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1
ATOMIC PHYSICS
(PHYS4011) LECTURE NOTES
Lecture notes based on a course given by Tom Kirchner.
The emphasis of the course is on solving atomic systems,
in particular the Hydrogen atom through perturbation theory.
Some relativistic quantum mechanics is introduced at the end
York University, 2011
Presented by:
LATEXNotes by:
TOM KIRCHNER
JEFF ASAF DROR
2011
YORK UNIVERSITY
2
CONTENTS
I. Introduction: the field-free Schrodinger hydrogen atom
A. Reduction to an effective one-body problem
B. Central-field problem for relative motion
C. Solutions of the Coloumb problem
D. Assorted Remarks
II. Atoms in electric fields: the Stark effect
A. Non-Degenerate perturbation theory (PT)
1. Comments
B. Degenerate Perturbation Theory
C. Effect on excited states: the linear Stark effect
3
3
3
3
5
7
8
10
12
14
III. Interaction of Atoms with Radiation
A. The semiclassical Hamiltonian
B. Time-Dependent Perturbation Theory
1. General Formulation
2. Comments
3. Discussion of 1st order result
4. Example: Slowly varying perturbation
5. Solution of TDSE up to 1st order
6. Comments
7. Example: Sudden Perturbation
8. Example:Periodic perturbation
C. Photoionization
1. Transitions into the continuum: Fermi’s golden rule (FGR)
2. Dipole Approximation
D. Outlook on Field Quantization
1. Construction of HF
2. Creation and Annihilation Operators
3. Interaction Between Photon Field and Electrons
4. The Transition Matrix Elements
5. Spontaneous Emission
18
18
21
21
23
23
24
25
25
26
26
28
28
30
31
32
34
35
36
38
IV. Brief Introduction to Relativistic Quantum Mechanics
A. Klien-Gordon Equation
1. Setting up a relativistic wave equation
B. Discussion of KG equation
C. Dirac Equation
1. Free Particles
2. Solutions of the free Dirac equation
3. Add Electromagnetic Potentials
4. The relativistic hydrogen problem
5. Nonrelativistic limit of the Dirac equations
43
43
43
43
44
44
47
49
49
51
Lecture 3 - Jan 09, 2011
3
I.
INTRODUCTION: THE FIELD-FREE SCHRODINGER HYDROGEN ATOM
A.
B.
Reduction to an effective one-body problem
Central-field problem for relative motion
Ĥrel =
p̂2
+ V (r);
2µ
Ĥrel φrel (r) = Erel φrel (r)
(I.1)
Note: omit hats from now on. We use the ansatz:
φrel (r) = R(r)Y`,m (φ, θ)
y(r)
=
Y`,m (φ, θ)
r
where y(r) ≡ rR(r). This gives the radial Schrodinger equation which given by
` (` + 1)
y` (r) = 0
y`00 (r) + − U (r) −
r2
where ≡
2µ
~2 Erel ,
U (r) =
2µ
~2 V
(I.2)
(I.3)
(I.4)
(r).
C.
Solutions of the Coloumb problem
Consider the potential
V (r) = −
Ze2
4πo r
(I.5)
For now on we will only consider Erel < 0 (we don’t consider the scattering problems of Erel > 0, only the bound
r
FIG. 1. The Hydrogen potential energy
states). The energy levels as well as the energy states are quantized.
→ n
Erel → En =
ψrel →
~2
~2 Z 2
2µ n = − 2µa2 n2
y (r)
ψn,`,m = n,`r Y`,m
≈ −13.6eV
Z2
n2
4
TABLE I. Some Hydrogen States and Their Properties. R ≡
n
1
2
2
2
2
~2
2µa
≈ 13.6eV
` m nr = n − ` − 1 state −En
0 0
0
1s RZ 2
RZ 2
0 0
1
2s
4
2
1 −1
0
2p−1 RZ
4
2
1 0
0
2p0 RZ
4
2
1 1
0
2p+1 RZ
4
where yn,` = An,` r`+1 e−κn r L2`+1
n−`−1 (2κn r) and κn =
√
−n =
Z
na ,
and L is the associated Laguerre polynomials. n is
2
0~
an integer and it’s called the principle quantum number. Lastly, a ≡ 4π
µe2 ≈ 0.53Å is called the Bohr radius.
There is a degeneracy in the energy since the energies are independent on `, m. For a given `, there are 2` + 1
states. For a given n we have the states ` = 0, 1, ..., n − 1. Hence the number of states is
n−1
X
2` + 1 = n2
(I.6)
`=0
Some states are summarized in table I The energy spectrum is shown in figure 2. Probability density is defined as
E
0
n=3
n=2
3s
2s
n=1
3p
2p
3d
1s
FIG. 2. The energy levels of Hydrogen
2
2
2
ρn,`,m (r) |φn,`,m | = Rn,`
(r) |Yn,` (φ, θ)|
(I.7)
If
1
Z
r
ρn,`,m (r)dr3 =
Z
0
∞
2
r2 Rn,`
(r)dr
z
Z
}|
{
2
|Y`,m | dΩ
(I.8)
5
then ρn,`,m dr3 is the probability to find the electron in the interval [r, r + dr]. We define the radial probability density
by
2
2
ρn,` (r) = r2 Rn,`
(r) |Y`,m (φ, θ)| dΩ
=r
2
(I.9)
2
Rn,`
(r)
(I.10)
Lecture 4 - Jan 11, 2012
Next we consider the momentum space representation of the Hydrogen atom. Recall
ψn,`,m (r) = hr|n, `mi
(I.11)
with Hrel |n, `mi = En |n, `, mi. Alternatively we can project our states onto momentum space
ψn,`,m (p) = hp|n, `mi
= hp| I |n, `, mi
Z
=
hp|ri hr|n, `, mi d3 r
<
Z
1
=
eip·r ψn,`,m (r)d3 r
3/2
<
(2π~)
Use k =
p
~
(I.12)
(I.13)
(I.14)
(I.15)
and
e
−ik·r
= 4π
∞
L
X
X
Bessel
z }| {
∗
(−i) jL (kr) YLM (Ωk ) YLM
(Ωr )
L
(I.16)
L=0 M =−L
where jL is the spherical Bessel function. Note that here the Ωr represents normal θ and φ while Ωk represents θ and
φ in spherical momentum coordinates. Using these relations
δL,` δM,m
ψn,`,m (p) =
4π
X
3/2
(2π~)
=
4π
(−i)
2
z
Z
r jL (kr) Rn,` (r)dr
}|
{
∗
YL,M
(Ωr )Y`,m (Ωr )dΩr
YL,M (Ωk )
(I.17)
0
L,M
`
3/2
L
∞
Z
Z
(−i)
0
(2π~)
= Pn,` (p)Y`,m (Ωp )
∞
r2 j` (kr)Rn,` (r)drY`,m (Ωk )
(I.18)
(I.19)
The probability density is
ρn,`,m (p) = |φn,`,m (p)|
2
(I.20)
2
(I.21)
While the radial probability density in momentum space is
ρn,` (p) = p2 |φn,`,m (p)|
Lecture 13th, 2012
D.
Assorted Remarks
1. Review Griffith’s chapters 4.1-4.3
2. Hydrogen-like ions are also solved (for Z ≥ 2). Energy scales like En ∝ Z 2 .
6
3. One can also look at a number of exotic systems using the same results as well. e.g. positronium(e+ , e− ),
muonium (µ+ , e− ), muonic atom (p, µ− ). When considering these systems the energy change is due to En ∝
m2
. For more on exotic systems consider C.T. book, volume I
µ = mm11+m
2
4. There are corrections to be looked at which we will consider in detail later
5. Thus far we have used SI units. In these units we have the Hamiltonian (for Hydrogen)
HSI = −
~2 2
e2
∇ −
2m
4π0 r
(I.22)
To eliminate some constants we introduce atomic units. To get rid of the constants we
• measure mass in me = 1a.u. (atomic unit).
• measure charge in units of e = 1a.u..
• measure angular momentum in units of ~ = 1a.u.
• measure permittivity of 4πo = 1a.u.
In short atomic units, a.u. are defined by me = e = ~ = 4πo = 1.
As a consequence of this the Hamiltonian in atomic units are given by
1
1
Ha.u. = − ∇2 −
2
r
• For length in atomic units we use Bohr’s radius, ao =
earlier definitions).
4πo ~2
m e e2
(I.23)
= 0.53Å = 1a.u. (as a consequence of our
2
• For energy in atomic units we consider the Hydrogen ground state. En=1 = − 2m~e a2 = −13.6eV = − 21 a.u.
o
(as a consequence of earlier definitions). Other units of energy may also be used:
1a.u.(of energy) = 27.2eV = 1 Hartree = 2 Rydberg
(I.24)
• For time in atomic units we use dimensional analysis in SI:
time =
distance × mass
distance2 × mass
distance
=
=
speed
momentum
angular momentum
(I.25)
Hence we define a unit of time:
to =
a2o me
= 2.4 × 10−17 s = 1a.u.
~
(I.26)
The Bohr-like revolution time of the electron around the proton in the Hydrogen ground state is
τ = 2πto
(I.27)
• We can now infer a velocity in atomic units:
2πao
vo
1
~
= 2.2 × 106 m/s =
c = 1a.u.
⇒vo =
me ao
137
τ = 2πto =
(I.28)
(I.29)
• The fine-structure constant is
α=
Lecture 6: January 16th, 2012
e2
~
1
=
=
=
4πo ~c
me ao c
137
1
c a.u.
(I.30)
7
II.
ATOMS IN ELECTRIC FIELDS: THE STARK EFFECT
From classical electromagnetism we know that a uniform electric field in the z direction with field strength F is
E = F k̂
(II.1)
φ(r) = −E · r
= −F z
(II.2)
(II.3)
(II.4)
W (r) = −eφ(r)
a = F ez
(II.5)
(II.6)
The electrostatic potential is
The potential energy of an electron is
We need to solve the stationary Schrodinger equation:
H |φα i = E |φα i
(II.7)
for (in atomic units)
1
1
H = − ∇2 − + F z
2
r
=
Ho + +W
(II.8)
(II.9)
To get a better intuition on the problem we sketch the potential energies for x = y = 0. This is shown in figure 3.
In principle there are no stationary state since “bound” electrons can always tunnel out of the potential well. This is
V
Fz(F>0)
z
Coloumb (-1/|z|) Eexcited
Coloumb (-1/|z|)
Eo
FIG. 3. The potential energies of the Stark effect
called ionization. In practice we only have ‘weak’ electric fields:
Flab ∼ 104 V /cm
F1s ∼
e2
= 5 × 109 V /cm
4πo a2o
8
Thus in practice the tunnel effect is unimportant for low-lying states. What does happen is the energy levels shift
and split.
A.
Non-Degenerate perturbation theory (PT)
1. Nondegerenate PT:General formulation
There are different types of perturbation theory.
Consider the Hamiltonian of the form
H |φα i = Eα |φα i
(II.10)
where we assume H = Ho + W and hφα |φβ i = δαβ . Assume further that
Ho |φoα i = Eα(0) |φoα i
D
(II.11)
E
is known and non degenerate with φoα |φoβ = δαβ . In other words that φα forms a basis. If we assume that W
is small then
W = λw
λ 1, λ ∈ <
(II.12)
We can now write
Ho + λw |φα (λ)i = Eα (λ) |φα (λ)i
(II.13)
Note that the eigenvalues and eigenvectors should depend on λ since the perturbation changes the system. We
can now use Taylor expansions:
1 d2 Eα dEα (0)
λ+
λ2 + ...
(II.14)
Eα (λ) = Eα +
dλ λ=0
2 dλ λ=0
d
|φα (λ)i = φ0α +
λ + ...
(II.15)
|φα i dλ
λ=0
Consider the derivative of the Schrodinger equation with respect to λ:
d
[(Ho + λw − Eα (λ)) |φα (λ)i] = 0
dλ
(Ho + λw − Eα (λ)) |φ0α (λ)i + (w − Eα0 (λ)) |φα (λ)i = 0
where
d
dλ
|φα (λ)i ≡ |φ0α (λ)i. Multiplying both sides of the equation by the bra hφβ (λ)|
φβ (λ)|H(λ) − Eα (λ)|φ0β (λ) + hφβ (λ)|w − Eα0 (λ)|φ(λ)i = 0
(II.16)
(II.17)
(II.18)
(II.19)
Assume that α = β
hφα (λ)|Eα (λ)
1
z }| {
z
}|
{
hφα | H(λ) + hφα (λ)|w|φα (λ)i − Eα0 (λ) hφα (λ)|φα (λ)i = 0
Eα0 (λ) = hφα (λ)| w |φα (λ)i
Eα0 (λ = 0) = hφoα | w |φoα i
(II.20)
(II.21)
(II.22)
Next we assume that α 6= β
hφβ (λ)| Eβ (λ) − Eα (λ) |φ0α (λ)i + hφβ (λ)| w |φα (λ)i = 0
(Eβ (λ) − Eα (λ)) hψβ (λ)|φ0α (λ)i + hφβ (λ)| w |φα (λ)i = 0
hφβ (λ)| w |φα (λ)i
hφβ (λ)|φ0α (λ)i =
Eα (λ) − Eβ (λ)
(II.23)
(II.24)
(II.25)
9
Consider the completeness relation and inserting inside the above equation gives
|φ0α (λ)i =
X
|φβ (λ)i hφβ (λ)|φ0α (λ)i
(II.26)
β
X
|φβ (λ)i
β
hφβ (λ)| w |φα (λ)i
Eα (λ) − Eβ (λ)
(II.27)
but this doesn’t include α = β. What about hφα (λ)|φ0α (λ)i?
d
hφα |φα (λ)i = 0
dλ
hφ0α (λ)|φα (λ)i + hψα(λ)|φ0α (λ)i = 0
∗
hφα (λ)|φ0α (λ)i
+ hψα(λ)|φ0α (λ)i
2 hφα (λ)|φ0α (λ)i
=0
=0
(II.28)
(II.29)
(II.30)
(II.31)
if hφα |φ0α i ∈ <. However the eigenstate of a Hermitian operator can always be transformed into a basis (by
taking linear combinations of them) such that they are real. Hence these overlaps are real. Which means that
the overlaps are zero.
|φ0 (λ)i =
X hφα (λ)| w |φα (λ)i
|φβ (λ)i
Eα (λ) − Eβ (λ)
(II.32)
β6=α
and hence
D X φ0β w φ0α 0
φ0β
|φα (λ = 0)i =
(0)
(0)
−
E
E
α
β6=α
β
(II.33)
d2
d 0
Eα (λ) =
E (λ)
dλ2
dλ α
(II.34)
One can go on to other orders as well
Lecture 7 - January 18th, 2012
We can express the perturbed energies by
dEα
1 d2 E λ+
λ2 + ...
dλ λ=0
2 dλ2 λ=0
E2
0 0 φα w φβ X
0 0
(0)
2
= E α + λ φα w φα + λ
+ ...
(0)
(0)
β6=α Eα − Eβ
E2
X φ0α W φ0β (0)
0
0
= E α + φα W φα +
+ ...
(0)
(0)
Eα − Eβ
β6=α
Eα (λ) = Eα(0) +
= Eα(0) + ∆Eα(1) + ∆Eα(2) + ...
0 0 E
0 X φα W φβ 0 φβ
|φα (λ)i = φα +
(0)
(0)
β6=α Eα − Eβ
(II.35)
(II.36)
(II.37)
(II.38)
(II.39)
(II.40)
10
1.
Comments
(0)
(0)
(a) These equations are non-defined only when Eα 6= Eβ ∀α, β. In other words non-degenerate systems.
(b) Convergence is difficult to check in perturbation theory but consistency checks can be done. One can check
that
E φ0 W φ0 α
β << 1
(II.41)
(0)
(0) Eα − Eβ If this is not fulfilled then it indicates that perturbation theory will likely fail.
(c) Full calculations for the energies, Eα , beyond first order are in general not possible (due to the infinite
sum).
(d) Griffith, 6.1; Liboff, B.1
2. Applications to H(1s) in an electric field
1
φ01s (r) = √ e−r
π
1
0
E1s
=−
2
W = Fz
We can calculate the first order energy correction:
(1)
∆E1s = φ01s F z φ01s
Z
F
e−2r zd3 r
=
π
Z
F
=
r cos θ sin θe−2r r2 drdθdφ
π
Z
Z
:Z0
F ∞ 3 −2r π 1 r e
=
sin 2θdθ dφ
02
π 0
=0
(II.42)
(II.43)
(II.44)
(II.45)
(II.46)
(II.47)
(II.48)
(II.49)
However this doesn’t tell us whether or not the rest of the correction orders are zero or non-zero. To get an idea
for the second order correction we check the consistency criterion. Consider the overlap of the contribution of
the 2p and 1s states. Note that this gives the smallest possible denominator (greatest correction).
D
D (0) φ F z |φ i φ(0) z φ0 2p 2p0
1s
1s =
(II.50)
1
1
(0)
(0)
E1s − En=2 − 2 + 8
≈ 2F
(II.51)
For the Stark effect the F ≈ 10−5 in atomic units. Hence we expect the second order effect to be small but non
zero. Next we estimate the full second order correction
D 2 X φ0β F z φ01s (2) (II.52)
∆E1s = (0)
(0)
β6=1s E1s − Eβ
To get an upper bound for this estimate we can replace the denominator by it’s minimum value which corresponds
to n = 2.
3
(0)
(0)
(0) (0) (II.53)
E1s − Eβ ≥ E1s − En=2 =
8
11
Thus we can write
8
X 2
(2) φ0β z φ01s ∆E1s = F 2
3
(II.54)
β6=1s
8 2 X 0 0 0 0 F
φ1s z φβ φβ z φ1s
3
β6=1s
X
0
8 2 0 0
0 = F φ1s z
φβ φβ z φ1s 3
=
(II.55)
(II.56)
β6=1s
but
X
φβ 0 φ0β = 1 − φ01s φ01s (II.57)
β6=2s
Hence
8
(2) ∆E1s = F 2 φ01s z 1 − φ01s φ01s z φ01s 3
(
)
0
:
8 2 0 2 0 0 0 0 0
= F
φ1s z φ1s − φ1s z φ1s φ
z φ1s
1s
3
=
8F 2 0 2 0 φ1s z φ1s
3
(II.58)
(II.59)
(II.60)
It turns out that the integral is easy to do
8F 2
(2) ∆E1s ≤
3
(II.61)
(2)
We also know that the energy correction is negative, ∆E1s < 0. One can also calculate the exact result:
9
(2)
∆E1s = − F 2
4
(II.62)
This is called the “quadratic Stark effect”. Since F is small this is a very small correction.
3. Interpretation
Consider a classical charge distribution in an electric field. The classical energy of a charge distribution is given
by
Z
U = ρ(r)φ(r)d3 r
(II.63)
Z
= −F ρ(r)zd3 r
(II.64)
= −F pz
(II.65)
where pz ≡ z component of the dipole moment. Connection to QM:
2
ρ(r) = − |φ1s (r)|
Z
pz = −
2
|φ1s (r)| zd3
= − hφ1s | z |φ1s i
D
E
(0)
(0)
(1)
(1)
= − φ1s + λφ2s + ... z φ1s + φ1s + ...
E
D
E
D
E
D
(0) (0)
(1) (0)
(0) (1)
= − φ1s z φ1s − λ φ1s z φ1s − λ φ1s z φ1s + O(λ2 ) + ...
(II.66)
(II.67)
(II.68)
(II.69)
(II.70)
12
The first term is zero. By choosing real eigenstates we can
E
ED
D
(0)
(0) (0) (0)
φβ F z φ1s
X φ1s z φβ
pz = −2
(0)
(0)
E1s − Eβ
β6=1s
E
(0) 2
X hφ1s | z φβ = −2F
+ O(λ2 )
(0)
E
−
E
(0)
β
1s
β6=1s
=−
=
2
(2)
∆E1s + O(λ2 )
F
(II.71)
(II.72)
(II.73)
9
F + O(F 2 )
2
(II.74)
In summary
(a)
E
D
(1)
(0) (0)
=0
∆E1s = 0 ⇐⇒ p(0)
z = − φ1s z φ1s
(II.75)
This is expected since a spherical charge distribution has no static dipole moment (and equivalently for a
spherical probability distribution)
(2)
(b) ∆E1s 6= 0 reflects that we have a nonzero induced dipole moment. This means that we have a nonzero
polarizability
αD =
B.
9
1 (1)
p =
F z
2
(II.76)
Degenerate Perturbation Theory
We have a Hamiltonian given by
H = Ho + λw
(II.77)
where
Ho φoα,j = Eα(0) φoα,j ,
j = 1, 2, ..., gα
H |φα,j i = Eα,j |φα,j i
(II.78)
(II.79)
Here we have assumed that after the perturbation the energy levels are non-degenerate (this is not always the case
but it is the case for the Stark effect). The states
o φα,j , j = 1, 2, ...gα
(II.80)
(0)
(0)
span the subspace (of Hilbert space) associated with Eα . Any linear combination is an eigenstate of Ho for Eα .
Effect of perturbation theory is shown in figure 4 Note that as the perturbation is increased we encounter a change
in state to some linear combination of states.
|φαj i → φ0α,j
(II.81)
We make the ansatz:
(1)
(2)
Eα,j (λ) = Eα(0) + λEα,j + λ2 Eα,j + ...
|φα,j (λ)i = φ0α,j + λ φ1α,j + ...
Inserting this into the Schrodinger equation:
o E
n
(1)
φ0α,j + λ φ(1) + ...
(Ho + λw) φ0α,j + λ φ1α,j + ... = Eα(0) + λEα,j
α,j
n Eo
n
Eo
(1)
(1)
Ho φ0α,j + λ w φ0α,j + Ho φα,j
+ O λ2 = Eα(0) φ0α,j + λ Eα(1) φ0α,j + Eα(0) φα,j
+ O λ2
(II.82)
(II.83)
(II.84)
(II.85)
13
0
(0)
E
E
1
E
2
E
E
E
3
4
g
FIG. 4. The effects of perturbation theory: degeneracy is lifted
By equation coefficients we know that
λ0 :
λ1 :
Ho φoα,j = Eα(0) φ0α,j
E
E
(1)
(1)
(1)
Ho φα,J + w |φα,j (0)i − Eα(0) φα,j − Eα,j |φα,j i = 0
The first equation just shows consistency. By adding on a bra to the second equation we get
E o (1)
(1) φβ,` Ho − Eα(0) φα,j + φoβ,` w − Eα,j φoα,j = 0
D
E (0)
(1)
Eβ − Eα(0) φoβ,` |φα,j + φoβ,` w φoα,j − Eα,j δαβ δ`,j = 0
(II.86)
(II.87)
(II.88)
(II.89)
Consider the case of α = β and ` = j
(1)
Eα,j = φoα,j w φoα,j
(II.90)
Consider the case of α = β and ` 6= j
φoβ,` w φoα,j = 0
(II.91)
This result is a property of the perturbation and it follows due to the ansatz we chose. However this is not true for
all perturbations. For example
hH(2s)| z |H(2po )i =
6 0
(II.92)
To ensure that this is always the case we need to diagonalize our perturbations. We modify our ansatz by
(1)
Eα,j = Eα(0) + λEα,j + ...
E
E
(1)
|φα,j i = φ̃oα,j + λ φα,j + ...
(II.93)
(II.94)
E P
E
D
E
0
0
gα
0 aα
with φ̃0α,j = k=1
such
that
φ̃
w
φ̃
φ
α,j = 0.
k,j
α,k
α,`
D
Insert the two ansatz into the Schrodinger equation, sort, and project on φ0α,` gα
X
k=1
E
(1) φ0α,` w − Eα,j φ̃0α,j = 0
(1) φ0α,` w − Eα,j φ0α,k aα
k,j = 0;
(II.95)
` = 1, 2, ...gα
(II.96)
14
This can be rewritten in matrix form as
 α
w11 − Eα,j
w12
w13
...
(1)

α
α
w
w
−
E
...
...

21
22
α,j

..
..
..
..

.
.
.
.

(1)
...
...
... wgα ,gα − Eα,j

aα
  a1,j
 α
  2,j
  ..
.

aα
gα ,j






0
0

.. 
. 
(II.97)
0
The condition for a nontrivial solution is
det (
)=0
(II.98)
(1)
and has gα roots Eα,j , j = 1, 2, ...gα .
The procedure for using degenerate perturbation theory is as follows
1. Solve “secular” (characteristic) equation and obtain eigenvalues
n
o
(1)
Eα,j , j = 1, ...gα
(II.99)
2. Insert eigenvalues into the matrix equations and obtain the expansion coefficients.
α
ak,j ; k, j = 1, ...gα
(II.100)
3. One can go further and check that
D
˜ w φ0 = E (1) δ
φ0α,`
α,j
α,j `,j
(II.101)
4. It’s possible to go on to calculate wavefunctions and 2nd order energy corrections but it’s tedious
C.
Effect on excited states: the linear Stark effect
1. Matrix elements
α
w`,k
= φ0α,` w φ0α,k
(II.102)
or for the Hydrogen atom (w = F z):
with φn`m (r) = Rn` (r)Y`m (Ω), z = r cos θ
r
φ0n`m z
0
φn`0 m0 =
4π
3
φ0n`m z φ0n`0 m0
q
4π
3 rY10
Z
∞
3
r Rn` (r)R
(II.103)
Z
n`0
(r)dr
Y`,m (Ω)Y10 (Ω)Y`0 ,m0 (Ω)dΩ
(II.104)
0
The radial integral is simple enough. We have done similar ones in the past. The angular integral can be done
in general using “Wiger-Eckant-Theorem”:
r
Z
p
4π
` L `0
` L `0
m
∗
Y`m
(Ω)YLM (Ω)Y`0 m0 (Ω)dΩ = (−1)
(2` + 1) (2`0 + 1)
(II.105)
−m M m0
0 0 0
2L + 1
Wigner’s 3j symbols
}|
Clebsch-Gordan Coefficients
{
z
}|
{
j1 j2 j3
j1 −j2 −m3
−1/2
with
= (−1)
(2j3 + 1)
× hj1 , m1 , j2 , m2 |j3 , −m3 i For more information on these
m1 m2 m3
topics refer to Liboff, chapter 9 or Cohen-T. Chapter 6 and 10.
z
15
Lecture 9 - January 25th, 2012 The selection rules can be written in terms of the Wigner 3j symbols.


m1 + m2 + m3 = 0
j1 j2 j3
6= ⇐⇒ and
(II.106)
m1 m2 m3

|j − j | ≤ j ≤ j + j (triangular condition)
1
2
3
1
2
j1 j2 j3
0 0 0


triangular condition
6 ⇐⇒ and
=

j + j + j = even
1
2
3
(II.107)
Applying these relations to our situation our integral is non zero only if
m = m0
∆` = ` − `0 = ±1
(II.108)
(II.109)
These are sometimes called the electric dipole selection rules (E1) (special case).
2. Linear Stark effect for H(n = 2)
Consider the degenerate stats
n
o
φo2s , φo2po , φo2p−1 , φo2p1
(II.110)
We consider the matrix eigenvalue problem:
X
(1) hφo2`m | w − En=2 φo2n,`0 ,m0 an=2
`,m,`0 ,m0 = 0
(II.111)
`0 ,m0 (n=2)
Consider
wi,j = hφo2`m | w φo2n,`0 m0
(II.112)
If i = j then wi,j = 0 because ∆` = 0. Further we have a symmetric matrix since the states are real. By using
the selection rules we see that


0 w12 0 0
0 0 0
w
w =  12
(II.113)
0
0 0 0
0
0 0 0
Hence the only potentially nonzero element is the w12 . Note that if the radial part is nonzero then the element
may still be zero. It’s easy to calculate the element explicitly:
(II.114)
w12 = hφo2s | z φo2po = −3a.u.
Hence we have the secular equation
−E (1) w12
0
0
w12 −E (1)
0
0
0
0
0
−E (1)
0
0
0
−E (1)
=0
This matrix is block diagonal and it’s easy to find the equation:
2 2
(1)
(1)
2
E
E
− w12 = 0
E (1) = {0, 0, w12 , −w12 }
(II.115)
(II.116)
(II.117)
Hence the first order energy corrections are
∆E (1) = {0, 0, −3F, 3F }
(II.118)
We are left with three lines. Two lines stay degenerate as represented by the zeroes however the originally one
line splits into three.
Next we calculate the mixing coefficients of the expansion, a`,m,`0 ,m0 . We insert the eigenvalues into our equation.
16
(1)
(a) First consider E1
(1)
= E2
=0
0 w12
 w12 0
 0
0
0
0

0
0
0
0

 
0
a2s
0   a2po  
=
0   a2p−1  
0
a2p+1

0
0
0
0
(II.119)
This is true only if a2s = a2po = 0. a2p−1 and a2p+1 are undetermined. The state that corresponds to this
state is any linear combination of the 3rd and 4th states. We choose
E
E (II.120)
φ̃E1 = φo2p−1
E (II.121)
φ̃E2 = φ2p−1
(1)
(b) Next we consider E3
= +w12 :

  
0
−w12 w12
0
0
a2s
0
0   a2p0   0 
 w12 −w12
=
 0
0
−w12
0   a2p−1   0 
0
0
0
0
−w12
a2p+1

(II.122)
This gives a2p−1 = a2p+1 = 0. The other two equations are
− w12 a2s + w12 a2po = 0
w12 a2s − w12 a2po = 0
(II.123)
(II.124)
This requires that a2s = a2po .
(c) The final eigenvectors for E4 = −w12 are given by a2s = −a2p0 , a2p−1 = a2p+1 = 0.
We have fixed the components of the eigenvectors (the energy corrections). We should fix the normalization of
the last two states (we include the previous states for concreteness):
E
1
φ̃E (1) = √ |φo2s i + φo2po
3
2
E
1
φ̃E (1) = √ |φo2s i − φo2po
4
2
E E
o
φ̃E (1) = φ2p−1
1 E E
φ̃E (1) = φo2p+1
(II.125)
(II.126)
(II.127)
(II.128)
2
3. Summary and interpretation
(a) Splitting of energy levels is shown in figure
D
(b) Note that φo2`,m z |φo2`m i = 0 hence the original states (φo2s , φ2p0 , φo2p±1 ) have no static dipole moment.
E E
However the Stark states φ̃E (1) , φ̃E (1) do have non zero static dipole moment
3
4
D
E
p0z,3 = − φE (1) z φE (1) )
3
= 3a.u.
pz,4 = −3a.u.
(c) The new states are shown in figure 6
(II.129)
3
(II.130)
(II.131)
17
F
m=0
E4
E
m=+/- 1
m=0
E3
FIG. 5. The linear Stark splitting of energy levels
x=y=0
x=y=0
E3
2s
z
2p0
z
x=y=0
E4
z
FIG. 6. The Stark states
(d) Note that we still have azimuthal symmetry.
[`z , w] = 0
(II.132)
Hence m is still a good quantum number
(e) The last question to consider is the effect on the n = 3 shell states. We would need to consider the
(3s, 3po , ..., 3d±2 )
(II.133)
18
This involves solving the secular equation of a 9 × 9 matrix.
Lecture 10, January 30th, 2012
(f) To solve the Stark effect on the n = 3 states one can decompose the matrix into blocks. To show this
method consider the L-shell problem one more time


−E (1) w12
0
0
 w12 −E (1)
0
0 

det 
(II.134)
(1)
 0
0
−E
0 
0
0
0
−E (1)
The matrix is block diagonal and we have 3 subspaces since the m states don’t mix (this can also by seen
by [`z , W ] = 0. For block diagonal matrices of the form:


A1 0 ...

.. 

(II.135)
det 
 0 A2 .  = det A1 det A2 ...
..
..
.
. ...
det A = 0 ⇐⇒ det Ai = 0;
∀i. For this case:
(1)
(1)
m = ±1
det −E (1) = 0 ⇐⇒ E1 = E1 = 0
m=0
det
−E (1) w12
w12 −E (1)
In the n = 3 shell we can subdivide the states by

0





−1

m= 1



−2



2
III.
= 0 ⇐⇒
E (1)
2
2
− w12
=0
3s, 3po , 3do
3p−1 , 3d−1
3p1 , 3d1
3d−2
3d2
(II.136)
(II.137)
(II.138)
INTERACTION OF ATOMS WITH RADIATION
Consider the scheme of Quantum Theory shown in figure 7.
A.
The semiclassical Hamiltonian
1. Consider a classical particle with charge q in an electromagnetic field. The force acting on the particle is simply
the Lorentzian force.
FL = q (E + v × B)
(III.1)
Introduce the electromagnetic potentials, A is the vector potential and φ is the scalar potential
E = −∇φ −
∂
A
∂t
B=∇×A
(III.2)
(III.3)
Inserting these equations gives
∂A
+ v × (∇ × A)
FL = q −∇φ −
∂t
(III.4)
19
FIG. 7. Scheme of Quantum Theory
Define a generalized potential energy,
U = q (φ − A · v)
(III.5)
L=T −U
m
= v 2 − qφ + qv · A
2
(III.6)
The Lagrangian is given by
(III.7)
The Lagrangian equations of motion
∂ ∂L
∂L
−
= 0 ⇐⇒ ma = FL
∂t ∂vi
∂xi
(III.8)
Hence the generalize potential energy yields the correct force. The Hamiltonian can now easily be extracted
H =p·v−L
(III.9)
Note that
p=
∂L
= mv + qA
∂v
(III.10)
is the canonical momentum. We can rearrange this equation for velocity:
v=
1
(p − qA)
m
(III.11)
We can now rewrite our Hamiltonian without v (we need to eliminate this variable to have a proper Hamiltonian):
p
1
q
2
· (p − qA) −
(p − qA) + qφ −
(p − qA) · A
m
2m
m
1
2
=
(p − qA) + qφ
2m
H=
2. Next consider the quantum mechanical Hamiltonian for an electron.
• To consider bound states we add another scalar potential, V due to a nucleus.
• The charge for the electron is q = −e
(III.12)
(III.13)
20
• Next we perform quantization (p → p̂ = ~i ∇ ). Hence
1
2
(p̂ + eA) − eφ + V
2m 1
~
~
Hψ(r, t) =
∇ + eA(r, t)
∇ + eA(r, t) φ(r, t) − eφ(r)ψ (r, t) + V (r)φ (r, t)
2m i
i
1
~e
~e
2 2
2 2
=
−~ ∇ φ + ∇ (Aψ) + A∇ψ + e A ψ − eφψ + V ψ
2m
i
i
H=
(III.14)
(III.15)
(III.16)
H
z
}|o
{
2
~2 2
e~
e~
e
= −
∇ +V ψ+
A · ∇ψ +
∇·A ψ+
A2 − eφ ψ
2m
mi
2mi
2m
(III.17)
W (t)
z
}|
{
e
e
e2 2
= Ho + A · p +
p·A+
A − eφ
m
2m
2m
(III.18)
where Ho is the Hydrogen Hamiltonian without electromagnetism and W (t) is the time dependent perturbation.
• Assume for the following the electromagnetic is a free electromagnetic field which is characterized by no
charges and no currents, ρ = 0; J = 0. Here its convenient to choose the Coloumb gauge, ∇ · A = 0. This
is useful because then we arrive at the equations
∇2 A −
1 ∂2A
=0
c2 ∂t2
(III.19)
and
φ=0
(III.20)
Note this doesn’t mean that there is no electric field (since E = − ∂A
∂t ). Consider the monochromatic
solution to the wave equation, a plane wave.
A(r, t) = Π̂ |Ac | cos (k · r − ωt + α)
(III.21)
One can perform two checks. Firstly
∇ · A = −Π̂ · k |Ac | sin (k · r − ωt + α) = 0
(III.22)
Hence Π̂ ⊥ k. Thus the vector potential is a transverse wave. The second check is the wave equation
1 ∂2A
c2 ∂t2
ω2
−k 2 A = − 2 A
c
∇2 A =
(III.23)
(III.24)
Thus we see that this is a solution given that k = ωc . Going back to the perturbation we have (due to the
Coloumb gauge p · A one term disappears and another disappears because we don’t have a scalar potential)
W (t) =
e
e2 2
A·p+
A
m
2m
(III.25)
Assume that A is weak. If this is the case we can neglect the second term:
W (t) ≈
e
A·p
m
(III.26)
21
Lecture 11 - February 1st, 2012
Test 1 is up to (and including) last lecture. In other words up to this point.
We can write the Hamiltonian due to an external field by
H = Ho + W (t)
(III.27)
where
Ho =
p2
+ V (r);
2m
W (t) =
e
e2 2
A(r, t) · p +
A (r, t)
m
2m
(III.28)
Note that since we have a time dependent Hamiltonian we cannot use the stationary Schrodinger equation. Thus we
use time dependent perturbation theory.
B.
Time-Dependent Perturbation Theory
1.
General Formulation
Consider a perturbation W (t) = λw(t) on a Hamiltonian which is stationary.
H(t) = Ho + λw(t)
(III.29)
The goal is to solve the time-dependent Schrodinger equation:
i~
∂
|ψ(t)i = H(t) |ψ(t)i
∂t
(III.30)
The SE is a initial value problem. We assume we know the state at some time. Assume for t ≥ to that
|φ(to )i = |φo i
W (t) = 0;
(III.31)
where Ho |ψj i = j |ψj i (j = 0, 1, ...). We assume that W is switched on at to and we look into how the system
develops in time.
Define |ψj (t)i = e−ij t/~ |φj i then inserting into the SE (before perturbation) by
i~
∂
|ψj (t)i = i~
∂t
∂ −ij t/~
e
|φj i
∂t
= j e−ij t/~ |φj i
= Ho |ψj i
(III.32)
(III.33)
(III.34)
Hence if |φi solves the SE then so does |ψi. We denote the |ψ(t)i by the solution after the perturbation has been
turned on. We expand these states as
|ψ(t)i =
X
cj |ψj (t)i
(III.35)
j
and insert into the time dependent SE


X
∂ X
i~
cj (t) |ψj (t)i = (Ho + W (t))
cj (t) |ψj (t)i
∂t
j
j
X
X
(i~ċj (t) + j ) e−ij t/~ |φj i = (Ho + W (t))
cj (t)e−ij t/~ |φj (t)i
j
(III.36)
(III.37)
j
(III.38)
22
We project these states onto the state hψk (t)| = hφk | eik t/~ :
δj,k
X
e
i
~ (k −j )t
z }| {
X i
hφk |φj i (i~ċj (t) + cj j ) =
e ~ (k −j )t cj (t) hφk | Ho + λw(t) |φj i
(III.39)
j
j
= c +
i~ċk (t) + ck (t)
k (t)
k
k
X
i
e ~ (k −j )t cj (t) hφk | λw(t) |φj i
(III.40)
j
i~ċk =
X
i
e ~ (k −j )t cj (t)λwkj
(III.41)
j
where wkj ≡ hφk | w |φj i. Note that thus far we have not made any approximations. These are sometimes called
coupled-channel equations. If we assume that W (t > tf ) = 0 then for t > tf the matrix elements on the right side
are 0 and hence
ċk (t > tf ) = 0
(III.42)
and the states are constant. The probabilities for transitions |φ0 i → hφk i (equivalently |ψ0 i → |ψk i ). The probability
for transition to state k is
2
2
2
pk = |ck |t>tf = |hψk |ψi|t>tf = |hφk |ψi|t>tf
(III.43)
P
To check the answer one can check that k pk = 1. This turns out to be true. We finally introduce perturbation
theory. Notice that the only thing we can expand is the expansion coefficients. We use a power series expansion
(0)
(1)
(2)
ck (t) = ck (t) + λck (t) + λ2 ck + ...
We insert this into the coupled channel equations
n
o
n
o
X i
(0)
(1)
(1)
(1)
i~ ċk + λċk + ... = λ
e ~ wkj (k −j )t cj + λcj + ...
(III.44)
(III.45)
j
consider zeroth order, λ0 :
(0)
i~ċk = 0
(III.46)
Next consider first order
(1)
i~ċk =
X
(0)
i
(1)
i
cj e ~ (k −j )t wkj
(III.47)
j
Lastly consider second order
(2)
i~ċk =
X
cj e ~ (k −j )t wkj
(III.48)
j
Notice that you can in principle get a solution for all orders using this method. Once you have first order you can get
second order and once you have second order you can get third order etc. In other words we solve these successively.
For first order
(0)
ck (t) = const = δk,0
(III.49)
Our initial condition was the ground state. Hence the coefficients are zero except the ground state coefficient. In
zeroth order we just stay in the ground (initial) state. Input this into higher orders:
X
i
(1)
i~ċk =
δj,0 e ~ (k −j )t wk,j
(III.50)
j
i
= e ~ (k −0 )t hφ| w(t) |φ0 i
Z
*
0 i t i ( − )t0
(1)
(1) ck (t) − e ~ k 0 hφk | w(t0 ) |φj i dt0
ck (t0 ) = −
~ to
Z
i t i (k −0 )t0
(1)
ck (t) = −
e~
hφk | w(t0 ) |φj i dt0
~ to
(III.51)
(III.52)
(III.53)
23
(1)
where ck (t0 ) = 0 since our initial conditions say that the system is in the ground state. In second order we have
(2)
i~ċk =
(1)
i
iX
~ j
Z
X
cj e ~ (k −j )t wk,j
(III.54)
j
(2)
i~ck = −
(2)
ck (t)
t0
00
i
i
e ~ (j −0 )t hφj | wj (t00 ) |φ0 i dt00 e ~ (k −j )t hφk | w(t) |φj i
(III.55)
to
Z 0
Z
1 X t 0 t 00 i (k −j )t0 i (j −0 )t00
dt
dt e ~
=− 2
e~
wk,j (t0 )wj,0 (t00 )
~ j t0
t0
(III.56)
Lecture 12 - February 6th, 2012
2.
Comments
1. ‘Exact’ calculations beyond 1st order are in general impossible (due to infinite sums)
2. Practical calculations of second order often rely on the ‘closure approximation ’. Notice that the second order
calculation is not an infinite sum if the j are constant (you can use the completeness relation formed from
wk,j (t0 )wj,0 (t00 ) = hφk | w(t0 ) |φj i hφj | w(t00 ) |φ0 i). Thus one can approximate that
j → ¯
(III.57)
where ¯ is an average energy value. This produces a second order correction of
(2)
ck (t)
1
=− 2
~
Z
t
0
Z
t0
t0
0
i
i
00
dt00 e ~ (k −¯)t e ~ (¯−0 )t hφk | w(t0 )w(t00 ) |φ0 i
dt
(III.58)
t0
3. The interpretation of the results are as follows. The first order result is called a direct transition since
w
|φ0 (t0 )i −
→ |φk (t)i
(III.59)
One the contrary for second order we have
w
w
|φ0 i −
→ |φj i −
→ |φk i
(III.60)
We call this a transition through a ‘virtual’ state (two steps). The |φj i state serves as an intermediate step. We
can see this pictorially as shown in figure 8.
3.
Discussion of 1st order result
(0)
(1)
ck (t) ≈ ck + λck
Z
i t iωk0 t0
= δk0 −
e
Wk0 (t0 )dt0
~ t0
(III.61)
(III.62)
0
where ωk0 ≡ k −
and Wk0 (t) = hφk | W (t) |φ0 i = λ hφk | w(t) |φ0 i is called a transition matrix element. The proba~
bility to transition to state k 6= 0 (from the initial, ground state) is
1st−order
P0→k
=
Z
2
1 t iwk0 t0
0
0
e
W
(t
)dt
k0
2
~
t0
(III.63)
24
t
Time
w(t')
t0
FIG. 8. Diagram of first order perturbation theory
For k = 0 we have an elastic collision (the particle stays in the initial state)
(0)
(1)
(2)
1st−order
P0→0
= c0 + λc0 + λ2 c0 + ...
(1) 2
(1)
(1)∗
(2)
(2)∗
≈ 1 + λ c0 + c0
+ λ2 c0 + c0 + c0
X
1st−order
=1−
P0→k
(III.64)
(III.65)
(III.66)
k6=0
This shows norm conservation even in first order.
4.
Example: Slowly varying perturbation
1. Slowly varying perturbation. An example of such a slowly varying perturbation is shown in figure 9. for k 6= 0
W(t)
t0
t
FIG. 9. Slowly varying perturbation
25
we have
i
ck (t) = −
~
t
Z
0
eiωk0 t Wk0 (t0 )dt0
(III.67)
t0


≈0
t
Z t


z
}|
{
0
i
1 iωk0 t0
1
=−
eiωk0 t W¯k0 (t0 ) dt0
e
Wk0 (t0 ) −

~  iωk0
iωk0 t0
t0
1
Wk0 (t)eiωk0 t
~ωk0
hφk | W (t) |φ0 i iωk0 t
=−
e
k − 0
≈−
(III.68)
(III.69)
(III.70)
2
p0→k = |ck |
(III.71)
2
=
|hφk | W (t) |φ0 i|
2
(k − 0 )
= 0(if for t > tf , W (tf ) = const)
(III.72)
(III.73)
This only works for a non-degenerate initial state (due to the energies in the denominator).
5.
Solution of TDSE up to 1st order
Consider the solution to the TDSE and insert in the 1st order coefficient
X
i
|ψ(t)i =
ck (t)e− ~ k t |φk i
(III.74)
k
i
= c0 (t)e− ~ 0 t |φ0 i +
X hφk | W (t) |φ0 i
k6=0
0 − k
i
e− ~ k t |φk i
Since we are using perturbation theory we require that c0 ≈ 1 in this case we have


|φ̃0 (t)i
}|
{
z
X hφk | W |φ0 i

 −i t
~ 0
|ψ(t)i ≈ 
|φk i
|φ0 i +
e
0 k


k6=0
(III.75)
(III.76)
E
φ̃0 (t) describes state of the system at time t up to 1st order corresponding to perturbed energy eigenvalue
(1)
0 (t) = 0 + hφ0 | W (t) |φ0 i
(III.77)
The system remains in the ground state of the total (instantaneous) Hamiltonian, H(t), at all times. This is often
called an adiabatic situation. The approximation of neglecting the time derivative of W is called the adiabatic
approximation.
6.
Comments
1. This argument can be generalized to strong perturbations (all orders). If perturbation varies slowly with time
the system is found in an eigenstate of the total Hamiltonian H(t) = H0 + W (t) at all times (“adiabatic
approximation”). Ref: D.Bohm, Quantum Theory, Chapter 20
2. Realizations of this formulism can found in
• Slow atomic collisions
26
• Stern-Gerlach experiment
Lecture 13, Feb 13th, 2012
7.
Example: Sudden Perturbation
Consider the following perturbation:
(
0
W (t) =
W
t ≤ t0
t > t0
(III.78)
The first order amplitude is
i
ck (t) = −
~
where ωk0 =
k −0
~
Z
t
ei(ω−ωk0 )t Wk0 (t0 )dt0
(III.79)
0
and Wk0 = hk| W (t) |0i. Thus
Z
i
ck (t) = − Wk0 eiωk0 dt0
~
Wk0 iωk0 t
e
−1
=−
~ωk0
(III.80)
(III.81)
2
2
P0→k (t) = |ck (t)| =
|Wk0 |
ωk0 {2 − 2 cos ωk0 t}
~2
(III.82)
2
=
where f (t, ωk0 ) =
ω
sin2 ( k0
2
2
ωk0
t
)
4 |Wk0 |
f (t, ωk0 )
~2
(III.83)
. The function f is independent of the particular perturbation. The perturbation is
shown in figure 10. Note that noticeable transitions only occur within ∆Ω =
2π
t .
This corresponds to an energy range
f
0.25
0.20
0.15
DΩ= 2tΠ
0.10
0.05
-15-10 -5
5 10 15
Ωk0
FIG. 10. The function f which determines the probability of transition
∆E =
2π~
t .
8.
Example:Periodic perturbation
Consider the potential below:
(
0
t ≤ t0 = 0
W (t) =
iωt
Be + B † e−iωt
t > t0 = 0
(III.84)
27
Note that W = W † . We consider the first order amplitude for a transition from state |ii to |f i. The amplitude is
cf i (t) = −
where ωf i =
i
~
Z
t
0
Wf i (t0 )eiωf i t dt0
(III.85)
0
f −i
~ .
i
cf i (t) = −
~
Z
hf | B |ii
t
e
i(ωf i +ωt0 dt0 )
†
Z
+ hf | B |ii
0
t
e
i(ωf i −ω)t0
0
dt
(III.86)
0
∗
we now define Bf i ≡ hf | B |ii and note that hf | B † |ii = hi| B |f i = Bif
. We can now write
cf i (t) = −
∗
Bif
Bf i
ei(ωf i +ω)t − 1 +
ei(ωf i −ω)t − 1
~ (ωf i + ω)
~ (ωf i − ω)
2
Pi→f (t) = |cf i (t)|
(III.88)
2
=
(III.87)
|Bf i |
2
~2 (ωf i + ω)
2
i(ωf i +ω)t
− 1 +
e
2
|Bif |
~2 (ωf i − ω)
2
i(ωf i −ω)t
− 1 + cross terms
e
(III.89)
The function is plotted in figure 11.
fi
=-
fi
=+
FIG. 11. The probability of a transition with oscillatory perturbation
We summarize the observations below
• Consider the case of ωf i = −ω ⇐⇒ f = i − ~ω. Hence we can only have a noticeable transition if the applied
frequency corresponds to the difference between energy of the two energy levels is ~ω. This is called stimulated
emission since this required a perturbation to emit energy. This is also called resonance de-excitation.
• For the second peak we have ωf i = ω ⇐⇒ f = i − ~ω. Hence we can only have a noticeable transition if we
have absorption of ~ω. This is also called resonant excitation.
Now we consider the connection to atom-radiation interaction. We found earlier that
e
W (t) = A(r, t) · p
m
(III.90)
with
A(r, t) = Π̂ |A0 | cos (k · r − ωt + α)
o
Π̂ n
=
A0 ei(k·r−ω)t + A∗0 e−i(k·r−ωt)
2
(III.91)
(III.92)
where A0 ≡ |A0 | eiα . Hence we have
W (t) =
o
e n
A0 ei(k·r−ωt) Π̂ · p + A∗0 e−i(k·r−ωt)
2m
Recall that we had W (t) = B † e−ωt + Beiωt thus we an make the identification
B=
e ∗ −ik·r
A e
Π̂ · p
2m 0
(III.93)
28
We need to check the criterion to avoid the overlap of resonances (neglect cross terms) this requires
2π
2π
ω ⇐⇒ t t
ω
∆ω =
(III.94)
Now we check the validity of first order time dependent perturbation theory.
t2 /4
Pi→f
2
}|
{ |B |2
4 |Bf i | z
fi
=
t2
f (t, ωji ± ω = 0) =
2
~
~2
To be valid this must be much less then 1. I.e.
|Bf i |2 2
~2 t
(III.95)
1. We combine these results to say that
2π
~
2π
=
ω
|ωf i |
|Bf i |
(III.96)
|f − i | |Bf i |
(III.97)
This is true only if
Hence the matrix element which causes the transition must be small when compared to the energy difference between
the states.
Lecture 14th, February 15th, 2012
C.
1.
Photoionization
Transitions into the continuum: Fermi’s golden rule (FGR)
Suppose that we don’t transition to a single state but into a band of states as shown in figure ?? We need to replace
f
f
+
-
i
FIG. 12. Jumping into an energy band
2
pi→f = |hφf |ψ(t(f ))i|
(III.98)
by
Z
f +∆
Pi→f =
pi→f (f 0 ) ρ (f 0 ) df
(III.99)
f −∆
with ρ(f 0 ) is the density of states. Free particle continuum states are
φf (r) = hr|pi =
1
(2π~)
i
3/2
e ~ p·r
(III.100)
29
hp|p0 i =
Z
hp|ri hr|p0 i d3 r
Z
0
i
1
e ~ (p −p) d3 r
=
3
(2π~)
(III.101)
r
= δ (p − p0 )
(III.102)
(III.103)
Consider
1 = hψ|ψi
Z
= hψ|pi hp|ψi d3 p
Z
2
= |φ(p)| d3 p
We can transfer this integral to an energy integral using f =
p2
2m .
d3 p = p2 dpdΩp = p2
(III.104)
(III.105)
(III.106)
We can use
dp
df dΩp
df
(III.107)
We use
1/2
dp
=
df
⇒
p = (2mf )
m
2f
1/2
(III.108)
This gives
d3 p = 2mf
q
=
r
m
df dΩp
2f
(III.109)
2m3 f df dΩp
(III.110)
We now go back to our normalization equation
Z
q
2
1 = |ψ(p)|
2m3 f df dΩp
(III.111)
pi→f (f )
Z
=
df
q
ρ(f )
2m3 f
zZ
}|
{
2
dΩp |ψ(p)|
Note that this was only for free particles. Hence for a free particle he density of states is
discuss photoionization we need to discuss
f +∆
Z
Pi→f =
pi→f (0f )ρ(0f )df
(III.112)
p
2m3 f . If we want to
(III.113)
f −∆
Recall our previous result for first order pi→f
Pi→f =
4
~2
Z
f +∆
|Bf i |
2
f t, ωf0 i + ω + f (t, ωf0 i − ω) ρ(0f )d0f
(III.114)
f −∆
If we assume that the interval range is small then ρ(0f ) and Bf i don’t vary much accross that interval. In this case.
Furthermore
the meximan hat (f) functions are only large at particular ω. If we are looking at absorption then only
f t, ωf0 i − ω is large across this band. In this case
abs
Pi→f
≈
4
2
|Bf i | ρ(f )
~2
Z
f +∆
f −∆
f (t, ωf0 i − ω)d0f
(III.115)
30
To carry out the integral we define ω̃ ≡ ωf0 i − ω and hence df = ~dω̃
abs
Pi→f
4
2
≈ |Bf i | ρ(f )
~
Z
sin2 ω̃t
2
dω̃
ω̃ 2
(III.116)
Technically the integral is from f −∆ to f +∆. However if we are centered on an absorption peak then contributions
from other parts of the function are small. Thus we may as well just extend the limits of the integral to ±∞, which
is a well known, analytic integal. With this we have
abs
Pi→f
=
2πt
2
|Bf i | ρ (f ) t
~
(III.117)
with f = i + ~ω One can easily show that we get a similar equation for stimulated emission:
SE
Pi→f
=
2πt
2
|Bf i | ρ (f ) t
~
(III.118)
d
Pi→f
dt
(III.119)
2π
2
|Bf i | ρ(f )
~
(III.120)
where f = i − ~ω. We define the transition rate by
Wi→f =
This is given by
Wi→f =
where f = i ± ~ω. This is called Fermi’s Golden Rule (FGR).
2.
Dipole Approximation
Typical situation is that the wavelength of the light used is large compared to the characteristic distance of atoms.
i.e. λ = 2π
k a0 . In this case we use the dipole approximation that says
eik·r = 1 + k · r + ...
≈1
(III.121)
(III.122)
Hence the field doesn’t change spatially across the atom. On Monday we said that
Bf i =
e ∗
A Π̂ hφf | e−ik·r p |φi i
2m 0
(III.123)
e ∗
A Π̂ · hφf | p |φi i
2m 0
(III.124)
Applying the dipole approximation we have
Bf i ≈
We now use a commutator relation:
p=
where Ho =
p2
2m
im
[H0 , r]
~
(III.125)
+ V . With this relation
hφf | p |φi i ≈
im
hφf | Ho r − rH0 |φi i
~
(III.126)
If we have φf and φi as eigenstates then we have
Dipole Matrix Elements
im
hφf | p |φi i =
(f − i )
~
z }| {
hφf | r |φi i
(III.127)
31
Thus we have (inserting back into previous equation)
Bfdip
i =
ie ∗
A (f − i ) Π̂ · hφf | r |φi i
2~ 0
(III.128)
For Π̂ = ẑ we have the standard selection rules, ∆m = 0, ∆` = ±1. We can now figure out the transition rates using
FGR. Using this one will find that
dip,ẑ
Wi→f
∝ cos2 θ
(III.129)
for φi = s-states. This is plotted in figure 13
z
FIG. 13. The transition probability as a function of θ for the 1s state
Lecture 15th - February 27th, 2012
D.
Outlook on Field Quantization
We need to find a Hamiltonian for atom and electromagnetic field
H = HA + HF + W
(III.130)
where HA is the Hamiltonian due to the atom, HF is the Hamiltonian due to the field, and W represents the interaction
of the atom with the field. We can write
H = Ho + W
(III.131)
where Ho = HA + HF . The steps toward a 1st-order PT treatment are
2
p
1. Determine H0 = HA + HF . We already know HA = 2m
+
†
we will use is that HF will be Hermitian, i.e. HF = HF .
e2
r .
However we don’t know HF . One requirement
2. Solve the eigenvalue problem of H0 . We already know the original eigenstates and energies:
HA |φj i = j |φj i
(III.132)
but we don’t know the states and energies below
HF |ρk i = ˜k |ρk i
(III.133)
32
If we denote |ψ` i as the full unperturbed states then we have
H0 |ψ` i = (HA + HF ) |φ` i |ρ` i
= (HA |φ` i) |ρ` i + |φ` i HF |ρ` i
= ` |ψ` i + ˜` |ψ` i = (` + ˜` ) |ψi
(III.134)
(III.135)
(III.136)
3. Determine W . We use the ansatz
W =
e
A·p
m
(III.137)
4. Obtain 1st - order transition rates (apply Fermi’s Golden Rule)
• Need
hψf | W |ψi i
1.
(III.138)
Construction of HF
The energy of a classical electromagnetic field in a vacuum of volume L3 is (denote this energy WEM )
Z
o
E 2 + c2 B 2 d3 r
WEM =
2
(III.139)
Recall for free electromagnetic waves (electric potential is zero) we have with the Coloumb gauge that
∇2 A −
1 ∂2A
=0
c2 ∂t2
(III.140)
Assume periodic boundary conditions for each side of cube
A (x, y, z = 0) = A (x, y, z = L)
A (x, y = 0, z) = A (x, y = L, z)
A (x = 0, y, z) = A (x = L, y, z)
This gives (kj = kx , ky , kz for j = 1, 2, 3)
1 = eikj L
(III.141)
and hence
kz =
2π
nz
L
(III.142)
The total vector potential is given by
A(r, t) =
X
Aλ (r, t)
(III.143)
λ
where
o
Π̂ n i(kλ ·r−ωλ t)
∗ −i(kλ ·r−ωλ t)
q
e
+
q
e
λ
λ
L3/2
n
o
λ is the mode index. Each mode is characterized by kλ , Π̂λ , ωλ . Π̂λ is the unit polarization vector, ωλ = ckλ ,
∂A
λ
λ
λ
λ
λ
λ
and kλ = 2π
L nx , ny , nz where nx , ny , nz ∈ Z. We can now calculate E = − ∂t and B = ∇ × A. By taking these
derivatives and inserting into the equation for WEM above we get (after a lot of manipulations)
X
WEM = 20
wλ2 qλ∗ qλ
(III.144)
Aλ (r, t) =
λ
33
We now perform a substitution amplitudes given by
√
Qλ ≡ 0 (qλ + qλ∗ )
√
Pλ ≡ i 0 ωL (qλ∗ − qλ )
This gives
1
qλ = √
2 0
Pλ
Qλ + i
ωλ
(III.145)
This gives
WEM = 20
X
x2λ
λ
=
1
40
Pλ2
2
Qλ + 2
ωλ
1X 2
Pλ + ωλ Q2λ
2
(III.146)
(III.147)
λ
We can now quantize this energy by promoting Pλ and Qλ to operators. We also demand them to be Hermitian and
follow the commutation relations for position and momentum. i.e.Pλ = Pλ† and Qλ = Q†λ and
[Qλ , Pλ0 ] = i~δλ,λ0
(III.148)
1X 2
Pλ + ωλ2 Q2λ = HF†
2
(III.149)
Since we have a quantized WEM we have HF :
HF =
λ
We know the eigenvalues of this Hamiltonian if it has the commutation relations of position and momentum. The
energies are
X
1
En1 ,n2 ,... =
~ωλ nλ +
(III.150)
2
λ
where nλ = 0, 1, 2, .... Lecture 16th - February 29th, 2012
Consider the following summary and discussion of our results.
1. For our expressions we have assumed that we have periodic boundary conditions. This forced us to have discrete
wavelengths inside our system. This enables us to change our integrals to sums
Z
X
d3 k →
λ
2. Note that
WEM =
1X 2
Pλ + ωλ2 Q2λ
2
(III.151)
λ
is independent on time (since the numbers Pλ and Qλ don’t change with time). i.e.
dWEM
dt
(III.152)
This means that the surface integral of the Poynting vector must be zero (from classical electrodynamics).
3. We quantized Pλ and Qλ by elevating them to Hermitian operators that obey the canonical commutation
relations
[Qλ , Pλ0 ] = i~δλ,λ0
(III.153)
34
4. Energy spectrum of our Hamiltonian is given by
E=
X
~ωλ
λ
1
nλ +
2
(III.154)
with nλ = 0, 1, 2, ....
5. The “conventional” but wrong interpretation is to associate
each mode with a particle in a parabolic potential
and then the eigenenergies in this mode Enλ = ~ω nλ + 12 are the ground and excited state energy levels
6. The alternate interpretation is to associate each mode with nλ particles (or quanta) in the same state. All these
quanta carry the same energy, ~ωλ (forgetting about the 12 ~ωλ term). Note that this only works because the
energy is linear in nλ otherwise. For example if you had 3 quanta (i.e. nλ = 3) we would have and energy of
~ω + ~ω + ~ω = 3~ω. This is the photon interpretation.
7. However we have not yet described the 21 ~ωλ factor. This is the zero-point energy. The full zero-point energy is
E0 =
X ~ωλ
2
λ
→∞
(III.155)
since this is an infinite sum. To make this finite we require a technique called renormalization. The zero-point
energy is the energy without any photons there. This relates to the idea that we have energy in a vacuum (which
causes effects such as spontaneous emission).
2.
Creation and Annihilation Operators
We introduce creation and annihilation operators given by
1
(ωλ Qλ + iPλ )
2~ωλ
1
b†λ = √
(ωλ Qλ − iPλ )
2~ωλ
bλ = √
These operators obey
h
i
bλ , b†λ0 = δλ,λ0
(III.156)
h
i
[bλ , bλ0 ] = b†λ , b†λ0 = 0
(III.157)
and
This gives
1
HF =
~ωλ
+
2
λ
X
1
=
~ωλ nλ +
2
X
b†λ bλ
(III.158)
(III.159)
λ
where nλ ≡ b†λ bλ is called the occupation number operator. The eigenvalue equation with HF =
1
(λ)
HF = |ψnλ i = ~ω nλ +
|ψnλ i
2
P
λ
(λ)
HF
is
(III.160)
where nλ = 0, 1, 2, ... and we have
D
E
ψnλ |ψn0λ = δnλ n0λ
(III.161)
35
The occupation number obeys
nλ |ψnλ i = nλ |ψnλ i
(III.162)
We use shorthand notation of |ψnλ i = |nλ i. One must be careful with this notation. Since we may write the following
equation
n̂λ |nλ + 1i = (nλ + 1) |nλ + 1i
(III.163)
Aˆwas used to differentiate the operator n̂λ from the number. We now consider the state
|nλ = 0i ≡ |0i
(III.164)
nλ |0i = |∅i
(III.165)
which we call the vacuum state. This gives
This is not the same as |0i! This is really the zero state. However since we have a ket on the left side we don’t want
to write just 0. One can show that
√
b†λ |nλ i = nλ + 1 |nλ + 1i
√
bλ |nλ i = nλ |nλ − 1i
In particular we have
bλ |0i = |∅i
(III.166)
Lecture 17th - March 5th, 2012
Here we generalize these results. We use the rule that if the eigenvalue of your system is the sum of a set of eigenvalues
then the resultant eigenvalues are the product of the eigenstates:
HF |n1 , n2 , ...i = E |n1 , n2 , ...i
(III.167)
P
P
where E = λ ~ωλ nλ + 21 = λ ~ωλ nλ + E0 and |n1 , n2 , ...i = |n1 i ⊗ |n2 i ⊗ ... are the product states. The operator
nλ counts the number of modes in mode λ:
n̂λ |n1 , n2 , ..., nλ , ...i = nλ |n1 , n2 , ..., nλ , ...i
√
b†λ |n1 , n2 , ..., nλ , ...i = nλ + 1 |n1 , n2 , ..., nλ + 1, ...i
√
bλ |n1 , n2 , ..., nλ , ...i = nλ |n1 , n2 , ..., nλ , ...i
3.
Interaction Between Photon Field and Electrons
In the beginning we discussed that we are using the semiclassical interaction with a perturbing Hamiltonian:
W =
e
A·p
m
(III.168)
with
A (r, t) =
o
X Πˆλ n
i(kλ ·r−ωλ t)
∗ −i(kλ ·r−ωλ t)
q
e
+
q
e
λ
λ
L3
(III.169)
λ
To quantize this operator we made the transformation given earlier:
r
1
Pλ
~
qλ = √
Qλ + i
=
bλ
2 0
ωλ
20 ωλ
(III.170)
36
Hence we have the vector potential in its new form given by
γ
{
z r}|
n
o
X
1
~
bλ ei(kλ ·r−ωλ t) + b†λ e−i(kλ ·r−ωλ t)
A (r, t) =
Πˆλ 3
L
20 ωλ
(III.171)
λ
This operator is time dependent. However we want an operator thats time independent. To achieve this we define
−iωλ t
bH
λ = bλ e
(III.172)
For this to make sense we need to show that
i~
d H H H
b = bλ , Hλ
dt λ
(III.173)
where we denote operators in the Heisenberg picture by superscript H.
Proof of statement above is shown below. The Hamiltonian in the Schrodinger interpretation is the same as in the
Heisenberg interpretation. Thus we have
h
i
1
(λ),H
H
−iωλ t
(III.174)
RHS = bλ , HF
=e
bλ , ~ω nλ +
2
= ~ωλ e−iωλ t [bλ , nλ ]
(III.175)
−iωλ t
(III.176)
= ~ωλ e
bλ
LHS = i~ω(−iωλ )bλ e−iωλ t = ~ωλ e−iωλ t bλ
= RHS
(III.177)
(III.178)
Hence our construction gave us an operator in the Heisenberg picture. Thus removing the time dependence is easy
and gives (where γ was defined above)
o
X n
A(r) = γ
Π̂ bλ eikλ r + b†λ e−ikλ ·r
(III.179)
λ
This is our quantized vector potential in the Schrodinger picture where the interaction is
W =
e
A·p
m
(III.180)
β
z
s
= −i
4.
hφf | W |ii =
}|
{
n
o
e 2 ~3
†
ikλ r
−ikλ ·r
Π̂
e
∇b
+
e
∇b
λ
λ
2ωλ 0 m2 L3
(III.181)
The Transition Matrix Elements
X
hφf | Wλ |φi i
(III.182)
D
hψf | ⊗ nf1 nf2 ... (Wλ ) ni1 ni2 ... ⊗ |ψi i
(III.183)
λ
=
X
λ
where |ψi are the electron states while |n1 n2 , ...i are the photon states. We have two parts to the equation the product
states are made up of a photon part and an electron part.
D
D
Xn
o
hφf | W |φi i = −iβ
hψf | eikλ ·r Π̂λ · ∇ |φi i ⊗ nf1 nf2 ... bλ ni1 ni2 ... + hφf | e−ikλ ·r Π̂λ ∇ |φi i ⊗ nf1 nf2 ... b†λ ni1 ni2 ...
λ
(III.184)
37
Note that we already know the electron matrix elements since we dealt with them earlier (we will get back to them
in more detail shortly). We now consider the photon matrix elements.
q
D
(III.185)
nf1 nf2 ...nfλ ... bλ ni1 ni2 ...niλ ... = (δnf ,ni δnf ,ni ...δnf ,ni −1 ...) niλ
1
1
2
2
λ
λ
This matrix element is highly selective. Furthermore we have
q
D
nf1 nf2 ...nfλ ... b†λ ni1 ni2 ...niλ ... = (δnf ,ni δnf ,ni ...δnf ,ni +1 ...) niλ + 1
1
1
2
2
λ
(III.186)
λ
Hence the condition to have a non zero transition elements are that the photon numbers don’t change by more then
one. Furthermore there can only be a single photon interaction at a time. The annihilation of a photon corresponds
to absorption of a photon, while the creation of a photon corresponds to an emission of a photon.
Summary:
The matrix elements
hφf | W |φi i =
6 0
(III.187)
are non-zero if and only if
1.
E
i i n1 n2 ... and nf nf ...
1 2
(III.188)
differ in exactly one mode (by one photon).
2. If we apply the dipole approximation then we have the dipole selection rules for the electronic part in the
non-zero mode (eikλ ·r ≈ 1). In this case we can use the typical selection rules of ∆` = ±1, ∆m = 0
3. Recall that we found in the constant perturbation or sudden approximation that the transition is small unless
energy is conserved. i.e.
Ei = Ef
(III.189)
H0 |φi i = (HA + HF ) |ψi i ni1 ni2 ...
= (HA |φi i) + |φi i HF ni1 ni2 ...
!
X
1
i
|φi i
= εi +
~ωλ0 nλ0 +
2
0
(III.190)
(III.191)
(III.192)
λ
This gives
Ef = εf +
X
λ0
~ωλ0
nfλ0
1
+
2
(III.193)
Energy conservation Ei = Ef implies that
εf = εi +
X
~ωλ0 niλ0 − nfλ0
(III.194)
λ0
εi ± ~ωλ
(III.195)
where the plus sign corresponds to absorption while the minus sign corresponds to emission.
Lecture 18 - March 7th, 2012
Note: Look carefully on last question of new assignment, it may be a test question
Recall we can write

p
−ik ·r

i niλ + 1
hψf | e λ Π̂λ · ∇ |ψip
hφf | W |φi i = −iβ × hψf | eikλ ·r Π̂λ · ∇ |φi i niλ

0
Note we can only have one of above.
(III.196)
38
5.
Spontaneous Emission
Thus is a special case of the term
− iβ hψf | e−ikλ ·r Π̂λ · ∇ |ψi i
q
niλ + 1
(III.197)
with niλ = nfλ − 1 = 0. Energy conservation needs to be fulfilled. In other words
εf = εi − i~ωλ
(III.198)
Fermi’s Golden rule says that
spon.emission
Wi→f
=
2π
2
|hφf | W |φi i| ρ(Ef )
~
where ρ is the density of states. We first need to find the density of states.
(
|ψi i : (discrete) Excited atomic state
Initial state =
|0i : No photon
(
Final state =
|ψf i : (discrete) atomic ground state
ε −ε
|1i : One photon with ω = i ~ f
(III.199)
(III.200)
(III.201)
We want to find the density of (photon) states. We look at the density of states with respect to k-space. Recall that
2π λ λ λ nx , ny , nz
L
(III.202)
∆N
∆nx ∆ny ∆nz
=
∆Vk
∆kx ∆ky ∆kz
(III.203)
kλ =
where nλi ∈ Z.
ρ(k) =
We use the relation between k and n shown in equation III.202. It’s easy to see that
3
2π
ρ(k) =
L
(III.204)
We can rewrite the differential as
d3 k = k 2 dkdΩ = k 2
dk
dEdΩ
dE
(III.205)
For photons we have
E = ~ω = ~ck
(III.206)
dE
dk
1
= ~c ⇐⇒
=
dk
dE
~c
(III.207)
Hence we have
With this relation we have
1
d k=
~c
3
E2
~2 ω 2
dEdΩ
(III.208)
39
We can now find the number differential
3 2
L
ω
dEdΩ
2π
~c3
ρ(E)dEdΩ
dN = ρd3 k =
(III.209)
(III.210)
Fermi’s Golden Rule says that
Spon.Emission
dWi→f
dΩ
3 2 2
e 2 ~3
2π L
ω
−ik·r
|
e
Π̂
·
∇
|φ
i
=
hψ
f
i
~ 2π
~c3 20 ωm2
L3
2 2
ω
e ~ −ik·r
=
|
e
Π̂
·
∇
|φ
i
hψ
f
i
8π 2 c3 0 m2
(III.211)
(III.212)
If we apply the dipole approximation then e−k·r ≈ 1 (this says that the size of the atoms is much smaller then the
wavelength of the light). This leaves us to consider
hψf | Π̂ · ∇ |ψi i =
i
hψf | Π̂ · p |ψi i
~
(III.213)
However one can show that
im
[HA , r] = p
~
if HA =
p2
2m
(III.214)
+ V (r). Using this relation we have
hψf | Π̂ · ∇ |ψi i = −
m
Π̂ · hψf | HA r − rHA |ψi i
~2
(III.215)
~ω
m z }| {
= − 2 (εf − εi ) Π̂ · hψf | r |ψi i
~
(III.216)
Putting this result together we have
Spon.Emission
dWi→f
dΩ
2
e2 ω 3
×
Π̂
·
hψ
|
r
|ψ
i
f
i 8π0 ~c3
α
z }| {
2
2 3
e ω
Π̂
·
r
=
if
8π 2 0 ~c3
=
(III.217)
(III.218)
We now sum over all polarizations thus we integrate. We choose our wave propagation direction, k, such that rif
is along the z axis and define the angle between these two axes as θ. The two linearly independent polarization
directions of the light must be perpendicular to the motion of the wave (⊥ k). If we choose one polarization to be
perpendicular to rif then this contribution is zero. This sets the other polarization direction to be
π
Π̂2 · rif = |r| cos
− θ = rif sin θ
(III.219)
2
Hence we have
Z 2
2
S.E
Wi→f =
|Π1 · rif | + |Π2 · rif | dΩ
(III.220)
Z
2
= α rif
sin2 θdΩ
(III.221)
8π 2
r
3 if
Thus the total transition rate in the dipolar approximation is
=α
S.E.
Wi→f
=
where rif = hψi | r |ψf i.
Discussion:
e2 ω 3
2
|rif |
3π0 ~c3
(III.222)
(III.223)
40
1. As an example consider a Hydrogen 2p → 1s transition
H(2p)
H(1s)
The lifetime is given by
1
dip
=
T2p→1s
S.E
W2p→1s
≈ 1.6 × 10−9 s
(III.224)
In a classical picture it takes the electron 10−16 s to circle around the nucleus. Thus this is a very long lifetime
with respect to this value.
2. We can consider a different decay of Hydrogen 2s → 1s It turns out that
1storder
T2s→1s
=∞
(III.225)
experiment
T2s→1s
= 0.1s
(III.226)
However experimentally we have
This state is metastable. We need second order perturbation theory (more then FGR) to do this. This corresponds to a two-photon process.
3. For an N -photon process one needs to consider N th order perturbation theory. The W operator is linear in b
and b† so in order to contribute a N photon process we need to combine more of these operators.
Lecture 19th - March 12th, 2012
Recall that the total transition rate in the dipole approximation is
s.e.
Wi→f
=
e2 ω 3 2
2
2
|xif | + |yif | + |zif |
3
3π0 ~c
(III.227)
The lifetime is simplify given by
Ti→f =
1
s.e.
Wi→f
(III.228)
Consider if we have an electron in the 3p state: φi = H(3p). This is shown in figure 14 The total decay rate is the
3p
2s
1s
FIG. 14. A decay from a Hydrogen 3p state
sum of the two rates:
W s.e. =
e2
3π0 ~c3
X
f (εf <εi )
3
ωif
|rif |
2
(III.229)
41
Here the rates are uncoupled (the rate of 3p → 2s doesn’t effect the rate of 3p → 1s). In practice of course this is not
the case.
Concluding remarks on photons:
Here we defined photons as the quanta of an electromagnetic field. The properties of the photon are
• Can be created or annihilated (hence they are not stable)
• Carry energy ~ωλ
• One can show that they carry momentum ~kλ . The reasoning is as follows. One could start from a classical
expression for momentum of the electromagnetic field:
Z
pEM = 0
(E × B) d3 r
(III.230)
V =L3
For our cube we know the vector potential (Aλ (r, t) =
P
λ
...) and hence we can find the electromagnetic fields
∂A
∂t
B=∇×A
E=−
we can then insert this result into the momentum and find the momentum of the fields:
X
pEM = 20
kλ ωλ qλ∗ qλ
(III.231)
λ
Quantizing this operator gives
pF =
X
~kλ b†λ bλ =
X
~kλ nλ
(III.232)
λ
Hence we have
pF |n1 , n2 , ...i =
X
~kλ nλ |n1 , n2 , ...i
(III.233)
λ
From this it is clear that momentum of each mode is ~kλ .
• Similarly one can show that photons also carry angular momentum (often called photon spin) given by ±~. This
is reasoned as follows. The angular momentum of an electromagnetic field is
Z
LEM = 0
r × (E × B) d3 r
(III.234)
L3
This gives a spin of ±~.
• Since one can show that the angular momentum is an integer the photons are bosons. Note that we didn’t find
any restriction for the number of photons that can be in a given mode. This is another way of showing that
photons are bosons.
• The photons move with the velocity of light v = c since that the velocity of electromagnetic waves in a vacuum.
By Einstein’s theory of special relativity we know that the photons have zero mass
• The photon treatment can be found in the following resources:
– Friedrich, Theoretical Atomic Physics
– Sakurai and Napolitano, Modern Quantum Mechanics
– Schiff, Quantum Mechanics
• The semiclassical atom-radiation interaction can be found in
– Liboff (Chap 13.5 - 13.9)
– Cohen-Tannoudiji (Chap 13)
42
As an aside we now go over the solution to the practice problem put online. Consider the harmonic oscillator
Hamiltonian:
H0 =
p2
m
+ ω02 x2
2m
2
(III.235)
W =
1
mω 2 x2 cos ωt
2
(III.236)
with the perturbation
i
~
c1storder
(t) = δk0 −
k
Z
t
0
eiωk0 t hk| W (t0 ) |0i dt
(III.237)
t0
We first need to find the matrix elements.
hk| W̃ |0i ≡ hk| x2 |0i
(III.238)
Here we use the following
r
x=
~
a + a†
2mω0
(III.239)
where a and a† are the annihilation and creation operators respectively.
x2 =
~
a2 + a† 2 + aa† a† a
2mω0
(III.240)
:0
~
:0
2 †
† †2
hk| W |0i =
hk|a |0i + hk| aa |0i + hk|a a |0i + hk| a |0i
2mω0 √
but a† |0i = |1i , a |1i = |0i , and a† |1i = 2 |2i. Thus
√
~ hk| W |0i =
hk|0i + 2 hk|2i
2mω0
√
~ =
δk0 + 2δk2
2mω0
We now summarize our results as
hk| W (t) |0i Thus
where we have ω20 = 2ω0 .
√
2 ω2
c2 (t) = −i
8 ω0
i (2ω0 + ω)
=
2~ mω 2 iωt
e + e−iωt
2mω0 4
Z
t
k6=0
2 ω2
8 ω0
1
(III.243)
e
0
ei2ω0 t
(III.244)
0
eiωt + e−iωt dt0
(III.245)
0
i(2ω0 +ω)t
−1 +
1
i (2ω0 − ω)
e
i(2ω0 −ω)t
−1
For the case of ω = 2ω0 (the resonance condition) we have (must go back to the integral to show this)
√
2
c2 (t) = −i
ω0 t
2
The probability is given by
P2 (t) =
Lecture 20th, March 19th, 2012
(III.242)
√
√
c2 (t) = −i
(III.241)
ω02 2
t
2
(III.246)
(III.247)
(III.248)
43
IV.
BRIEF INTRODUCTION TO RELATIVISTIC QUANTUM MECHANICS
A.
1.
Klien-Gordon Equation
Setting up a relativistic wave equation
Recall the non-relativistic case. We start from a classical expression for the energy of a free particle
E−
p2
=0
2m
(IV.1)
We then quantize this expression by promoting the energy and momentum parts to operators:
∂
∂t
p2 → −i~2 ∇2
E → i~
(IV.2)
(IV.3)
This gives
∂
i~ + ~2 ∇2 ψ = 0
∂t
(IV.4)
The relativistic case is as follows. We start with the relativistic energy momentum relation:
E 2 = p2 + m2
(IV.5)
We then quantize this equation using the prescription above
2
2 2 2
2 4
2 ∂
+~ c ∇ −m c ψ =0
−~
∂t2
(IV.6)
or in natural units (KG equation)
−
B.
∂2
2
2
+
∇
−
m
ψ=0
∂t2
(IV.7)
Discussion of KG equation
(i) KG equation is a second order PDE in both space and time
(ii) KG equation is Lorentz covariant (this means that this equation behaves under LT the way that it should)
(iii) Time development is determined from two initial conditions (since it’s second order in time) for the wavefunction
and for the time deriviative. i.e.,
ψ(t0 ) =
∂ψ(t0 )
∂t
(IV.8)
However this is not compatible with one of the postulates of quantum mechanics which says that to get the full
wavefunction we just need to act on the wavefunction at a certain time with the time evolution operator.
Û
ψ(t0 ) −
→ ψ(t)
(IV.9)
(iv) Another problem arises with the continuity equation. One can derive the probability current in the KG equation:
∂ρ
= −∇ · J = 0 ,
∂t
∂ψ ∗
i~
where ρ = 2mc
ψ ∗ ∂ψ
is the probability density and J =
2
∂t − ψ ∂t
current. However ρ is not positive definite and can be negative!
(IV.10)
i~
2m
(ψ∇ψ ∗ − ψ ∗ ∇ψ) is the probability
44
(v) Consider the following trial solution of the KG equation:
ψ(r, t) = A sin (k · r − ωt)
∂2ψ
= −Aω 2 sin (k · r − ωt) = −ω 2 ψ
∂t2
∇2 ψ = −k 2 ψ
(IV.11)
(IV.12)
(IV.13)
substitution into the KG equation gives
~2 ω 2 ψ = ~2 c2 k2 + m2 c4 ψ
(IV.14)
E 2 = c2 p2 + m2 c4
(IV.15)
This gives
with E = ~ω and p = ~k. However this solution doesn’t solve the Schrodinger equation! The time dependent
Schrodinger equation cannot be solved by a real solution. However this is not true for the KG equation.
Subsitution into the current and probability densities given above for this solution gives
J=0
ρ=0
(IV.16)
(IV.17)
This solution is clearly problematic. On the other hand we can try another solution for the KG equation:
ψ(r, t) = Aei(k·r−ωt)
(IV.18)
E 2 = p2 c2 + m2 c4
(IV.19)
with the same energy momentum relation:
The Klien Gordon equation does not seem to have any restriction on energy thus we can have
p
E = ± p2 c2 + m2 c4
(IV.20)
(vi) Adding the Coloumb potential to the free KG equation and solving gives solutions that don’t agree with experiments. This is what bothered Schrodinger. This is because the KG equation does not take spin into account.
(vii) In 1934 the KG equation was recognized as correct wave equation for spin 0 particles.
C.
1.
Dirac Equation
Free Particles
Dirac wanted an equation first order in space and time. He hoped this would remove some of the problems with
the KG equation. Dirac tried the an satz was
i~
∂ψ
= Hψ
∂t
He had a set of requirements that his equation was to fulfill.
(i) Compatibility with the relativistic energy relation: E 2 = p2 c2 + m2 m4
(ii) Covariant with respect to Lorentz transformations.
(iii) An equation that is consistent with the continuity equation and probability interpretation
(iv) He didn’t want to invent any new quantization rules
(IV.21)
45
We have our energy momentum relation
E 2 = p2 c2 + m2 c4
(IV.22)
E = pc + mc2
(IV.23)
If we had the equation (which is not true!)
then we are first order and space and time. Dirac’s idea was to write
H = α · pc + βmc2
.
(IV.24)
Dirac realized that this an satz can fulfill the energy momentum relation if α and β are matrices. Trying to fulfill
the requirements listed above gives the restrictions on α and β. All we know so far is that we have three α matrices
and 1 β matrix. We call these N × N matrices. Since these are matrices we require wavefunctions with a number of
components equal to the dimensionality of the matrices. These wavefunctions are called ‘spinor wavefunctions’. We
can denote this spinor wavefunction as follows


ψ1 (r, t)
 ψ2 (r, t) 


(IV.25)
Ψ=

..


.
ψN (r, t)
where N is the dimensionality of the matrices. The first requirement says that each component ψi (r, t) must fulfill
the KG equation.
Lecture 21 - March 21st, 2012
We now go through Dirac’s wishlist in order to find conditions for α and β: First each component of ψi (r, t) must
fulfill the Klien Gordon equation. Consider the Dirac equation:
∂
(IV.26)
i~ − H Ψ = 0
∂t
iterating the Dirac equation gives
d
d
i~ − H
i~ − H Ψ = 0
dt
dt
2
d
d
H Ψ=0
−~2 2 + H 2 − 2i~
dt
dt
2
d
−~2 2 + H 2 − 2H 2 Ψ = 0
dt
− ~2
−~2
(IV.27)
(IV.28)
(IV.29)
d2
= H 2Ψ
dt2
0
(IV.30)
∂2
Ψ = cα · p + βmc2 cα · p + βmc2 Ψ
2
∂t


!
X
X
~
d
~
d
2
2
= c
αj
+ βmc  c
αk
+ βmc Ψ
i j
dxj
i
dxk
k




2
3 X
X
∂
~mc
d
= −~2 c2
αj αk
+
(αj β − βαj )
+ β 2 m2 c4 Ψ


∂xj ∂xk
i
dxj
j
j,k




3 X
X αj αk + αk αj ∂ 2
~mc
d
+
(αj β − βαj )
+ β 2 m2 c4 Ψ
= −~2 c2


2
∂xj ∂xk
i
dxj
j
j,k
(IV.31)
46
Recall the Klien Gordan equation says that
− ~2
∂2
ψ = −~2 c2 ∇2 + m2 c4 ψ
2
∂t
(IV.32)
The left hand sides are trivially equal to one another. Thus we require certain conditions. Clearly there are no first
order derivatives in the KG equation. Thus we require
αj β + βαj ≡ {αj , β} = 0
.
(IV.33)
2
The KG equation only has second derivatives in the form of a Laplacian. Thus all the cross terms such as ∂x∂1 x2 must
be zero. This can be done by introducing a Kronecker delta. However the term must be 1 for each second derivative
∂2
of a given variable (e.g. ∂x
2 ):
1
αj αk + αk αj = 2δjk
(IV.34)
β2 = 1
(IV.35)
Lastly we require
This conditions must be fulfilled for all j, k = 1, 2, 3. Further requirements(one that wasn’t mentioned in Dirac’s
wishlist but important) was that the Dirac Hamiltonian should be Hermitian:
H = H†
(IV.36)
This requires
αj = α†
;
β = β†
.
(IV.37)
Hence they have real eigenvalues. Notice our requirements tell us that
αj2 = 1 and β 2 = 1
Thus the eigenvalues must be ±1 for all four of these matrices. Consider the equation
αj β + βαj = 0 .
If we multiply this equation on the right with β we we have
αj β 2 = −βαj β
αj = −βαj β
Tr(αj ) = −Tr(βαj β)
Tr(αj ) = −Tr(αj )
(IV.38)
where in the last step we used β 2 = 1. Hence the trace of αj must be zero. The same can be shown for β by
multiplying by αj instead of β in the beginning. Thus we have another condition on αj and β, αj and β are traceless.
Since the matrices are tracelss the sum of the eigenvalues is zero (easy to prove). Since the eigenvalues are ±1 and
the only way to have a sum of 0 we need to have an even dimension! We need 4 anticommuting matrices.
The simplest choice for the dimension would be N = 2. However we already know that the Pauli matrices make
up three independent anticommuting matrices and there cannot be a fourth. Hence we need to go up to a larger
dimension. We choose the next lowest dimension, namely N = 4. Matrices are obey all the conditions we have above
are








0 0 0 1
0 0 0 −i
0 0 1 0
1 0 0 0
0 0 1 0
0 0 i 0 
 0 0 0 −1 
0 1 0 0 
αx = 
; αy = 
; αz = 
; β=
(IV.39)
0 1 0 0
0 −i 0 0 
1 0 0 0 
0 0 −1 0 
1 0 0 0
i 0 0 0
0 −1 0 0
0 0 0 −1
47
Note that we can rewrite these as
0 σx
0 σy
σz 0
1 0
αx =
; αy =
; αz =
; β=
σx 0
σy 0
0 σz
0 −1
(IV.40)
We now take a look at the Dirac equation more explicitly:



ψ1
ψ1
 ψ 
∂ ψ  i~  2  = c (αx px + αy py + αz pz ) + βmc2  2 
ψ3
∂t ψ3
ψ4
ψ4

(IV.41)
These are four coupled equations because α have all off-diagonal terms. Working out these equations gives (when you
multiply by α and β matrices)
∂
~c
∂
∂
∂
i~ ψ1 =
−i
ψ4 +
ψ3 + mc2 ψ1
(IV.42)
∂t
i
∂x
∂y
∂z
∂
~c
∂
∂
∂
i~ ψ2 =
+i
ψ3 −
ψ4 + mc2 ψ2
(IV.43)
∂t
i
∂x
∂y
∂z
~c
∂
∂
∂
∂
−i
ψ2 +
ψ1 − mc2 ψ3
(IV.44)
i~ ψ3 =
∂t
i
∂x
∂y
∂z
∂
~c
∂
∂
∂
i~ ψ4 =
+i
ψ2 − mc2 ψ4
ψ1 −
(IV.45)
∂t
i
∂x
∂y
∂z
2.
Solutions of the free Dirac equation
To solve them we use the ansatz
ψj (r, t) = uj ei(k·r−ωt)
(IV.46)
with E = ~ω and p = ~k. This gives four linearly independent solutions. The two solutions


1
 0 

=
; u(2) = 
χ1 
χ2

u(1)

0
1 
χ01 
χ02
(IV.47)
correspond to an energy of
p
E = + p2 c2 + m2 c4
(IV.48)
The factors χ1 , χ2 , χ01 , χ02 are kinematic factors. In the limit of v c these factors are zero. The two other solutions,

 0 
φ1
φ1
 φ2 
 φ02 
(4)
=
; u =
1 
0 
0
1

u(3)
(IV.49)
p
vc
correspond to negative energy, E = − p2 c2 + m2 c4 with φ1 , φ2 , φ01 , φ02 −−−→ 0. These negative solutions are difficult
to interpret. Dirac’s interpretation is shown in figure 15. Dirac says that the negative energy levels are all occupied
by negative energy electrons. Since electrons are fermions they cannot all pile in these negative energy states. These
particles make up the vacuum. Thus we use a photon with high enough energy we can excited these particles and
create a hole in the Dirac sea. Dirac inferred that these holes are observable as antiparticles called the positrons.
These are holes in the negative energy spectrum and hence have positive energy.
This interpretation says that the vacuum is a a fully occupied Dirac sea with no electrons with energy E > mc2 .
48
E
mc
Filled States
2
0
-mc2
Enegy Gap (2 x 511 keV)
Hole
Filled States ("Dirac Sea")
FIG. 15. The Dirac Interpretation of the solutions to the Dirac equation. He says that all the negative energy levels are filled.
However we cannot observe these particles
Lecture 22 - March 26th, 2012
As an aside recall that Dirac’s problem with the KG equation was that it didn’t give physical answers for the
continuiuty equation. For the Dirac equation one can show that
2
2
2
2
ρ = Ψ† Ψ = |ψ1 | + |ψ2 | + |ψ3 | + |ψ4 | > 0
†
Jk = cΨ αk Ψ
(IV.50)
(IV.51)
for k = 1, 2, 3. The Dirac equations gives strictly positive probabilities as expected and parallels the probabilities in
the Schrodinger equation.
We now come back to the Dirac interpretation. The process of pair creation is given by
γ → e+ + e−
(IV.52)
To conserve energy and momentum this process must occur in the presence of a nucleus. Annihilation can produce
photons:
e+ + e− → 2γ
(IV.53)
The consequence of the Dirac interpretation is that particle number is not conserved in this theory. Not only photons
can be created and annihilated but also photons can. This suggests that this is a many-particle theory.
Recall that we have 4 energy solutions. This is called a doublet structure. This represents the spin degree of
freedom:
u(1) , u(3)
(2)
u
(4)
,u
“spin up”
“spin down”
At this point from our analysis its not clear that this spin has anything to do with angular momentum.
Now consider the components in the u vectors labeled χ. These are non-vanishing ‘small’ components which
implies that e+ and e− are intrinsically connected. For small velocity (v c) the Schrodinger (Pauli) equation can
be recovered from the Dirac equation.
49
3.
Add Electromagnetic Potentials
We now add EM potentials to the Dirac theory. This is often called the “minimal coupling prescription”. Recall
that the get from the free Schrodinger equation to one with potentials we add a potential:
p2
∂
ψ=
ψ
∂t
2m
(IV.54)
∂
1
2
ψ=
(p + eA) − eφ
∂t
2m
(IV.55)
i~
In EM field we have
i~
We can extract a recipe from this.
p → p + eA
∂
∂
i~ → i~ + eφ
∂t
∂t
Recall the free Dirac equation is
i~
∂ψ
= α · pc + βmc2 ψ
∂t
(IV.56)
Using our recipe we have
∂ψ
= α · (p + eA) + βmc2 − eφ ψ
(IV.57)
∂t
One can show that this equation is also Lorentz covariant which says that it has the same form in all inertial reference
frames.
i~
4.
The relativistic hydrogen problem
This corresponds to a given choice for the potential in the Dirac equation
A = 0;
φ=
Ze
4π0 r
.
(IV.58)
However this is not Lorentz covariant since we chose a nonconvariant form for the potential (if you have a boost then
the vector potential would be nonzero). This introduces a small error. To solve the Dirac equation we use the usual
an satz
Ψ(r, t) = Φ(r)e−iEt/~
(IV.59)
this gives
cα · p + βmc2 −
Ze2
4π0 r
Φ(r) = EΦ(r)
Lecture 23 - March 28th, 2012
The bound-state eigenenergies for the bound states can be obtained and written as
" s
#
2
(Zα)
2
En,j = mc − 1 +
2
(n − δj )
q
2
2
1
1
where δj ≡ j + 2 −
.
j + 12 − (Zα) and α = ma~0 c ≈ 137
n = 1, 2, 3, ...
is still the principle quantum number and
j=
is the total angular momentum quantum number.
Discussion
1 3 5
1
, , , ..., n −
2 2 2
2
(IV.60)
(IV.61)
50
(i) δj ≥ 0 for Zα ≤ 1
(ii) En,j 6= En,j 0 this accounts for fine-structure.
2
(iii) One can expand En,j in powers of the small parameters (Zα) . This gives
4
2
En,j = mc
2
(Zα )
(Zα )
−
1−
2n2
2n3
1
j+
1
2
3
−
4n
!
(IV.62)
The first term (0th order) is the rest energy mc2 . The second term (1st order) we have the non relativistic
binding energy since
~
ma20
(IV.63)
~ Z2
2ma20 n2
(IV.64)
mc2 α2 =
and
En(2) = −
The third term (2nd order) gives a relavistic correction which is j dependent (fine structure). The fine structure
is shown in figure ??:
Continuum
E=mc2
3s,3p,3d
n=3
3d 5/2
3p3/2 , 3d 3/2
3s1/2 , 2p1/2 , 3d1/2
2s,2p
n=2
2p3/2
2s1/2 , 2p1/2
10^-4 eV
1s
n=1
1s1/2
Schrodinger
1.8 x 10^-4 eV
Dirac
Notice that in Schrodinger’s theory our reference frame was zero. On the other hand with Dirac we have relativity
and we use E = mc2 as our new reference frame.
Further corrections of bound state hydrogen problem:
(i) Hyperfine structure coupling of magnetic moments of the electron to the nuclear magnetic moment. e.g.
Continuum
E=mc2
'trilet'
n=1
1s1/2
6 x 10-6eV
'singlet'
(ii) QED effects (the Lamb shift) e.g.
51
Continuum
E=mc2
2s 1/2
2p 1/2
n=2 2s 1/2,2p1/2
5.
4.4 x 10 -6eV
Nonrelativistic limit of the Dirac equations
The starting point is the stationary Dirac equation:
cα · p + βmc2 + V (r) Φ = EΦ(r)
(IV.65)
we write the 4-spinor as

φ1
φ 
= 2 
χ1
χ2

Φ=
We can write αj =
0 σj
σj 0
φ
χ
(IV.66)
with σj as the Pauli matrices. Further we have β =
1 0
0 −1
. Inserting into the
Dirac equation gives
c
0 σ
σ 0
·p
φ
χ
=
1 0
E − V (r) − mc2
Φ(r)
0 −1
(IV.67)
This gives two coupled matrices for φ and χ:
cσ · pχ = E − V (r) − mc2 φ
cσ · pφ = E − V (r) + mc2 χ
Isolating the second equation for χ gives
χ=
c
σ · pφ
E − V (r) + mc2
Inserting this equation into the top equation gives
c2
σ·p
σ · p φ = E − V (r) − mc2 φ
2
E − V (r) + mc
(IV.68)
(IV.69)
Lecture 24th, 2012
Isolating for φ we have
φ=
1
σ·p
(E − V (r) − mc2 )
c2
σ
·
p
φ
E − V (r) + mc2
We now make a weak relativistic approximation. We define
ε = E − mc2 mc2
Further we assume
V (r) mc2
(IV.70)
52
We now expand consider the term in the left hand side of equation IV.69:
1
1
ε−V
c2
≈
=
1
−
ε−V
ε + 2mc2 − V (r)
2m
2mc2
2m 1 + 2mc
2
(IV.71)
This gives
1
ε−V
σ · p φ = (ε − V ) φ
σ·p 1−
2m
2mc2
1
~
ε−V
∇V
2
(σ · p) + σ ·
σ · p φ = (ε − V ) φ
1−
2m
2mc2
i
2mc2
(IV.72)
(IV.73)
Now for Pauli matrices we can write
(σ · A) (σ · B) = ∇ (A · B) + iσ · (A × B)
(IV.74)
In our case we have A × B = p × p = 0. Further we have that the gradient of a function is (given no radial dependence
∇V =
1 dV
dr
r dr
(IV.75)
Inserting in these relations we have
ε−V
~ 1 dV σ · rσ · p 1
2
1−
p
+
φ = (ε − V ) φ
2m
2mc2
i r dr
2mc2
~
ε−V
1
dV
dV
p
~
1−
+
(r
·
p)
+
σ
·
L
φ = (ε − V ) φ
2m
2mc2
i 4m2 c2 r dr
4m2 c2 r dr
(IV.76)
(IV.77)
where we have used equation IV.74 We define these terms T1 , T2 and T3 respectively.
Interpretation of Terms
As preparation consider a term acting on a non-relativistic wavefunction
p2
ε − mc2 − V (r) ψ =
ψ
2m
This is not the term we have. In our case we have
2
2 2
ε−V
p2
1
p
p
T1 φ = 1 −
≈
−
2mc2 2m
2m 2mc2 2m
p2
p4
=
−
2m 8m3 c2
(IV.78)
(IV.79)
Here we see the relativistic correction of kinetic energy. The second term T2 is not Hermitian. However it can easily
be made Hermitian by considering Hermitian average
T̄2 ≡
=
1
T2 + T2†
2

(IV.80)
† 


 
1
 ~ 1 dV r · p − ~  1 dV r ·p 


2 i r dr
i |r {z
dr }  
(IV.81)
F(r)
Now
†
(F · p) = p† F† (r)
†
=p ·F
=p·F
†
(IV.82)
(IV.83)
(IV.84)
53
With this we have
T̄2 =
1
8m2 c2
~ dV
~
1 dV
r·p− p·r
i dr
i
r dr
(IV.85)
By inserting p = ~i ∇ and doing some algebra (exercise) we get
~2
∇2 V ≡ HD
8m2 c2
T̄2 =
(IV.86)
This term is called the Darwin hamiltonian. Note that
1
∇2
= −4πδ(r)
r
(IV.87)
Thus we finally have
HD =
Ze2 ~2
δ(r)
8m2 c2 0
(IV.88)
This represents “Zitterbewgung” or trembling motion. Lastly consider the third term
~ 1 dV
σ·L
4m2 c2 r dr
1 1 dV
=
S·L
2m2 c2 r dr
(IV.89)
T3 =
(IV.90)
with S = ~2 σ. This is a spin orbit coupling term. This is the reason we call spin-orbit coupling a relativistic effect.
The weak relativistic limit of the Dirac equation takes the form
H |φi = ε |φi
(IV.91)
p4
Ze2
+ − 3 2 + HD + T3
4π0 r
8m c
(IV.92)
with
H = f racp2 2m −
= H0 + W
(IV.93)
You can account for W in first order perturbation theory. We obtain
4
∆E (1) = hW i =
mc2 (Zα)
2
n3
1
j+
1
2
−
3
4n
(IV.94)