x3 , . . .. However, the commutation
rules would not be as simple in this case.
One could think, that the second quantisation leads to different physical results. This is not the case. Both formalisms are completely equivalent.
However, second quantisation has several practical advantages, especially
when dealing with many-particle systems. Moreover, second quantisation is
more flexible for phenomena without a constant number of particles (e.g.
annihilation and creation of photons).
9.2
Transformation of operators to second quantisation
The procedure to quantise fields described above consisted in starting
from the Lagrange and Hamilton function (total energy), quantise the fields
Ψ, and to expand them in terms of orthonormal solutions of the field equation with the coefficients (amplitudes) being operators. Based on this approach, arbitrary observables can be translated into second quantisation.
9.2.1
Single particle operators
In first quantisation the Hamiltonian of N identical, non interacting particles is the sum of the Hamiltonians of each particle
X
Ĥ =
H(r̂ i , p̂i )
(9.18)
i
132
CHAPTER 9. SECOND QUANTISATION
In second quantisation, this is replaced by the quantized Hamiltonian of
the Schrödinger field. As discussed in (9.5), the latter is composed of the
Expectation value of the first-quantisation Hamiltonian
HSQ = hΨ|Ĥ|Ψi ,
where the wave function Ψ is quantized according to (9.7). This procedure can
be applied in the same way to arbitrary, so-called single-particle operators.
Single-particle operators are the ones, which have the form (9.18) in first quantisation, i.e. they are composed of an independent sum over all particles of the same
operator acting on just one of the particles. Terms describing correlations between the particles are not described by single-particle operators, and will
be discussed below. We apply the same procedure to an arbitrary operartor Ô (generically expressed as a function of x̂ and p̂ = −i~∇). e.g. Ô = x̂.
The expectation value is
Z
hÔi = Ψ∗ (x) Ô Ψ(x) d3 x .
Using the same procedure as for Ĥ this leads to
Z
(Ô)SQ = Ψ̂† (x) Ô Ψ̂(x) d3 x .
(9.19)
As in (9.9), we can expand the field operator into eigenfunctions ϕn (x) of
Ĥ (or in arbitrary orthogonal functions χn (x), as in (9.15)) and get
X
Ψ̂(x) =
ϕn (x)b̂n
n
(Ô)SQ =
XZ
ϕ∗n (x)
n,m
|
(Ô)SQ =
X
Ô ϕm (x) d x b†n bm
{z
}
3
On,m
On,m b†n bm .
(9.20)
n,m
This result provides an alternative expression for arbitrary (single-particle) operators
in second quantisation. The Hamiltonian and - as just shown - arbitrary
single-particle operators are expressed in terms of creation and destruction operators as
X
ÔSQ =
On,m b†n bm .
(9.21)
n,m
with On,m = hϕn |Ô|ϕm i =
Z
dx ϕ∗n (x)Ôϕm (x) .
(9.22)
9.2. TRANSFORMATION OF OPERATORS TO SECOND QUANTISATION133
Physical meaning of the field operator Ψ̂
The particle density ρ(x) in x in first quantisation becomes the particle
density operator in second quantisation.
ρ̂(x) ≡ Ψ̂† (x)Ψ̂(x)
(9.23)
Then one can show that the operator Ψ̂† (x) creates a particle at the position x. This means that
|xi ≡ Ψ̂† (x) |0i
(9.24)
becomes a state with one particle in x (with a delta-wave function). In
order to proof this, we show that |xi is an eigenstate of the density (9.23)
Proof:
ρ̂(y) |xi =Ψ̂ † (y)Ψ̂ (y)Ψ̂† (x) |0i
= Ψ̂† (y) Ψ̂† (x)Ψ̂(y) + δ(x − y) |0i = δ(x − y)Ψ̂† (y) |0i
= δ(x − y)Ψ̂† (x) |0i = δ(x − y) |xi
Equivalence between FQ and SQ for single-particle operators
As discussed above, second quantisation is just an alternative formalism
to deal with many-particle systems in an efficient way. Of course, the requirement is that first (FQ) and second (SQ) quantisation should give the
same physical results. In other words, expectation values of corresponding
operators must coincide in the two formalisms.
To evaluate expectation values of operators in FQ one has first to symmetrize (or antisymmetrize) the wave function, as discussed in Chap. 6,
which for more than two or three particles becomes quite laborious.
In second quantisation (anti)-symmetrisation is not necessary, as it is automatically taken care by states of the form (9.12).
We will not prove systematically the equivalence of FQ and SQ here. This
would be a lengthy procedure. We refer to the abundant literature. However, we will illustrate this for simple examples. This gives us the opportunity to exercise a bit with creation and destruction operators.
We consider an arbitrary single-particle operator Ô, e.g. the position operator x̂. For a particle with wave function ϕν (x), the expectation value of
Ô in first quantisation is given by
Z
hÔi = ϕ∗ν (x) Ô ϕν (x) dx .
134
CHAPTER 9. SECOND QUANTISATION
In second quantisation, the state has the following form
|νi = b†ν |0i
and the expectation value of the operator (9.20) becomes
X
hÔi =
On,m h0|bν b†n bm bν† |0 i
(9.25)
n,m
To evaluate this, we first deal with the right hand side of (9.25):
bm b†ν |0i = b†ν bm |0i + [bm , b†ν ] |0i = δm,ν |0i
| {z } | {z }
0
δm,ν
Another way to obtain to this result is to interpret the state b†ν |0i as a “single
particle in level ν”. bm destroys a particle in level m: if m 6= ν this results in
“nothing” (0), otherwise, if m = ν, we arrive at the vacuum state (state without
particles).
The left hand side is evaluated by using the hermitian conjugate, which
results in a similar expression as above:
†
h0| bν b†n = bn b†ν |0i = (δn,ν |0i)† = δn,ν h0|
In total, (9.25) results in (cf. also (9.21))
Z
X
On,m δn,ν δm,ν h0|0i = Oν,ν = ϕ∗ν Ô ϕν (x) dx .
hÔi =
n,m
This is also what is expected in first quantisation. First and second quantisation lead to the same result.
More interesting is the case of a state with more than one particle in second
quantisation. Here, several levels are occupied, e.g. for two particles:
|ν, µi =
1 † †
b b |0i .
Zνµ ν µ
The normalisation constant Zνµ can be determined by√analogy with a system
of harmonic oscillators. For ν = µ, we have Zνν = 2. For ν 6= µ the state
corresponds to having two “oscillators” in the first excited state. Here the
normalisation constant is Zνµ = 1 · 1 = 1. The normalisation constant can
be also determined explicitly by carry out the scalar product of the state
with itself. (see here): Sec. A.25 The expectation value of an arbitrary
single-particle operator (9.21) in this state becomes
1 X
hOi = 2
Onm h0|bµ bν b†n bm bν† bµ† |0 i
(9.26)
Zνµ nm
9.2. TRANSFORMATION OF OPERATORS TO SECOND QUANTISATION135
As before, we first consider the right-hand side of the expectation value.
We want to end up having the annihilation operator bm on the right, in
order to destroy the vacuum state.
bm b†ν b†µ |0i = b†ν bm b†µ |0i + δm,ν b†µ |0i
= b†ν b†µ bm |0i +δm,µ b†ν |0i + δm,ν b†µ |0i
| {z }
=
=0
†
δm,µ bν |0i
+ δm,ν b†µ |0i .
(9.27)
This result can be understood in the following way: bm should destroy a particle
in level m. If there is no particle in m, the result is 0, otherwise the other particle
in the initial state remains.
The left-hand side is again evaluated by using the hermitian conjugate of
the right-hand side:
h0|bµ bν b†n = h0|bµ δn,ν + h0|bν δn,µ .
Combining the two we get (cf. (9.27))
h0|bµ bν b†n bm b†ν b†µ |0i = δn,ν δm,µ h0|bµ b†ν |0i + δn,ν δm,ν h0|bµ b†µ |0i
+δn,µ δm,µ h0|bν b†ν |0i + δn,µ δm,ν h0|bν b†µ |0i
Due to h0|bm b†n |0i = δnm , this becomes
h0|bµ bν b†n bm b†ν b†µ |0i = δn,ν δm,µ δν,µ + δn,ν δm,ν + δn,µ δm,µ + δn,µ δm,µ δν,µ
= 2δn,ν δm,ν δν,µ + δn,ν δm,ν + δn,µ δm,µ .
For ν = µ this results in (9.26) with (9.28)
h0|bµ bν b†n bm b†ν b†µ |0i = 4δn,ν δm,ν .
so that the eigenvalue is
4
hOi = Oν,ν = 2Oν,ν .
2
For ν 6= µ wir get (9.26) mit (9.28)
h0|bµ bν b†n bm b†ν b†µ |0i = δn,ν δm,ν + δn,µ δm,µ ,
i.e.
hOi = Oν,ν + Oµ,µ
(9.28)
136
CHAPTER 9. SECOND QUANTISATION
These are results we would obtain in first quantisation for two particles
in the symmetric states
|Ψi = |ϕν i1 ⊗ |ϕν i2
bzw
1
|Ψi = √ |ϕν i1 ⊗ |ϕµ i2 + |ϕµ i1 ⊗ |ϕν i2
2
To evaluate this in first quantisation, we shuld have used the respective
operators in first quantisation for two particles
Ô = Ô1 + Ô2 .
(e.g. x̂ = x̂1 + x̂2 ).
Summary: single-particle operators
In general, for a system with N identical particles a single-particle operator
Ô has the form
SINGLE - PARTICLE OPERATORS
for Ô =
N
X
O(x̂i , p̂i )
in first quantisation
i=1
ÔSQ =
X
On,n0 b†n bn0
in second quantisation
(9.29)
n,n0
with
On,n0 = hϕn | Ô|ϕn0 i =
Z
ϕ∗n (x)O(x, −i~∇)ϕn0 (x)d3 x .
The many-body basis states are given in (9.12). These bosonic states are
automatically symmetrised.
9.2.2
Two-particle operators
In addition to the single-particle operators that describe the kinetic energy and the interaction with external potentials, two-particle operators
9.2. TRANSFORMATION OF OPERATORS TO SECOND QUANTISATION137
also play an important role. Two-particle operators describe the interaction between particles expressed as a sum of interactions between pairs of
particles. One of the most important examples is the Coulomb interaction.
We start from the classical form of the interaction between two charge densities ρ1 (x) and ρ2 (x0 )
Z Z
ρ1 (x)V (x, x0 )ρ2 (x0 )d3 xd3 x0
Z Z
1
=
ρ(x)V (x, x0 )ρ(x0 )d3 xd3 x0 − self.energy ,
2
Ĥint =
where we use the total charge density ρ(x) = ρ1 (x) + ρ2 (x), as the two
charges are indistinguishable. The factor 1/2 compentsates for the double counting and the self.energy term subtracts the interaction energy
ρ1 (x)ρ1 (x0 ) of a charge with itself. We now replace the particle density
with the corresponding operator in second quantisation.
ρ̂(x) = Ψ̂† (x)Ψ̂(x)
(9.30)
and get
Ĥint
1
=
2
Z Z
Ψ̂† (x)Ψ̂(x)V (x, x0 )Ψ̂† (x0 )Ψ̂(x0 )d3 xd3 x0 − self.energy , .
(9.31)
The subtraction of the self energy term can be taken into account by requiring that the interaction vanishes if less than two particles are present,
i.e.
Ĥint |0i = 0
Ĥint b†l |0i = 0 .
In second quantisation this requirement is naturally achieved by “normal
ordering” creation and destruction operator (i.e., all destruction operators are
moved to the right). We obtain:
138
CHAPTER 9. SECOND QUANTISATION
I NTERACTION PART OF THE H AMILTONIAN IN SECOND QUANTISATION
IN TERMS OF FIELD OPERATORS ( CF. (9.31))
for Ĥint
(Ĥint )SQ
N
1 X
=
V (x̂i , x̂j )
in FQ
2 i6=j=1
Z Z
1
=
Ψ̂† (x)Ψ̂† (x0 )V (x, x0 )Ψ̂(x0 )Ψ̂(x)d3 xd3 x0
2
(9.32)
in SQ .
By introducing an arbitrary single-particle basis ϕn (x) as in (9.9) and by
using (9.32) we obtain
I NTERACTION PART OF THE H AMILTONIAN IN SECOND QUANTISATION
IN ARBITRARY SINGLE - PARTICLE BASIS
1 X
Vn b†n1 b†n2 bn3 bn4
2 n ,n ,n ,n
Z Z1 2 3 4
ϕ∗n1 (x)ϕ∗n2 (x0 )V (x, x0 )ϕn3 (x0 )ϕn4 (x)d3 xd3 x0
Vn =
Ĥint =
(9.33)
= hn1 | hn2 | V̂ |n4 i |n3 i .
9.3
Second quantisation for fermions
Up to now we used a form that correctly reproduces the particle statistics
for bosons. To describe fermions we have to use different creation and
destruction operators. We consider a two fermion state:
1
|Ψi = √ |ϕ1 i1 ⊗ |ϕ2 i2 − |ϕ2 i1 ⊗ |ϕ1 i2 .
2
9.3. SECOND QUANTISATION FOR FERMIONS
139
In second quantisation, this state has the form
|Ψi = c†1 c†2 |0i .
To assure the correct symmetry, we clearly require
c†i c†j = −c†j c†i
and {c†i , c†j } = 0
with {A, B} := AB + BA
(Anti-commutator) .
For fermions we have also to consider spin (σ = ±1 this means the sign
of one component, e.g. of z) . In general, the following anti commutation
relations hold (without proof) .
A NTI - COMMUTATION RELATIONS FOR FERMIONS
{c†i,σ , c†j,σ0 } = 0
{ci,σ , cj,σ0 } = 0
{ci,σ , c†j,σ0 } = δi,j δσ,σ0 .
(9.34)
(9.35)
(9.36)
The Pauli principle is accounted for, since (c†l,σ )2 |0i = 0.
From now on we can include the spin σ in the total quantum number n, unless
we need to specify spin explicitly. In other words, n describes both the
orbital quantum numbers as well as the spin projection: nσ → n. In this
way, single-particle operators have the same form for both bosons and
fermions. (cf. (9.29))
S INGLE - PARTICLE OPERATORS FOR FERMIONS IN SQ
Ô =
X
On,n0 c†n cn0
n,n0
On,n0 = hϕn | Ô|ϕn0 i
(9.37)
140
CHAPTER 9. SECOND QUANTISATION
Operators are often independent of spin, so that On,n0 is diagonal in spin
and only depends on the orbital part of n, n0 . I.e., with explicit introduction
of spin indices
Onσ,n0 σ0 = δσ,σ0 On,n0
(9.38)
The interaction has the same form as for bosons (cf. (9.33)).
T WO PARTICLE OPERATORS FOR FERMIONS
Hint =
1 X
2 n ,n ,n ,n
1
2
3
Vn c†n1 c†n2 cn3 cn4 .
4
Again, matrix elements (Vn ) are the same as for bosons (cf. (9.33)). Note: the
sum over ni in principle includes a sum over 4 spins. However, in general
the pairs n1 , n4 and n2 , n3 have the same spin, because those appear in the
same density ρ̂(x) = Ψ† (x)Ψ(x). If we include spin explicitly, we have
Vn1 σ1 ,n2 σ2 ,n3 σ3 ,n4 σ4 = δσ1 ,σ4 δσ2 ,σ3 Vn1 ,n2 ,n3 ,n4
(9.39)
For the sake of completeness, we here present one- two-particle operators
for fermions in second quantisation with explicit indication of spin for the
case in which the operators do not act on spin:
9.3. SECOND QUANTISATION FOR FERMIONS
1- AND 2- PART.
Ô =
X
OPER . FOR FERMIONS IN
141
SQ WITH SPIN EXPLICIT
On,n0 c†n,σ cn0 ,σ
n,n0 ,σ
Z
On,n0 =
ϕ∗n (x)O(x, −i~∇)ϕn0 (x)d3 x .
1 X X
Vn c†n1 ,σ c†n2 ,σ0 cn3 ,σ0 cn4 ,σ
(9.40)
2 n ,n ,n ,n σ,σ0
1 2 3 4
Z Z
Vn1 ,n2 ,n3 ,n4 =
ϕ∗n1 (x)ϕ∗n2 (x0 )V (x, x0 )ϕn3 (x0 )ϕn4 (x) d3 x d3 x0
Hint =
= hn1 | hn2 | V̂ |n4 i |n3 i
Field operators for fermions have similar properties as for bosons:
{Ψ̂σ0 (x0 ), Ψ̂†σ (x)} = δσ0 ,σ δ(x0 − x)
{Ψ̂σ0 (x0 ), Ψ̂σ (x)} = {Ψ̂†σ0 (x0 ), Ψ̂†σ (x)} = 0
and, similarly to (9.24), Ψ̂†σ (x) creates a fermion with spin σ at the position
x. With the help of these operators one can express single-particle operators and interaction terms in the same way as for bosons: ( (9.19) and
(9.32))
9.3.1
Useful rules for (anti) commutators
When working in second quantisation one often has to carry out commutators or anticommutators between operators consisting of product of
creation and destruction operators. In this case, the (possibly repeated) application of these relation can be useful. The proof is here: Sec. A.26
[a b, c] =
= a[b, c] + [a, c]b
= a{b, c} − {a, c}b
(9.41)
142
CHAPTER 9. SECOND QUANTISATION
[c, a b] =
= [c, a]b + a[c, b]
= {c, a}b − a{c, b}
(9.42)
9.4. SUMMARY: FOCK SPACE
9.4
143
Summary: Fock space
A Fock space describes a system with different particle numbers. A natural
set of basis states is given by the occupation-number representation.
|N1 , N2 , · · · , Nn , · · · i
(9.43)
The subscript indices represent a single-particle state or level. E.g.:
1 → (n = 1, l = 0, m = 0, σ =↑) 2 → (n = 1, l = 0, m = 0, σ =↓)
3 → (n = 2, l = 0, m = 0, σ =↑) 4 → (n = 2, l = 0, m = 0, σ =↓)
5 → (n = 2, l = 1, m = −1, σ =↑) 6 → (n = 2, l = 1, m = −1, σ =↓)
7 → (n = 2, l = 1, m = 0, σ =↑) 8 → (n = 2, l = 1, m = 0, σ =↓)
9 → (n = 2, l = 1, m = +1, σ =↑) 10 → (n = 2, l = 1, m = +1, σ =↓)
···
The capital Nn indicates how many particles there are in level n. Clearly
Nn = 0, 1 for fermions, due to Pauli principle, and Nn = 0, 1, 2, · · · for
bosons.
An arbitrary state of the Fock space is a linear combination of (9.43)
|Ψi =
X
N1 ,N2 ,··· ,Nn ,···
CN1 ,N2 ,··· ,Nn ,··· |N1 , N2 , · · · , Nn , · · · i .
(9.44)
Second quantisation allows us to express these states with the help of creation
and annihiltaion operators:
|N1 , N2 , · · · , Nn , · · · i
N1 N2
Nn
1
=√
· · · b†1
b†2
· · · b†n
· · · |0i
N1 !N2 ! · · · Nn !
(9.45)
The vacuum |0i is the state without particles and has the property (9.13),
and the operators bi obey the commutation (9.10a) for bosons or anticommutation rules (9.36) for fermions. Moreover, they are the hermitian conjugate of each other:
(9.46)
(bn )† = b†n
Single-and two-particle operators can be written in both case in the form
(9.29) and (9.33).
Corresponding “bra” states can be obtained by a useful rule of hermitian
conjugation:
†
hΨ|  = † |Ψi
(9.47)
144
CHAPTER 9. SECOND QUANTISATION
so that from (9.45)
hN1 , N2 , · · · , Nn , · · ·| = (|N1 , N2 , · · · , Nn , · · · i)†
1
1
1
√
=√
··· √
· · · h0| · · · (bn )Nn · · · (b2 )N2 (b1 )N1
N1 ! N2 !
Nn !
(9.48)
(9.47) together wit (9.13) implies that the “left vacuum” is annihilated by
creation operators:
h0| b†n = 0
∀n
(9.49)
9.5
Unitary transformations
It may sometimes
be
out
convenient to carry
a transformation from a set
of operators bn , b†n to another set cn , c†n . This is allowed as long as
the cn , c†n obey the same commutation or anticommutation relation as the
bn , b†n . It can be shown that this is the case if the transformation is done
with the help of a unitary matrix U :
X
ck =
Uk,n bn
(9.50)
n
Using its unitarity (U −1 = U † ) the inverse reads
X
∗
bn =
Uk,n
ck
(9.51)
k
Due to (9.46), the † operators are transformed as
X
∗
c†k =
Uk,n
b†n
(9.52)
n
As already discussed in Sec. 9.1.2 the transformation (9.51) corresponds to
replacing operators (b†n ) creating particle on one basis set of orbitals (ϕn (x))
to new operators (c†k ) creating particle on another basis set (say χk (x)).
From (9.50), we easily get the (anti)commutation rules:
X
X
X
†
∗
[bn , b†m ]± =
Uk,n Un,p
= (U U † )k,p = δkp
[ck , c†p ]± =
Uk,n
Up,m
|{z} | {z }
n
n
m
†
Um,p
δnm
(9.53)
i.e. the correct ones. Here, we used a common notation [· · · ]± where the +
stays for anticommutators and − for commutators.
9.5. UNITARY TRANSFORMATIONS
9.5.1
145
Application: tight-binding hamiltonian
Consider a chain of L lattice sites n = 0, 1, · · · , L − 1 with periodic boundary conditions n + L ≡ n (for example, these could be points on a ring, e.g.
Benzene, or a very long chain, for which boundary conditions don’t matter). b†n creates a particle (fermion or boson) on the lattice site n (this could
be, for example, an orbital on that site). The tight-binding hamiltonian
reads
X
X †
(9.54)
Ĥ = ε
b†n bn − t
bn+1 bn + b†n bn+1 .
n
n
and describes particles on orbitals with energies ε, and an amplitude t for
each particle for moving to the left or to the right orbital (“hopping”).
This model is easily solved by discrete Fourier transformation (9.50) with
1
Uk,n = √ e2π i k n/L
L
(9.55)
Notice that also the indices k = 0, · · · , L − 1 have periodic boundary conditions: Uk+L,n = Uk,n . In this way, it can be seen that U is indeed unitary:
(U U † )k,p =
X
∗
Uk,n Up,n
=
n
1 X 2π i (k−p) n/L
e
= δk,p
L n
(9.56)
All indices have to be understood from 0 to L − 1.
In this way, (9.54) becomes (we use (9.51))
Ĥ = ε
XX
n
p,k
|
∗
Up,n Uk,n
c†p ck − t
{z
X
†
∗
∗
Up,n+1 Uk,n
+ Up,n Uk,n+1
cp ck .
(9.57)
n,p,k
}
δp,k
Now
X
∗
Up,n+1 Uk,n
=
n
1 2π i p/L X 2π i (k−p) n/L
e
e
= e2π i p/L δk,p
L
n
similarly
X
∗
Up,n Uk,n+1
= e−2π i p/L δk,p
n
So that (9.57) becomes
Ĥ =
X
k
(ε − 2t cos(2πk/L)) c†k ck
(9.58)
146
CHAPTER 9. SECOND QUANTISATION
which is in diagonal form.
The operators c†k create particles with (crystal) momentum Pk = 2πk/L
and energy εk ≡ (ε − 2t cos(2πk/L)). The mode with lowest energy is the
one with k = 0, so the bosonic ground state with N particles is simply
N
B
GN = √1
c†0 |0i
N!
all particles are in the lowest level: it is a Bose-Einstein condensate. Its energy
is N (ε − 2t).
For fermions one has to start filling levels according to Pauli principle.
Introducing spin explicitly (which does not affect the Hamiltonian), the
ground state with two particles is
|G2 i = c†0↑ c†0↓ |0i .
Since the levels with k = 1 and k = L − 1 ≡ −1 have the same energy (see
(9.58)), the state with three fermions is four-fold degenerate:
|G3 i = c†±1σ c†0↑ c†0↓ |0i
σ =↑↓ .
For N particles the ground state is obtained by filling the levels up to the
Fermi energy (see solid-state physics)
Y
|GN i =
c†k,σ |0i .
|2πk/L|<PF ,σ
9.5.2
Field operators
The transformation (9.15), whose inverse is given in (A.42) is a continuous
version of (9.50). While operators such as b†n , c†k create particles on discrete
levels, field operators Ψ† (x) create particles on continuous single-particle
states, in this case on an eigenstate of the position operator x̂, as discussed
around (9.24).
9.6
Heisenberg time dependence for operators
Creation and annihilation, as well as field operators discussed above can
be assigned a time dependence according to the usual Heisenberg representation , which for an arbitrary operator (in both first or second quantisation) reads:
∂
d
i~ Ô(t) = [Ô(t), Ĥ] + i~ Ô(t)
(9.59)
dt
∂t
9.6. HEISENBERG TIME DEPENDENCE FOR OPERATORS
147
For time-independent Hamiltonians and operators, the formal solution
reads, as usual
Ô(t) = ei Ĥ t/~ Ô e−i Ĥ t/~ .
(9.60)
Since operators can be quite generally expressed as polynomials in the
b, b† and/or Ψ̂, Ψ̂† operators, it is easy to show that the time dependence
can be pulled over to these operators. For example, take the interaction
operator (9.32) (we omit the subscript SQ ):
1
2
i Ĥ t/~
Ĥint (t) = e
Z Z
1
ei Ĥ
2
×e
i Ĥ t/~
1
=
2
t/~
Z Z
0 −i Ĥ
i Ĥ t/~ †
Ψ̂† (x) e|−i Ĥ t/~
{ze
} Ψ̂ (x )e
t/~
t/~
=
V (x, x0 )
1
0
−i Ĥ t/~ i Ĥ t/~
e
|
Ψ̂(x )e
Z Z
Ψ̂† (x)Ψ̂† (x0 )V (x, x0 )Ψ̂(x0 )Ψ̂(x)d3 xd3 x0 e−i Ĥ
Ψ̂(x)e−i Ĥ t/~ d3 xd3 x0
{z
}
Ψ̂(x,t)
Ψ̂† (x, t)Ψ̂† (x0 , t)V (x, x0 )Ψ̂(x0 , t)Ψ̂(x, t)d3 xd3 x0 ,
(9.61)
i.e. the time dependence of Ĥint (t) is obtained by inserting the time dependence in the constituent field operators.
9.6.1
Time dependence of field operators
For a noninteracting Hamiltonian (9.11a) (we use U (x) instead of V (x) for
the potential:
~2 0 2
0
d x Ψ̂ (x)Ψ̂ (x ) −
∇ + U (x ) Ψ̂(x0 )
2m
~2 0 2
†
0
0
− Ψ̂ (x ) −
∇ + U (x ) Ψ̂(x0 )Ψ̂ (x)
2m
Z
~2 0
2
3 0
†
0
†
0
0
=
d x Ψ̂(x)Ψ̂ (x ) ∓ Ψ̂ (x )Ψ̂(x) −
∇ + U (x ) Ψ̂(x0 )
2m
|
{z
}
d
i~ Ψ̂ (x) =
dt
Z
3
0
†
0
δ(x−x0 )
= −
2
~
∇2 + U (x) Ψ̂(x) ,
2m
(9.62)
where the − sign is for bosons and + for fermions. We now
also
include
the interaction, (9.32) so that the total Hamiltonian is ĤSQ + Ĥint
SQ
. One
148
CHAPTER 9. SECOND QUANTISATION
obtains (without proof)
Z
~2 2
d
∇ + U (x) Ψ̂(x) +
d3 x0 ρ̂(x0 )V (x, x0 )Ψ̂(x) (9.63)
i~ Ψ̂(x) = −
dt
2m
Both expressions are formally analogous to the Schödinger equation in
first quantisation.
9.6.2
Time dependence of creation and annihilation operators
The Heisenberg time dependence for c, c† operators described by the noninteracting Hamiltonian is given by (9.16)
i~
X
X
d
cn (t) =
hk,m [cn (t), c†k (t)cm (t)] =
hn,m cm (t)
dt
|
{z
}
m
k,m
(9.64)
δn,k cm (t)
For the case of a diagonal hamiltonian (9.11a), for which hn,m → δn,m En
cn → bn the solution is simple
bn (t) = e−iEn t/~ bn (0)
(9.65)
For the more general case (9.64) the solution can be formally obtained by
introducing a column vector


c1


c ≡  ... 
cn
together with the matrix h = {hn,m }. Then (9.64) becomes a matrix equation
d
i~ c(t) = h · c(t)
dt
with formal solution
c(t) = e−ih t/~ c(0)
These expressions are used for Green’s functions in many-body solid state
physics.
9.7. MOMENTUM SPACE
9.7
149
Momentum space
For translation invariant systems it may be convenient to carry out the
transformation (9.15) where the ortonormal basis set consists of eigenfunction of momentum, i.e. plane waves. The basis is, in principle, continuous,
so that the sum in (9.15) should be replaced by an integral. However, it is
often common practice to restrict the space to a cubic box of edge L with periodic boundary condition. In the end, physical results are obtained by letting
L → ∞.
In a periodic box, the normalized eigenfunction of the momentum operator p̂
are
1
Ω = L3 = volume
(9.66)
ϕk = √ ei k·x
Ω
with
k=
2π
(n1 , n2 , n3 ) ni = −∞, · · · , +∞
L
(9.67)
Then (9.16) becomes
Ĥ =
X
hk0 ,k c†k0 ck
k,k0
hk0 ,k
1
=
Ω
Z
3
i(k−k0 )·x
d xe
~2 k2
+ U (x)
2m
=
~2 k2
U (k − k0 )
δk,k0 +
(9.68)
2m
Ω
where we have used two useful formulas
Z
1
0
d3 xei(k−k )·x = δk,k0
Ω
Z
0
d3 xei(k−k )·x U (x) = U (k − k0 )
(9.69)
(9.70)
is the Fourier transform of U (x) (wit abuse of notation).
The interacting term (9.33) can be obtained by first carying out the integral.
We consider the case of a potential which depend on distance only:
Z Z
ϕ∗k1 (x)ϕ∗k2 (x0 )V (x − x0 )ϕk3 (x0 )ϕk4 (x) d3 xd3 x0
Z Z
1
= 2
d3 x d3 r e−i(k1 −k4 )·x e−i(k2 −k3 )·(x−r) V (r)
Ω
1
= δk1 +k2 −k3 −k4 V (k2 − k3 )
Ω
(9.71)
150
CHAPTER 9. SECOND QUANTISATION
where we have transformed x0 = x − r, used (9.69) and again introduced
the Fourier transform of V , analogously to (9.70).
In this way, by taking k3 = k2 −q and exploiting the δ leading to k4 = k1 +q
it is convenient to write (9.33) (or (9.40)) as
Ĥint =
1 X X
V (q) c†k1 ,σ c†k2 ,σ0 ck2 −q,σ0 ck1 +q,σ ,
2Ω k ,k ,q σ,σ0
1
(9.72)
2
where for convenience we have introduces spin (for the case of Fermions,
k1
ä
k1 + q
σ
V (q)
σ′
k2 − q
k2
Figure 9.1: Schematic representation of an interacting process
otherwise the expression is the same for bosons). The interaction term
describes a process in which two particles with momenta k2 − q and k1 +
q come together and exchange momentum q so that total momentum is
conserved.
The amplitude for the process is given by the Fourier transform of the
interaction evaluated at the transferred momentum q. On the other hand,
the external potential given by U in (9.68) can be written as (again with
spin, when we have Fermions)
ĤU =
1 XX
U (q) c†k−q,σ ck,σ
Ω k,q σ
(9.73)
and describe a transfer of momentum q from the potential center, so momentum is not conserved.
9.8. FREE FERMIONS
151
U (q)
k
k−q
Figure 9.2: Scattering by a potential
9.8
Free fermions
For free fermions we have (9.68) with U = 0:
XX
Ĥ =
εk n̂k,σ
k
(9.74)
σ
2 2
~k
n̂k,σ ≡ c†k,σ ck,σ
2m
As mentioned in Sec. 9.5, the ground state |F i (state with minimal energy)
is the one in which all levels k up to a certain energy εF (Fermi energy), or,
~2 k2
equivalently with |k| smaller than the Fermi momentum kF with εF = 2mF
are occupied (n̂k,σ = 1) and the other ones are empty (n̂k,σ = 0):
Y Y †
ck,σ |0i .
(9.75)
|F i =
εk ≡
|k|≤kF
σ
These states can be represented by a filled sphere, the Fermi sphere. The
ä
kF
Figure 9.3: Fermi sphere
total number of particles is given by the sum of the k points within the sphere
times 2 for spin:
Z
X
d3 k
2Ω 4π 3
Np = 2
1 = 2Ω
=
k
(9.76)
3
(2π)3 3 F
|k|≤kF (2π)
|k|≤kF
152
CHAPTER 9. SECOND QUANTISATION
R d3 k
P
where we have used k · · · = Ω (2π)
3 · · · cf. here: Sec. A.28. We, thus
have the relation between kF and the particle density np :
kF3 = 3π 2 np
(9.77)
Excited states are obtained by removing a particle from within the Fermi
sphere (i.e., creating a hole) and putting it outside: particle-hole excitations:
c†k1 ,σ ck2 ,σ0 |F i
|k1 | > kF , |k2 | ≤ kF
(9.78)
the corresponding excitation energy is
ä
kF
Figure 9.4: Particle-hole excitation.
Eex,k1 σ,k2 σ = Ek1 σ,k2 σ − EF = εk1 − εk2
(9.79)
These excitations play an important role in solid state physics. For example, at finite temperature particle-hole excitations are created up to an energy Eex,k1 σ,k2 σ ≈ kB T .
9.8.1
Single-particle correlation function
One quantity of interest is the single-particle correlation function
G(x, t; x0 , t0 ) ≡ hF| Ψ̂σ (x, t)Ψ̂†σ (x0 , t0 ) |F i .
(9.80)
Physically, this describes the amplitude that after creating a particle at x0
at time t0 , this is found at time t in x. It thus describes particle transport
from x0 to x.
We evaluate this quantity for the case of free fermions. First, we transform
the field operators according to (9.9) with the ϕn replaced by the eigenfunctions (9.66) and the b by c.
X
1
(9.81)
Ψ̂σ (x) =
ck,σ √ eik·x
Ω
k
9.8. FREE FERMIONS
153
Then
Ψ̂σ (x, t)Ψ̂†σ (x0 , t0 ) =
1 X ik·x −ik0 ·x0
e e
ck,σ (t)c†k0 ,σ (t0 )
Ω 0
(9.82)
k,k
Since the Hamiltonian (9.74) is diagonal in the c, their Heisenberg time
dependence is given by the same form as in (9.65)
ck,σ (t) = e−iεk t/~ ck,σ .
(9.83)
We are left with the evaluation of expectation values of operators in (9.80)
(cf. (9.82))
hF| ck,σ c†k0 ,σ |F i
clearly, if one creates a particle at k0 , this same particle has to be annihilated
again in order to get the same state, therefore k = k0 . Moreover, k0 must be
outside of the Fermi sphere: |k0 | > kF . This gives for the above expectation
value
hF| ck,σ c†k0 ,σ |F i = δk,k0 θ(|k| − kF )
(9.84)
Using (9.82), (9.83), and (9.84) we obtain for (9.80)
1 X
0
eik·(x−x ) e−iεk τ /~
G(x, t; x0 , t0 ) = G(|x − x0 |, t − t0 ) =
| {z } | {z }
Ω
|k|>kF
τ
r
| {z }
R
1
=
(2π)3
Z
1
2π 2 r
Z
2
d3 k
|k|>kF (2π)3
−iεk τ /~
k dk e
Z
2π
k>kF
1
0
d cos θ eik|x−x | cos θ
| −1
{z
}
2 sin kr
kr
=
∞
dk k sin kr e−iεk
τ /~
(9.85)
kF
For τ = 0 the integral can be evaluated analytically yielding the equal-time
correlation function
G(r, 0) =
3np z cos z − sin z
2
z3
z ≡ rkF
(9.86)
this describes decaying oscillations with wavevector kF (Friedel oscillations) produced by adding a particle.
Another interesting information is the behavior of (9.85) for large τ . In this
case, it can be seen that the leading contribution to the k integral comes by
expanding εk around kF : εk ≈ εkF + ~vF (k − kF ), where vF is the so-called
Fermi velocity. Due to the sinkr, one has in the exponent terms of the form
e−iεkF
τ /~ −i(vF τ ∓r)(k−kF ) ±irkF
e
e
,
154
CHAPTER 9. SECOND QUANTISATION
G(r, 0)
kF r
ä
Figure 9.5: Equal time correlation function (9.86)
where ± come from the sin. Due to the integral over k, the term in the
center oscillates fast, and cancels out unless the argument vF τ ∓ r ≈ 0.
This describes particles moving with the Fermi velocity vF . On top of this,
there are again spatial oscillations with momentum kF and time oscillations with frequency εkF /~.
9.8.2
Pair distribution function
The evaluation of the pair distribution function illustrates how fermions
affect each other even without interaction. We start from the state Ψ̂σ (x) |F i
in which a particle has been removed from the Fermi sphere at position x.
Its normalisation is given by
hF| Ψ̂†σ (x)Ψ̂σ (x) |F i = hF| ρ̂σ (x) |F i = np /2 .
(9.87)
Here, we used the density operator ρ̂σ (x) (with given spin) cf. (9.23), and
since the filled Fermi sphere is translation invariant, its expectation value,
the average density is x-independent. We now determine the probability
density n2p gσ,σ0 (r) to find a particle with spin σ 0 at a distance r from the fist
one. This is given by the expectation value of ρ̂σ0 (x + r) in this state. We
get:
n 2
p
gσ,σ0 (r) = hF| Ψ̂†σ (x)Ψ̂†σ0 (x + r)Ψ̂σ0 (x + r)Ψ̂σ (x) |F i
(9.88)
X
1
= 2
e−i(k1 −k4 +k2 −k3 )·x e−i(k2 −k3 )·r hF| c†k1 ,σ c†k2 ,σ0 ck3 ,σ0 ck4 ,σ |F i
Ω k ,k ,k ,k
2
1
2
3
4
We now distinguish between two cases:
• σ = −σ 0
In this case, it is clear that in (9.88) for each spin separately the particles that has been destroyed must be created with the same k, i.e. we
9.8. FREE FERMIONS
155
get k1 = k4 and k2 = k3 . Then, the result is independent of r and is
given by

2
2 X 
gσ,−σ (r) = 
1
=1
(9.89)
np Ω
|k|<kF
• σ = σ0
In this case, we have two terms contributing in the sum (9.88). The
first one with k1 = k4 and k2 = k3 , whose contribution is the same
as before. The second one has k1 = k3 and k2 = k4 . Its contribution
(exchange contribution) is
n 2
p
2
ex
gσ,σ
(r) =
1 X −i(k2 −k1 )·r
e
hF| c†k1 ,σ c†k2 ,σ0 ck1 ,σ ck2 ,σ |F i
Ω2 k ,k
| {z }
1
2
−ck2 ,σ ck1 ,σ
In the expectation value hF| · · · |F i (with the minus sign) one first
destroys a particle in k1 then in k2 , then creates the one in k2 then in
k1 . This gives 1 provided |k1 | and |k2 | < kF . In this way, the sums
over k1 and k2 are independent of each other, we thus get
2
X
1
i k1 ·r ex
e
gσ,σ (r) = − Ω |k1 |<kF
The sum is like the one in (9.85) for τ = 0 and one integrates from 0
to kF instead, yielding −G(r, 0). In total, thus
ex
gσ,σ (r) = 1 + gσ,σ
(r) = 1 −
9
(sin z − z cos z)2
6
z
z = kF r
(9.90)
This pair distribution function is suppressed at low distances. This
is a pure exchange effect depending on the asymmetry of the wave
function. It signals the fact that the probability to find two fermions
with the same spin at a distance r . k1F is suppressed.
156
CHAPTER 9. SECOND QUANTISATION
gσ,σ
ä
kF r
Figure 9.6: Equal spin pair distribution function gσ,σ (9.90)
Chapter 10
Quantisation of the free
electromagnetic field
The quantisation of the electromagnetic field is also necessary in order to
end up with a description in terms of photons. Like in the case of the
elastic string of Chap. 8 one starts from a contiuous object and ends up
with discrete oscillators describing particles.
The procedure for quantisation of the electromagnetic field is similar to
the case of the Schrödinger field in Chap. 9, there are, however, some important differences leading to technical issues.
• The dynamical variable is the vector potential A, which, being a vector, has three components.
• In the Coulomb gauge, ∇ · A = 0 so the three components of the
vector are not independent from each other
• A is a real field, with the consequence that Fourier components with
opposite k are not independent
10.1
Lagrangian and Hamiltonian
The electric and the magnetic field can be determined from a scalar and a
vector potential
1 ∂A
− ∇φ
c ∂t
B =∇×A
E=−
157
(10.1)
158CHAPTER 10. QUANTISATION OF THE FREE ELECTROMAGNETIC FIELD
We use Gaussian units and c is the speed of light. For simplicity, we use
units in which c = 1. We use the Coulomb gauge for which
φ=0
∇·A=0.
(10.2)
The homogeneous Maxwell equations can be eliminated by using (10.1). The
inhomogeneous Maxwell equations reduce (by using (10.2)) to the wave
equation for A:
1 ∂2
∇2 A − 2 2 A = 0 .
(10.3)
c ∂t
We now show that the Lagrange function
Z
2
Z
µ
2
3 µ
A − (∇ × A) = d3 r
E2 − B2
(10.4)
L= d r
2
2
.
which gives the conjugate momentum density Πn
Πn =
.
δL
.
δ An
= µ An = −µEn
(10.5)
leads to the correct equation of motion (10.3). This is proven here: Sec.
A.27 Here, we introduced the factor µ, from which, of course the equations of motions do not depend, but which in the end has to be fixed to
µ=
1
4π
(10.6)
in order to reproduce the correct energy of the electromagnetic field in
Gauss’ units.
As usual, we obtain the Hamilton function H from the Legendre transformation
2
Z
Z
Π
2
3
3 µ
+ (∇ × A)
H = d rΠ·A−L= d r
2 µ2
Z
µ
= d3 r
E2 + B2
(10.7)
2
.
which, gives the total energy of the electromagnetic field (c.g.s units).
Quantisation cannot be yet applied, since not all variables are independent,
due to (10.2). In other words, commutation rules in the form
[Ân (r), Π̂m (r 0 )] = i~ δn,m δ(r − r 0 )
would contradict (10.2). (See here: Sec. A.29 ).
NO!
(10.8)
10.2. NORMAL MODES
10.2
159
Normal modes
To make progress, we now carry out a discrete Fourier transformation. (See
Sec. 9.7) More specifically, we consider a cubic “box” with volume Ω =
L3 and periodic boundary conditions, and we expand A in terms of the
orthonormal functions √1Ω eik·r
1 X ik·r
A(r) = √
e
Ak
Ω k
(10.9)
The inversion rules are
1
Ak = √
Ω
Z
d3 r e−ik·r A(r) .
The same transformation applies for Π(r)
1 X ik·r
Π(r) = √
e
Πk .
Ω k
(10.10)
(10.11)
The periodic boundary conditions make the k discrete:
k=
2π
(n1 , n2 , n3 )
L
ni integers.
(10.12)
Moreover, since A(x) and Π(x) are real fields,
Ak = A∗−k
10.3
Πk = Π∗−k
(10.13)
Quantisation
In analogy to (8.16), we would expect second quantisation to be generated
by introducing the commutators [Ânk , Π̂mk0 ] = i~ δnm δkk0 (here, n, m are
components of the vectors). However, there are two issues: First, due to
the fact that the Fourier basis is complex, one has to require instead
[Ânk , Π̂m−k0 ] = i~ δnm δkk0
not yet correct!
(10.14)
in order to obtain local commutation rules in r space: details here: Sec.
A.28
1 X ik·r −ik0 ·r0
e e
[Ânk , Π̂m−k0 ]
[Ân (r), Π̂m (r 0 )] =
Ω 0
kk
1 X ik·(r−r0 )
= i~ δnm
e
= i~ δnm δ(r − r 0 )
not yet correct!
Ω k
160CHAPTER 10. QUANTISATION OF THE FREE ELECTROMAGNETIC FIELD
Notice that a change in sign of k0 in the commutators, would have introduced a change in sign in r 0 , which is not correct.
This is not yet the whole story: these commutation rules are still incorrect,
as discussed in Eq. 10.8. They are in contradiction with the transversality
condition (10.2) (See here: Sec. A.29). To enforce this condition, we express Ak and Πk in terms of two real units vectors (polarisations) uk,s with
s = ±1, which are mutually orthogonal and orthogonal to k:
k · uk,s = 0
uk,s · uk,s0 = δss0
X
X
Âk =
q̂k,s uk,s
Π̂k =
p̂k,s uk,s
s
(10.15)
(10.16)
s
and now assume canonical commutation rules between the coefficients
in the form
[q̂k,s , p̂−k0 ,−s0 ] = i~ δs,s0 δk,k0
(10.17)
We use the conventions, coming from (10.13)
uk,s = u−k,−s = u∗k,s
†
q̂k,s = q̂−k,−s
p̂k,s = p̂†−k,−s
(10.18)
The Hamiltonian (10.7) becomes (again, we don’t show the “ˆ” explicitly)
µ X Π−k · Πk
+ (k × A−k ) · (k × Ak )
H=
2 k
µ2
1 X
=
Π−k · Πk + µ2 k 2 A−k · Ak ,
(10.19)
2µ k
where in the last line we have exploited the fact that k · Ak = 0.
Inserting (10.16) in one of the terms in H yields
X
X
Π−k · Πk =
u−k,s1 p−k,s1 · uk,s2 pk,s2 =
uk,−s1 · uk,s2 p−k,s1 pk,s2
s1 s2
=
X
s1 s2
p−k,−s pk,s
s
where we have exploited the orthogonality of the u and (10.18). Similarly
X
A−k · Ak =
q−k,−s qk,s
s
The Hamiltonian (10.19), thus becomes
H=
1 XX
p−k,−s pk,s + µ2 k 2 q−k,−s qk,s
2µ k s
(10.20)
10.3. QUANTISATION
161
The treatment is similar to the case of the harmonic oscillator, with slight
variations. One introduces creation and destruction operators (remember
that k ≡ |k|)
1
(µk q̂k,s + i p̂k,s )
2~µk
1
=√
(µk q̂−k,−s − i p̂−k,−s ) = (ak,s )†
2~µk
ak,s = √
a†k,s
The inverse of this expression is
s
~ q̂k,s =
ak,s + a†−k,−s
2µk
r
~µk p̂k,s = −i
ak,s − a†−k,−s
2
(10.21)
(10.22)
These operators obey usual commutation relations (easy to show by using
(10.17))
[ak,s , a†k0 ,s0 ] = δkk0 δss0
(10.23)
with all other commutators being zero. In terms of these, (10.20) acquires
the familiar form, where we reintroduce the speed of light c, so that ωk =
ck:
XX
~ωk a†k,s ak,s + const.
(10.24)
H=
k
s
Here, the (infinite) constant comes, as for the usual harmonic oscillator,
from commutations of p and q and describes the zero-point energy.
(10.23) together with (10.24) describe again a set of independent harmonic
oscillators. Each one of these oscillator describes a photon mode with
wavevector k and polarisation vector uk,s . The “creation” operator a†k,s
increases by 1 the number of photons in this mode, while ak,s decreases it
by 1.
The commutators of the fields are evaluated here: Sec. A.30
162CHAPTER 10. QUANTISATION OF THE FREE ELECTROMAGNETIC FIELD
Chapter 11
Interaction of radiation field with
matter
Starting from the results of the previous chapters, we will treat here the
interaction between radiation field and matter, and apply the formalism
to two cases of interest:
(i) Absorption and emission of radiation
(ii) Bremsstrahlung
In both cases, the electromagnetic field will be treated in second quantisation
according to the results of Sec. 10. The electron Hamiltonian will be treated
in first quantisation for case (i), this means that we will consider just one
electron at a time. We will use the full second quantisation (for both radiation
and electrons) formalism for case (ii).
11.1
Free radiation field
The Hamiltonian of the free electromagnetic (radiation) field follows from
the corresponding Hamilton function, that we derived in (10.24).
163
164CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
H AMILTONIAN OF THE FREE ELECTROMAGNETIC FIELD
Hem =
X
~ωk a†k,s ak,s
(11.1a)
k,s
ωk = |k| c .
11.2
Electron and interaction term
The Hamiltonian of a single electron in an electromagnetic field is given in
(7.1).
2
q
1 p − A(r, t) + V (r)
H=
(11.2)
2m
c
We consider the contribution of a static field V (r), plus an electromagnetic
field in the Coulomb gauge described by the vector potential A. For simplicity, we neglect electron spin.
We split (11.2) in a term describing an electron in an external potential (for
example in an atomic orbital)
Hel =
1 2
p + V (r)
2m
(11.3)
and the electron-photon interaction term
q
q2
A(r, t)2 .
Hel−ph = − A(r, t) · p +
2
mc
2mc
|
{z
} |
{z
}
H0
(11.4)
H 00
The two terms in (11.4) are termed paramagnetic (H 0 ) and diamagnetic (H 00 ),
respectively. Notice that despite of the fact that r and p are noncommuting
operators, the order in the product A(r, t) · p doesn’t matter due to the
Coulomb gauge ∇ · A = 0.
We will consider Hel−ph (11.4) as a perturbation of the unperturbed Hamiltonian H0 ≡ Hel + Hem . Since radiation is treated in second quantisation,
we have to replace in Hel−ph the vector potential A with its second quanti-
11.3. TRANSITION RATE
165
sation expression (10.9) with (10.16), (10.22) and µ = 1/(4π), leading to
X 2π~c2 1/2 X
†
ik·r
ak,s + a−k,−s
uk,s e
A(r) =
Ωωk
s
k
X γk X
†
−ik·r
ik·r
√
,
(11.5)
=
uk,s ak,s e
+ ak,s e
Ω
s
k
γk ≡
2π~c2
ωk
1/2
(11.6)
where we have used the symmetry (10.18), and reintroduced the speed of
light c.
The paramagnetic interaction term in (11.4), thus becomes
q X γk †
0
ik·r
−ik·r
√
H =−
ak,s e
+ ak,s e
uk,s · p
(11.7)
mc k,s
Ω
11.3
Transition rate
We evaluate the rate for a transition between an initial state |ii and a final
state |f i under the influence of the perturbation H 0 within Fermi’s golden
rule (5.27). We neglect contributions coming from H 00 : we will discuss why
later.
First, the states |ii and |f i are eigenstates of the unperturbed hamiltonian
H0 ≡ Hel + Hem
These are product states of the form
|mi = |electron statem i ⊗ photon statem
(11.8)
m = i, f .
(11.9)
Here, |electron statem i is an eigenstate of Hel , which we assume discrete,
for example an atomic or molecular orbital, with energy εm and denote it
as
|εm i.
photon state
is an eigenstate of the many-body photon Hamiltonian
m
Hem , and can be written in the occupation number representation as |{nm
ks }, ∀(k, s)i,
m
(See Sec. 9.43). Here nks is the number ofphotons with momentum k and
polarisation s in the state photon statem .
The total energy Em of the state |mi is, thus given by (neglecting the zeropoint energy)
X
E m = εm +
~ωk nm
(11.10)
ks .
k,s
166CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
Fermi’s golden rule (5.27) for the transition rate, thus becomes
Wi→f =
2π
~
2
f
εf ⊗ {nks } H 0 εi ⊗ {niks } δ(Ef − Ei )
From the expression (11.7) it is clear that H 0 can create or destroy a single
photon. Therefore, |f i can differ from |ii by having one photon more (photon emission) in a given mode, or one less (photon absorbtion). We consider
these two processes separately.
11.3.1
Photon emission
The final state will have just one additional photon in a single mode, say
(k, s), i.e.
nfk0 s0 = nik0 s0
nfks = niks + 1
(k0 , s0 ) 6= (k, s)
(11.11)
For the energy difference we, thus, have
Ef − Ei = εf − εi + ~ωk
The matrix element of (11.7) is given by
0 q X γk0
0 √ uk0 ,s0 · εf pe−ik ·r εi {nfks }a†k0 ,s0 {niks } . (11.12)
f H i = −
mc k0 ,s0 Ω
Due to (11.11), the photon matrix element is given by
f † i q i
{nks }ak0 ,s0 {nks } = nks + 1 δk,k0 δs,s0
In total, using (11.6), we have for the emission rate
e
=
Wi→f
4π 2 q 2
i
uk,s · εf pe−ik·r |εi i2 (11.13)
δ(ε
−
ε
+
~ω
)(n
+
1)
f
i
k
ks
Ωm2 ωk
This term describes the transition from a higher εi to a lower εf = εi − ~ωk
electronic state upon emitting a photon. This can occur spontaneously, i.e.
without the presence of other photons i.e. niks = 0 “spontaneous emission”. The presence of an external field niks > 0 with that resonance frequency enhances the emission rate by a factor proportional to its intensity
“stimulated emission”.
11.3. TRANSITION RATE
11.3.2
167
Photon absorption
For a phonon absorption process the discussion is similar. In this case, the
final state will have one photon less in k, s, i.e.
nfk0 s0 = nik0 s0
nfks = niks − 1
(k0 , s0 ) 6= (k, s)
(11.14)
The energy difference
Ef − Ei = εf − εi − ~ωk
The matrix element is given by
q X γk0
0
√ uk0 ,s0 · εf peik ·r |εi i {nfks }ak0 ,s0 {niks } (11.15)
hf| H 0 |ii = −
mc k0 ,s0 Ω
| √ {z
}
niks δk,k0 δs,s0
Finally, the absorption rate becomes
e
Wi→f
=
2
ik·r
4π 2 q 2
i pe
|ε
i
δ(ε
−
ε
−
~ω
)
n
u
·
ε
i
f
i
k
k,s
f
ks
Ωm2 ωk
(11.16)
In this case the absorption rate is proprotional to niks , i.e. absorption takes
place only if photons with that frequency are present, which is somewhat
obvious.
11.3.3
Electric dipole transition
An expansion in multipoles of the charge corresponds to an expansion in
powers of a0 /λ. Here, a0 is the typical spatial extension of the charge (e.g.
Bohr’s radius) and λ the wavelength of radiation. For a0 λ one can
restrict to the lowest order, i.e. the electric dipole term. This corresponds
to neglecting the ik · r term in the exponent of (11.13) and (11.16) which is
of the order of a0 /λ.
The remaining matrix element of the momentum can be then transformed
into
im im εf p |εi i =
εf [Hel , r] |εi i =
εf Hel r − rHel |εi i
~
~
im
= (εf − εi ) εf r |εi i .
~
I.e. it is proportional to the dipole matrix element.
(11.17)
168CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
Selection rules
The dipole matrix element (11.17) is nonzero only between certain pair of
states. This leads to the so called dipole selection rules. For example, the
two states |εf i |εi i must have different parity under reflection r → −r,
and, in case of atomic orbitals, must have the quantum number l differing
by ±1, i.e.
lf = li ± 1
(11.18)
This can be understood by the fact that a photon has angular momentum
j = 1. In the case of photon absorption, the initial state consisting of the
electron and one photon has, according to the general rules of addition of
angularm momenta (4.12), a total angular momentum jtot,i where li − 1 ≤
jtot,i ≤ li + 1. In the final state the photon has been absorbed, so angular
momentum conservation fixes lf to be equal to jtot,i . Considering the fact
that lf = li is excluded since it has the same parity, this gives the selection
rule (11.18). The same discussion holds upon exchanging final and initial
state for the case of photon emission.
This discussion neglects the contribution of the spatial part of the photon
wave function to jtot . This is correct within the electric dipole approximation
in which k is neglected, i.e. the photon wave function is a constant. Higher
orders in k, i.e. higher orders in the multipole expansion, allow for transitions between more distant angular momenta.
11.3.4
Lifetime of an excited state
In the absence of external radiation there can be only spontaneous photon
emission acompanied
an excited state i into a state with
Pby ae decay from
dPi
lower energy. Since f Wi→f = − dt describes the rate of decrease of the
average occupation (population) of the excited state, this is equal to the
inverse lifetime τ1 .
From (11.13), and in the electric dipole approximation (11.17), we obtain
in the absence of photons in the initial state (nk,s = 0)
X
1 X X 4π 2 q 2
2
uk,s · εf r |εi i2 (11.19)
=
δ(ε
−
ε
+
~ω
)(ε
−
ε
)
f
i
k
f
i
τ
Ω~2 ωk
ε
s
k
f
Where we have explicitly written the sum over the final states f as a sum
over the photon momentum k the polarisation s, and the electron final
states εf .
11.3. TRANSITION RATE
169
In the infinite volume limit, we can replace (cf. (A.44))
Z
Z
1X
1
1
3
−→
d k=
k 2 dk d cos θ dϕ .
3
3
Ω k
(2π)
(2π)
The sum over the photon polarisation in (11.19)
2
X
uk,s · εf r |εi i
| {z }
s dif
(11.20)
can be carried out in the following way: Without restriction we can take
uk,s=−1 perpendicular to dif , so that there is only a contribution from s = 1.
Let us call α the angle between uk,1 and dif . Then, due to the fact that uk,1
and k are orthogonal, the angle between k and dif is θ = π2 − α. One obtains
for (11.20)
|dif |2 cos2 α = |dif |2 sin2 θ
Here, again,
dif ≡ εf r |εi i
(11.21)
is the dipole matrix element between initial and final state.
For a fixed final state εf in (11.19), we carry out the integration by taking
dif in the z direction. Therefore, (11.19) becomes
Z
4π 2 q 2
1 X
k 2 dk d cos θ dϕ
=
δ(εf − εi + ~c k)(εf − εi )2 |dif |2 sin2 θ
3
2
τ
(2π) ~ ck
εf
Z
X q2
=
|dif |2 (εi − εf )3 sin2 θd cos θ
4
3
~ c
ε
f
2
X q2 4
2 (εi − εf ) (εi − εf )
|d
|
=
if
~c 3c2
~2
~
|{z}
ε
(11.22)
f
≈1/137
~/τ describes the energy uncertainty ∆εi (lifetime broadening) of the level
εi . Accordingly: the ratio ∆εi /(εi − εf ) describes how sharp a certain transition is.
(ε −ε )
Let us estimate this ratio it from (11.22): i ~ f is a typical frequency, whose
inverse is the typical time for an electron to go around an orbit. dif is the
(ε −ε )
typical radius of an orbit, so that βe = dif i ~ f /c is a typical speed of the
electron in units of the speed of light.
2
In total, ∆εi /(εi − εf ) ∼ βe2 α, where α is the fine structure constant q~c ≈
1/137.
170CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
11.4
Nonrelativistic Bremsstrahlung
As we know from classical electrodynamics, accerated charges emit radiation. If two particles scatter with each other, they will emit radiation.
We would like to treat this process quantum-mechanically. Therefore we
consider the scattering of an electron with a nucleus described by a static potential V (r), an take into account the interaction with radiation as well. For
simplicity, we neglect the electron spin, since it does not change during
scattering and has no effect on the result. As we consider scattering of a
single electron, the result is independent of particle statistics and is also
valid for bosons.
The Hamiltonian initially consists of three parts:
H = Hkin + Hem + Hint ,
the kinetic energy of the electron, the contribution from the free electromagnetic field (11.1a) and the interaction between charge and field.
11.5
Kinetic energy of the electron
We now rewrite the kinetic part of the Hamiltonian of the electrons in second quantisation (cf. (9.11a), (9.9) or (9.37)). We use as eigentfunctions
ϕn (x) plane waves with periodic boundary conditions in a cubic box of
the length L (cf. Sec. 9.7)
P LANE WAVE BASIS
1
ϕq (r) = √ eiqr
Ω
2πnα
;
n α ∈ N0
qα =
L
X
ϕq (r)ϕ∗q (r 0 ) = δ(r − r 0 )
(11.23b)
ϕ∗q (r)ϕq0 (r) d3 r = δq,q0
(11.23d)
(11.23a)
(11.23c)
q
Z
The completeness and orthogonality of plane waves follow from the fact that
11.6. INTERACTION BETWEEN CHARGE AND RADIATION FIELD171
they are eigenfunctions of the momentum operator. It also follows that
the kinetic energy is diagonal in this basis. As in (9.11a) we get
K INETIC ENERGY FOR PLANE WAVES
Hkin =
X
Eq c†q cq
(11.24a)
q
2 2
Eq =
11.6
~q
.
2m
(11.24b)
Interaction between charge and radiation field
We use once again c = 1. To derive the interaction between radiation field
and electron, we start from the expression of the total electron Hamiltonian
(7.1). As described in chapter 9 we use the energy expectation value and
transform the electron wave function into a field operator. The result is the
kinetic energy plus the interaction
2
1 ~
∇ − eA(r) + V (r) Ψ(r)
Hkin + Hint = d rΨ (r)
2m i
Z
Z
−~2 ∇2
e~
3
†
= d rΨ (r)
Ψ(r) −
d3 rΨ† (r)A(r) · ∇Ψ(r) + . . .
2m
im
|
{z
}|
{z
}
Hkin
H0
Z
Z
e2
3
†
d3 rΨ† (r)A2 (r)Ψ(r) . (11.25)
. . . + d rΨ (r)V (r)Ψ(r) +
2m
|
{z
} |
{z
}
Z
†
3
H 00
HV
We evaluate again the interaction term H 0 (11.25), now with the electron
field in second quantisation. We use again (11.5) for the vector potential.
For a more efficient notation, we define the operators
aτk,s
(
a†k,s
=
ak,s
für
für
τ = +1
τ = −1
172CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
and use them to rewrite (11.5) in a more compact form
X 2π~ 1/2
uk,s aτk,s e−iτ kr .
A(r) =
Ωk
k
(11.26)
s=±1
τ =±1
Please remember that ∇·A = 0. Therefore, ∇(·Af (r)) = A · ∇f (r). Thereby,
the Hamiltonian representing the interaction H 0 in (11.25) becomes (we
reintroduce c, so that ωk = ck)
1/2
X 2π~c2
τ
−iτ kr
· ∇Ψ(r)
uk,s ak,s e
d rΨ (r)
Ωωk
k,s,τ
1/2
Z
e~ X 2π~c2
τ
3
†
−iτ kr
=−
uk,s ak,s ·
d rΨ (r)e
∇Ψ(r) (11.27)
imc ks,τ
Ωωk
e~
H =−
imc
0
Z
3
†
For the expression in square brackets we introduce the expansion of the
field operators in plane waves (cf. (9.9) with (11.23a)), as we did for the
kinetic energy.
d3 r −iqr −iτ kr iq0 r
e
e
∇e
Ω
q,q0
Z 3
X
d r i(q0 −q−τ k)r
†
0
=i
cq cq 0 q
e
Ω
0
q,q
X
c†q cq0 q0 δq−(q0 −τ k)
=i
[. . .] =
X
c†q cq0
Z
q,q0
=i
X
q0 c†q0 −τ k cq0 .
(11.28)
q0
This result is plugged in the square brackets in the Hamiltonian (11.27)
and leads to
1/2
e~ X 2π~c2
0
q · uk,s c†q−τ k cq aτk,s
H =−
mc kq
Ωωk
s,τ
e~ X
=−
mc kqs
2π~c2
Ωωk
1/2
†
†
†
q · uk,s cq+k cq ak,s + cq cq+k ak,s .
For the last transformation, we used k · uk,s = 0.
(11.29)
11.7. DIAMAGNETIC CONTRIBUTION(ADDENDUM)
6
qf =
q f = qi + k
I
k, s
qi
173
qi − k@ k, s
@
6
I
@
@
qi
Figure 11.1: Feynman diagramms for the two Scattering events of the interaction term of the Hamiltonian (11.29).
The interaction term consists of two parts, shown in Fig. (372) with Feynman diagramms. The picture on the left represents the first part of the
Hamiltonian, in which an electron with momentum qi and a photon with
momentum k and polarisation s are annihilated and a new electron with
mometum qf = qi + k is created. The second term describes the inverse
process - first, an electron is annihilated and then, a photon and an electron
are created.
11.7
Diamagnetic contribution(Addendum)
Now we will evaluate the so-called diamagnetic part H 00 in (11.25). As we
will see, this part does not contribute to the Bremmstrahlung. Once again
we use (11.26) for A,
Z
e2
d3 rΨ† (r)A2 (r)Ψ(r)
H =
2mc2
2π~c2 X
1
0
0 0
A2 =
uk,s · uk0 ,s0 aτk,s aτk0 ,s0 e−i(τ k+τ k )·r
√
Ω
ωk ωk0
0
00
k,k
s,s0
τ,τ 0
174CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
1
π~e2 X
0
uk,s · uk0 ,s0 aτk,s aτk0 ,s0 . . .
H =
√
0
mΩ
ωk ωk
0
00
k,k
s,s0
τ,τ 0
Z
...
†
3
d rΨ (r)e
−i(τ k+τ 0 k0 )r
Ψ(r)
The term [. . .] is the same as in (11.27), with the only difference that τ k →
(τ k + τ 0 k0 ). Accordingly, we use the result in (11.28)
[. . .] =
X
c†q cq+τ k+τ 0 k0
q
H 00 =
1
π~e2 X
0
uk,s · uk0 ,s0 aτk,s aτk0 ,s0 c†q cq+τ k+τ 0 k0 .
√
mΩ
ωk ωk0
0
q,k,k
s,s0
τ,τ 0
D IAMAGNETIC TERM
π~e2 X
1
uk,s · uk0 ,s0 a†k,s a†k0 ,s0 c†q cq+k+k0
H =
√
mΩ
ωk ωk0
0
00
q,k,k
s,s0
+ ak,s ak0 ,s0 c†q cq−k−k0 + a†k,s ak0 ,s0 c†q cq+k−k0
†
†
+ ak,s ak0 ,s0 cq cq−k+k0
(11.30)
The two lower processes describe scattering of an electron with a photon.
For those, the total momentum is conserved. This is Compton scattering,
where one photon is scattered off a (free) electron.
11.8
Potential term
Finally, we are left with the term representing the potential in (11.25). In
second quantisation the expression is derived using (9.37):
11.8. POTENTIAL TERM
c†q cq−k−k0 ak,s ak0 ,s0
6
175
c†q cq+k+k0 a†k,s a†k0 ,s0
6
qi
?
k, s
qi − k − k 0 k0 , s0
qi
6 k, s
0 qi + k + k
- k0 , s 0
c†q cq+k−k0 a†k,s ak0 ,s0
c†q cq−k+k0 ak,s a†k0 ,s0
6
qi
6 k, s
0 qi + k − k
k0 , s0
6
qi
?
k, s
qi − k + k0 - k0 , s 0
Figure 11.2: The contributions to H 00 ((11.30)). The lower two terms contribute to Compton scattering. The upper terms describe emission and absorption of two photons.
176CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
P OTENTIAL TERM
HV =
X
V (q, q0 )c†q cq0
q,q 0
V (q, q0 ) =
Z
ϕ∗q (r)V (r) ϕq0 (r) d3 r
Z
1
0
=
eir(q−q ) V (r) d3 r = V (q − q0 )
Ω
(11.31)
For the Coulomb potential for atomic number Z we have
Ze2
r
4πZe2
V (q − q0 ) = −
.
Ω |q − q0 |2
V (r) = −
(11.32)
Derivation of the Fourier transformation here: Sec. A.31 .
11.9
Transistion amplitude
11.9.1
First order
We now split the Hamiltonian, which at this point consists of the following
terms,
H = Hkin + Hem + HV + H 0 + H 00 ,
|
{z
} |
{z
}
H0
Hint
into an unperturbed term H0 and a perturbation term Hint . Eigenvectors
of H0 have the form (Electron spin is neglected)
|Ψi = c†q1 c†q2 · · · a†k1 ,s1 a†k2 ,s2 · · · |0i .
We are interested in processes where an electron is scattered and a photon
is emitted. That means
Initial state:
Final state:
|ii = c†qi |0i
|f i = c†qf a†k,s |0i .
(11.33a)
11.9. TRANSISTION AMPLITUDE
177
We use Fermis Golden Rule ((5.27) with (5.33)), to determine the transition
rate
2π
|Mf i |2 δ(Ef − Ei )
~
X
Hf n Hni
Mf i = Hf i +
Ei − En + iO+
n
Wi→f =
(11.33b)
In principle, H 0 (11.29) in first order already could describe the process we
are interested in. The matrix element would then be
1/2
e~ X 2π~c2
0
q · uk0 ,s0
Mf i =hf |H |ii = −
mc 0 0
Ωωk0
k qs
†
†
†
× h0 |cqf ak,s cq+k0 cq ak0 ,s0 + cq−k0 cq ak0 ,s0 cq† i |0 i
(11.34)
Obviously, only the second summand contributes and we require
k0 = k,
s0 = s,
q = qi ,
q f = qi − k
Therefore, we have
e~
Mf i = −
mc
2π~c2
Ωωk
1/2
qi · uk,s δqf −qi +k .
Nevertheless, one can show, that in this process energy and momentum
cannot be simultaneously conserved. Proof here: Sec. A.32 .
11.9.2
Second order
Therefore, first order perturbation theory does not contribute to the bremsstrahlung
process and we proceed to second order, i.e. the second term in (11.33b).
(The Compton term (11.30) needs a photon in the initial state or two photons in the initial/final state and does not contribute either.
We consider the matrix elements (of H 0 (11.29) and HV (11.31))
0
0
Hf n Hni = HV f n + H f n
HV ni + H ni
= HV f n HV ni + H 0 f n H 0 ni + H 0 f n HV ni + HV f n H 0 ni .
The first two matrix elements vanish. In the first term, no photons are created
by HV , thus the final state cannot be reached. In the second term, a photon
178CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
qf @ k, s
@
6
I
@
@
I
@
@
qf @
6
q = qf + k
t
t
@
q = qf + k
V (qf + k − qi )
qi
V (qf + k − qi )
k, s
qi
Figure 11.3:
Second order processes for bremsstrahlung (cf. (11.29),
(11.31)). On the left, with intermediate state n1 and on the right with intermediate state n2 .
exists in an intermediate state, but applying H 0 once more, it either annihilates this photon or creates a second one. Both possible results do not
correspond to the final state. Therefore, we are left with
Hf n Hni = H 0 f,n1 HV n1 ,i + HV f,n2 H 0 n2 ,i .
The corresponding processes are shown in (11.3). Here, we have two possible intermediate states
|n1 i = c†q |0i
|n2 i = c†q a†k,s |0i ,
(11.35)
were we still have to sum over q, k, s.
The contributing matrix elements (cf. (11.29), (11.31), (11.33a), (11.35)) are
1
1
e~ 2π~c2 2 X
hf |H |n1 i = −
√ q0 · uk0 ,s0
mc
Ω
ωk0
k 0 q0 s0
†
†
†
× h0|akf ,sf cqf cq0 +k0 cq0 ak0 ,s 0 + cq0 cq0 +k0 ak0 ,s 0 c†q |0i
0
1
e~ 2π~c2 2 X
1
†
†
=−
√ q0 · uk0 ,s0 h0|akf ,sf cqf cq0 cq0 +k0 ak0 ,s 0 c†q |0i
0
mc
Ω
ω
k
k 0 q0 s0
11.9. TRANSISTION AMPLITUDE
179
1
1
e~ 2π~c2 2 X
†
=−
√ q0 · uk0 ,s0 h0|cqf cq0 cq0 +k0 c†q |0i
0
mc
Ω
ωk
k 0 q0 s0
× h0|akf ,sf a†k0 ,s0 |0i
1
e~ 2π~c2 2 X
1
=−
√ q0 · uk0 ,s0 δq0 ,qf δq0 +k0 ,q δk0 ,kf δs0 ,sf
mc
Ω
ωk0
k 0 q0 s0
1
e~ 2π~c2 2 1
qf · ukf ,sf δq,qf +kf = hf |H 0 |n1 i
(11.36)
=−
√
mc
Ω
ωkf
1
e~ 2π~c2 2 X
1
hn2 |H |ii = −
√ q0 · uk0 ,s0
mc
Ω
ωk0
k 0 q0 s0
†
†
†
× h0|ak,s cq cq0 +k0 cq0 ak0 ,s 0 + cq0 cq0 +k0 ak0 ,s 0 c†qi |0i
0
1
e~ 2π~c2 2 X
1
†
†
=−
√ q0 · uk0 ,s0 h0|ak,s cq cq0 cq0 +k0 ak0 ,s0 c†qi |0i
0
mc
Ω
ω
k
k0 q0 s0
1
1
e~ 2π~c2 2 X
†
=−
√ q0 · uk0 ,s0 h0|cq cq0 cq0 +k0 c†qi |0i
0
mc
Ω
ωk
k 0 q0 s0
× h0|ak,s a†k0 ,s0 |0i
1
e~ 2π~c2 2 X
1
=−
√ q0 · uk0 ,s0 δq0 ,q δq0 +k0 ,qi δk0 ,k δs0 ,s
mc
Ω
ωk0
k 0 q0 s0
1
e~ 2π~c2 2 1
=−
√ q · uk,s δq,qi −k
mc
Ω
ωk
1
e~ 2π~c2 2 1
=−
(11.37)
√ qi · uk,s δq,qi −k = hn2 |H 0 |ii .
mc
Ω
ωk
For the last step we used that k · uk,s = 0. We need the matrix elements of
180CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
HV (11.31)
hn1 |HV |ii = h0|cq HV c†qi |0i
X
=
V (q0 − q00 )h0|cq cq† 0 cq00 c†qi |0i
q0 ,q00
=
X
q0 ,q00
V (q0 − q00 )δq0 ,q δq00 ,qi
= V (q − qi ) .
This matrix element has to be multiplied with H 0 f,n1 (11.36) and we get
H 0 f,n1 HV n1 ,i
1
e~ 2π~c2 2 1
qf · uk,s V (qf + k − qi ) δq,qf +k .
=−
√
mc
Ω
ωkf
The contributions to Mf,i are therefore (cf. (11.35))
(1)
Mf i
H 0 f,n1 HV n1 ,i
=
(11.38)
Ei − En1 (q) + iO+
q
1
e~ 2π~c2 2 1
V (qf + k − qi )
=−
qf · uk,s
√
mc
Ω
ωk f
Ei − En1 (qf + k) + iO+
X
En1 (q) :=
~2 q2
2m
Due to energy conservation, we require
Ei = Ef =
~2 qf 2
+ ~ωkf
2m
The energy denominator in (11.38) is therefore
E1 := Ef − En1 (qf + kf ) + iO+
~2 (qf 2 − (qf + kf )2 )
=
+ ~ωkf + iO+
2m
~2 (2qf · kf + k2f )
= ~ωkf −
+ iO+
2m
~2 k2f
= ~ωkf − ~vf · kf −
+ iO+
2m
k f · vf
~ |kf |
= ~ωkf 1 −
−
+ iO+ ≈ ~ωkf
|kf | c
2mc
11.9. TRANSISTION AMPLITUDE
181
where the last approximation is valid in the nonrelativistic limit vf c.
Moreover, we have
hf |HV |n2 i = h0|cqf akf ,sf HV c†q a†k,s |0i
X
=
V (q0 − q00 )h0|cqf akf ,sf cq† 0 cq00 c†q a†k,s |0i
q0 ,q00
=
X
q0 ,q00
V (q0 − q00 )δq0 ,qf δq00 ,q δk,kf δs,sf
= V (q − qf )δk,kf δs,sf .
This matrix element has to be multiplied with H 0 n2 ,i (11.37) and we get
HV f,n2 H 0 n2 ,i
1
e~ 2π~c2 2 1
=−
qi · ukf ,sf V (kf + qf − qi ) δq,qi −kf δk,kf δs,sf .
√
mc
Ω
ωk f
The contribution to Mf i is
(2)
HV f,n2 H 0 n2 ,i
Ei − En2 (q, k) + iO+
q,k,s
1
e~ 2π~c2 2 1
V (kf + qf − qi )
=−
qi · ukf ,sf
√
mc
Ω
ωkf
Ei − En2 (qi − kf , kf ) + iO+
(11.39)
Mf i =
X
The energy denominator is therefore
E2 := Ei − En2 (qi − kf , kf ) + iO+
~2 qi 2 − (qi − kf )2
− ~ωkf
=
2m
~2 k2f
~2 qi · kf
= −~ωkf +
−
m
2m kf · vi ~ |kf |
= −~ωkf 1 −
+
+ iO+ ≈ −~ωkf
|kf | c
2mc
182CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
The matrix element we were looking for is (cf. (11.38),(11.39))
(1)
(2)
Mf i = Mf i + Mf i
1
−4πZe2
e~ 2πc2 2
=−
ukf ,sf · qf − qi
mc ~Ωωk3 f
Ω(kf + qf − qi )2
1
4πZe2
2πe2 2
=−
ukf ,sf · vf − vi
~Ωωk3 f
Ω(kf + qf − qi )2
| {z }
=:∆v
The photon momentum is much smaller than the one of the electron (k =
ωk /c), so that we can omit kf in the central term and replace there qf −qi →
m∆v. Finally, we get
Mf i =
11.10
4πZe2 ~2
Ωm2 (∆v)2
21
2πe2
~Ωωk3 f
ukf ,sf · ∆v
(11.40)
Cross-section
The total cross-section σtotal defines an effective scattering surface: Particles going through that surface are scattered by the bremsstrahlung process:
σtotal =
Iscatt
number of scattered particles/time
≡
flux (or current density) of incident particles
Jincid
If Np is the total number of incident particles in volume Ω, then
Jincid =
Np
vi
Ω
vi = particle velocity
The probability per unit time that one particle
Pis scattered, is given by the
transition rate summed over all final states,
f Wi→f . The total current of
scattered particles is thus given by
X
Iscatt = Np
Wi→f
f
In total, we thus have
σtotal
Np
Iscatt
=
=
Jincid
P
f
Np
Ω
Wi→f
vi
(11.41)
11.10. CROSS-SECTION
183
using (11.41), (11.40) and (5.27) we get
σtotal
Ω X 2π
~2 qf 2
2
=
|Mf i | δ
+ ~ωkf − Ei
vi k s q ~
2m
f f f
2π X Ω 16π 2 Z 2 e4 ~4
2πe2
=
~ ksq vi Ω2 m4 (∆v)4
~Ωωk3
~2 q2
+ ~ωk − Ei
2m
4(2π)4 ΩZ 2 e6 ~4 X |uk,s · ∆v|2 ~2 q2
+
~ω
−
E
=
δ
k
i
4
~2 vi Ω3 m4
2m
|∆v| ωk3
ksq
4(2π)4 Z 2 e6 ~2 X |uk,s · ∆v|2 ~2 q2
=
δ
+ ~ωk − Ei
4 3
2
4
vi Ω m
2m
|∆v| ωk
ksq
× |uk,s · ∆v|2 δ
Trasforming the sum into an integral, we get (cf. (A.44))
Z
Z
1
1X
1
3
−→
dq =
q 2 dq dΩq ,
Ω q
(2π)3
(2π)3
where dΩq is the differential spherical angle.
For the cross-section the sum over wavevectors of the photon as well as
of the electron is transformed into an integral and we get an expression
independent of the volume.
Z
Z
|uk,s · ∆v|2
4Z 2 e6 ~2 X
2
2
q
dq
dΩ
k
dkdΩ
σtotal =
q
k
vi m4 (2π)2 s
|∆v|4 ωk3
~2 q2
×δ
+ ~ωk − Ei .
2m
We use the relation k = ωk /c and get
Z
4Z 2 e6 ~2 X
dω |uk,s · ∆v|2 ~2 q2
2
σtotal = 3
q
dqdΩ
dΩ
δ
+
~ω
−
E
.
q
k
i
c vi m4 (2π)2 s
ω
2m
|∆v|4
(11.42)
Note that the polarisation vector uk,s in only dependent on the direction
of k, but not on its absolute value.
We consider now low energy photons, that means that the energy of the
photons is much smaller than the energy of the electrons. Due to the δ
184CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
function (qf2 = qi2 ), we have elastic scattering. This results in
⇒
|vf | = |vi |
|∆v|2 = vi2 − 2vi vf cos(θ) + vf2
= vi2 2 1 − cos(θ)
θ
= 4vi2 sin2 ( ) ,
2
where θ is the scattering angle. (11.42) gives
XZ
4Z 2 e6 ~2
dω |uk,s · ∆v|2 2m 2
2
q
dqdΩ
dΩ
σtotal =
δ q − qi2
q
k
4 θ
3
4
2
2
4
16c vi m (2π) s
ω vi sin 2 ~
Z
Z 2 e6 ~2 X
dω |uk,s · ∆v|2 m δ q − qi
2
=
q dqdΩq dΩk
16c3 vi m4 π 2 s
ω vi4 sin4 2θ
~2 qi
Z
Z 2 e6 ~2 mqi2 X
dω |uk,s · ∆v|2
=
dΩ
dΩ
q
k
16c3 m4 π 2 vi ~2 qi s
ω vi4 sin4 2θ
| {z }
m2
~3
X
Z 2 e6
=
16c3 m2 π 2 ~ s
Z
dω |uk,s · ∆v|2
dΩq dΩk
ω vi4 sin4 2θ
We can directly read the differential cross-section out of this expression.
D IFFERENTIAL CROSS - SECTION OF THE BREMSSTRAHLUNG
|uk,s · ∆v|2
vi4 sin4 ( 2θ )
e2 |∆v⊥ |2
Z 2 e4
= 2 4 4 θ
m vi sin ( 2 ) 16π 2 c3 ~ω
d3 σ
Z 2 e6
=
dΩk dΩqf dωk
16c3 m2 π 2 ~ω
P
s
(11.43)
Here, ∆v⊥ is the part of ∆v which is perpendicular to k.
The first factor in (11.43) is the formula for Rutherford scattering and the
second one gives the probability density for a photon with energy ~ω to
be found in the spatial angle unit dΩk .
We conclude with some comments
11.10. CROSS-SECTION
185
• Z 2 dependence ⇒ To generate X-Rays, we need elements with high
atomic numbers.
At the same time, the melting point has to be high, because a large
amount of energy is left behind in the substrate in the form of heat.
Therefore, tungsten is often used as X-Ray source.
• We have a continous spectrum. However, there is a maximum photon energy.
Eph
~ω max
~ck max
2π~c
λmin
≤ Ei
= Ei = O(eV)
= Ei = O(eV)
= (eV)
2π~c
O(eV)
hc
=
∼ 10−6 m
eV
λmin =
Duane and Hunt’ Law
186CHAPTER 11. INTERACTION OF RADIATION FIELD WITH MATTER
Chapter 12
Eine kurze Einführung in die
Feynman’schen Pfadintegrale
Ausgangspunkt ist die Übergangsamplitude hx 0 , t0 |x, ti, ein Teilchen im
Zustand x 0 zur Zeit t0 anzutreffen, wenn es zur Zeit t am Ort x präpariert
worden ist. Mit dem Zeitentwicklungsoperator für einen nicht explizit zeitabhängigen
Hamilton-Operator erhalten wir
hx 0 , t0 |x, ti = hx 0 | e−iH
t−t0
~
|xi
.
(12.1)
Wenn wir den Propagator hx 0 , t0 |x, ti kennen, können wir daraus bequem
berechnen, wie sich ein beliebiger Zustandsvektor mit der Zeit entwickelt
t−t0
ψ(x 0 , t0 ) = hx 0 | e−iH ~ |ψ(t)i
Z
t−t0
= hx 0 | e−iH ~ |xi ψ(x, t) d3 x
Z
= hx 0 , t0 |x, ti ψ(x, t) d3 x
.
Wir gehen von dem Einteilchen-Hamilton-Operator
H = H0 + H1 =
p2
+ V (x)
2m
(12.2)
aus. Da die beiden Beiträge nicht kommutieren gilt
e−i
∆t
H
~
= e−i
∆t
H0
~
e−i
∆t
H1
~
+ O((∆t)2 )
,
mit ∆t = (t0 − t). Um die Faktorisierung dennoch durchführen zu können,
muß dafür gesorgt werden, daß der Vorfaktor von H (hier ∆t) kleine wird.
187
188CHAPTER 12. EINE KURZE EINFÜHRUNG IN DIE FEYNMAN’SCHEN PFADINTEG
Das erreicht man durch ∆t → δ ≡ ∆/N
∆t
∆t
∆t
−iH N ~
N~ . . . e
hx 0 , t0 |x, ti = hx 0 | |e−iH N ~ e−iH{z
} |xi
.
N Faktoren
schieben wir den Einheitsoperator 11 =
RVor jede Exponentialfunktion
3
|xi ihxi | d xi mit passendem Index „i” ein und erhalten
Z
Z
N
Y
δ
0 0
,
hx , t |x, ti = . . . d3 x1 . . . d3 xN −1 hxi−1 |e−iH ~ |xi i
i=1
mit den Definitionen x0 = x 0 und xN = x und δ = ∆t
Unter Ausnutzung
N
von (12.2) wird hieraus
Z
Z
N Y
p2 δ
δ
0 0
3
3
−i 2m
−iV
(x
)
i
~ |x i e
~
+ O δ2
hx , t |x, ti =
. . . d x1 . . . d xN −1
hxi−1 |e
i
i=1
=
Z Y
3
d xi
i
Y
N
i=1
hxi−1 |e
2 δ
~
p
−i 2m
P
δ
|xi i e−i i V (xi ) ~ + O δ 2 .
p2 δ
(12.3)
Wir verarbeiten nun die Matrixelemente hxi−1 |e−i 2m ~ |xi i dadurch, daß
wir den Einheitsoperator ausgedrückt in Impulseigenzuständen einfügen
Z
p2 δ
p2 δ
−i 2m
~
hxi−1 |e
|xi i = d3 q hxi−1 |e−i 2m ~ |qi hq|xi i
Z
q2 δ
.
= d3 q e−i 2m ~ hxi−1 |qihq|xi i
Mit
i
hx|qi = (2π~)−3/2 e ~ q·x
wird daraus
2 δ
~
p
−i 2m
Z
q2 δ
i
|xi i = (2π~)
d3 q e−i 2m ~ e− ~ q·(xi −xi−1 )
Z
1 δ
2 2m
−3
= (2π~)
d3 q e−i 2m ~ (q + δ q·(xi −xi−1 ))
Z
1 δ
m
1 δ m
2
2
−3
= (2π~)
d3 q e−i 2m ~ (q+ δ (xi −xi−1 )) ei 2m ~ ( δ (xi −xi−1 ))
Z
2
δ xi −xi−1
1 δ 02
im
−3
3 0 −i 2m
q
2
~
δ
~
= (2π~)
dq e
e
|
{z
}
hxi−1 |e
−3
( π2m~
)3/2
iδ
=
m
iδ2π~
−3/2
e
im
2
δ
~
xi −xi−1
δ
2
.
189
Somit vereinfacht sich (12.3) zu
hx 0 , t0 |x, ti
Z
Z Y
= ...
−3/2
m
iδ2π~
d 3 xi
ei
P
i
m xi −xi−1 2
(
) −V
2
δ
(xi )
δ
~
(12.4)
+ O δ2 .
i
Im Limes N → ∞ geht das Argument der Exponentialfunktion über in
Z t0
X m xi − xi−1
m
2
(
) − V (xi ) δ −→
ẋ(t)2 − V (x(t)) dt
N →∞ t
2
δ
2
i
Z t0
=
L[x(t)] dt = S[x(t)]
.
t
Damit erhalten wir den von Feynman vorgeschlagenen alternativen Zugang zur Quantenmechanik über Pfadintegrale
F EYNMAN ’ SCHES P FADINTEGRAL
0
0
hx , t |x, ti =
Z
Z
x0
x
x0
x
D[x(t)] = lim
N →∞
D[x(t)] eiS[x(t)]/~
Z
3
d x1 . . .
Z
3
(12.5)
d xN −1
m
iδ2π~
−3N/2
.
Die Übergangsamplitude von x zur Zeit t nach x 0 zur Zeit t0 ist die Summe
(Integral) über alle Pfade. Gemittelt wird der Phasenfaktor, dessen Argument die Wirkung zu gegebenem Pfad ist. Der wesentliche Unterschied
zur klassischen Physik ist die Tatsache, daß nicht nur der Pfad minimaler Wirkung beiträgt. Die Wirkung ist das Zeit-Integral der LagrangeFunktion.
Man beachte die große Ähnlichkeit zu den Grundprinzipien der QM, die
wir im Zusammenhang mit dem Doppelspaltexperiment abgeleitet haben.
Geht man von der allgemeinen Gültigkeit des Pfadintegral-Formalismus
(12.5) aus, ist es einfach, den Einfluß eines Vektorpotentials (Magnetfeld)
auf die Wahrscheinlichkeitsamplitude eines bestimmten Pfades anzugeben.
Man erhält lediglich aufgrund von (siehe Mechanik)
q
1
L = mv 2 − qΦ(x) + A · v
2
c
(12.6)
190CHAPTER 12. EINE KURZE EINFÜHRUNG IN DIE FEYNMAN’SCHEN PFADINTEG
einen zusätzlichen Phasenfaktor, die sogenannte
P EIERLS -P HASE
q
ϕPeierls = ei c~
12.1
R t0
t
A· dx
dt
dt
q
= ei c~
Rx0
x
A·dx
(12.7)
Aharonov-Bohm-Effekt
Ein Experiment, in dem die Peierls-Phase beobachtet werden kann, ist
der Aharonov-Bohm-Effekt. Hierbei werden Teilchen durch einen Doppelspalt geschickt, hinter dem sich in einem räumlich beschränkten Gebiet, wie in Abbildung (12.1) dargestellt, eine magnetische Spule befindet.
Innerhalb der Spule liegt ein konstantes B-Feld der Stärke B0 vor und außerhalb ist das B-Feld Null. Wir betrachten die beiden in der Abbildung skizzierten
Pfade, die zum Pfadintegral beitragen werden. Auf diesen Pfaden liegt kein
B-Feld vor. Das Vektorpotential hingegen ist im Bereich außerhalb der Spule
im Abstand r vom Spulenzentrum gegeben durch
A=
B0 R2
eϕ
2r
.
Wir werden allerdings für die folgenden Überlegungen den tatsächlichen Wert von A nicht benötigen. Es sei Ψ0α für α = 1, 2 die Wahrscheinlichkeitsamplitude für den Pfad α bei Abwesenheit der Spule. Der zusätzlich B-Feld-abhängige Phasenfaktor Faktor ist gemäß (12.7)
R
e
−i ~c
Pfadα Adx
e
Die Summe der Amplituden der beiden Pfade ergibt
Ψ1 + Ψ2 = Ψ01 e
=e
=e
e
−i( ~c
)
e
−i( ~c
)
R
e
)
−i( ~c
R
R
Pfad1
Pfad2
Pfad2
Adx
e
)
−i( ~c
+ Ψ02 e
e
−i( ~c
)(
Adx
(Ψ01 e
Adx
(Ψ01 e−i( ~c )
e
R
R
Pfad1
H
C
Pfad2
−
Adx
R
Adx
Pfad2 )Adx
+ Ψ02 )
+ Ψ02 )
.
Der Vorfaktor fällt bei der Berechnung von Erwartungswerten heraus.
Nicht so das Linienintegral auf einem geschlossenen Weg um das eingeschlossene
12.2. QUANTEN-INTERFERENZ AUFGRUND VON GRAVITATION191
B-Feld. Die Elektrodynamik (Satz von Stokes) liefert
I
Z
Adx = BdS
.
C
Das hintere Integral ist ein Oberflächenintegral, das den magnetischen
Fluß Φ durch die von C umschlossene Fläche angibt. Offensichtlich können wir nun die Kurve C deformieren, solange wir nicht den Bereich B = 0
verlassen. Mit der Konstanten für das elementare Flußquantum
Φ0 := 2π
c~
e
erhalten wir schließlich
Ψ1 + Ψ2 = e
e
)
−i( ~c
R
Pfad2
Adx
(Ψ01 e
−i2π ΦΦ
0
+ Ψ02 )
.
Wenn also Φ/Φ0 = n eine ganze Zahl ist, addieren sich Ψ01 und Ψ02 konstruktiv
und bei Φ/Φ0 = n+1/2 destruktiv. Daß heißt, daß bei Variation des B-Feldes
eine oszillierendes Signal am Schirm beobachtet wird. Wir haben somit ein
Ergebnis abgeleitet, in das in Zwischenschritten das nicht eindeutig festgelegte (eichabhängige) Vektorpotential eingeht. Das Endergebnis hingegen hängt nur von B ab und ist eichinvariant. Das obige Resultat hat weitere
interessante Konsequenzen. Wenn geladene Teilchen auf geschlossenen
Bahnen umlaufen, z.B. in geeignet geformten Spulen, so muß der eingeschlossene
magnetische Fluß in Einheiten von Φ0 quantisiert sein, damit die Wellenfunktion eindeutig ist. Diese Quantisierung wurde erstmals 1961 mit supraleitenden Spulen in einem homogenen Magnetfeld nachgewiesen. Da in den
supraleitenden Spulen Cooper-Paare die elementaren Objekte bilden und
diese die Ladung 2e haben, wurde als Flußquant der Wert Φ0 /2 gefunden.
12.2
Quanten-Interferenz aufgrund von Gravitation
Wir werden hier untersuchen, wie man den Pfadintegral-Formalismus nutzen
kann, um ein überraschendes Interferenz-Experiment zu beschreiben, daß
sensitiv genug ist, Interferenz aufgrund des Einflußes der Erdgravitation auf
Neutronen zu beobachten. Man beachte, daß der Unterschied zwischen der
elektromagnetischen und der Gravitationskraft zwischen Elektronen und Neutronen 1040 beträgt. Man verwendet einen annähernd monochromatischen Neutronenstrahl, der ähnlich wie im obigen Doppelspalt-Experiment
192CHAPTER 12. EINE KURZE EINFÜHRUNG IN DIE FEYNMAN’SCHEN PFADINTEG
in zwei Pfade zerlegt wird, die anschließend wieder zusammenlaufen.
Einer der Teilstrahlen verläuft auf einem Weg, der eine höhere potentielle
Energie im Schwerefeld der Erde hat. Die Apparatur ist auf einer ebenen Platte angebracht, die um eine Achse um einen beliebigen Winkel
ϕ verkippt werden kann, so daß der Beitrag der potentiellen Energie zur
Wirkung durch
∆S = T · mg gl1 sin(ϕ)
beschrieben werden kann. l1 ist hierbei die Breite der Platte und T die Zeit,
die das Neutron benötigt, die Länge l2 der Anordnung zu durchlaufen.
Wir können T über T = mt l2 /p durch den Impuls ausdrücken. Der Impuls
wiederrum hängt mit der de Broglie Wellenlänge über p = h/λ zusammen.
Der Unterschied der Wirkung auf den beiden Pfaden ist damit
∆S =
mg mt gl1 l2 λ sin(ϕ)
2π~
.
◦
Für Neutronen mit λ = 1.4A wurde die Interferenz als Funktion von ϕ
experimentell ermittelt. Man findet eine Periode von O(5◦ ). Da ~ im Nenner vorkommt geht die Periodenlänge im klassischen Grenzfall ~ → 0
gegen Null und die Interferenz-Oszillationen sind nicht mehr beobachtbar. Ein weiterer interessanter Punkt ist die Tatsache, daß hier beide Typen
von Massen, träge mt und schwere mg eingehen. Dieses Experiment ermöglicht
einen Test der Gleichheit dieser Massen auf mikroskopischem Gebiet. Man
findet die Gleichheit auch hier bestätigt.
12.2. QUANTEN-INTERFERENZ AUFGRUND VON GRAVITATION193
Figure 12.1: Meßanordnung zum Aharonov-Bohm-Effekt. Durch den
Kreis verläuft der magnetische Fluß senkrecht zur Bildebene. Außerhalb
des Kreises verschwindet das Magnetfeld.
194CHAPTER 12. EINE KURZE EINFÜHRUNG IN DIE FEYNMAN’SCHEN PFADINTEG
Appendix A
Details
A.1
Proof of eq. 1.7
back to pag. 15 Let us call the left and right side of (1.7) l(ϕ) and r(ϕ),
respectively. In addition, Differentiating both sides w.r.t. ϕ yields




0 −1 0
− sin ϕ − cos ϕ 0
0  e−iϕAz r0 (ϕ) =  cos ϕ
− sin ϕ 0  (A.1)
l0 (ϕ) =  1 0
0 0
0
0
0
0
Multiplying the left expression from the right by l(ϕ)−1 r(ϕ) yields the right
term, i.e.
l0 (ϕ)l(ϕ)−1 r(ϕ) = r0 (ϕ) ⇒ log(l(ϕ))0 = log(r(ϕ))0 .
This equation, with the initial conditions
l(0) = r(0)
gives
l(ϕ) = r(ϕ)
A.2
Evaluation of (1.9)
back to pag. 15 The calculation is simplified by the fact that novanishing
terms only come from mixed products Ax Ay . We thus get
(1 + iAy ϕ)(1 + iAx ϕ)(1 − iAy ϕ)(1 − iAx ϕ) + · · · =
= 1 + ϕ2 (−Ay Ax + Ay Ax + Ax Ay − Ay Ax ) + · · · = 1 + ϕ2 [Ax , Ay ] + · · ·
195
196
A.3
APPENDIX A. DETAILS
Discussion about the phase ωz in (1.14)
back to pag. 16 In the expansion of the r.h.s. of (1.14) we also expand the
phase ωz , since this is also “small” like ϕ (it vanishes when ϕ → 0). We
obtain instead of (1.16)
1−
ϕ2
ωz2
+ i( Jz + ωz 11) .
2
~
(A.2)
If this has to be equal to (1.15) up to O(ϕ2 ), it follows that ωz = O(ϕ2 ).
Therefore we have
[Jx , Jy ] = i~(Jz + dz 11)
(A.3)
with dz = ϕ~2 ωz = const.. This constant can be absorbed in the definition:
Jeα = Jα + dα so that (1.17) holds for the Jeα operators.
This means that the rotation operators Jα are not uniquely defined: there
is an arbitrary constant, which by convention is set to zero. The choice of
the constant does not affect physical results.
A.4
Transformation of the components of a vector
back to pag. 18
The projection of a vector A on the direction defined by the unit vector n is
given by An := n · A. A rotation transforms n into n0 = R n. Accordingly
the projection of the operator is transformed after the same rotation into
A0n = n0 · A. This gives
A0n = (Rβα nα )Aβ .
In order to determine the α-component A0α we set n = eα , i.e.
A0α = Rβα Aβ = RT A α
(A.4)
(A.5)
This shows that the components of a vactor transform according to RT =
R−1 .
A.5
Commutation rules of the orbital angular momentum
back to pag. 25
A.6. FURTHER COMMUTATION RULES OF J
197
We start by proving (1.20) with A = r:
[Lα , rβ ] = εαγρ [rγ pρ , rβ ]
= εαγρ rγ [pρ , rβ ]
| {z }
(A.6)
−i~δρβ
= −i~εαγβ rγ
= i~εαβγ rγ
.
The proof for A = p is analogous.
We now evaluate the commutator [Lx , Ly ].
[Lx , Ly ] = [Lx , rz px − rx pz ] = i~(−ry px + rx py ) = i~Lz
(A.7)
together with its cyclic permutations, this proves that L satisfies (1.17).
A.6
Further Commutation rules of J
back to pag. 19
[J 2 , Jβ ] = [Jα Jα , Jβ ]
= Jα [Jα , Jβ ] + [Jα , Jβ ]Jα
= i~ εαβγ Jα Jγ + εαβγ Jγ Jα
.
After renaming α ↔ γ in the last term it follows
[J 2 , Jβ ] = i~ εαβγ Jα Jγ + εγβα Jα Jγ
= i~ εαβγ Jα Jγ − εαβγ Jα Jγ = 0
.
[J+ , J− ] = [Jx + iJy , Jx − iJy ]
= [Jx , Jx ] + [Jy , Jy ] +i[Jy , Jx ] − i[Jx , Jy ]
| {z } | {z }
=0
=0
= −2i[Jx , Jy ] = −2i · i~Jz = 2~Jz
[Jz , J± ] = [Jz , Jx ] ± i[Jz , Jy ]
= i~εzxy Jy ± i(i~)εzyx Jx
= i~Jy ± ~Jx = ±~(Jx ± iJy )
[J 2 , J± ] = 0
This follows immediately from (1.21).
198
A.7
APPENDIX A. DETAILS
Uncertainty relation for j = 0
back to pag. 23
The uncertainty relation implies in general
| (∆Jx )2 | · | (∆Jy )2 | ≥
~
·j
2
.
For j = 0 one has
|h[Jx , Jy ]i| = ~|hJz i| = ~2 · |m| = 0
⇒
(∆Jx )2 · (∆Jy )2 can become zero.
(A.8)
(A.9)
(A.10)
A.8 J × J
back to pag. 24
(J × J )α = εαβγ Jβ Jγ
1
= (εαβγ Jβ Jγ + εαγβ Jγ Jβ )
2
1
= εαβγ [Jβ , Jγ ]
2
1
= εαβγ i~εβγδ Jδ
2
i~
= 2 · δαδ Jδ
2
= i~Jα
A.9
Explicit derivation of angular momentum operators and their eigenfunctions
back to pag. 27
Using (1.50) the components of the angular momentum operator can be
expressed as

∂
∂
− z ∂y
) 
Lx = ~i (y ∂z
~
∂
∂
∂
~
∂
∂
Ly = i (z ∂x − x ∂z )
⇒ L± =
(y ∓ ix) − z(
∓i )

i
∂z
∂y
∂x
~
∂
∂
Lz = i (x ∂y − y ∂x )
A.9. EXPLICIT DERIVATION OF ANGULAR MOMENTUM OPERATORS AND THEIR EIGENFU
With the spherical coordinates
x = r · sin θ cos ϕ
y = r · sin θ sin ϕ
z = r · cos θ
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
∂θ
∂θ ∂x ∂θ ∂y ∂θ ∂z
∂
∂
∂
+ r cos θ sin ϕ
− r sin θ
= r cos θ cos ϕ
|
{z
} ∂x | {z } ∂y
∂z
cot θ·x
cot θ·y
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
∂ϕ
∂ϕ ∂x ∂ϕ ∂y ∂ϕ ∂z
∂
∂
= − r sin θ sin ϕ
+ r sin θ cos ϕ
| {z } ∂x | {z } ∂y
y
x
∂
∂
−y
=x
∂y
∂x
Thus
Lz =
~ ∂
i ∂ϕ
(A.11)
We now show that
L± = ~e±iϕ (±
∂
∂
+ i cot θ )
∂θ
∂ϕ
(A.12)
Proof:
∂
∂
∂
± r cos θ sin ϕ
∓ r sin θ
∂x
∂y
∂z
∂
∂
+i · r · cot
· sin θ} cos ϕ
− i · r · cot
θ{z
· sin θ} sin ϕ )
| θ{z
|
∂y
∂x
cos θ
cos θ
~e±iϕ (±r cos θ cos ϕ
L± =
=
~e
±iϕ
∓r · sin θ
∂
∂
+ (ir · cos θ cos ϕ ± r · cos θ sin ϕ)
∂z
∂y
∂
−(ir · cos θ sin ϕ ∓ r cos θ cos ϕ)
∂x
200
APPENDIX A. DETAILS

=
~e±iϕ ∓r · sin θ
∂
∂
+ ir · cos θ (cos ϕ ∓ i sin ϕ)
{z
} ∂y
|
∂z
e∓iϕ

∂
±r · cos θ (cos ϕ ∓ i sin ϕ) 
{z
} ∂x
|
e∓iϕ
=
=
=
∂
∂
∂
~ ∓e r · sin θ
+ iz
±z
∂z
∂y
∂x


∂
∂
∂
∓ i )
~ (∓ r · sin θ cos ϕ −i r sin θ sin ϕ) + iz(
|
{z
} | {z } ∂z
∂y
∂x
x
y
∂
∂
∂
~(−i) (y ∓ ix) − z(
∓i )
∂z
∂y
∂x
∓iϕ
q.e.d
Firs we determine the ϕ-part of the eigenfunctions Ylm (θ, ϕ). From Eqs.
(1.40b) and (1.51) it follows
∂ m
Yl (θ, ϕ) = im Ylm (θ, ϕ)
∂ϕ
1
⇒ Ylm (θ, ϕ) = √ eimϕ Θm
l (θ)
2π
(A.13)
Θm
l is still an arbitrary function which depends on θ only. Instead of (1.48)
we now use (1.31), i.e. L+ |l, m = li = 0 and obtain the equation
∂ m=l
Y
(θ, ϕ)
∂θ l
+
i
cos θ ∂ m=l
Y
(θ, ϕ) =
sin θ ∂ϕ l
|
{z
}
ilYll (θ,ϕ)
(A.13)
=
⇒
⇒
cos θ m=l
eimϕ ∂ m=l
!
√
Θl (θ) − l
Θl (θ) = 0
sin θ
2π ∂θ
∂ l
cos θ l
Θl (θ) = l ·
Θ (θ)
∂θ
sin θ l
Θll (θ) = Cl · sinl θ
A.9. EXPLICIT DERIVATION OF ANGULAR MOMENTUM OPERATORS AND THEIR EIGENFU
cos θ l
∂
sinl θ = l · cos θ sinl−1 θ = l ·
Θ (θ)
∂θ
sin θ l
Proof:
The normalisation gives
Z
2
dΩ Ylm=l (θ, ϕ) = |Cl |2
Zπ
sin2l θ sin θ dθ
0
2
= |Cl | ·
Zπ
sin2l+1 θ dθ
0
√
= |Cl |2 π
= |Cl |2
Γ(l + 1)
Γ(l + 1 + 12 )
(l!2l )2 · 2 !
=1
(2l + 1)!
Convention
(−1)l
Cl = l
2 l!
r
(2l + 1)!
2
From this it follows
(−1)l
Ylm=l (θ, ϕ) = l
2 l!
r
(2l + 1)! ilϕ l
e sin θ
4π
The eigenfunction associated to smaller values of the magnetic quantum
number m are obtained by repeated application of equation (1.43)
1
|l, m − 1i = [l(l + 1) − m(m − 1)]− 2
1
= [(l + m)(l − m + 1)]− 2
L−
~
L−
~
⇒
m=l:
1
|l, l − 1i = [(2l) · 1]− 2
L−
~
|l, li
|l, mi
|l, mi
202
APPENDIX A. DETAILS
m=l−1:
− 21
|l, l − 2i = (2l)
− 12
[(2l − 1) · 2]
L−
2
|l, li
~
..
.
− 21
[l, l − ni = [(2l)(2l − 1) . . . (2l − n + 1)n!]
=
(2l − n)!
(2l)!n!
12 or with (n = l − m)
|l, mi =
m
Yl (θ, ϕ) =
(l + m)!
(2l)!(l − m)!
21 L−
n
~
n
L−
~
|l, li
(l + m)!
(2l)!(l − m)!
−iϕ
−e
12 L−
~
l−m
|l, li
l−m
∂
∂
Yll (θ, ϕ)
( − i cot θ )
∂θ
∂ϕ
Yll (θ, ϕ) is of the form Cl eiϕl · sinl θ =: Cl eiϕl f (θ).
L−
to Yll yields
A single application of
~
∂
−iϕ iϕl
− Cl e e
− i · cot θ · (+il) f (θ)
∂θ
∂
i(l−1)ϕ
= −Cl e
+ cot θ · l f (θ)
∂θ
|
{z
}
ξ
Conjecture:
ξ =
|l, li
1 d
(sinl θ · f (θ))
l dθ
sin θ
Proof
1
d
d
l
l
ξ =
( sin θ) f (θ) + sin (θ) f (θ)
dθ
sinl θ dθ
1
d
l
l
=
l · cot θ sin θ f (θ) + sin θ f (θ)
dθ
sinl θ
d
=
+ l · cot θ f (θ)
dθ
A.9. EXPLICIT DERIVATION OF ANGULAR MOMENTUM OPERATORS AND THEIR EIGENFU
q.e.d.
Thus, ξ can also be expressed as
ξ = −
⇒
(
d
(sinl θ · f (θ))
l−1
sin θ d(cos θ)
1
·
1
L− l
d
)Yl (θ, ϕ) = Cl ei(l−1)ϕ l−1 ·
(sinl θ · f (θ))
~
sin θ d(cos θ)
We have started from the general form
Yll (θ, ϕ) = Cl eilϕ f (θ)
The application of
and:
L−
ist therefore
~
eilϕ → ei(l−1)ϕ
1
d
f (θ) → f˜ =
(sinl θ · f (θ))
l−1 dcos θ
sin θ
A second application leads to
(
L− 2 l
1
d
) Yl = Cl ei(l−2)ϕ l−2
(sinl−1 f˜)
~
sin θ dcos θ
= Cl ei(l−2)ϕ
d2
sinl θ · f (θ)
sinl−2 θ d(cos θ)2
1
We recognize the regularity
(
L− n l
1
dn
) Yl = Cl ei(l−n)ϕ l−n
sinl θ · f (θ)
~
sin θ d(cos θ)n
.
We now insert the special form f (θ) = sinl θ and obtain
S PHERICAL HARMONICS
(−1)l
Ylm (θ, ϕ) = l
2 l!
s
(2l + 1)(l + m)! eimϕ
1
dl−m
√ ·
·
sin2l θ (A.14)
m
l−m
2(l − m)!
sin
θ
d(cos
θ)
2π
Ylm are termed SPHERICAL HARMONICS.
204
A.10
APPENDIX A. DETAILS
Relation betwen L2 and p2
back to pag. 34
L
2
r×p
=
α
· r×p
α
= εαβγ εαβ 0 γ 0 rβ pγ rβ 0 pγ 0
= (δββ 0 δγγ 0 − δβγ 0 δγβ 0 )rβ pγ rβ 0 pγ 0
= rβ pγ rβ pγ − rβ pγ rγ pβ
| {z }
|{z}
rβ pγ −i~δβγ
pβ rγ +i~δβγ
= r 2 p 2 − i~ r · p − r̂β p̂γ p̂β r̂γ − i~ rp
= r 2 p 2 − 2i~ r · p − rβ pβ · pγ rγ
|{z}
(A.15)
rγ pγ −3i~
L
A.11
2
2
2
2
= r p + i~ r · p − (r · p)
Proof that σ ≤ 0 solutions in (2.17) must be
discarded
back to pag. 41
The second solution of (2.17), σ = −l, must be discarded because it is too
singular around r = 0, as we show below.
The wave function is
ψ(r) =
χ(r)
Y (θ, ϕ)
r
(A.16)
Its norm is
Z
2
3
|ψ(r)| d r =
Z
Z∞
2
|Y (θ, ϕ)| dΩ ·
|
{z
} 0
=1
Z∞
=
χ(r)2 dr
.
0
Its contribution around r = 0 gives (for c0 6= 0 )
Z
|c0 |2 r2σ dr
χ(r)2 2
r dr
r2
A.12. DETAILS OF THE EVALUATION OF THE RADIAL WAVE FUNCTION FOR HYDROGEN2
which diverges for σ ≤ −1.
For σ = l = 0 the norm does not diverge, however the kinetic energy does.
Z
~2
Ekin =
|∇ψ(r)|2 d3 r
.
(A.17)
2m
R3
The angular part of the wave function Y00 =
again the behavior for r → 0
∇ψ(r) ∼ ∇(
√1
4π
is a constant. We consider
c0
er
)∼− 2
r
r
⇒ ||∇ψ(r)||2 ∼
|c0 |2
r4
Therefore the integral (A.17) diverges.
A.12
Details of the evaluation of the radial wave
function for hydrogen
back to pag. 48
Aus dem Verhältnis der Entwicklungskoeffizienten (2.19) und (2.27) wird
somit
2γ(µ + l + 1) − 2 aZ0
cµ+1
=
cµ
(µ + l + 2)(µ + l + 1) − l(l + 1)
µ+1 +l −n
= 2γ
(µ + l + 2)(µ + l + 1) − l(l + 1)
.
Es ist sinnvoll, dimensionslose Entwicklungskoeffizienten
dµ := (2γ)−µ cµ
einzuführen, mit
µ+l+1−n
dµ+1
=
dµ
(µ + l + 2)(µ + l + 1) − l(l + 1)
µ+l+1−n
= 2
l + l(3 + 2µ) + (1 + µ)(2 + µ) − l2 − l
µ+l+1−n
=
2l(1 + µ) + (1 + µ)(2 + µ)
l−n+µ+1
=
.
(2l + 2 + µ)(1 + µ)
(A.18)
206
APPENDIX A. DETAILS
Die Koeffizienten sind mit den Abkürzungen ν = n − l − 1 und α = 2l + 2
−ν
d0
α·1
−ν(−ν + 1)
−ν + 1
d1 =
d0
=
(α + 1) · 2
α(α + 1) · 1 · 2
−ν + 2
−ν(−ν + 1)(−ν + 2)
=
d2 =
d0
(α + 2) · 3
α(α + 1)(α + 2) · 1 · 2 · 3
..
.
d1 =
d2
d3
dµ =
−ν(−ν + 1) . . . (−ν + µ − 1) 1
d0
α(α + 1) . . . (α + µ − 1)
µ!
.
(A.19)
Diese Koeffizienten in die Reihe
Rnl (r) =
n−l−1
X
µ
cµ r =
µ=0
n−l−1
X
dµ (2γr)µ
µ=0
eingesetzt führen mit d0 = 1 auf die entartete Hypergeometrische Funktion
ν
X
−ν(−ν + 1) . . . (−ν + µ − 1) (2γr)µ
Rnl (r) = 1 F1 (−ν, α; 2γr) :=
α(α + 1) . . . (α + µ − 1)
µ!
µ=0
.
Es handelt sich um orthogonale Polynome vom Grade ν mit den Eigenschaft
1 F1 (−ν, α; 0)
=1
1 F1 (0, α; r) = 1
(A.20)
(A.21)
.
Die entartete Hypergeometrische Funktion hängt mit dem orthogonalen
Laguerre-Polynom1 Lνα−1 vom Grad ν zusammen
Lνα−1 (r)
ν+α−1
=
1 F1 (−ν, α; 2γr)
ν
.
Das führt uns zum Endergebnis (2.28)
1
Es gibt in der Literatur zwei unterschiedliche Definitionen der Laguerre-Polynome.
Zwischen der hier verwendeten Definition Lα
n und der anderen Definition L̃ besteht die
(−1)α α
α
Beziehung Ln = (n+α)! L̃n+α
A.13. POTENTIAL OF AN ELECTRON IN THE GROUND STATE OF A H-LIKE ATOM207
A.13
Potential of an electron in the ground state
of a H-like atom
back to pag. 58
The potential V (R) produced in a point R by an electron with wave func 3 12
e−Z r (see (3.3)) centered in the origin of the coordinates
tion ψ(r) = Zπ
is given by
Z
1
−V (R) = d3 r |ψ(r)|2
.
|R − r|
The radial component of the electric field in R is given by Gauss’ law:
E(R) =
QR
R2
with the total charge contained between 0 and R:
Z R
−QR = 4π
r2 |ψ(r)|2 = 1 − (1 + 2RZ(1 + RZ))e−2RZ
0
The potential V (R) is given by
Z R
1
E(R0 )d R0 =
−V (R) =
1 − (1 + ZR)e−2ZR
R
∞
A.14
(A.22)
Löwdin orthonormalisation
back to pag. 56
Let us start from a non-orthonormal, yet linear independent basis
hψi |ψj i = Sij
∗
Sij is invertible and hermitian: Sji
= Sij .
An orthonormal basis is given by the vectors |ϕα i obtained as (Einstein
sum convention)
−1
(A.23)
|ϕα i = |ψj i Sjα2
Proof:
−1∗
−1
−1
−1
1
1
hϕβ |ϕα i = Siβ 2 hψi |ψj i Sjα2 = Sβi 2 Sij Sjα2 = (S − 2 SS − 2 )αβ = δαβ
1
1
That S − 2 SS − 2 = 11 can be shown by introducing the unitary matrix U that
diagonalizes S:
U † SU = Sdiag
208
APPENDIX A. DETAILS
Then
1
1
1
−1
1
−1
2
2
S − 2 SS − 2 = U U † S − 2 U U † SU U † S − 2 U U † = U Sdiag
Sdiag Sdiag
U † = U 11U † = 11 .
−1
−1
2
2
Sdiag
Sdiag Sdiag
is a diagonal matrix with diagonal elements of the form
−1
−1
sn 2 sn sn 2 = 1, and thus equal to 11.
A.15
Some details for the total S
back to pag. 72
Generators of rotations
Let us consider a product state of two single spin states
|Φ1 i1 ⊗ |Φ2 i2 .
Let us carry out a rotation by an angle α around, say, the z-axis, then
|Φ01 i = e−iα
|Φ02 i = e−iα
are the rotated spin states, and
z
S1
~
z
S2
~
|Φ1 i
|Φ2 i
|Φ01 i1 ⊗ |Φ02 i2
the rotated product state. Now, it is easy to show that the latter equals
e−iα
I.e.,
Sz
~
Sz
~
|Φ1 i1 ⊗ |Φ2 i2 .
is the generator of rotation in the product space.
Commutators
Although the above result should be sufficient, we now explicitly prove
the commutation relations of S:
[S α , S β ] =
=
=
α
β
[S , S ] =
[(S1α + S2α ), (S1β + S2β )]
[S1α , S1β ] + [S2α, S2β ]
αβγ S1γ + S2γ
αβγ S γ
.
(A.24)
A.16. PROOF OF WIGNER ECKART’S THEOREM FOR VECTORS
A.16
209
Proof of Wigner Eckart’s theorem for vectors
back to pag. 81
The idea consists in showing the proportionality
hj, m| V̂ M |j1 , m1 i ∝ hj1 , j2 = 1, j, m| |j1 , m1 i |j2 = 1, m2 = Mi
(A.25)
To do this, we go back to the definition of the Clebsch-Gordan coefficients
(4.9b), which describe the coefficients of the expansion of the |j1 , j2 , j , mi
in the product basis.
C(j, m|j1 , m1 ; j2 , m2 ) = hj1 , j2 , j, m| |j1 , m1 i |j2 , m2 i
(A.26)
We rephrase in another language the procedure we carried out in Sec. 4.6.2:
We insert the operator Jˆ+ = Jˆ1+ + Jˆ2+ in (A.26) and apply it once to the right
and once to the left (in this case it acts as a Jˆ− ) and set the results equal
α+ (j1 , m1 ) hj1 , j2 , j, m| |j1 , m1 + 1i |j2 , m2 i
+ α+ (j2 , m2 ) hj1 , j2 , j, m| |j1 , m1 i |j2 , m2 + 1i
= α− (j, m) hj1 , j2 , j, m − 1| |j1 , m1 i |j2 , m2 i
where
α± (j, m) =
p
j(j + 1) − m(m ± 1)
(A.27)
(A.28)
this gives a recursion equation for the coefficients of |j1 , j2 , j, m − 1i in
terms of the ones for |j1 , j2 , j, mi. Starting from the one with the largest 2
m, i.e. |j1 , j2 , j, ji, we get all the others down to |j1 , j2 , j, −ji.
The commutation rules for vector operators (1.20) lead for (4.23) to commutation rules similar to (1.24), i.e
√
√
[Jˆ+ , V̂+1 ] = 0
[Jˆ+ , V̂0 ] = 2~ V̂+1
[Jˆ+ , V̂−1 ] = 2~ V̂0
which can be summarized to
[Jˆ+ , V̂M ] = α+ (1, M ) ~ V̂M +1
(A.29)
where the α+ (1, M ) are the (A.28) with j = 1 and a + . For simplicity, we
now use ~ = 1.
We now put both sides of (A.29) between two states ((A.25)):
ha, j, m| [Jˆ+ , V̂M ] |b, j1 , m1 i = α+ (1, M ) ha, j, m| V̂M +1 |b, j1 , m1 i
2
These can be determined by requiring that J − gives zero
210
APPENDIX A. DETAILS
expanding the commutator one gets
α− (j, m) ha, j, m − 1| V̂M |b, j1 , m1 i − α+ (j1 , m1 ) ha, j, m| V̂M |b, j1 , m1 + 1i
= α+ (1, M ) ha, j, m| V̂M +1 |b, j1 , m1 i
i.e.
α− (j, m) ha, j, m − 1| V̂M |b, j1 , m1 i =
(A.30)
α+ (j1 , m1 ) ha, j, m| V̂M |b, j1 , m1 + 1i + α+ (1, M ) ha, j, m| V̂M +1 |b, j1 , m1 i
This has the same form as (A.27) with j2 = 1, m2 = M . Remember, there
we said that (A.27) is sufficient to get all coefficients for a given fixed set
j, j1 , j2 starting from just one. Since (A.30) is formally the same, it means
that within a given set of (a, j; b, j1 ) the matrix elements of all three components
of V̂ between all pairs of m, m1 are proportional to the hj1 , m1 | hj2 = 1, m2 = M | |j1 , j2 = 1 , j , mi.
This gives (4.24), where ha, j| |V̂ | |b, j1 i is the proportionality constant and
can be determined by evaluating just one of the matrix elements on the
left, i.e. for one set of m, M, m1 . Then one automatically has all of them.
(4.24), compared with (4.9b) suggests that V̂M behaves similarly to a state
with angular momentum quantum numbers j2 = 1, m2 = M . This also
implies that the only nonzero matrix elements are such that |j − j1 | ≤ 1
and m = m1 + M .
A.17
Proof of the projection theorem
back to pag. 83
Consider that with the representation (4.23) the scalar product of two vectors reads
X
 · B̂ =
(−1)M ÂM B̂−M
(A.31)
M
We evaluate
hn, j, m1 | V̂ · Ĵ |n, j, m2 i =
=
X
(−1)M
X
m0
M
X
M
(−1)M hn, j, m1 | V̂M Jˆ−M |n, j, m2 i
hn, j, m1 | V̂M |n, j, m0 i hn, j, m0 | Jˆ−M |n, j, m2 i
since Ĵ does not change n, j. From (4.33) this becomes
X
X
= γ(n, j)
(−1)M
hn, j, m1 | JˆM |n, j, m0 i hn, j, m0 | Jˆ−M |n, j, m2 i
M
m0
2
= γ(n, j) hn, j, m1 | Ĵ |n, j, m2 i
A.18. A REPRESENTATION OF THE DELTA DISTRIBUTION
211
i.e.
γ(n, j) =
hn, j, m1 | V̂ · Ĵ |n, j, m2 i
(A.32)
2
hn, j, m1 | Ĵ |n, j, m2 i
which gives (4.34).
A.18
A representation of the delta distribution
back to pag. 92
Die Funktion ist auf Eins normiert:
Z
∆t (ω) dω = 1.
Das Verhalten von ∆t (ω) ist in Abbildung (A.1) als Funktion von ω zu
festem t wiedergegeben. Man erkennt, daß ∆t (ω) bei ω = 0 konzentriert
ist, eine Breite proportional zu 1t und eine Höhe proportional zu t hat.
Diese Funktion verhält sich im Limes t → ∞ also wie die Delta-Funktion.
Figure A.1: Plot der Funktion ∆t (ω) (durchgezogen) und der Einhüllenden ω22tπ
fi
(gestrichelt).
Z
lim
t→∞
f (ω) ∆t (ω) dω = f (0)
,
212
APPENDIX A. DETAILS
vorausgesetzt, die Test-Funktion f (ω) hat die Eigenschaft limω→∞
Für diese Klasse von Funktionen gilt (5.26)
lim ∆t (ω) = δ(ω)
f (ω)
ω
= 0.
.
t→∞
Integral evaluation of (6.7)
A.19
back to pag. 108 Wir wollen die Berechnung des Zwei-Zentren-Integrals
(6.7) explizit vorführen. Da die Dichten rotationssymmetrisch sind, ist es
sinnvoll, Kugelkoordinaten zu verwenden
∆E
(1)
Z
=
32π 2
Z
∞
dx1 x21
e
−x1
∞
Z
0
dx2 x22
e
−x2
Z
Z
dΩ1
dΩ2
0
1
|~x1 − ~x2 |
.
Zur Auswertung des inneren Winkelintegrals legen wir die z-Achse in
Richtung von ~x1 . Man erhält dann ein vollständiges Differential
Z
1
dΩ2
= 2π
|~x1 − ~x2 |
Z
π
d cos(θ) p
0
Z
1
1
x21 + x22 − 2x1 x2 cos(θ)
1
dξ p 2
x1 + x22 − 2x1 x2 ξ
ξ=+1
q
2π
2
2
= −
x1 + x2 − 2x1 x2 ξ x1 x2
= 2π
−1
ξ=−1
=
2π
x1 x2
x 1 + x2 − x1 − x2 Da es nun
R keine Winkelabhängigkeit mehr gibt, liefert das zweite Winkelintegral dΩ2 = 4π. Wenn wir schließlich die verbleibenden Radialanteile
berechnen, erhalten wir
∆E
(1)
Z
Z ∞
Z8π 2 ∞
−x1
−x2
x1 − x 2 dx
x
e
dx
r
e
x
+
x
−
=
1
1
2
2
1
2
32π 2 0
Z 0x1
Z ∞
Z ∞
Z
−x1
2 −x2
−2Zx2
=
dx1 x1 e
2
dx2 x2 e
+ 2x1
dx2 x2 e
4 0
0
x1
Z x1
Z
Z ∞
Z ∞
−x1
2 −x2
−2Zx2
=
dx1 x1 e
dx2 x2 e
+x1
dx2 x2 e
2 0
0
x1
|
{z
}
|
{z
}
2−e−x1 (2+2x1 +x21 )
e−x1 (1+x1 )
A.20. EXAMPLE OF EVALUATION OF MATRIX ELEMENTS IN PRODUCT STATES213
=
=
=
=
Z
Z ∞
2
2
−x1
−x1
2 − e (2 + 2x1 + x1 − x1 − x1
dx1 x1 e
2 0
Z ∞
Z ∞
Z
1 ∞
−t
−2t
2 −2t
Z
dt t e −
dt t e −
dt t e
2 0
0
0
Z ∞
Z
Z ∞
1 ∞
1
−t
−x
2 −x
Z
dt t e −
dx x e −
dx x e
4 0
16 0
0
1
1
1 1
5
Z Γ(2) − Γ(2) − Γ(3) = Z 1 − − − ) = Z
4
16
4 8
8
.
Wir haben somit (6.8).
A.20
Example of evaluation of matrix elements
in product states
back to pag. 111 As an example, we determine the last matrix element in
(6.10).
1
h2lm|1 ⊗ h100|2
|100i1 ⊗ |2lmi2
|r̂ 1 − r̂ 2 |
To do this, we insert the identity
Z
I = dr 1 dr 2 |r 1 i1 |r 2 i2 hr 1 |1 hr 2 |2
Z
dr 1 dr 2 h2lm|1 ⊗ h100|2
1
|r 1 i1 |r 2 i2 hr 1 |1 hr 2 |2 |100i1 ⊗ |2lmi2
|r̂ 1 − r̂ 2 |
Since |r 1 i1 |r 2 i2 is an eigenstate of
Z
1
|r̂ 1 −r̂ 2 |
dr 1 dr 2 h2lm|1 ⊗ h100|2 |r 1 i1 |r 2 i2
Z
=
1
hr 1 |1 hr 2 |2 |100i1 ⊗ |2lmi2
|r 1 − r 2 |
dr 1 dr 2 h2lm|1 |r 1 i1 h100|2 |r 2 i2
Z
=
this becomes
dr 1 dr 2 ϕ∗2lm (r 1 )ϕ∗100 (r 2 )
1
hr 1 |1 |100i1 hr 2 |2 |2lmi2
|r 1 − r 2 |
1
ϕ100 (r 1 )ϕ2lm (r 2 )
|r 1 − r 2 |
which corresponds to the exchange term in (6.11)
214
APPENDIX A. DETAILS
A.21
Some proofs for Chap.(8)
back to pag. 122
Equivalence of (8.5) and (8.4)
Using a discrete version of the derivative u0i =
we have for the r.h.s. of (8.5) 3
ui −ui−1
∆x
together with (8.6),
1 ∂L
∂`i
1 ∂`i
1 ∂`i+1
∂`
d ∂`
=
+
−
⇒
−
0
0
∆x ∂ui
∂ui ∆x ∂ui ∆x ∂ui+1 ∆x→0 ∂u dx ∂u0
(A.33)
where `i ≡ `(ui , u0i , u̇i , xi , t). This gives (8.4). An alternative derivation is by
partial integration.
Canonical commutation rules for the p̂m , q̂m
Z
L
Z
[q̂m , p̂ ] =
L
dx0 Um (x) Um0 (x0 ) [û(x), π̂(x0 )]
dx
m0
0
0
Z
L
= i~
Z
dx
0
Z
0
L
dx0 Um (x) Um0 (x0 ) δ(x − x0 )
L
dx Um (x) Um0 (x)
= i~
0
= i~δm,m0 .
(A.34)
Hamiltonian in terms of p̂m , q̂m .
L
1
0
0
Un (x)Um (x)p̂n p̂m + KLUn (x)Um (x) q̂n q̂m dx
ρ
0
Z
Z
1 X L
KL X L 0
0
=
Un (x)Um (x) dx p̂n p̂m +
U (x)Um
(x) dx q̂n q̂m
2ρ n,m 0
2 n,m 0 n
|
{z
}
|
{z
}
1X
Ĥ =
2 n,m
Z
2δ
kn
n,m
δn,m
=
3
1
2
X
n
1 2
p̂ + KL kn2 q̂n2
ρ n
.
Strictly speaking for all i but the last one.
A.22. PROOF OF (9.3)
215
We have used the fact that
Z L
0
Un0 (x)Um
(x) dx = δn,m
(A.35)
0
proof:
Z
L
L
0
Un0 (x)Um
(x)
dx =
0
=
Un0 (x)Um (x)
0
|
}
kn2
{z
Z
=0
L
−
Z
0
L
Un00 (x) Um (x) dx
| {z }
2 U (x)
=−kn
n
Un (x)Um (x) dx = kn2 δn,m .
0
A.22
Proof of (9.3)
back to pag. 128 First we transform the Lagrangian a little bit. We exploit
the fact that
∇ (Ψ∗ ∇Ψ − (∇Ψ∗ )Ψ) = Ψ∗ ∇2 Ψ − (∇2 Ψ∗ )Ψ .
After integration over the volume the first term gives a surface integral
(Gauss’ law), which vanishes in an infinite volume. Then (9.1) becomes
Z ~2
2 ∗
∗
∗
(∇ Ψ )Ψ − V (x)Ψ Ψ d3 x .
(A.36)
L=
i~Ψ Ψ̇ +
2m
So that
π(x, t) =
δL
= i~Ψ∗ (x, t)
δ Ψ̇
δL
~2 2 ∗
= −V (x)Ψ∗ (x, t) +
∇Ψ
δΨ
2m
Lagrange II:
−i~Ψ̇∗ (x, t) = V (x)Ψ∗ (x, t) −
A.23
~2 2 ∗
∇Ψ
2m
Gauge transformation for the wave function
back to pag. 117
216
APPENDIX A. DETAILS
For simplicity, we work in units in which ~ = c = 1. Using (7.9), we can
rewrite (7.11) as
1
i∂t Ψ0 =
(p − qA)2 Ψ0 + qϕ Ψ0
(A.37)
2m
with
p = p − q∇χ
i∂ t = i∂t + q∂t χ
(A.38)
Applying these to the transformed wave function (7.10), we observe that
p Ψ0 = −i∇Ψ0 − Ψ0 q∇χ = Ψ 0 q∇χ + eiqχ(r,t) (−i∇Ψ)−Ψ 0 q∇χ
= e iqχ(r,t) p Ψ
(A.39)
similarly
i ∂t Ψ0 = −Ψ0 q ∂t χ + eiqχ(r,t) i∂t Ψ + Ψ0 q ∂t χ = e iqχ(r,t) i ∂t Ψ
(A.40)
In other words,
p eiqχ(r,t) = eiqχ(r,t) p
∂t eiqχ(r,t) = eiqχ(r,t) ∂t
(A.41)
Therefore, when applying the differential operators p and ∂ t , they “move” the
eiqχ(r,t) to the left and “lose” the · · ·. Therefore, (A.37) becomes
1
2
iqχ(r,t)
iqχ(r,t)
(p − qA) Ψ + qϕΨ
e
i∂t Ψ = e
2m
which is equivalent to (7.6).
A.24
Commutators and relation for the Schrödinger
field quantisation
back to pag. 129
We consider a more general case, namely (9.15), of which (9.9) is a special case. First of all one can invert the (9.15)
R in the usual way by scalar
multiplication with the wave functions: e.g. χ∗m (y) dy · · · . The first term
gives
Z
ĉm =
χ∗m (y) dy Ψ̂(y)
the other one is the hermitian conjugate
Z
†
ĉn = χn (z) dz Ψ̂† (z)
(A.42)
A.25. NORMALISATION OF TWO-PARTICLE STATE
217
so that the commutation rules become
Z
†
[cm , cn ] =
dy dz χ∗m (y)χn (z)δ(y − z) = δm,n
The above proof have simply made use of the fact that the χm are a complete, orthogonal basis set, so this holds obviously for the bn and (9.9) as
well.
We now consider the Hamiltonian (9.11a), using (9.15). But this is easily
generalizable for arbitrary single particle operators:
Z
~2 2
†
∇ + V (x) Ψ̂(x) dx =
ĤZQ = Ψ̂ (x) −
2m
X X Z
~2 2
†
3
∗
=
ĉn
ĉm d x χn (x) −
∇ + V (x) χm (x)
2m
n
m
|
{z
}
hn,m
For the case in which χn are replaced by the ϕn , the eigenfunctions of the
Hamiltonian, then clearly hn,m = En δn,m .
A.25
Normalisation of two-particle state
back to pag. 134 Wir wollen zunächst die Normierung des Vektors
√ bestimmen. Für ν = µ kennen wir die Normierung schon Zνν = 2!. Für
ν 6= µ erhalten wir
2
Zνµ
= h0|bµ bν b†ν b†µ |0i .
Wir bringen den Vernichter bν ganz nach rechts, da dann bν |0i verschwindet.
bν b†ν b†µ |0i = b†ν bν b†µ |0i + b†µ |0i
µ6=ν
= b†ν b†µ bν |0i +b†µ |0i = b†µ |0i .
| {z }
=0
Nun multiplizieren wir noch von links mit dem Vernichter bµ und bringen
ihn ebenfalls nach rechts
bµ bν b†ν b†µ |0i = bµ b†µ |0i = b†µ bµ |0i +|0i = |0i
| {z }
=0
und die gesuchte Normierung ist
2
Zνµ
= h0|0i = 1 .
218
APPENDIX A. DETAILS
A.26
Commutations rules
back to pag. 141
[a b, c] = abc − cab
a[b, c] + [a, c]b = abc − acb + acb − cab = abc − cab
a{b, c} − {a, c}b = abc + acb − acb − cab = abc − cab
which proves (9.41).
(9.42) is proven by observing that all terms are just minus the ones occurring in (9.41)
A.27
Lagrangian for electromagnetic field
back to pag. 158
First, observe that
2(∇ × A)2 =
X
ij
(∂i Aj − ∂j Ai )2 = 2
X
ij
(∂i Aj ) (∂i Aj − ∂j Ai ) =
The first equality is because in the sum over i, j only terms i 6= j contribute.
Consider the case i = 1, j = 2, this gives (∇×A)2z , the same for i = 2, j = 1,
which gives a factor 2. Then, similarly 2, 3 gives the x component, and
3, 1 the y component. The second equality is obtained by expanding and
interchanging indices
One can, thus, write
X
X
(∇ × A)2 =
∂i (Aj (∂i Aj − ∂j Ai )) −
Aj ∂i (∂i Aj − ∂j Ai )
ij
ij
The first term on the r.h.s. is a divergence, which, integrated over a volume, can be transformed into a surface integral. Taking the surface at infinity, this term can be made to vanish. The second term can be simplified
due to te fact that from (10.2) ∂i Ai = 0. Thus the only contribution left is
X
−
Aj ∂i2 Aj
ij
giving for (10.4)
L=µ
Z
1
d3 r
2
.
2
A +
!
X
i
A i ∇2 A i
(A.43)
A.28. INTEGRALS VS SUMS
219
We thus have the two terms of the Lagrange equation (cf. (10.5))
..
d
Πi = µ A i
dt
δL
= µ∇2 Ai .
δAi
Setting then equal gives indeed (10.3)
A.28
Integrals vs sums
back to pag. 159
We use the fact that
1 X ik·(r−r0 )
e
= δ(r − r 0 )
Ω k
ik·r
which is due to the completeness of the basis e√Ω .
Alternatively a convenient proof is obtained in the infinite volume Ω → ∞
limit.
Here, one can convert the sum over k into an integral. The discrete k are
due to the finite “box” as k = 2π
(n1 , n2 , n3 ) (cf. (10.12)). Therefore in the
L
L → ∞ limit, each point k can be replaced by the small cube of volume
3
around it. Thus,
∆k 3 = 2π
L
∆k
3
X
k
··· →
Z
d3 k · · ·
which gives
1X
1
··· =
Ω k
Ω∆k 3
Z
1
d k ··· =
Ω
3
L
2π
3 Z
3
d k ··· =
Z
Therefore, one has
1 X ik·(r−r0 )
e
→
Ω k
Z
d3 k ik·(r−r0 )
e
= δ(r − r 0 )
(2π)3
d3 k
··· .
(2π)3
(A.44)
220
APPENDIX A. DETAILS
A.29
Commutators and transversality condition
back to pag. 158 The commutation rules (10.8) and (10.14) are in contradiction with the transversality condition (10.2).
To see this, just take the divergence
[∂n · Ân (r), Π̂m (r 0 )] = i~ δn,m ∂n δ(r − r 0 )
| {z }
∇·A=0
The left-hand side vanishes identically, while the right hand side does not.
This holds equivalently for (10.14), which is its Fourier transform. Multi0
and sum over n, m:
ply (10.14) with kn km
[
X
kn Ânk ,
n
|
A.30
X
0
Π̂m−k0 km
] = i~
m
{z
k·Âk =0
X
nm
0
kn km
δnm δkk0 = i~ |k|2 δkk0
}
Commutators of fields
back to pag. 161
The completeness of the triplet {uk,−1 , u,k1 , kk } gives
X
unk,s umk,s +
s
kn km
= δn,n
k2
Therefore, the commutator
[Ânk , Π̂m−k0 ] =
X
unk,s um−k0 ,−s0 [q̂ks , p̂−k0 −s0 ]
ss0
=
X
s
A.31
unk,s umk,s = δn,m −
kn km
.
k2
Fourier transform of Coulomb potential
back to pag. 176
A.32. (NO) ENERGY CONSERVATION FOR THE FIRST-ORDER PROCESS (11.34)221
V (∆q) =
=
=
=
Z ix∆q
−Ze2
e
d3 x
Ω
r
Z
Z 1 Z ∞
−Ze2 2π
eir|∆q|ξ
dϕ
dξ
r2 dr
Ω
r
0
−1
0
Z 1
2 Z ∞
−2πZe
drr
dξeir|∆q|ξ
Ω
0
−1
2 Z ∞
ir|∆q|
−2πZe
e
− e−ir|∆q|
drr
Ω
ir|∆q|
0
Z
−2πZe2 ∞
ir|∆q|
−ir|∆q|
=
dr e
−e
iΩ|∆q| 0
−1
−2πZe2 −1
−
=
iΩ|∆q| i|∆q| −i|∆q|
−2πZe2
=
iΩ|∆q|2
Der Beitrag von der oberen Integrationsgrenze führt zu einer unendlich
stark oszillierenden Funktion, die verschwindet bereits, wenn man über
ein infinitesimales ∆q integriert. Integrale über q werden in der weiteren
Rechnung noch auftreten und somit ist V (q) bewiesen.
A.32
(No) energy conservation for the first-order
process (11.34)
back to pag. 177 To show that momentum and energy cannot be simultaneously conserved in the process (11.34), let us work in the reference frame
in which the initial electron is at rest q i = 0. Then, the initial energy Ei = 0
is also zero. By momentum conservation, then q f = −k. Since k cannot be
zero, the final energy Ef > 0, which is incompatible with energy conservation.