p. 54
MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II�
Instructor: Prof. Andrei Tokmakoff
Interaction of Light with Matter
We want to derive a Hamiltonian that we can use to describe the interaction of an electromagnetic field with charged particles: Electric Dipole Hamiltonian. Semiclassical: matter treated quantum mechanically Field: classical Brief outline of electrodynamics: See nonlecture handout. Also, see Jackson, Classical Electrodynamics, or Cohen-Tannoudji, et al., Appendix III. > Maxwell’s Equations describe electric and magnetic fields (E, B ).
> For Hamiltonian, we require a potential.
> To construct a potential representation of E and B , you need a vector potential A (r , t
) and a
scalar potential ϕ (F,t ).
>
A and ϕ are mathematical constructs that can be written in various representations (gauges).
We choose a gauge such that ϕ = 0 (Coulomb gauge) which leads to plane-wave description of E
and B :
∂ 2 A (r ,t )
−∇ A (r,t ) + ∈0 µ0
=0
∂t
2
∇⋅ A = 0
This wave equation allows the vector potential to be written as a set of plane waves: i (k ⋅r −ω t )
ˆe
A (r , t ) = A0 ∈
ˆe
+ A*0 ∈
−i (k ⋅r −ω t )
(oscillates as cos ωt)
ˆ 0 ⇒ k⊥∈
ˆ where ∈
ˆ is the polarization direction of the vector potential.
since ∇ ⋅ A = 0, k ⋅ ∈=
E=−
∂A
i k ⋅ r −ωt )
ˆe(
= iωA 0 ∈
+ c.c.
∂t
B = ∇× A = i ( k× ∈) A 0 e
b̂ k
i ( k ⋅ r −ωt )
+ c.c
(oscillates as sin ωt)
p. 55
ˆ ⊥ n̂
so we see that k̂ ⊥ ∈
ˆ is the direction of the electric field polarization and ∈
n̂ is the direction of the magnetic field polarization. ˆ
∈
b̂
k
We define
1
2 E0
1
2 B0
= iωA0
= i k A0
(
E0
B0
=
ˆ
E ( r , t ) = E 0 ∈sin
( k ⋅ r − ωt )
B ( r , t ) = B0 bˆ sin ( k ⋅ r − ωt )
ω
k
=c
)
p. 56
Hamiltonian for radiation field interacting with charged particle
We will derive a Lagrangian for charged particle in field, then use it to determine classical
Hamiltonian, then replace classical operators with quantum.
Start with Lorentz force on a charged particle:
F = q ( E + v × B)
(1)
where r is the velocity. In one direction (x ) , we have:
Fx = q ( E x + yBz − zBy )
(2)
The generalized force for the components of the force in the x direction in Lagrangian Mechanics
is:
Fx = −
∂U d  ∂U 
+ 

∂x dt  ∂x 
(3)
U is the potential. Using our relationships for E and B in terms of A and ϕ in eq. (2) and
working it into the form of eq. (3), we can show that:
U = qϕ − q r ⋅A
(4)
See CTDL, app. III, p. 1492. Confirm by plugging into (3).
Now we can write a Lagrangian
L = T −U
= 12 mr 2 + q r ⋅A − qϕ
(5)
Now the Hamiltonian is related to the Lagrangian at:
H = p ⋅ r − L
= p ⋅ r − 12 m r 2 − q r ⋅A − q
ϕ
p=
∂L
= m r + qA ⇒
∂r
r=
1
m
(6)
( p − qA )
Now substituting (7) into (6), we have:
(7)
p. 57
p ⋅ (p − qA )−
H=
1
m
H=
1
2m
1
2m
(p − qA )2 − mq (p − qA )⋅ A + qϕ
[p − qA (r , t )]2 + qϕ (r , t )
This is the classical Hamiltonian for a particle of charge q in an electromagnetic field. So, in the
Coulomb gauge (ϕ = 0) , we have the Hamiltonian for a collection of particles in the absence of a
field:

 p2
H0 = ∑  i + V0 (ri )

i  2mi
and in the presence of the field:
H=∑
i
(
1
2mi
( p − q A ( r ))
i
i
i
2
+ V0 ( ri )
)
Expanding:
H = H0 − ∑
i
qi
( pi ⋅A + A ⋅ pi
) +
2mi
qi
∑ 2m
i
2
A
i
Generally the last term is considered small—energy of particles high relative to amplitude of
potential—so we have:
H = H0 + V (t )
V (t ) = ∑
i
qi
(pi ⋅A + A ⋅ pi )
2mi
Now we are in a position to substitute the quantum mechanical momentum for the classical:
p = −i ∇
V (t ) = ∑
i
Matter: Quantum; Field (A): Classical
i
qi (∇i
⋅A + A ⋅∇i )
2mi
Notice ∇ ⋅ A = (∇ ⋅ A ) + A ⋅ ∇ (chain rule), but we are in the Coulomb gauge (∇ ⋅ A = 0), so
∇ ⋅ A = A ⋅∇
p. 58
V (t ) = ∑
i
i qi
A ⋅∇ i
mi
= −∑
i
qi
A ⋅ pi
mi
For a single charge particle our interaction Hamiltonian is
V (t ) =
−q.
A⋅p
m
Using our plane-wave description of the vector potential:
V (t ) =
−q
m
i(k ⋅r −ωt )
 A ∈⋅
ˆ
+ c.c.
 0 p e

Electric Dipole Approximation
If the wavelength of the field is much larger than the molecular dimension (λ → ∞ )( k → 0), then
e ik ⋅ r →1. If r0 is the center of mass of a molecule: eik ⋅r = e ik ⋅r eik ⋅(r − r )
i
0
i
0
= e ik ⋅r0 [1 + ik ⋅(ri − r0 ) + … ]
For UV, visible, infrared—not X-ray— k ri − r0 <<1, set r0 = 0
ei k ⋅ r →1 .
We do retain higher-order terms to describe higher order interactions with the field.
Retain second term for quadrupole transition moment: charge distribution interacting with
gradient of electric field and magnetic dipole.
p. 59
Electric Dipole Hamiltonian
[
Using A0 =
]
−q
ˆ p e− iωt + c.c.
A0 ∈⋅
m
V (t ) =
iE0
2ω
−iqE
0
ˆ p e − iωt − ∈⋅
ˆ p e+ iωt 
∈⋅
2mω

V(t) =
−qE
0
ˆ p ) sin ωt
(∈⋅
mω
−q
=
( E(t) ⋅ p )
mω
V(t) =
Electric Dipole Hamiltonian
or for a collection of charge particles (molecules):

E
q
ˆ i )  0 sin ωt
V ( t ) = −  ∑ i (∈⋅p

ω
 i mi
Harmonic Perturbation: Matrix Elements
For a perturbation V (t ) = V0 sin ωt the rate of transitions induced by field is
wk =
π
2
Vk
2
[δ (Ek − E
− ω ) + δ (Ek − E + ω )]
Let’s look at the matrix elements for the E.D.H.
Vk = k V0
=
qE0 ˆ
k ∈⋅ p
mω
Evaluate the bracket k p
kp
=
using [r, H0 ]=
m
k rH0 − H0 r
i
= imω k k r
∴Vk = iqE 0
ωk
ˆ r
k ∈⋅
ω
i p
m
p. 60
or for a collection of particles
Vk = iE 0
= iE 0
ωk
ω
ωk
ω


ˆ  ∑ q i ri 
k ∈⋅
 i

ˆ
k ∈⋅µ
dipole moment
So we can write the electric dipole Hamiltonian as
V ( t ) = − µ ⋅ E(t)
So the rate of transitions between quantum states induced by the electric field is
wk =
≈
π
2
π
2
E0
2
2 ωk
2
ω
ˆ
k µ ⋅∈
ˆ
E0 2 k µ ⋅ ∈
2
2
[δ (Ek − E − ω )+ (E k − E
[δ (ω k
− ω )+ δ (ω k + ω )]
+ ω )]