MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Andrei Tokmakoff
p. 13
QUANTUM DYNAMICS1
The motion of a particle is described by a complex wavefunction ψ ( r , t ) that gives the
probability amplitude of finding a particle at point r at time t . If we know ψ (r ,t0 ) , how does it
change with time?
?
ψ (r ,t0 ) → ψ (r,t )
t > t0
We will use our intuition here (largely based on correspondence to classical mechanics)
We start by assuming causality: ψ (t0 ) precedes and determines ψ (t ) .
Also assume time is a continuous parameter: lim ψ (t
) = ψ (t0 )
t →t 0
Define an operator that gives time-evolution of system.
ψ (t ) = U(t, t 0 )ψ (t 0 )
This “time-displacement operator” is similar to the “space-diplacement operator”
ψ ( r ) = eik (r −r0 ) ψ ( r0 )
which moves a wavefunction in space.
U does not depend on ψ . It is a linear operator.
if ψ ( t 0 ) = a1 ϕ1 ( t 0 ) + a 2 ϕ ( t 0 )
ψ ( t ) = U ( t, t 0 ) ψ ( t 0 )
= a1U ( t, t 0 ) ϕ1 ( t 0 ) + a 2 U ( t, t 0 ) ϕ2 ( t 0 )
= a1 ( t ) ϕ1 + a 2 ( t ) ϕ2
From Merzbacher, Sakurai, Mukamel
p. 14
while ai (t ) typically not equal to ai (0) ,
∑ a (t ) = ∑ a (t )
n
n
n
0
n
Properties of U(t,t0)
Time continuity: U(t , t ) = 1
Composition property:
U(t2 , t0 ) = U(t 2 , t1 )U(t1 , t0 )
(This should suggest an exponential form).
Note: Order matters!
ψ (t2 ) = U(t 2 , t1 )U(t1 , t0 ) ψ (t0 )
= U(t 2 , t1 ) ψ (t1 )
∴U(t , t0 )U(t0 , t ) = 1
∴ U −1 ( t, t 0 ) = U ( t 0 , t ) inverse is time-reversal
Let’s write the time-evolution for an infinitesimal time-step, δt.
lim U ( t 0 + δt, t 0 ) = 1
δt →0
We expect that for small δt , the difference between U(t0 , t0 ) and U ( t 0 + δt, t 0 ) will be linear
in δt . (Think of this as an expansion for small t):
U ( t 0 + δt, t 0 ) = U ( t 0 , t 0 ) − iΩδt
Ω is a time-dependent Hermetian operator. We’ll see later why the expansion must be complex.
Also, U ( t 0 + δt, t 0 ) is unitary. We know that U-1U = 1 and also
U † ( t 0 + δt, t 0 ) U ( t 0 + δt, t 0 ) = (1 + iΩ†δt ) (1 − iΩδt ) ≈ 1
We know that U ( t + δt, t 0 ) = U ( t + δt, t ) U ( t, t 0 ) .
p. 15
Knowing the change of U during the period δt allows us to write a differential equation for the
time-development of U(t , t0 ). Equation of motion for U :
lim U ( t + δt, t 0 ) − U ( t, t 0 )
d U ( t, t 0 )
=
δt → 0
δt
dt
=
U
 ( t + δt, t ) −1 U ( t, t 0 )
δt → 0
δt
lim
The definition of our infinitesimal time step operator says that
U ( t + δt, t ) = U ( t, t ) − iΩδt = 1 − iΩδt . So we have:
∂U (t,t 0 )
∂t
= −iΩU (t,t 0 )
You can now see that the operator needed a complex argument, because otherwise probability
amplitude would not be conserved (it would rise or decay). Rather it oscillates through different
states of the system.
Here Ω has units of frequency. Noting (1) quantum mechanics says E = ω and (2) in classical
mechanics Hamiltonian generates time-evolution, we write
Ω=
i
H
Ω can be a function of time!
∂
U (t,t 0 ) = HU (t,t 0 )
∂t
eqn. of motion for U
Multiplying from right by ψ (t0 ) gives
i
∂
ψ = Hψ
∂t
p. 16
We are also interested in the equation of motion for U † . Following the same approach and
recognizing that U† (t, t 0 ) acts to the left:
ψ ( t ) = ψ ( t 0 ) U † ( t, t 0 )
we get
−i
∂ †
U ( t, t 0 ) = U † ( t, t 0 ) H
∂t
Evaluating U(t,t0): Time-Independent Hamiltonian
Direct integration of i ∂U ∂t = HU suggests that U can be expressed as:
i
U(t , t0 ) = exp  − H(t − t0 )


Since H is an operator, we will define this operator through the expansion:
2
−iH
 iH

 −i   H ( t − t 0 ) 
exp  − ( t − t 0 )  = 1 +
+…
( t − t0 ) +  
2


 
2
(NOTE: H commutes at all t .) You can confirm the expansion satisfies the equation of motion for U . For the time-independent Hamiltonian, we have a set of eigenkets:
H n = En n
∑
n
n =1
n
So we have
U ( t, t 0 ) = ∑ exp −
 iH ( t − t 0 ) /  n n
n
= ∑ n exp −
 iE n ( t − t 0 ) /  n
n
So,
p. 17
ψ ( t ) = U ( t, t 0 ) ψ ( t 0 )
 −i

= ∑ n n ψ ( t 0 ) exp  E n ( t − t 0 ) 


n
cn ( t 0 )
= ∑ n cn ( t )
c n ( t ) = c n ( t 0 ) exp  −iωn ( t − t 0 ) 
n
Expectation values of operators are given by
A(t) = ψ(t) A ψ(t)
= ψ ( 0 ) U † ( t, 0 ) AU ( t, 0 ) ψ ( 0 )
For an initial state ψ (0) = ∑ cn (0 ) n
n
A = ∑ c*m m m e+ iωm t m A n e − iωn t n n c n
n,m
=
∑c
c A mn e−ωnm t
*
m n
n,m
=
∑c (t)c (t)A
*
m
n
mn
n,m
What is the correlation amplitude for observing the state k at the time t ?
ck ( t ) = k ψ ( t ) = k U ( t, t 0 ) ψ ( t 0 )
= ∑ k n n ψ ( t0 ) e
− iωn ( t − t 0 )
n
Evaluating the time-evolution operator: Time-Dependent Hamiltonian
If H is a function of time, then the formal integration of i ∂U ∂t = HU gives
 −i t

U ( t, t 0 ) = exp  ∫ H ( t ′ ) dt ′
t0


Again, we can expand the exponential in a series, and substitute into the eqn. of motion to
confirm it; however, we are treating H as a number.
p. 18
i
U(t, t0 ) = 1 −
∫
t
t0
H(t′ )dt′ +
2
1  −i 
2!  
∫
t
t0
dt′′H(t ′)H (t′′) + …
NOTE: This assumes that the Hamiltonians at different times commute!  H ( t ′ ) , H ( t ′′ )  = 0
This is generally not the case in optical + mag. res. spectroscopy. It is only the case for special
Hamiltonians with a high degree of symmetry, in which the eigenstates have the same symmetry
at all times. For instance the case of a degenerate system (for instance spin ½ system) with a
time-dependent coupling.
Special Case: If the Hamiltonian does commute at all times, then we can evaluate the time-
evolution operator in the exponential form or the expansion.
U ( t,t0 ) = 1 −
i
∫
t
t0
H ( t ′ ) dt ′ +
1  −i 
 
2!  
2
∫
t
t0
t
dt ′∫ dt ′′H ( t ′ ) H ( t ′′ ) +…
t0
If we also know the time-dependent eigenvalues from diagonalizing the time-dependent
Hamiltonian (i.e., a degenerate two-level system problem), then:
U ( t,
0
 −i t

j exp  ∫ ε j ( t ′ ) dt′ j
t0


)=∑
j
More generally: We assume the Hamiltonian at different times do not commute. Let’s proceed
a bit more carefuly:
Integrate
∂
U ( t,
∂t
To give:
U ( t, t 0 ) = 1 −
0
)=
i
−i
∫
t
t0
H ( t ) U ( t,
0
dτ H ( τ ) U ( τ,
)
0
)
This is the solution; however, U(t , t0 ) is a function of itself. We can solve by iteratively
substituting U into itself.
First Step:
U ( t, t 0 ) = 1 −
i
 i
∫t0 dτH ( τ ) 1−
t
∫
τ
t0
dτ′H ( τ′ ) U ( τ′,
 −i  t
 −i 
= 1 +   ∫ dτH ( τ )  
  t0
 
2
∫
t
t0
0
) 

τ
dτ ∫ dτ′H ( τ ) H ( τ′ ) U ( τ′,
t0
0
)
p. 19
Next Step:
 −i  t
U (t, t 0 ) = 1 +   ∫ dτH ( τ )
  t0
 −i 
+ 
 
2
 −i 
+  
 
3
∫
t
∫
t
t0
t0
τ
dτ ∫ dτ′H ( τ ) H ( τ′ )
t0
τ
τ′
t0
t0
dτ ∫ dτ′∫ dτ′′H ( τ ) H ( τ′ ) H ( τ′′ ) U ( τ′′, t 0 )
From this expansion, you should be aware that there is a time-ordering to the interactions. For
the third term, τ ′′ acts before τ ′, which acts before τ : t 0 ≤ τ ′′ ≤ τ ′ ≤ τ ≤ t .
Notice also that the operators act to the right.
This is known as the (positive) time-ordered exponential.
 −i t

 −i t

U (t, t 0 ) ≡ exp +  ∫ dτ H ( τ )  = Tˆ exp  ∫ dτ H ( τ ) 
t0
t0




 −i 
= 1+ ∑  

n =1 
∞
n
∫
t
t0
τ
t
t0
t0
dτn∫ dτn … ∫ dτ1
H ( τn ) H ( τn −1 )…H ( τ1 )
Here the time-ordering is: t 0 → τ 1 → τ 2 → τ 3 .... τ n → t
t0 →
τ ′′ → τ ′ → τ
…
Compare this with the expansion of an exponential:
n
1  −i 
1+ ∑  

n =1 n! 
∞
∫
t
t0
t
dτn … ∫ dτ1 H ( τn ) H ( τ n −1 )…H ( τ1 )
t0
Here the time-variables assume all values, and therefore all orderings for H(τ i ) are calculated.
The areas are normalized by the n! factor. (There are n! time-orderings of the τ n times.)
We are also interested in the Hermetian conjugate of U(t,t0 ), which has the equation of motion
p. 20
∂ †
+i
U ( t, t 0 ) = U † ( t, t 0 ) H ( t )
∂t
If we repeat the method above, remembering that U† (t,t 0 ) acts to the left:
ψ ( t ) = ψ ( t 0 ) U † ( t, t 0 )
then from U † ( t, t 0 ) = U † ( t 0 , t 0 ) +
i
∫
t
t0
dτU † ( t, τ ) H ( τ ) we obtain a negative-time-ordered
exponential:
i
U † ( t, t 0 ) = exp − 


dτ H ( τ ) 
t0

∫
t
∞
i
= 1+ ∑  

n =1 
Here the H(τ i ) act to the left.
n
∫
t
t0
τn
τ2
t0
t0
dτn ∫ dτn −1 … ∫ dτ1H ( τ1 ) H ( τ2 )…H ( τn )