Download PART 1 Identical particles, fermions and bosons. Pauli exclusion

O. P. Sushkov
School of Physics, The University of New South Wales, Sydney, NSW 2052, Australia
(Dated: March 1, 2016)
PART 1
Identical particles, fermions and bosons.
Pauli exclusion principle.
Slater determinant.
Variational method.
He atom.
Multielectron atoms, effective potential.
Exchange interaction.
2
Identical particles and quantum statistics.
Consider two identical particles.
2 electrons
2 protons
2
12
C nuclei
2 protons
. . . .
The wave function of the pair reads
ψ = ψ(~r1 , ~s1 , ~t1 ...; ~r2 , ~s2 , ~t2 ...)
where
r1 , s1 , t1 , ... are variables of the 1st particle.
r2 , s2 , t2 , ... are variables of the 2nd particle.
r - spatial coordinate
s - spin
t - isospin
. . . - other internal quantum numbers, if any.
Omit isospin and other internal quantum numbers for now.
The particles are identical hence the quantum state is not changed under
the permutation :
ψ (r2 , s2 ; r1 , s1 ) = Rψ(r1 , s1 ; r2 , s2 ) ,
where R is a coefficient.
Double permutation returns the wave function back, R2 = 1, hence R = ±1.
The
spin-statistics theorem claims:
* Particles with integer spin have R = 1, they are called bosons (Bose - Einstein
statistics).
* Particles with half-integer spins have R = −1, they are called fermions (Fermi
- Dirac statistics).
3
The spin-statistics theorem can be proven in relativistic quantum mechanics. Technically
the theorem is based on the fact that due to the structure of the Lorentz transformation
the wave equation for a particle with spin 1/2 is of the first order in time derivative (see
discussion of the Dirac equation later in the course).
At the same time the wave equation for a particle with integer spin is of the second order
in time derivative.
An example: The vector potential in electrodynamics is to some extent equivalent to the
wave equation of photon. The photon has spin S = 1. Maxwell’s equation for the vector
~ reads
potential A
h
1 ∂2
c2 ∂t2
i
2 ~
~
− ∇ A = 4π~j.
here ~j is electric current. The equation contains the second time derivative.
Comment: Do not mix permutation with parity, these are different operations.
permutation : ψ (r1 , s1 ; r2 , s2 ) → ψ(r2 , s2 ; r1 , s1 )
space reflection: ψ (r1 , s1 ; r2 , s2 ) → ψ(−r1 , s1 ; −r2 , s2 )
4
Example: Statistics influence rotational spectra of diatomic molecules.
Consider rotational spectrum of Carbon2 molecule that consists of two
A
12
12
C isotops.
C nucleus has spin S = 0, hence it is a boson.
12
C
r1
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
12
C
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
r2
FIG. 1: Rotation of Carbon2 molecule
The wave function of the molecule reads
Ψ(1, 2) = U [(~r1 + ~r2 )/2] V ~r1 − ~r] ϕ1 , ϕ2
Here ϕ1 and ϕ2 are spin wave functions of the first and the second nucleus respectively.
U is the wave function of the center of mass motion.
V is the wave function of the relative motion.
Spin of the nucleus is zero, S = 0. Hence ϕ1 = ϕ2 = 1.
V (~r1 − ~r2 ) = χ(| r~1 − r~2 |)Ylm (~
r1 − r~2 )
where Ylm is spherical harmonic.
Let us perform permutation of the particles
V (2, 1) = χ (|~r2 − ~r1 |) Ylm (~r2 − ~r1 ) = (−1)l χ(| ~r1 − ~r2 |)Ylm (~r1 − ~r2 ) = (−1)l V (1, 2) .
Here I have used the exact mathematical relation (see 3d year quantum mechanics)
Ylm (−~r) = (−1)l Ylm (~r) .
Requirement of Bose statistics: ψ (2, 1) = ψ (1, 2)
Hence only even values of l are allowed in the rotational spectrum of C2 consisting of two
12
C isotops.
5
In a molecule consisting of two identical
12
C isotops only even values of l are allowed.
In a molecule consisting of two different isotops, say 12 C and 13 C or 12 C and 14 C all values
of l are allowed.
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
l=4
l=4
l=3
12
l=2
l=2
l=0
l=1
l=0
12
C C
12
14
C C
FIG. 2: Rotational spectra of C2 molecule. Left: two identical nuclei each with spin S=0. Right:
two distinguishable nuclei each with spin S=0.
6
Two noninteracting fermions in an external potential.
Consider any external potential: Coulomb field of a nucleus, a potential well, etc.
Let ϕa and ϕb be single particle states in the potential.
For an infinite potential well these are simple standing wave as it is illustrated in the Fig.
below
ϕb
ϕa
FIG. 3: Two lowest single particle orbital quantum states in an infinite square well potential.
Let us put two electrons with parallel spins in the potential.
Requirement of Fermi statistics: ψ (1, 2) = −ψ (2, 1).
Therefore, the many-body (in this case “many” = 2) wave function is
1
Ψ (1, 2) = √ [ϕa (r1 ) ϕb (r2 ) − ϕa (r2 ) ϕb (r1 )] | ↑i1 | ↑i2
2
(1)
If ϕa = ϕb then Ψ ≡ 0.
Thus, one cannot put two fermions in the same single-particle quantum state.
This is Pauli exclusion principle.
If spins of the electrons are opposite then the single particle states are different, ϕa↑ 6= ϕa↓ ,
even if the coordinate states are identical. So, such two-electron state is possible.
1
Ψ (1, 2) = ϕa (r1 ) ϕa (r2 ) √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ]
2
(2)
7
Slater determinant
Let us introduce spin in the definition of the single particle orbitals, and let us enumerate
these orbitals by index i : In these notations orbitals of the previous example are
ϕ1 (r, s) = ϕa↑ ≡ ϕa (r) | ↑i
ϕ2 (r, s) = ϕa↓ ≡ ϕa (r) | ↓i
ϕ3 (r, s) = ϕb↑ ≡ ϕb (r) | ↑i
ϕ4 (r, s) = ϕb↓ ≡ ϕb (r) | ↓i
Hence the wave function (1), page 6 can be written as determinant
1
1 ϕ1 (1) ϕ1 (2) Ψ (1, 2) = √ [ϕ1 (1) ϕ3 (2) − ϕ1 (2) ϕ3 (1)] ≡ √ 2
2 ϕ3 (1) ϕ3 (2) Similarly, the wave function (2), page 6 also can be written as determinant.
1
1 ϕ1 (1) ϕ1 (2) Ψ (1, 2) = √ [ϕ1 (1) ϕ2 (2) − ϕ1 (2) ϕ2 (1)] ≡ √ 2
2 ϕ2 (1) ϕ2 (2) I repeat the meaning of the notation ϕi (x)). Here the index i enumerates single particle
orbitals and x shows coordinates (spatial, spin, etc.) of a particle.
Using these notations we can write the many-body wave function for arbitrary number of
noninteracting fermions
ϕ (1) ϕ (2) .... ϕ (N) 1
1
1
ϕ (1) ϕ (2) .... ϕ (N) 2
2
2
1 Ψ (1, 2, ...N) = √ ....
.... .... .... N! ....
.... .... .... ϕN (1) ϕN (2) .... ϕN (N) This form was suggested by Slater and it is called Slater determinant.
8
Permutation of two particles is equivalent to permutation of two columns.
For Example
ϕ (2) ϕ (1) .... ϕ (N) 1
1
1
ϕ (2) ϕ (1) .... ϕ (N) 2
2
2
1 Ψ (2, 1, ...N) = √ ....
.... .... .... = −Ψ (1, 2, ...N) .
N! ....
.... .... .... ϕN (2) ϕN (1) .... ϕN (N) So, the Fermi statistics requirement is automatically satisfied.
If two orbitals in the determinant coincide: i = j; then the determinant vanishes because
there are two identical lines. This describes the Pauli exclusion principle.
Slater determinant is a very convenient form for the wave function. Unfortunately this form
is exact only for noninteracting fermions.
9
Interacting fermions. Variational solution for He atom ground state
Hamiltonian of He atom
e2
p~2 2 Ze2 Ze2
p~1 2
−
+
+
−
2m 2m
r1
r2
|r~1 − r~2 |
Here we neglect all relativistic/magnetic effects. The relative magnitude of these effects is
Ĥ =
∼ v 2 /c2 ∼ (Zα)2 = (Z/137)2 ≪ 1.
Schrodinger equation
Hψ (1, 2) = Eψ (1, 2)
can be solved exactly numerically but:
1) The solution is very involved technically.
2) There is no exact solution for three (Li atom) and more electrons.
So, we need an approximate but relatively simple method which can be propagated to
multi-electron systems.
Variational method
Energy of the system reads
D
E Z
Ψ|Ĥ|Ψ = Ψ∗ ĤΨd3 r1 d3 r2 d3 r3 ...
The wave function is normalized
hΨ|Ψi =
Z
Ψ∗ Ψd3 r1 d3 r2 d3 r3 ... = 1
Let us find minimum of energy with respect to variation of Ψ∗ .
To account for the normalization constraint let us use the Lagrange multiplier method
with λ being the Lagrange multiplier
δ
δΨ∗
(x)
[hΨ∗ H|Ψi − λ hΨ∗ |Ψi] = 0
⇒ Ĥψ − λψ = 0 ⇒ Ĥψ = λψ ⇒ λ = E.
So, we end up with usual Schrodinger equation. In this case the number of variational
parameters = ∞, Ψ∗ (x) at each point x. The variational method does not bring anything
new.
10
For practical applications we choose a finite number of parameters and hence the variational
method gives an approximate answer.
Example He-like ion ground state
Hydrogen-like ion: single electron in Coulomb field of a nucleus with charge Z
(see 3d -year quantum mechanics.)
The Hamiltonian:
Z
11
00
00
11
00
11
00
11
FIG. 4: Hydrogen-like ion
H=
p2
Ze2
−
2m
r
The electron ground state energy:
ǫ=−
Z 2 me4
2h̄4
The electron wave function:
ϕ(r) = Ae−Zr/aB
s
Z3
A=
πa3B
aB is Bohr radius, and A is the normalization constant.
Atomic units:
Eatomic =
E
me4 /h̄4
me4
= 27.2eV
h̄4
ratomic = r/aB
h̄2
aB =
≈ 0.53Å = 0.53 10−8 cm
me2
11
Below I omit the subscript “atomic”. The electron Hamiltonian, energy, and wave function
in atomic units are
p2 Z
∆ Z
− =− −
2
r
2
r
Z2
ǫ=−
2
ϕ(r) = Ae−Zr
r
Z3
A=
π
H=
Now consider He-like ion. He-like ion Hamiltonian in atomic units reads
Ĥ = −
∆1 ∆2
Z
1
Z
−
− − +
2
2
r1 r2 |r1 − r2 |
Let us consider the two electron variational wave function of the following form
Ψ = ϕ (1) ϕ (2) Φs
1
Φs = √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ]
2
where Φs is the spin wave function corresponding to total spin zero. For electron orbital ϕ
we use the hydrogen-like ansatz,
ϕ (r) =
s
3
Zef
f −Zef f r
e
,
π
where Zef f is a variational parameter.
According to the variational method we have to calculate energy of the system and then
minimize it with respect to variation of Zef f .
Z
1 1
Z Ψ
E = hΨ|H|Ψi = − hΨ|∆1 + ∆2 |Ψi − Ψ + Ψ + Ψ 2
r1 r2
|~r1 − ~r2 Z
Φ†s ϕ∗ (2) ϕ∗ (1) ∆1 ϕ (1) ϕ (2) Φs d3 r1 d3 r2
Z
Z
† ∗
3
∗
3
ϕ (1) ∆1 ϕ (1) d r1
ϕ (2) ϕ (2) d r2
= Φs |Φs
Z
=
ϕ∗ (1) ∆1 ϕ (1) d3 r1
hΨ|∆1 |Ψi =
12
Assignment problem:
Z
2
ϕ∗ (1) ∆1 ϕ (1) d3 r1 = Zef
f
Remember that
∆=
1 d 2d
r
r 2 dr dr
Z
Z Z
Ψ Ψ =
Φ†s ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ∗ (2) Φs d3 r1 d3 r2
r1
r1
Z
Z
† Z
∗
3
∗
3
= Φs |Φs
ϕ (2) ϕ (2) d r2
ϕ (1) ϕ (1) d r1
r1
Z
Z
=
ϕ∗ (1) ϕ (1) d3 r1
r1
Assignment problem:
Z
Ψ ϕ∗ (1)
Z
ϕ (1) d3 r1 = ZZef f
r1
Z
1
1
Ψ =
Φ† ϕ∗ (2) ϕ∗ (1)
ϕ (1) ϕ (2) Φs d3 r1 d3 r2
|r~1 − r~2 |
|r1 − r2 |
Z
1
∗
ϕ (1) ϕ (2) d3 r1 d3 r2
= hΦs |Φs i ϕ∗ (2) ϕ∗ (1)
|r~1 − r~2 |
Z
1
=
ϕ∗ (2) ϕ∗ (1)
ϕ (1) ϕ (2) d3 r1 d3 r2
|r~1 − r~2 |
Assignment problem:
Z
1
ϕ (1) ϕ (2) d3 r1 d3 r2
ϕ∗ (2) ϕ∗ (1)
|r~1 − r~2 |
Z
6
Zef
1
f
= 2
e−2Zef f r1 e−2Zef f r2
r12 dr1 dΩ1 r22 dr2 dΩ2
π
|r1 − r2 |
5
= Zef f
8
13
Altogether the energy is
5
2
E = hΨ|H|Ψi = Zef
f − 2ZZef f + Zef f
8
It is minimum at Zef f = Z −
5
.
16
The physical energy which is the minimum energy is
2 5 E = hΨ|H|Ψi = − Z −
16 Z=2
27
→−
16
2
In electron volts
27
E=−
16
2
27.2 = −77.38eV .
Experiment: E = −78.9eV , so the simple variational solution works remarkably well.
14
Multi-electron atom and effective self-consistent potential
A particular electron
Averaged electron cloud
Z
11
00
00
11
00
11
00
11
FIG. 5: A cartoon of multielectron atom. A particular electron which we consider is shown by
the black line. Other electrons which together with nucleus produce the effective potential for the
“black” electron are shown by red.
The effective potential method reduces (approximately) the many-body problem to a single
particle problem.
H many-body → Hsp , “sp” stands for single-particle
Hsp =
p2
+ Vef f (r)
2m
(3)
For a neutral atom

 r << aB : V = − Ze2
ef f
Z
r
 r >> aB : Vef f = − e2
r
Veff
r
−e2
r
−Ze2
r
FIG. 6: A sketch of the effective potential of a neutral atom.
(4)
15
The effective potential is different from the simple Coulomb potential of a point-like nucleus.
Therefore the hydrogen degeneracy of states with the same principal quantum numbers is
lifted.
Energy levels in a Coulomb potential
3s
3p
2s
2p
Energy levels in the effective atomic potential
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
3d
1s
3s
3p
2s
2p
1s
FIG. 7:
In the periodic table of elements the levels are filled from down to up.
3d
16
Exchange interaction
Consider two lowest excitations of He atom.
The experimental spectrum is as follows
Spectroscopic
Energy
notation
1s2s
1
S0
———————
20.62eV
1s2s
3
S1
———————
19.82eV
ground state
1s2
1
S0
———————
The spectroscopic notation is
0.eV
2S+1
ˆ ~ˆ ~ˆ
LJ , J~ = L
+S
For example 1 S0 means that
Letter S ⇒ L = 0
Lef t superscript = 1 ⇒ 2S + 1 = 1 ⇒ S = 0 spin
Right subscript = 0 ⇒ J = L + S = 0 + 0 = 0
The states |1s2s, 1 S0 > and |1s2s, 3 S4 > differ by total spin only,

0
S=
1
Question:
The interaction is spin independent (Coulomb).
Why the states have different energies?
17
Single particle orbitals
ϕ1 (r) ≡ ϕ1s (r) =
q


Z 3 ef f −Zef f r
e
π
ϕ2 (r) = ϕ2s (r) − some f unction with one node 
We do not need an explicit form of ϕ1 (r) and ϕ1 (r) for our discussion.
The two electron wave functions (Slater determinants) are
1
1
|1s2s, 1 S 0 i = √ [ϕ1 (r1 )ϕ2 (r2 ) + ϕ1 (r2 )ϕ2 (r1 )] × √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ]
2
2
1
|1s2s,3 S1 i = √ [ϕ1 (r1 )ϕ2 (r2 ) − ϕ1 (r2 )ϕ2 (r1 )] × | ↑i1 | ↑i2
2
Reminder: Spin wave functions of two electrons are
1
|S = 0, Sz = 0i = √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ]
2
|S = 1, Sz = 1i = | ↑i1 | ↑i2
1
|S = 1, Sz = 0i = √ [| ↑i1 | ↓i2 + | ↓i1 | ↑i2 ]
2
|S = 1, Sz = −1i = | ↓i1 | ↓i2
Let us calculate contributions to energies that come from the Coulomb interaction between
electrons
E1 S0
1 1
S0
=
S0 |r~1 − r~2 Z
1
1
=
[ϕ∗1 (1)ϕ∗2 (2) + ϕ∗1 (2)ϕ∗2 (1)]
[ϕ1 (1)ϕ2 (2) + ϕ1 (2)ϕ2 (1)]
2
|~r1 − ~r2 |
Z
Z ∗
|ϕ1 (r1 )|2 |ϕ2 (r2 )|2 3 3
ϕ1 (1)ϕ1 (2)ϕ∗2 (2)ϕ2 (1) 3 3
=
d r1 d r2 +
d r1 d r2
|~r1 − ~r2 |
|~r1 − ~r2 |
1
ϕ1
r1
ϕ1
ϕ1
r1
ϕ2
ϕ2
r2
ϕ2
ϕ2
r2
ϕ1
FIG. 8: Feynman-like diagrams for the direct and the exchange interaction
18
E3 S1
1 3
S1
=
S1 ~r1 − ~r2 Z
1
1
[ϕ∗1 (1)ϕ∗2 (2) − ϕ∗1 (2)ϕ∗2 (1)]
[ϕ1 (1)ϕ2 (2) − ϕ1 (2)ϕ2 (1)]
=
2
|~r1 − ~r2 |
Z ∗
Z
ϕ1 (1)ϕ1 (2)ϕ∗2 (2)ϕ2 (1) 3 3
|ϕ1 (1)|2 |ϕ2 (2)|2 3 3
d r1 d r2 −
d r1 d r2
=
|r~1 − r~2 |
|~r1 − ~r2 |
3
ϕ1
r1
ϕ1
ϕ1
r1
ϕ2
ϕ2
r2
ϕ2
ϕ2
r2
ϕ1
FIG. 9: Feynman-like diagrams for the direct and the exchange interaction
Thus, the energy is
ϕ1
r1
ϕ1
ϕ1
r1
ϕ2
ϕ2
r2
ϕ2
ϕ2
r2
ϕ1
E
exchange contribution
direct contribution
Due to statistics the exchange term depends on total spin in spite of the fact that the
interaction is spin independent.
Comparing with experimental data on He we find:
ϕ1
r1
ϕ2
2
0.76 eV
ϕ2
r2
ϕ1
Magnetism in solids is due to exchange interaction.