MIT Department of Chemistry�
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Andrei Tokmakoff
p. 33
PERTURBATION THEORY
Given a Hamiltonian
H(t ) = H0 + V (t )
where we know the eigenkets for H0
H0 n = En n
we often want to calculate changes in the amplitudes of n induced by V (t ) :
ψ (t ) = ∑ cn (t
) n
where
n
ck (t ) = k ψ (t ) = k U (t,t 0 )ψ (t0 )
In the interaction picture, we defined
bk (t ) = k ψ I
= e +i ω k r ck (t )
which contains all the relevant dynamics. The changes in amplitude can be calculated by solving
the coupled differential equations:
∂
−i
bk =
∂t
∑e
− iωnk t
Vkn ( t ) b n ( t )
n
For a complex system or a system with many states to be considered, solving these equations isn’t practical. Alternatively, we can choose to work directly with UI (t , t0 ), and we can calculate bk (t ) as:
bk = k U I (t,t0 )ψ (t 0 )
where
 −i
U I ( t, t 0 ) = exp + 

∫
t
t0

VI ( τ ) dτ 

Now we can truncate the expansion after a few terms. This is perturbation theory, where the
dynamics under H0 are treated exactly, but the influence of V (t ) on bn is truncated. This works
well for small changes in amplitude of the quantum states with small coupling matrix elements
relative to the energy splittings involved.
bk ( t ) ≈ bk ( 0 ) ; V
Ek − En
p. 34
Transition Probability
Let’s take the specific case where we have a system prepared in
, and we want to know the
probability of observing the system in k at time t , due to V (t ) .
Pk ( t ) = b k ( t )
b k ( t ) = k U I ( t, t 0 )
2
 i
b k ( t ) = k exp +  −

= k
−
i
∫
t
t0
 −i 
+ 
 
∫

dτ VI ( τ ) 

t
t0
dτ k VI ( τ )
2
∫
t
t0
dτ2
∫
τ2
t0
dτ1 k VI ( τ2 ) VI ( τ1 )
+…
using
k VI (t )
bk (t ) = δ k −
i
= k U0† V (t ) U0
t
∫t
0
−i
+ ∑  
m
dτ 1 e− iω
2
t
∫t
0
τ
k 1
= e −i ω k t Vk (t )
Vk (τ 1 )
τ2
“first order”
dτ 2 ∫ dτ 1 e− iω mk τ 2 Vkm (τ 2 ) e −i ω m τ1 Vm (τ1 ) + …
t0
“second order”
This expression is usually truncated at the appropriate order. Including only the first integral is
first-order perturbation theory.
If ψ 0 is not an eigenstate, we only need to express it as a superposition of eigenstates, but
remember to convert to ck (t ) = e −ω k t bk (t ).
Note that if the system is initially prepared in a state
, and a time-dependent perturbation is
turned on and then turned off over the time interval t = −∞ to + ∞ , then the complex amplitude in
the target state k is just the Fourier transform of V(t) evaluated at the energy gap ω k .
p. 35
Example: First-order Perturbation Theory
Vibrational excitation on compression of harmonic oscillator. Let’s subject a harmonic oscillator
to a Gaussian compression pulse, which increases the frequency of the h.o.
H =
p2
x2
+ k(t )
2m
2
A′ = δk 0 = A / 2πσ
k ( t ) = k 0 + δk ( t )
H = H0 + V ( t ) =
 ( t − t 0 )2 
δk ( t ) = A′ exp  −

2


2
σ


 ( t − t 0 )2 
p2
x2
A′x 2
+ k0
+
exp  −

2


2m
2
2
2
σ


H0
H0 n = En n
k 0 = mΩ 2
V( t )
1

H 0 = Ω  a †a + 
2

1

En = Ω  n + 
2

If the system is in 0 at t0 = −∞ , what is the probability of finding it in n at t = ∞ ?
bn ( t ) =
for n ≠ 0:
=
−i
−i
∫
t
t0
A′ n x 2 0
dτ Vn 0 ( τ ) eiωn− τ
∫
+∞
−∞
dτ eiωn 0 τ e −τ
2
2σ2
ωn 0 = nΩ
bn ( t ) =
−i
A′ n x 2 0
∫
+∞
−∞
dτ einΩτ−τ
∫
+∞
−∞
2
/ 2σ2
exp ( ax 2 + bx + c ) dx =
(
−π
2
exp c − 14 ba
a
)
p. 36
b n ( t ) ==
−i
2 2
A n x 2 0 e−2n σ Ω
2
/4
What about matrix element?
x2 =
(a + a )
mΩ
† 2
=
( aa + a a + aa
†
mΩ
†
+ a †a † )
First-order perturbation theory won’t allow transitions to n = 1, only n = 0 and n = 2 .
Generally this wouldn’t be realistic, because you would certainly expect excitation to v=1
would dominate over excitation to v=2. A real system would also be anharmonic, in which case,
the leading term in the expansion of the potential V(x), that is linear in x, would not vanish as it
does for a harmonic oscillator, and this would lead to matrix elements that raise and lower the
excitation by one quantum.
However for the present case,
2 x2 0 = 2
mΩ
So,
b2 =
2 2
− 2i
A e−2σ Ω
mΩ
2
P2 = b 2 =
2 A 2 −4σ2Ω2
e
m 2Ω2
A = δk 0 2πσ
Significant transfer of amplitude occurs when the compression pulse is short compared to the
vibrational period.
1
<< Ω
σ
Validity: First order doesn’t allow for feedback and bn can’t change much from its initial value.
for P2 ≈ 0
A << mΩ
p. 37
First-Order Perturbation Theory
A number of important relationships in quantum mechanics that describe rate processes come from
1st order P.T. For that, there are a couple of model problems that we want to work through:
(1) Constant Perturbation
ψ (t0 ) =
. A constant perturbation of amplitude V is applied to t0 . What is Pk ?
V (t ) = θ (t − t 0 ) V
V(t)
0 t < 0
=
V t ≥ 0
t
t0
To first order, we have:
bk = δ k −
i
t
∫t
i ω k (τ − t0 )
dτ e
Vk
0
Vk independent of time
k U †0 V U 0
= δk +
−i
= δk +
−Vk
exp(iω k (t − t0 ))− 1
Ek − E
Vk
t
∫t
= Ve
iωk ( t−t 0 )
i ω k (τ −t 0 )
dτ e
0
[
]
using e i∅ − 1 = 2iei 2 sin ∅ 2
∅
iω t −t /2
− 2iVk e k ( 0 )
+
sin (ω k (t − t0 ) / 2)
Ek − E
= δk
For k ≠
Pk = bk
we have
2
=
4Vk
2
Ek − E
2
sin 2 12 ω k (t − t0 )
or setting t0 = 0 and writing this as we did in lecture 1:
p. 38
Pk =
V2
2
2 sin (∆t / )
∆
or Pk =
V2t 2
2
where ∆ =
sinc 2 ( ∆t / 2
Ek − El
2
)
Compare this with the exact result:
Pk =
V2
sin 2
V2 + ∆2
(
)
∆2 + V2 t /
Clearly the P.T. result works for V << ∆. (…not for degenerate systems)
to k as a function of the energy level splitting ( E k − E ) :
The probability of transfer from
2
Vkl t 2 /
2
Area scales linearly
with time.
2π / ( t − t 0 )
−4 π
t − t0
−2 π
t − t0
0
2π
t − t0
4π
t − t0
Ek − El
Time-dependence:
Ek=El
(exact solution for Ek=El )
Pk(t)
Ek-El ≥ Vkl
0
π /Vkl
0
Ek-El >> Vkl
t
p. 39
Time dependence on resonance (∆=0):
expand sin x = x −
x3
+ …
3!

V 2  ∆t ∆ 3 t 3
Pk = 2 
−
+ …
3
∆ 
6

=
V 2
2
2
t 2
This is unrealistic, but the expression shouldn’t hold for ∆=0.
Long time limit: The sinc2(x) function narrows rapidly with time giving a delta function:
sin 2 ( ax 2 ) π
lim
= δ(x)
t →∞
ax 2
2
lim Pk (t ) =
t→ ∞
2π Vk
2
δ (Ek − E )(t − t 0 )
A probability that is linear in time suggests a transfer rate that is independent of time! This
suggests that the expression may be useful to long times:
∂P ( t ) 2π Vk
wk (t) = k
=
∂t
2
δ ( Ek − E
)
This is one statement of Fermi’s Golden Rule, which describes relaxation rates from first order
perturbation theory. We will show that this will give long time exponential relaxation rates.
p. 40
(2)
Harmonic Perturbation
Interaction of a system with an oscillating perturbation turned on at time t0 = 0. This
describes how a light field (monochromatic) induces transitions in a system through dipole
interactions.
V (t ) = V cos ωt = − µE0 cosω t
observe
Vk (t ) = Vk cosω t
V
= k eiω t + e −i ωt
2
[
V(t)
τ
t
t0
To first order, we have:
bk = k ψ I ( t ) =
−i
∫
t
t0
dτ Vk ( τ ) eiωk τ
=
−iVk
2
=
i ω +ω t
i ω +ω t
i ω −ω t
i ω −ω t
−Vk  e ( k ) − e ( k ) 0 e ( k ) − e ( k ) 0 
+


ωk + ω
ωk − ω
2 

∫
t
t0
i ω +ω τ
i ω −ω τ
dτ e ( k ) − e ( k ) 
iθ
Setting t0 → 0 and using eiθ − 1 = 2ie 2 sin θ 2
i ( ωk −ω) t / 2
i ω +ω t / 2
sin ( ωk − ω) t / 2 e ( k ) sin ( ωk + ω) t / 2 
−iVk  e
bk =
+


ωk − ω
ωk + ω


Notice that these terms are only significant when
ω ≈ ω k : resonance!
]
p. 41
First Term
Second Term
max at: ω = + ω k
ω = −ω k
Ek > E
Ek < E
Ek = E + ω
Ek = E − ω
Absorption
k
Stimulated Emission
l
(resonant term)
l
(anti-resonant term)
k
For the case where only absorption contributes, Ek > E , we have:
Vk
2
Pk = b k =
or
2
E 02 µ k
( ωk
( ωk
2
− ω)
2
2
− ω)
2
sin 2  12 ( ωk − ω) t 
sin 2  12 ( ωk − ω) t 
The maximum probability for transfer is on resonance ω k = ω
2
Vkl t 2 / 4
2
2π / t
-2
-1
0
1
2
ω − ωkl / 2π
p. 42
We can compare this with the exact expression:
Vk
2
Pk = b k =
2
( ωk
2
− ω) + Vk
2
2
sin 2  21

Vk
2
2
+ ( ωk − ω) t 

which points out that this is valid for couplings Vk that are small relative to the detuning
∆ω = ( ωk − ω) .
Limitations of this formula:
By expanding sin x = x −
x3
+ … , we see that on resonance ∆ω = ωk − ω → 0
3!
2
lim
V
Pk ( t ) = k 2 t 2
∆ω → 0
4
This clearly will not describe long-time behavior, but the expression is not valid for ∆ω =0.
Nontheless, it will hold for small Pk , so
t <<
2
Vk
(depletion of 1 neglected in first order P.T.)
At the same time, we can’t observe the system on too short a time scale. We need the field to
make several oscillations for it to be a harmonic perturbation.
t>
1
ω
≈
1
ωk
These relationships imply that
Vk << ω k
2π / t << ωkl