Chapter 8. Approximation Methods – Part II
8.1 Time-Independent Perturbation (Degenerate)
Let’s begin with a brief review. Hamiltonian H = H 0 + H ′ , where H ′ is much weaker than H 0 .
′ = n (0) H ′ m (0) .
We already know: H 0 n (0) = En(0) n (0) and H nm
The Problem: solving the equation
H n = En n ,
using the basis of
(8.1)
{ n } . How to solve it exactly?
(0)
The matrix method. n = ∑ cnm m (0) , and we want to find cnm by solving Eq. (8.1).
m
En n = En ∑ cnm m (0) ,
(8.2)
H n = (H 0 + H ′)∑ cnm m (0) = ∑ cnm (Em(0) + H ′) m (0) .
(8.3)
m
m
m
Combining Eqs. (8.1), (8.2), and (8.3), we find that
0 = ∑ cnm (Em(0) − En + H ′) m (0) .
(8.4)
m
Multiplying Eq. (8.4) on the left by l (0) , we obtain the fundamental equation of the matrix
method:
′ .
0 = (El(0) − En )cnl + ∑ cnmH lm
(8.5)
m
This is an eigenstate equation, exactly equivalent to finding the eigenvalues and eigenvectors
of the matrix for H = H 0 + H ′ using the basis of
{ n } , with matrix elements
(0)
′ , by solving the determinantal equation
H nm = En(0)δnm + H nm
det(H −λI ) = 0 .
(8.6)
The problem with using the matrix method is the matrix often has an infinite dimensionality
since the number of eigenstates
{ n } is often infinite. Usually it includes a summation over
(0)
all bound and continuum states, and the latter are continuous and infinite in number. Potentials
121
without continuum states, such as the harmonic oscillator, have an infinite number of bound
states. So the above determinant has an infinite number of rows and columns.
In practice, an accurate solution can be obtained for the states lowest in energy by using a
finite matrix by truncating states after N terms, and the determinant is evaluated for N ×N .
Convergence is tested by increasing the value of N. Modern computers make this calculation
possible in many realistic cases.
However, diagonalizing a large matrix ( N ~ 1, 000 −1, 000, 000 ) is non-trivial. The First- and
second-order time-independent perturbation theory can give accurate estimation of En and n ,
if H ′ is much weaker than H 0 .
En = En(0) + En(1) + En(2) + 
(8.7)
n = n (0) + n (1) + n (2) + 
(8.8)
′ ,
En(1) = n (0) H ′ n (0) = H nn
(8.9)
Then Eq. (8.1) leads to
n (1) = ∑
m≠n
′
H mn
E
(0)
n
En(2) = ∑
m≠n
−E
(0)
m
′
H nm
m (0) ,
(8.10)
.
(8.11)
2
En(0) − Em(0)
The condition for safely applying Eqs. (8.9), (8.10), and (8.11) is
′
En(0) − Em(0)  H nm
(8.12)
i.e., there is no degeneracy. How to treat the degenerate case, where Eq. (8.12) doesn’t hold any
more? Specifically, when En(0) = Em(0) ,
′
H nm
En(0) − Em(0)
diverges. We need to derive the degenerate
′ , we’ll have to employ the
perturbation theory. In practice, when En(0) − Em(0)  H nm
degenerate perturbation theory.
Assume that E1(0) = E 2(0) =  = E g(0) = E , and these g degenerate states
1(0) , 2(0) , , g (0) form a sub-space D. The first-order equation,
122
(H
0
)
(
)
− En(0) n (1) = En(1) − H ′ n (0) ,
(8.13)
still holds, but we cannot find the correct solution to n (1) by multiplying m (0) on the left of
(
)
(
)
the above equation, m (0) H 0 − En(0) n (1) = m (0) En(1) − H ′ n (0) . Since when n,m ∈ D the
left hand side of the above equation will vanish, then the components of n (1) in the sub-space D
will just disappear; however, in the exact solution, these contributions cannot be neglected. In
addition, without correctly considering the degenerate sub-space D, the first- and second-order
corrections to energy will also be wrong.
g
Instead, one chooses a state k as a linear combination of m (0) ∈ D : k = ∑C km m (0) ,
m=1
then one could get rid of the degeneracy by solving the matrix equation in the sub-space.
Multiplying k on the left of Eq. (8.13), we obtain
(
)
k H 0 − En(0) n (1) = k (λ − H ′) n (0) .
(8.14)
The left hand side is still zero, but now this equation becomes an eigenstate problem for k in
the degenerate sub-space using the basis of n (0) :
0 = k (λ − H ′) n (0) .
Using the completeness in D:
g
∑m
(0)
(8.15)
m (0) = I , we show that it is an eigenstate problem:
m=1
0 = k (λ − H ′) n (0) ⇒ n (0) H ′ k = n (0) λ k = λ n (0) k
g
And n (0) H ′ k = n (0) H ′I k = ∑ n (0) H ′ m (0) m (0) k ⇒
m=1
g
∑
n (0) H ′ m (0) m (0) k = λ n (0) k
(8.16)
m=1
This is equivalent to the normal form of the eigenstate equation H ′ k = λ k , and the
eigenvalues λ are the first-order corrections to En : En(1) = λn , with n = 1, 2, ..., g , and the
123
eigenstates kn replaces n (0) for higher-order corrections to eigenstate n . The second-order
correction in energy becomes
E
(2)
n
=
∑E
m≠D
′
H nm
(0)
n
2
− Em(0)
.
(8.17)
Question: why there is no contribution from the subspace-D? Similarly, the first-order correction
in eigenstate becomes
n (1) =
∑E
′
H mn
(0)
n
−E
Summary of the degenerate perturbation theory:
m∉D
(0)
m
m (0) .
(8.18)
1) Identify degenerate unperturbed eigenkets and construct the g ×g matrix H ′ if the
degeneracy is g .
2) Diagonalize the g ×g matrix by solving the secular equation.
3) Roots of the secular equation are the first-order energy shifts. Eigenkets of the g ×g
matrix are the zero-order kets for Hamiltonian H (Eigenkets of H 0 ).
4) For higher-orders (2nd or higher for energy and 1st or higher for eigenstates), use the
corresponding non-degenerate perturbation theory except excluding all contributions
from the unperturbed degenerate subspace D.
Here we assume H ′ can be diagonalized, so that we can safely apply the second-order
corrections to energy using the non-degenerate perturbation. However, in some very rear cases
diagonalization of the H ′ matrix in the subspace D doesn’t work as we expect. If so, one has to
construct the second-order degenerate perturbation theory to correctly compute En(2) , as is the
case in Problem 2 of HW5.
124
8.2 Linear Stark Effect
A hydrogen atom is subjected to a uniform electric field E , and the interaction Hamiltonian H ′
is given by (Note: here e is the electron charge, i.e.,
H ′ = −eE ⋅ r = −e | E | r cos θ = −e | E | z ,
(8.19)
where we take the z-axis to be along the direction of E, and θ the angle between r and E.
Now we need to evaluate the matrix elements of H ′ using the usual basis nlm (Note: here
we omit spin which doesn’t interact with static electric field.), i.e., n ′l ′m ′ z nlm . The selection
rule for electric dipole transition
⎧
⎪ l′ = l ±1
unless ⎪
⎨
⎪
⎪
⎩ m′ = m
n ′l ′m ′ z nlm = 0
(8.20)
This follows from the Wigner-Eckart theorem and parity consideration.
For the n = 1 level, there is no degeneracy and E1s(1) = −e | E | 100 z 100 = 0 , thus no (1storder) energy shift to the ground state of a hydrogen atom in a uniform electric field.
For the n = 2 level, the degeneracy is four: 2s : 200 ; 2pz : 210 ; 2p± : 21 ± 1 , and
2s
2pz
H′ =
2s 2pz
2p+
2p−
0 M
M* 0
0
0
0
0
0
0
0
0
0
0
0
0
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
2p+
2p−
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
(8.21)
The only non-zero matrix element
M = −e | E | 2s z 2pz = −e | E | ∫ φ*2s (r)r cos(θ)φ2p (r)d 3r .
z
(8.22)
Here denoting a 0 as Bohr radius, we write the eigenfunctions for H atam:
φ2s (r) =
φ2p (r) =
z
1
8πa
3
0
(1 −r / 2a 0 )e
r cos(θ)
32πa
5
0
e
−r/2a0
125
.
−r/2a0
,
(8.23)
(8.24)
The integral in Eq. (8.22) has two
parts. The angular part gives
4π / 3 while the radial part is
−9a 0 / 4π . Combining both
integrals we find that
M = 3ea 0 | E | .
The matrix [Eq. (8.21)] has
block-diagonal form with a 2×2
part in the upper left corner and
two blocks with single elements.
|E|
Fig. 8.1: Stark effect for n = 2 states. Left: Energy levels in nonrelativistic theory; Right: Relativistic corrections are included.
The four eigenvalues are
λ = ±M, 0, 0 , and the corresponding eigenstates are ± =
1
2
( 2s
)
± 2pz , 2p+ and 2p− ,
(1)
(1)
respectively. The energy shifts for these 4 eigenstates are E ±(1) = ±M , E 2p
= 0 and E 2p
= 0,
+
−
as schematically plotted in Figure 8.1, in comparison with relativistic results.
8.3 Quadratic Stark Effect and Polarizability of H Atom
The energy shift for state k of a H atom in an electric filed E is
ΔEk = E
(1)
k
+E
(2)
k
2
2
+  = −e | E | ξk,k +e | E |
∑E
j ≠k
| ξkj |2
(0)
k
− E j(0)
+
(8.25)
Notes: (1) here k and j are collective indices representing (n,l,m) and (n ′, l ′, m ′) ; (2) Matrix
elements ξkj = k z j ; (3) For degenerate case, summation over j excludes the degenerate
states of k and Ek (1) = λk .
The first-order correction to energy gives the linear Stark effect, while the second term the
quadratic effect. Because ξk,k = 0 , the linear Stark effect always comes from energy splitting of
degenerate states. The quadratic Stark effect is much weaker than the linear Stark effect for weak
electric field, except for the 1s state where the linear stark effect is zero.
126
The quadratic Stark effect of an atom is associated with its polarizability α defined in terms
of the energy shift of the atomic state:
1
ΔE = − α | E |2 .
2
(8.26)
Eqs. (8.25) and (8.26) show that for the ground state
∑
α = −2e 2
2
k z 1s
∞
E1s(0) − Ek(0)
k≠(100)
.
(8.27)
Here the sum over k includes not only all bound states with n > 1 but also the continuum states
with positive energy. The summation is very difficult to evaluate!
A simple estimation of the sum in Eq. (8.27) assumes that the denominators in Eq. (8.27)
were a constant, then
∞
∑
k≠(100)
k z 1s
2
∞
2
= ∑ k z 1s
k
∞
= ∑ 1s z k k z 1s = 1s z 2 1s .
(8.28)
k
Since x 2 = y 2 = z 2 = r 2 / 3 , and
φ1s (r) =
1
πa
3
0
e
−r/a0
,
(8.29)
we obtain r 2 = a 02 with a 0 Bohr radius. Denominators in summation Eq. (8.27) are not equal,
and
e2
(1 −1 / 4) .
2a 0
(8.30)
16 3
a 0 ≈ 5.33 a 03 .
3
(8.31)
−E1s(0) + Ek(0) ≥ −E1s(0) + E 2s(0) =
As a result, we obtain the upper limit for polarizability α
α < 2e
2
a 02
1
3 e
(1 −1 / 4)
2a 0
2
=
Question: how would you find a good lower limit for α ?
Experimental measurements give α = 4.5 a 03 , so the crude estimation is not bad! The
remaining part of this section provides an exact derivation of the polarizability of atomic
hydrogen. Hydrogen is the only atom for which the polarizability can be derived exactly. Very
127
accurate numerical values are known for many atoms and ions. The exact result for hydrogen has
been known since 1926.
An important physical point is that there are no rigorous bound states for an atom in an
electric field. Figure 7.2 shows a graph of the potential energy along the z-direction for the
potential energy
e2
(8.32)
−e | E | z .
|z |
A bound electron can escape the barrier from
V(z) = −
point A to B (How to compute its transition rate?)
For small fields, the tunneling rate is
infinitesimally small, but for larger fields it
becomes an important process, thus a strong DC
Fig. 8.2: Potential energy V(z) for an
electron of a H atom in an electric field.
electric field can strip electrons from atoms.
Here we focus on small electric fields. The trick to obtaining the exact result is to avoid
evaluating the direct summation in Eq. (8.27); instead, the first-order change in the eigenfunction
is defined by an inhomogeneous differential equation. The exact solution to this equation gives
an exact result for the polarizability. Similar methods are applied to other atoms, but the
equations must be solved numerically.
The induced polarization in the atom is
Pz = ∫ ψ*1s (r) ez ψ1s (r)d 3r ,
(8.33)
where the perturbed wavefunction for the ground state
ψ1s (r) = φ1s (r) + ψ1s(1)(r) + ψ1s(2)(r) + 
(8.34)
and φ1s (r) is the unperturbed wavefunction, i.e., φ1s (r) = ψ1s(0)(r) . Because
ψ1s(1)(r) = ∑
k≠1s
k (−e | E | z) 1s
E1s − Ek
φk (r) ,
(8.35)
ψ1s(1)(r) ∝ e | E | , and we can easily verify that ψ1s(2)(r) ∝ e 2 | E |2 . Note: We omitted the obvious
superscript (0), i.e., here k = k (0) , 1s = 1s (0) , Ek = Ek(0) , and E1s = E1s(0) .
128
We define
ψ1s(1)(r) = e | E | φ1s(1)(r)
(8.36)
then
2
Pz = e ∫ z φ1s (r) +e | E | φ1s(1)(r) +O(| E |2 ) d 3r
= 2e | E | ∫ φ1s (r)zφ (r) d r +O(| E | ).
2
(1)
1s
3
2
(8.37)
Polarizability α describes the linear relation between polarization and field strength,
α=
Pz
|E|
= 2e 2 ∫ φ1s (r)zφ1s(1)(r) d 3r .
(8.38)
The exact value of α can be obtained if we can find exact solution to φ1s(1)(r) . Operate on both
sides of Eq. (8.35) by (H 0 − E1s ) , we find the equation for φ1s(1)(r) :
k (−e | E | z) 1s
(H 0 − E1s )ψ1s(1)(r) = ∑
E1s − Ek
k≠1s
(Ek − E1s )φk (r)
(8.39)
= e | E | ∑ r k k z 1s = e | E | zφ1s (r).
k≠1s
Using Eq. (8.36), finally we reach the equation
(H 0 − E1s )φ1s(1)(r) = zφ1s (r) ,
(8.40)
⎡ 2
e2
e 2 ⎤⎥ (1)
2
⎢−
∇
−
+
φ (r) = zφ1s (r) .
⎢ 2m
r
2a 0 ⎥⎥⎦ 1s
⎢⎣
Define ρ = r / a 0 , solving Eq. (8.41) one gets
φ1s(1)(r) = φ1s(1)(ρ, θ) =
cos(θ)
2E 0 πa 0
(8.41)
ρ (1 + ρ / 2)e −ρ ,
(8.42)
where E 0 = e 2 / 2a 0 , the Rydberg energy.
Insert this result into the integral (8.38) for the polarizability. The radial integral is put into
dimensionless form:
α=
=
2e 2a 04
2
∫ ρ (1 + ρ / 2)e
4
2 4
0
2(e / 2a 0 ) π a
−2ρ
dρ
∫ cos (θ)dΩ
2
3
0
2a 27 4π 9 3
= a0
π 16 3
2
So the exact theoretical value of α agrees perfectly with the measured value.
129
(8.43)
8.4 Zeeman Effect and Paschen-Back Effect
We discuss H (or H-like) atoms in a weak uniform magnetic field B, and the resulting energy
shifts are called the Zeeman effect. The Hamiltonian

H B = −µ ⋅ B .
 

Here the magnetic moment µ = µL + µS , where
(8.44)
µ

µL = − 0 L ,

µ
µ

µS = − 0 gS ≈ −2 0 S ,


where the Bohr magneton µ0 =
(8.45)
(8.46)
−e
= 5.788×10−5 eV/T with −e = 1.602×10−19 C. Thus
2mec
HB =
µ0

(L + 2S)⋅ B =
µ0B

(Lz + 2Sz ) .
(8.47)
Here we assume a weak uniform magnetic field B is along z-direction with strength B.
⎡ e 2B 2 2
⎤
2 ⎥
Omitting the quadratic term ⎢⎢
(x
+
y
)
2
⎥ , the total Hamiltonian H = H 0 + H LS + H B ,
⎢⎣ 8mc
⎥⎦
where
H0 =
H LS =
p2
p2
e2
+VC (r) =
− ,
2me
2me r
(8.48)
ζ(r)
L ⋅S,
2
(8.49)
with
ζ(r) =
 2 1 dVC (r)
.
2me2c 2 r dr
(8.50)
For an orbital state φi (r) = r i , first-order perturbation theory gives
i H LS i =
ζi
2
L⋅S ,
(8.51)
with spin-orbital energy parameter
ζi = i ζ(r) i = ∫ | φi (r) |2 ζ(r)d 3r .
130
(8.52)
We study the effects of weak magnetic field (energy splitting is smaller than spin-orbital
coupling) using the eigenkets of H 0 + H LS , i.e., the common eigenstates of these four operators:
J2 , Jz, L2, and S2, designated as j,m j ;l,s , instead of those for L2, Lz, S2, and Sz designated as
l,ml ;s,ms = l,ml ⊗ s,ms . Here the total angular momentum J = L + S . Let’s simplify the
notations for a fixed l -value (and s = ½): j,m ≡ j,m j ;l,s and ml ,ms ≡ l,ml ;s,ms .
Energy shift due to spin-orbital coupling
ΔE LS ≈ jm H LS jm =
ζnl

2
jm L ⋅ S jm =
⎤
ζnl ⎡
⎢ j(j + 1)−l(l + 1)− 3 ⎥ .
2 ⎢⎣
4 ⎥⎦
(8.53)
Energy shift due to magnetic field B:
ΔE B ≈ H B =
µ0B

Lz + 2Sz =
µ0B

J z + Sz .
(8.54)
Here J z = m . jm Sz jm is more difficult to evaluate, since jm is NOT an eigenket of
Sz , but ml ,ms is, so we write jm as a linear combination of basis ml ,ms . We know that
j = l ± 1 / 2 (since j =| l −s |, | l −s | +1, , l + s ) and m = m j = ml + ms , thus
j = l ± 1 / 2,m = c+ ml = m −1 / 2,ms = 1 / 2 + c− ml = m + 1 / 2,ms = −1 / 2 , (8.55)
where c+ and c− are determined to be
c+ = ±
l ±m +1/ 2
2l + 1
and
c− =
l m +1/ 2 .
2l + 1
(8.56)
The expectation value of Sz :
Sz = jm Sz jm =

m
.
|c+ |2 − |c− |2 = ±
2
2l + 1
(
)
(8.57)
We obtain Lande’s formula for the energy shift due to magnetic field B:
⎛
1 ⎞⎟ .
⎟⎟
ΔE B ≈ µ0Bm ⎜⎜1 ±
⎜⎝
2l + 1 ⎟⎠
(8.58)
Figures 8.3 & 8.4 show Zeeman effect in the n = 2 multiplet of H atom. The energy splitting
( 3ζ / 2 ) between 2p3/2 and 2p1/2 is 4.5×10−5 eV, compared with µ0 = 5.788×10−5 eV/T
(Question: what is the effective magnetic field generated by spin of an electron?)
131
Fig. 8.3: Fine Structure and Zeeman effect of H atom
for the
states.
Fig. 8.5: p-state energy splitting of
Hg in a magnetic filed.
Figure 8.5 shows that the energy splitting
is linear at both weak (µ0B  ζ) and
strong (µ0B  ζ) magnetic field, the
Paschen-Back effect. Here the magnetic
interaction H B dominates over spinorbital H LS , so that we choose the basis
Fig. 8.4: Energy splitting by spin-orbital
coupling (left) and a weak magnetic field (right).
set ml ,ms ≡ l,ml ;s,ms , which
diagonalizes H B :
H B = µ0B ml ,ms (Lz + 2Sz ) ml ,ms = µ0B (ml + 2ms ) .
(8.59)
Here the original 2(2l + 1) degeneracy in a H atom is now reduced by H B to states with the
same ml + 2ms . Now we need to evaluate L ⋅ S with respect to ml ,ms to find the energy
shift due to H LS :
L ⋅ S = ml ,ms ⎡⎢LzSz + (L+S− + L−S+ ) / 2⎤⎥ ml ,ms =  2mlms .
⎣
⎦
(8.60)
We obtain the total energy shift under a strong magnetic field:
ΔE ≈ E (1)m ,m = µ0B(ml + 2ms ) + ζmlms .
l
s
132
(8.61)
8.5 Time-Dependent Hamiltonian and Quantum Dynamics
8.5.1 Review: Quantum Dynamics With Time-Independent Hamiltonian
Last semester we discussed quantum dynamics for a time-independent Hamiltonian,
H 0 φn = En φn .
(8.62)
ψ(0) = ∑ cn φn ,
(8.63)
At t = 0 , the initial state
n
then at an arbitrary time t > 0 ,
ψ(t) =U(t) ψ(0) = e
−iH 0t/
ψ(0) = ∑ cne
−iEnt/
φn .
n
The probability of finding the system in state φf at time t is φf ψ(t)
2
(8.64)
=|c f |2 , which is time
independent. (Question: if so, how do you know ψ(t) changes by measurement?)
Eq. (8.64) suggests that when t = 0 , if a particle stays at one eigenstate of H 0 , φi , it will
always stay at the same state, i.e., the transition probability to any other eigenstate φf ( f ≠ i )
is zero. What can induce a transition from φi to φf ?
We can add a time-independent perturbation H ′ so that the matrix element H if′ = i H ′ f
is not zero. The total Hamiltonian H = H 0 + H ′ , whose eigenstates Φn can be obtained
exactly using the matrix method or approximately using perturbation theory. We denote
H Φn = εn Φn ,
(8.65)
ψ(0) = ∑ cn Φn ,
(8.66)
and Φn = ∑ dn,m φm . When t = 0 ,
m
n
then at an arbitrary time t > 0 ,
ψ(t) =U(t) ψ(0) = e −iHt/ ψ(0) = ∑ cne
n
−iεnt/
Φn .
(8.67)
One can easily figure out the probability for transition from φi (at t = 0 ) to φf at time t
with i ≠ f . The key is to expand φi and φf as linear combinations of
133
{ Φ }.
n
Considering a weak perturbation H ′ , one can use the time-independent perturbation theory
to find the Transition Probability: Pi→f (t) = φf ψ(t)
2
for ψ(0) = φi . Using the first-order
perturbation theory,
φn ≈ Φn + ∑
m≠n
′
H mn
En − Em
Φm
(8.68)
one can obtain
Pi→f (t) =
H fi′
Ei − E f
(e
iω fit
)
2
−1 ,
(8.69)
where ω fi = (E f − Ei ) /  .
8.5.2 The Fundamental Equation for Time-Dependent Hamiltonian
So far we have only considered time-independent Hamiltonian; however, there are many
quantum-mechanical systems of importance with time dependence. For example, a laser with
angular frequency ω , which generates an electric field
E = E0 cos(ωt) .
(8.70)
H = H 0 + H ′(t) .
(8.71)
Here we consider a Hamiltonian
At t = 0 , the initial state ψ(0) = ∑ cn (0) φn , then what is ψ(t) at an arbitrary time t > 0 ?
n
Inspired by Eq. (8.64), we can write down
ψ(t) = ∑ cn (t)e
−iEnt/
n
φn ,
(8.72)
and cn (t) can be determined by solving the time-dependent Schrödinger Equation
i
∂
ψ(t) = H ψ(t) .
∂t
(8.73)
Plugging Eq. (8.72) into (8.73), one obtains the equations for cn (t) :
i
∂
∂t
⇒
⎡
⎤
⎡
⎤
⎢ ∑ cm (t)e −iEmt/ φm ⎥ = ⎡⎢H 0 + H ′(t)⎤⎥ ⎢ ∑ cm (t)e −iEmt/ φm ⎥
⎦⎢ m
⎢⎣ m
⎥⎦ ⎣
⎥⎦
⎣
i ∑ cm (t)e
m
−iEmt/
φm = ∑ cm (t)H ′(t)e
m
134
−iEmt/
φm .
(8.74)
(8.75)
Multiplying both sides of Eq. (8.75) by φn on the left, we reach the fundamental equation:
′ (t)e iωnmtcm (t) ,
icn (t) = ∑ H nm
(8.76)
m
where cn (t) = dcn (t) / dt and ωnm = (En − Em ) /  . Explicitly in matrix form:
⎡
⎢
⎢
⎢
i ⎢
⎢
⎢
⎢
⎢⎣
⎡
c1 ⎤⎥ ⎢ H 11′
⎥ ⎢
′ e iω21t
c2 ⎥ ⎢ H 21
⎢
=
⎥
c3 ⎥⎥ ⎢⎢ H ′ e iω31t
⎢ 31
⎥
 ⎥⎦ ⎢⎣

H 12′ e
iω12t
′
H 22
′e
H 32
H 13′ e
iω13t
′e
H 23
iω23t
iω32t
′
H 33


⎤
 ⎥ ⎡⎢ c1
⎥⎢
 ⎥⎥ ⎢ c2
⎢
⎥
 ⎥ ⎢⎢ c3
⎥⎢
 ⎥⎦ ⎢⎣ 
⎤
⎥
⎥
⎥
⎥.
⎥
⎥
⎥
⎥⎦
(8.77)
This is the coupled differential equation for obtaining the probability of finding state φn at time
t , given the initial condition of ψ(0) = ∑ cm (0) φm .
m
8.5.3 Dynamics of Two-Level Systems: Rabi Oscillation
You can imagine that exact solution to Eq. (8.76) [or (8.77)] is extremely difficult, and in most
cases we have to use the time-dependent perturbation theory, as we’ll discuss later on. There is
an exactly solvable problem of enormous practical importance, namely a two-level system with a
sinusoidal oscillating H ′ . Specifically,
⎡ E
0 ⎤⎥
⎢
H0 = ⎢ 1
⎥,
⎢ 0 E2 ⎥
⎣
⎦
with eigenstates 1 and 2 . Without losing generality, we assume E 2 > E1 . And
⎡
0
H ′(t) = ⎢⎢
−iωt
⎢⎣ γe
⎤
γe iωt ⎥
⎥,
0 ⎥⎦
(8.78)
(8.79)
where constants γ and ω are real and positive. Question: what realistic systems and problems
can be modeled by this quantum system?
Initially at t = 0 , the system stays in the lower level of H 0 , i.e., ψ(0) = 1 , so that
c1(0) = 1 and c2(0) = 0 . Then what are the probabilities of finding the system staying in the
lower level and higher level at time t > 0 , respectively? Obviously, we need to determine c1(t)
and c2(t) by solving Eq. (8.76), and then the probabilities are |c1(t) |2 and |c2(t) |2 , respectively.
135
I. I. Rabi solved this problem in 1927, so the solutions are called Rabi’s formula:
|c1(t) |2 = 1− |c2(t) |2
|c2(t) |2 =
⎡
⎤
4γ 2
2 ⎢ t
2
2
2⎥
sin
4γ
+

(ω
−
ω
)
,
21 ⎥
⎢ 2
4γ 2 +  2(ω − ω21 )2
⎣
⎦
(8.80)
where ω21 = (E 2 − E1 ) /  . Define the oscillatory frequency
Ω=
1
4γ 2 +  2(ω − ω21 )2 ,
2
(8.81)
we study the most important case of resonant oscillation when ω ≈ ω21 , i.e., ω ≈ E 2 − E1 .
At the exact resonance, ω = ω21 , Ω = γ /  , c1(t) = cos(Ωt) and c2(t) = sin(Ωt) , as plotted
in Figure 7.6. From t = 0 to t = π / 2γ , the two-level system absorbs energy, |c1(t) |2
decreases and |c2(t) |2 increases. At t = π / 2γ only the upper level is occupied. Then from
t = π / 2γ to t = π / γ , the system loses (emits) energy, |c1(t) |2 increases and |c2(t) |2
decreases until t = π / γ , when only the lower level is occupied. This absorption-emission
cycle (e.g., optical absorption and emission) is repeated indefinitely.
When ω is away from resonance, the absorption-emission cycle still takes place, but the
maximum of |c2(t) |2 is less than 1 and |c2(t) |2 doesn’t go all the way to 0. Figure 8.7 plots
|c2(t) |2 max as a function of ω , which has a resonance peak of 1 at ω = ω21 and its full width at
half maximum is 4γ /  . Question: Figure 8.7 reveals what fundamental principle in quantum
theory?
Fig. 8.6: Resonance absorption-emission cycle.
136
Fig. 8.7: Maximum transition
probability as a function of .
8.6 Time-Dependent Perturbation Theory
8.6.1 The General Theory
Exact solutions to the fundamental equation (8.76) are impossible in most realistic situations. As
in the case of time-independent problem, we can obtain approximate (and exact) solutions
through perturbation methods.
Let’s specify the problem. H = H 0 + H ′(t) , with H 0 φn = En φn . Initial condition:
ψ(0) = φi , i.e., cn (0) = δni . What is ψ(t) = ∑ cn (t)e
−iEnt/
φn ? Here we need to determine
n
cn (t) . Rewriting the exact quantum dynamics equation,
′ (t)e
icn (t) = ∑ H nm
iωnmt
cm (t) ,
m
(8.82)
we want to obtain cn (t) through perturbation expansion,
cn (t) = cn(0)(t) +cn(1)(t) +cn(2)(t) + 
(8.83)
We start from the zeroth-order by ignoring H ′ :
ic(0)
(t) = 0 ⇒ c (0)
(t) = c (0)
(0) = 0
fi
fi
fi
(for i ≠ f ) .
(8.84)
Here we use two subscripts i and f to denote that the system starts from φi at t = 0 , and the
probability of finding the system stays in state φf , which is different from the initial state (We
are interested in transition), is
Pi→f (t) =|c fi (t) |2 =|c (1)
(t) +c (2)
(t) + |2 .
fi
fi
(8.85)
Following exactly the same procedure for time-independent perturbation theory, we plug the
zeroth-order solutions into Eq. (8.82) to obtain the first-order equation:
ic(1)
(t) = H fi′ (t)e
fi
iω fit
c (1)
(t) = −
fi
⇒
i t
iω t ′
H fi′ (t ′)e fi dt ′
∫
 0
(8.86)
Plugging the first-order solutions into Eq. (8.82), the second-order solutions are obtained:
ic(2)
(t) = H fi′ (t)e
fi
iω fit (1)
fi
c (t) ⇒
2
⎛ i⎞
c (t) = ⎜⎜− ⎟⎟⎟
⎜⎝  ⎟⎠
(2)
fi
t
∑∫ ∫
n
0
t′
0
H fn′ (t ′)e
137
iω fnt ′
H ni′ (t ′′)e
iωnit ′′
dt ′′ dt ′
(8.87)
Keeping doing this leads to the nth-order solutions:
ic(n)
(t) = H fi′ (t)e
fi
iω fit (n−1)
fi
c
(t) ⇒
n
⎛ i⎞
c (t) = ⎜⎜⎜− ⎟⎟⎟
⎝  ⎟⎠
(n)
fi
∑ ∫
m1 ,m2 ,.mn−1
t
0
t1
dt1 ∫ dt2  ∫
0
tn−1
0
dtn
(8.88)
iω fm t1
iωm m t2
iωm i tn ⎤
⎡ ′
⎢H fm1 (t1 )e 1 H m′ 1m2 (t2 )e 1 2 H m′ n−1i (t2 )e n−1 ⎥ .
⎣
⎦
This is called the Dyson’s series. Last semester we discussed it in Quantum Dynamics [Eq. (3.6)
on page 37], and the solution for the time-evolution operator U(t,t0 ) to the time-dependent
Schrödinger equation,
i
∂
U(t,t0 ) = H(t)U(t,t0 ) ,
∂t
(8.89)
is
∞
n
U(t,t0 ) = 1 + ∑ (−i /  )
n=1
∫
t
t0
t1
dt1 ∫ dt2  ∫
t0
tn−1
t0
dtnH(t1 )H(t2 )H(tn ).
(8.90)
We can readily generalize Eq. (8.86) for an arbitrary initial state, ψ(0) = ∑ cn (0) φn , then
n
c (0)
(t) = c f (0) , and the first-order perturbation:
f
c (1)
(t) = −
f
t
2
i
cn (0) ∫ H ′fn (t ′)exp(iω fnt ′)dt ′
∑
0
 n≠f
(8.91)
In this class we mostly just consider the first-order time-dependent perturbation, which is a
(t) | 1 (Question: Why?), i.e., it is valid for small transition
good approximation if |c (1)
fi
(t) | 1 , the time
probability when the initial state has not been significantly depleted. When |c (1)
fi
integral
∫
t
0
H fi′ (t ′)e
iω fit ′
dt ′   . Since the integral depends on the action time t , its relation to
the characteristic time tc   / E f − Ei defines three most important cases (limits):
(1) t  tc , the adiabatic limit;
(2) t  tc , the sudden perturbation;
(3) t ~ tc , steady and weak perturbation → Fermi’s Golden Rule for Transition Rate.
138
8.6.2 The Adiabatic Limit
Adiabatic limit is when the perturbation is very long (slow) compared to tc . An example is when
the alpha-particle goes by the atom very slowly. The important feature of the adiabatic limit is
that the system evolves very slowly in time. The atom can slowly polarize, and depolarize, as the
alpha goes by, but the electrons never change their atomic states.
If t  tc , one can switch on from t = −∞ until t = 0 , and time-dependent perturbation
reduces to time-independent perturbation. The Hamiltonian can be written as
H(t) = H 0 +e t/τ H ′ ,
(8.92)
for −∞ < t ≤ 0 and τ > 0 . The adiabatic limit is τ → ∞ , and we can show that
lim τ→∞ c (1)
(t = 0) = −
fi
H ′fi
i 0
t ′/τ
′
′
′
H
e
exp(iω
t
)d
t
→
,
fi
! ∫−∞ fi
Ei − E f
(8.93)
which is exactly the same as that from the first-order time-independent perturbation. When
H 0 → H 0 + H ′ adiabatically, φn → ψn , with H 0φn = En(0)φn and H ψn = En ψn , and the firstorder perturbation in wavefunction: ψn(1) = ∑
m≠n
′
H mn
E
(0)
n
− Em(0)
φm . However, when E f = Ei ,
tc → ∞ , it is impossible to adiabatically switch on.
If a system is switched on and off adiabatically, we expect the system will go back to the
original state before the perturbation. Example: A 1D charged simple harmonic oscillator (
H 0 = p 2 / 2m + mω 2x 2 / 2 ) in the ground state at t = −∞ , with H ′(t) = −eεx exp(−t 2 / τ 2 ) .
Question: what is the probability that the oscillator is in the state n ( n ≠ 0 ) at time t = ∞ ?
Solution: using the first-order time-dependent perturbation theory [Eq. (8.86)]
cn(1)0 (∞) = −
2
2
i ∞
−eε) n x 0 e −t /τ exp(inωt)dt .
(
∫
 −∞
Here the operator x = x 0(a + a † ), where x 0 =
(8.94)

, and n x 0 ≠ 0 only when n = 1 .
2mω
Therefore, cn 0(t) = 0 when n > 1 . For n = 1 , Eq. (8.94) becomes
(1)
c1,0
(∞) =
ieεx 0

∫
∞
−∞
e −t
2
/τ 2
exp(iωt)dt =
139
ieεx 0 π

(
)
τ exp −ω 2τ 2 / 4 .
(8.95)
The probability that the oscillator is in n-th state ( n ≠ 0 ) at time t = ∞ is 0 except for n = 1:
2
P0→1(∞) = cn(1)(∞) =
πe 2ε2 2
τ exp −ω 2τ 2 / 2 .
2mω
(
)
(8.96)
(1) τ → ∞ , adiabatically switch on and off, P0→1(∞) = 0 ; the SHO goes back to the ground state.
(2) τ → 0 , suddenly switch on-and-off, P0→1(∞) = 0 ; Why no transition?
(3) τ =
1
πe 2ε2 e −1/2
, the maximum transition probability P0→1(∞) =
.
2mω ω 2
ω
8.6.3 Sudden Perturbation
Sudden approximation is when the perturbation is very short compared to tc. An example is when
the alpha-particle goes rapidly by the atom. Then it is very likely that some electrons in the atom
are excited to other orbital states.
When t < 0, H = H i ; t > 0 , H = H f . Here both H i and H f are time-independent:
H = H i + θ(t)(H f − H i ) ,
⎧1
⎪
⎪
⎪
⎩0
where the step function θ(t) = ⎪⎨
(8.97)
(t > 0)
. When t < 0, ψi (t) ; t > 0, ψf (t) . Continuity at
(t < 0)
t = 0 : ψi (0) = ψf (0) if H f and H i are both finite, because
ψf (τ / 2)− ψi (−τ / 2) = −
i τ/2
H(t)ψ(t)dt → 0 when τ → 0 .
 ∫−τ/2
(8.98)
For time-independent H i and H f : H i φn(i ) = En(i )φn(i ), and H f φn(f ) = En(f )φn(f ) , and the time
evolution:
ψi (t) = ∑ an exp(−En(i )t / )φn(i )
and
n
ψf (t) = ∑ bn exp(−En(f )t / )φn(f ) .
(8.99)
n
Continuity ψi (0) = ψf (0) ⇒
∑a φ
n
(i )
n n
= ∑ bmφm(f )
m
⇒ bm = ∑ an φm(f ) φn(i )
(8.100)
n
Therefore, if a particle initially stays at φn(i ) , then bm,n = φm(f ) φn(i ) , i.e., the probability of
finding the particle stays in state of φm(f ) is
2
Pn→m = bm,n = φm(f ) φn(i )
140
2
(8.101)
Example: A one-electron H-like atom is in its ground state, whose nuclear charge changes
suddenly from Z to Z + 1 (the β decay). Question: what is the probability that it is still in the
ground state after the β decay?
Solution: Since the change in potential has the spherical symmetry, the angular wave
functions won’t change, i.e., they are still described by the quantum numbers l and m. Thus only
radial wave functions need to be considered:
3/2
⎛ Z ⎞⎟
R1s (r) = 2 ⎜⎜⎜ ⎟⎟
⎝⎜a 0 ⎟⎠⎟
⎛ Zr ⎞⎟
exp ⎜⎜⎜− ⎟⎟ .
⎜⎝ a 0 ⎟⎟⎠
(8.102)
Eq. (8.100) suggests that probability of remaining in the ground state:
P = R1sZ+1 R1sZ
3/2
Z +1
1s
R
Z
1s
R
⎛ Z ⎞⎟
= 2 ⎜⎜⎜ ⎟⎟
⎜⎝a 0 ⎟⎟⎠
2
3/2
⎛ Z + 1 ⎞⎟
⎜⎜
⎟
⎜⎜ a ⎟⎟⎟
⎝ 0 ⎠
∫
∞
0
2
.
3/2
⎛ Zr ⎞⎟
⎛ (Z + 1)r ⎞⎟
⎡
⎤
1
2
⎜
⎜
⎥
⎟⎟ r dr = ⎢1 −
exp ⎜⎜− ⎟⎟ exp ⎜⎜−
2⎥
⎢
⎜⎝ a 0 ⎟⎟⎠
⎜⎝
a 0 ⎟⎟⎠
(2Z
+
1)
⎣
⎦
(8.103)
⇒
3
⎡
⎤
1
⎥ .
P = ⎢⎢1 −
2⎥
(2Z
+
1)
⎣
⎦
3
(8.104)
3
⎛8⎞
⎛ 24 ⎞
For Z = 1 , P = ⎜⎜⎜ ⎟⎟⎟ = 0.702 ; for Z = 2 , P = ⎜⎜⎜ ⎟⎟⎟ = 0.885 .
⎝ 9 ⎟⎠
⎝ 25 ⎟⎠
8.6.4 Weak and Steady Perturbation Over Time Period t ~ tc : Fermi’s Golden Rule
Golden Rule is when the perturbation is weak but steady. The perturbation can cause the atom or
particle to change its state. The rate of change is such a useful formula that Fermi called it the
‘‘Golden Rule.’’ One example: classical electromagnetic field E(t) = E 0 exp[i(kx − ωt + φ)] .
In this case,
H ′(t) = H ′ exp(−iωt) ,
(8.105)
the first-order TD perturbation theory leads to
c (1)
(t) = −
fi
i t
′
H ′fie −iωt exp(iω fit ′)dt ′ ,
∫
0
!
where H fi′ = φf H ′ φi and ω fi = E f − Ei .
141
(8.106)
i(ω −ω)t
t i(ω −ω)t ′
i
i
e fi
−1
fi
′
′
′
c (1)
(t)
=
−
H
e
d
t
=
−
H
fi
fi ∫0
fi


i(ω fi − ω)
=
=
H fi′ 1 − cos[(ω fi − ω)t ]−i sin[(ω fi − ω)t ]
ω fi − ω

H fi′ 2sin[(ω fi − ω)t / 2]
ω fi − ω

e
i[(ω fi −ω)t−π ]/2
Set Ω = ω fi − ω = ω f − ωi − ω , we obtain
2
Fig. 8.8: A plot of
2 ⎡ sin(Ωt / 2) ⎤
1
⎥ .
Pi→j (t) = c (t) = 2 H fi′ ⎢
⎢ Ω/2 ⎥

⎣
⎦
(1)
fi
2
Figure 8.8 shows that when Ω → 0 , Pi→j (t) has the maximum value of
.
(8.107)
2
1
′
H
t 2 for a
fi
2

fixed t ; and it decreases rapidly as Ω increases. We define transition rate
Ri→f (t) ≡
2 2sin(Ωt)
d
1
.
Pi→f (t) = 2 H fi′
dt
Ω

(8.108)
The above rate is time-dependent; however, we really care about the rate for t → ∞ :
Ri→f ≡ limt→∞ Ri→f (t) =
2
2
1
2sin(Ωt)
1
H fi′ limt→∞
= 2 H fi′ 2πδ(Ω)
2
Ω


Ri→f =
⇒
2
2π
H fi′ δ(E f − Ei − ω)

(8.109)
This is the famous Fermi’s Golden Rule (FGR), though Dirac should get the full credit.
(1) The δ function enforces energy conservation.
(2) The transition rate R is NOT infinite as apparently suggested by FGR. In all the practical
applications, the function will be integrated over some parameters to give finite rate.
(3) One-to-many transition: φf is degenerate: Ri→f =
2
2π
H fi′ δ(E f − Ei − ω)
∑
 f ≠i
(4) If final states are continuous with density of states ρ(E f ) , which is the number of states per
unit energy at E f , FGR becomes Ri→f =
Ri→f =
2
2π
H fi′ δ(E f − Ei − ω)ρ(E)dE ⇒
∫

2
2π
H fi′ ρ(E f ) , with E f = Ei + ω .

(5) When ω = 0 , FGR gives the scattering rate between two states of the same energy.
142
(8.110)
8.7 Photoelectric Effect in Hydrogen Atom
A realistic example of employing FGR is photoelectric effect. A hydrogen atom is in its ground
state, and the incident electromagnetic wave is described by
A(r,t) = A0 cos(k ⋅ r − ωt) .
(8.111)
If ω is high enough, the electron can be knocked out to be free. Question: what is the
differential cross section of the photoelectric effect in H atom?
We use FGR, Ri→f =
2
2π
H fi′ δ(E f − Ei − ω) , and we need to

(1) evaluate H fi′ = f H ′ i and (2) treat the δ -function. Here the initial state is i = 100 ,
and φ1s (r) = r 100 =
1
exp(−r / a 0 ) , the ground state of H atom. The final state:
πa 03
f = p f , a free particle with momentum p f , whose wavefunction is,
φf (r) = r p f =
1
(8.112)
exp(ip f ⋅ r / !) .
V
Note: Here we use the big-box normalization, and the volume of the big box is V. See page 47 in
QM I lecture notes. This is often more convenient.
Next we examine the perturbation Hamiltonian (m and e are the electron mass and charge.)
H ′(t) = −
e
e
A⋅p = −
cos(k ⋅ r − ωt)A0 ⋅ p
mc
mc
e ⎡ i(k⋅r−ωt )
=−
e
+e −i(k⋅r−ωt ) ⎤⎥ A0 ⋅ p ,
⎢
⎣
⎦
2mc
(8.113)
where e i(k⋅r−ωt ) corresponds to light absorption, while e −i(k⋅r−ωt ) corresponds to light emission.
Since emission is irrelevant to the photoelectric effect,
H ′(t) = −
e ik⋅r
e A0 ⋅ pe −iωt ≡ H ′e −iωt .
2mc
(8.114)
The time-independent part of the perturbation matrix elements
H fi′ = N ∫ e
where N ≡ −
e 1
2mc V
1
πa 03
−ip f ⋅r/ ik⋅r
e A0 ⋅(−i∇) e
.
143
−r/a0
d 3r ,
(8.115)
In order to simplify the evaluation of the matrix element in Eq. (8.115), we employ the
electric dipole approximation. Since the wave length ( λ ) of incident wave is much larger than
Bohr radius a 0 , ka 0  1 ⇒ e ik⋅r ≈ 1 . Physically, the H atom sees a homogeneous
electromagnetic wave. Then
H ′(t) = −
e
iωe
A0 ⋅ pe −iωt , which is equivalent to −
A0 ⋅ re −iωt .
2mc
2c
(8.116)
i
p2
+V(r) ⇒ ⎡⎢ r,H 0 ⎤⎥ = i∇pH 0 = p ⇒
Proof: H 0 =
⎣
⎦
m
2m
p=
m⎡
r,H 0 ⎤⎥ .
⎦
i ⎣⎢
(8.117)
Then the relevant matrix elements
m
m
f ⎡⎢ r,H 0 ⎤⎥ i =
f (rH 0 − H 0 r) i
⎣
⎦
i
i
m
= (Ei − E f ) f r i = imω f r i .
i
f pi =
H ′(t) = −
(8.118)
iωe
A0 ⋅ re −iωt is often called the electric dipole approximation, because electric
2c
field
E=−
and electric dipole moment
1 ∂A iω
= A0e −iωt ,
c ∂t
2c
(8.119)
d = er ,
iωe
H ′(t) = −d ⋅ E = −
A0 ⋅ re −iωt .
2c
(8.120)
(8.121)
We can use either A0 ⋅ p or A0 ⋅ r to evaluate H fi′ . Here we use the former and integrate by parts,
H fi′ = N ∫ e
I ≡∫e
−ip f ⋅r/
e
−ip f ⋅r/
dr = ∫
−r/a0
∞
0
A0 ⋅(−i∇) e
π
∫ ∫
0
dr = NA0 ⋅ p f ∫ e
−r/a0
−ip f ⋅r/
e
−r/a0
dr .
2π −ip r cos(θ)/
−r/a0 2
f
0
e
e
r dr sin(θ)dθdφ
−r/a
⎛e ipf r/ −e −ipf r/ ⎞⎟
− ∞ ipf r/
∂e 0
−ip f r/
⎜⎜
−r/a0 2
⎟
= 2π ∫ ⎜
r dr = 2π
⎟⎟e
∫0 (e −e ) ∂(1 / a ) dr
0 ⎜
⎟
ip
r
/

ip
⎝
⎠
f
f
0
∞ ip r/
2πi
∂
8π
−ip r/ −r/a
=
(e f −e f )e 0 dr =
.
∫
p f ∂(1 / a 0 ) 0
a 0[(1 / a 0 )2 + (p f / )2 ]2
∞
144
(8.122)
We compute the photoelectric transition rate using
FGR: Ri→f =
2
2π 2 2
N I A0 ⋅ p f δ(E f − Ei − ω) .

This corresponds to the probability of finding an
emitted electron with momentum exactly equal to p f ,
which is of little practical interest. How can you
determine an exact momentum?!
Instead, we want to know the probability of finding
emitted electrons in a finite solid angle dΩ with
Fig. 8.9: Photoelectric Effect.
momentum in the range of p f and p f + dp f , as shown
in Figure 8.9, which is proportional to differential cross section (dσ / dΩ) and thus can be
measured experimentally. There are many degenerate states in a finite solid angle dΩ with
momentum in the range of p f and p f + dp f , and we use the density of states ρ(E) to replace the
δ -function:
Ri→dΩ =
where ρ(E) =
2
2π 2 2
N I A0 ⋅ p f ρdΩ (E) ,

(8.123)
dN
(Note: this is different from what on Page 48 in QM I notes, why?).
dE
Consider a free particle in a big cubic box with lattice L , ki L = 2πni , with i = x,y,z .
L3
V
dN = dnxdnydnz =
dk =
k 2dkdΩ
3
3
(2π)
(2π)
E =  2k 2 / 2m ⇒ dE =
2
kdk ⇒
m
dN
mkV
mpV
ρdΩ (E) =
= 2
dΩ = 3
dΩ .
3
dE
 (2π)
 (2π)3
(8.124)
(8.125)
The transition rate
Ri→dΩ =
=
2 mp V
2π 2 2
f
N I A0 ⋅ p f
dΩ
3

 (2π)3
4a 03e 2p f A0 ⋅ p f
2
π 4mc 2[1 + (p f a 0 / )2 ]4
dΩ,
with p f = 2m(Ei + ω) . Here Ei is energy for the initial state (ground state).
145
(8.126)
Transition rate Ri→dΩ depends on the angle between the polarization field A0 and the
outgoing electron’s momentum p f . Using the coordinate system plotted in Figure 8.9, we write
A0 ⋅ p f = A0 p f cos θ . Then the total ionization rate Rionization = Ri→all is obtained by integrating
Ri→dΩ over all solid angles:
Ri→all = ∫
=
=
4a 03e 2p f A0 ⋅ p f
2
π 4mc 2[1 + (p f a 0 / )2 ]4
4a 03e 2p f3A02
π 4mc 2[1 + (p f a 0 / )2 ]4
16a 03e 2p f3A02
3 4mc 2[1 + (p f a 0 / )2 ]4
dΩ
∫∫ cos
2
θd(cos θ)dφ
(8.127)
.
Each ionization absorbs energy ω , so the energy absorption rate is
dE abs
dt
= ω ⋅Ri→all .
(8.128)
The incident light beam’s energy flux density (the Poynting vector, the rate of energy transfer
per unit area) is
S=
c
E×B ,
4π
(8.129)
where
E=−
1 ∂A
ω
= − A0 sin(k ⋅ r − ωt) ,
c ∂t
c
(8.130)
B = ∇×A = −(k ×A0 )sin(k ⋅ r − ωt) ,
(8.131)
c
ω2 2 2
E×B =
A sin (k ⋅ r − ωt) k̂
4π
4πc 0
(8.132)
thus
S=
The time average over a cycle is
ω2 2
S=
A .
8πc 0
(8.133)
The total absorption cross section σ of an atom is defined as a perfectly absorbing disk of
area σ , which will absorbs energy at the rate
dE abs
dt
= σ ⋅S = σ
146
ω2 2
A .
8πc 0
(8.134)
Comparing Eqs. (8.128) and (8.134), we obtain the (total) photoelectric cross section
σ=
=
ω ⋅Ri→all
A02ω 2 / (8πc)
128πa 03e 2p f3
3mcω 3[1 + (p f a 0 / )2 ]4
(8.135)
.
Similarly, we associate a differential cross section dσ / dΩ , with energy (outgoing electrons)
flowing into a solid angle dΩ :
ω ⋅R
dσ
1
= 2 2 i→dΩ
dΩ A0 ω / (8πc) dΩ
=
32a 03e 2p f3 cos2 θ
mcω 3[1 + (p f a 0 / )2 ]4
147
(8.136)
.