MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Andrei Tokmakoff
p. 28
THE RELATIONSHIP BETWEEN U(t,t0) AND cn(t)
For a time-dependent Hamiltonian, we can often partition H = H0 + V (t )
H0 : time-independent; V (t ) : time-dependent potential. We know the eigenkets and eigenvalues of H0 : H0 n = En n
We describe the initial state of the system (t = t 0 ) as a superposition of these eigenstates:
ψ ( t 0 ) = ∑ cn n
n
For longer times t , we would like to describe the evolution of ψ in terms of an expansion in
these kets:
ψ (t ) = ∑ cn (t ) n
n
The expansion coefficients ck (t ) are given by
ck (t ) = k ψ (t ) = k U (t,t 0 )ψ (t0 )
Alternatively we can express the expansion coefficients in terms of the interaction picture wavefunctions bk (t ) = k ψ I (t )
(This notation follows Cohen-Tannoudji.) Notice
ck ( t ) = k ψ ( t ) = k U 0 U I ψ ( t 0 )
= e−iωk t k U I ψ ( t 0 )
= e−iωk t b k ( t )
2
2
so that bk (t ) = ck (t ) . Also, bk (0 ) = ck (0) . It is easy to calculate bk (t ) and then add in the extra oscillatory term at the end. p. 29
Now, starting with
∂ψ I
i
∂t
= VI ψ I
we can derive an equation of motion for bk
∂ bk
= k VI U I ψ I ( t 0 )
∂t
i
inserting
∑n
ψ I (t0 ) = ∑ bn n
n
= ∑ k VI n n U I ψ I ( t 0 )
n =1
n
n
= ∑ k VI n b n ( t )
n
i
∂b k
= ∑ Vkn ( t ) e− iωn k t b n ( t )
∂t
n
This equation is an exact solution. It is a set of coupled differential equations that describe how
probability amplitude moves through eigenstates due to a time-dependent potential. Except in
simple cases, these equations can’t be solved analytically, but it’s often straightforward to
integrate numerically.
Exact Solution: Resonant Driving of Two-level System
Let’s describe what happens when you drive a two-level system with an oscillating potential.
V ( t ) = V cos ωt = Vf ( t )
This is what you expect for an electromagnetic field interacting with charged particles: dipole
transitions. The electric field is
E ( t ) = E 0 cos ωt
For a particle with charge q in a field E , the force on the particle is
F = qE
which is the gradient of the potential
p. 30
Fx = −
∂V
= qE x
∂x
⇒ V = −qE x x
qx is just the x component of the dipole moment µ . So matrix elements in V look like:
k | V(t) |
= −qE x k | x |
cos ωt
More generally,
V = −E ⋅ µ .
So,
V ( t ) = V cos ωt = − E 0 ⋅µ cos ωt .
Vk ( t ) = Vk cos ωt = − E 0 ⋅µ k cos ωt
We will now couple our two states k +
starts in
with the oscillating field. Let’s ask if the system
what is the probability of finding it in k at time t ?
The system of differential equations that describe this situation are:
i
∂
b k ( t ) = ∑ b n ( t ) Vkn ( t ) e−ωnk t
∂t
n
= ∑ b n ( t ) Vkn e − iωnk t × 12 ( e− iωt + e+ iωt )
n
i ω −ω t
i ω +ω t
i b k = 12 b Vk e ( k ) + e ( k )  + 12 b k Vkk eiωt + e− iωt 
= (1) and (2)
i ω −ω t
i ω +ω t
i b = 12 b V eiωt + e− iωt  + 12 b k V k e ( k ) + e ( k ) 
or
= (3) and (4)
 e − i( ωk +ω) t + e− i( ωk −ω) t 


We can drop (2) and (3). For our case, Vii = 0 . We also make the secular approximation (rotating wave approximation) in which the i ω k + ω )t
nonresonant terms are dropped. When ω k ≈ ω , terms like e ±i ωt or e (
rapidly and so don’t contribute much to change of cn .
oscillate very
p. 31
So we have:
bk =
−i
2
b Vk e (
i ωk −ω) t
b =
−i
2
bk V k e
(1)
− i ( ωk −ω) t
(2)
Note that the coefficients are oscillating out of phase with one another.
Now if we differentiate (1):
 b Vk ei( ωk −ω) t + i ( ωk − ω) b Vk ei( ωk −ω) t 


bk =
−i
2
b =
2i
− i ω −ω t
bk e ( k )
Vk
(3)
Rewrite (1):
(4)
and substitute (4) and (2) into (3), we get linear second order equation for b k .
2
V
b k − i ( ωk − ω) b k + k 2 b k = 0
4
This is just the second order differential equation for a damped harmonic oscillator:
ax + bx + cx = 0
x=e
− ( b 2a ) t
( A cos µt + B sin µt )
µ=
 4ac − b 2 
1
2
(remember b k (0)=0 and b (0)=1)
With a little more work, we find
Pk = b k ( t ) =
Vk
2
ΩR =
P = 1 − bk
1
2a
2
1
2
Vk
2
 Vk

2
2
sin 2 Ω r t
+
2
( ωk
− ω)
+
2
( ωk
2
− ω) 

2
1
2
p. 32
Amplitude oscillates back and forth between the two states at a frequency dictated by the coupling. Resonance: To get transfer of probability amplitude you need the driving field to be at the same frequency as the energy splitting. Note a result we will return to later: Electric fields couple states, creating coherences! On resonance, you always drive probability amplitude entirely from one state to another. 1
Pkl
on resonance
ωkl − ω = Vkl
0.5
0
0
large detuning
Efficiency of driving between
t
π / Vkl
and k states drops off with detuning.
Pmax
2 Vkl /
ω kl