Time-dependent perturbation theory

1. Introduction

Let's consider a system with the following Hamiltonian:

H(t) = H0 + H1(t).

Where H0 is the time-independent unperturbed Hamiltonian 
whose eigenvalues and eigenstates are known exactly. H1(t) 
represents a small time-dependent external perturbation. 

The eigenstates of H0 are written as:

H0|ψn> = En|ψn>

If the system is in any of these eigenstates then, it remains 
in this state for ever in the absence of an external perturbation. 

If the considered system is initially in some eigenstate of the 
unperturbed Hamiltonian, that is an exact eigenstate of this
Hamiltonian, then it will be found, after a perturbation, in some 
other eigenstate at a certain time. 

The time-dependent perturbation theory is used to determine 
transistions between the unperturbed energy eigenstates of a 
system.

2. Expression of the time-dependent amplitude

At a time t = 0, the state of the system is:
|ψ(0)> = Σ cn|ψn>

At a time t = t, the state of the system is:
|ψ(t)> = Σ cn exp{- iEnt/ℏ}|ψn>

The probability of finding the system in state |ψs> 
at time "t" is:
Ps(t) = |<ψs|ψ(t)>|2 = |<ψs|Σ cn exp{- iEnt/ℏ}|ψn>|2 = 
|Σ cn exp{- iEnt/ℏ}<ψs|ψn>|2  = |Σ cn exp{- iEnt/ℏ}δ(s,n)|2 = 
|cs|2 = Ps(0)

The probability to find an unperturbed system in a 
certain state does not depend on time.



2. Perturbed Hamiltonian

The system is now perturbed, its Hamiltonian is:

H = H0 + H1(t)

The probability Ps(t) of finding the perturbed 
system in a certain state must now depend on time, that is 
Ps(0) = |cs(t)|2. Therefore, 
at a time t = t, the state of the system is:
|ψ(t)> = Σ cn(t) exp{- iEnt/ℏ}|ψn>

The time-dependent Schrodinger equation is:

iℏ∂|ψ(t)>/∂t = H(t)|ψ(t)> = [H0 + H1(t)]|ψ(t)>

We have:
iℏ∂|ψ(t)>/∂t = 
iℏΣ [∂cn(t)/∂t exp{- iEnt/ℏ} - (iEn/ℏ)cn(t) exp{- iEnt/ℏ}]|ψn> = 
iℏ∂Σ [∂cn(t)/∂t exp{- iEnt/ℏ} |ψn> + Σ [En cn(t) exp{- iEnt/ℏ}]|ψn>
    (2.1)

And

[H0 + H1(t)]|ψ(t)> = H0|ψ(t)> + H1(t)|ψ(t)> = 
H0Σ cn(t) exp{- iEnt/ℏ}|ψn> + H1(t)|Σ cn(t) exp{- iEnt/ℏ}|ψn> = 
Σcn(t) exp{- iEnt/ℏ}  En|ψn> + Σ cn(t) exp{- iEnt/ℏ} H1(t)|ψn>     (2.2)

Equating (2.1) and (2.2) yields:
iℏΣ ∂cn(t)/∂t exp{- iEnt/ℏ} |ψn> = Σ cn(t) exp{- iEnt/ℏ} H1(t)|ψn>

The projection over a bra <ψm| gives:

iℏ Σ ∂cn(t)/∂t exp{- iEnt/ℏ} <ψm|ψn> = 
Σ cn(t) exp{- iEnt/ℏ} <ψm|H1(t)|ψn>

Using <ψm|ψn> = δ(m,n), and Σ(n)<ψm|ψn> = 1 
(The unperturbed eigenstates are orthogonal), we find:

iℏ∂cm(t)/∂t exp{- iEmt/ℏ} = 
Σ(n) cn(t) exp{- iEnt/ℏ} <ψm|H1(t)|ψn>

Or:
iℏ ∂cm(t)/∂t = Σ(n) cn(t) exp{- i(En - Em)t/ℏ} <ψm|H1(t)|ψn>

Let's write:
ωmn = (Em - En)t/ℏ and Hmn (t) = <ψm|H1(t)|ψn>
We have then:
iℏ∂cm(t)/∂t = Σ(n) cn(t) exp{ iωmn} Hmn(t)     (2.3)


Let's recap:

iħ∂cm(t)/∂t = Σ cn(t) exp{- i(En - Em)t/ħ} <ψm|H1(t)|ψn>
ωmn = (Em - En)t/ħ 
Hmn(t) = <ψm|H1(t)|ψn>
iħ∂cm(t)/∂t = Σ(n) cn(t) exp{ iωmnt} Hmn(t)

3. Two level time-dependent system - Rabi oscillations

The simplified example of a quantum system is the one 
with only two stationary states (levels) |ψ(0)1>
and |ψ(0)2> with energies E(0)1
and E(0)2 respectively. The equation (2.3) becomes:

iℏ∂c1(t)/∂t = Σ(1,2) cn(t) exp{ iω1nt} H1n(t) = 
c1(t) exp{ iω11t} H11(t) + c2(t) exp{ iω12t} H12(t) 

Similarly,
iℏ∂c2(t)/∂t = Σ(1,2) cn(t) exp{ iω2nt} H2n(t) = 
c1(t) exp{ iω21t} H21(t) + c2(t) exp{ iω22} H22(t) 

Recal:
Hmn(t) = <ψm|H1(t)|ψn>
We assume that the elements of the time-dependent 
perturbation are zero between aa eigebstate and itself, i.e.:
H11(t) = <ψ1|H1(t)|ψ1> = 0, and 
H22(t) = <ψ2|H1(t)|ψ2> = 0,
We assume also the off-diagonal elements of the time-dependent 
perturbation are equal to a constant, i.e.:
H12(t) = <ψ1|H1(t)|ψ2> = ħV
H21(t) = <ψ2|H1(t)|ψ1> = H*12(t) = ħV*
(H1(t) is Hermitian operator)

The two equations for c1 and c2 becomes 
a system of two differential equations:

i∂c1(t)/∂t =  c2(t) exp{ iω12t} V12(t) 
i∂c2(t)/∂t = c1(t) exp{ iω21t} V21(t)  


We will use : V12(t)V21(t)  = V12(t)V*12(t) = |V|2, 
and V12(t) = V*12(t) = |V|.

The initial condition we set is that at t = 0 the system is 
definitely in state 1, that is: c1(0) = 1 and c2(0) = 0.

We have:
c1(t)  = (i/|V|) exp{ - iω21t} ∂c2(t)/∂t 
Then:
∂c1(t)/∂t = (i/|V|) exp{ - iω21t}[∂2c2(t)/∂t2 - iω21 ∂c2(t)/∂t]

Substituting this equation in the first one of the system yields:

(-1/V21(t)) exp{ - iω21t} [∂2c2(t)/∂t2 - iω21 ∂c2(t)/∂t] = 
c2(t) exp{ iω12t} V12(t) 

With ω21 = - ω12, it becomes:

∂2c2(t)/∂t2 - iω21 ∂c2(t)/∂t + c2(t) |V|2 = 0


This differential equation is of the form:

ay" + by' + cy = 0, 
a = 1
b = - iω21
c = + |V|2
Then:
Δ = b2 - 4ac; = - (ω221 + 4|V|2) < 0
In this case the roots and of the auxiliary equation 
(ar2 + br + c = 0)
are complex numbers:
r1 = α + i β
r2= α - i β
Where 
α  = - b/2a 
β= (- Δ)1/2/2a

The solution of this equation  is:
y = C1 exp{r1x} + C2 exp{r2x} = exp{αx} (A cos βx + B sin βx)


The solution of our equation is:
c2(t)  = exp{(iω21/2)t} 
(A cos[(- Δ)1/2/2]t + B sin [(- Δ)1/2/2]t)

We know that c1(0) = 1, and c20) = 0 , so A = 0,
c2(t)  = exp{(iω21/2)t} B sin [(- Δ)1/2/2]t

We have c1(t)  = (i/|V|) exp{ - iω21t} ∂c2(t)/∂t 
First:
∂c2(t)/∂t = (iω21/2) exp{(iω21/2)t} B sin [(- Δ)1/2/2]t 
+ [(- Δ)1/2/2] exp{(iω21/2)t} B cos [(- Δ)1/2/2]t 
= exp{(iω21/2)t} B {(iω21/2)  sin [(- Δ)1/2/2]t 
+ [(- Δ)1/2/2] cos [(- Δ)1/2/2]t }
Therefore:

c1(t)  = (iB/|V|) exp{(-iω21/2)t}{(iω21/2) sin [(- Δ)1/2/2]t 
+ [(- Δ)1/2/2] cos [(- Δ)1/2/2]t }

Using c1(0) = 1, we get:
1 = (iB/|V|) {[(- Δ)1/2/2] }, then:
B = -iV21(0)) /[(- Δ)1/2/2]
Then:
c2(t)  = exp{(iω21/2)t} B sin [(- Δ)1/2/2]t
c2(t)  = -i|V| exp{(iω21/2)t}  sin [(- Δ)1/2/2]t/[(- Δ)1/2/2]

Therefore:

c1(t)  =  exp{(- iω21/2)t}{(iω21/2)  sin [(- Δ)1/2/2]t 
+ [(- Δ)1/2/2] cos [(- Δ)1/2/2]t }/[(- Δ)1/2/2]
= exp{(- iω21/2)t} [cos ((- Δ)1/2/2)t + (iω21)  sin [(- Δ)1/2/2]t /2[(- Δ)1/2/2]]

To recap:

β = [(ω221 + 4|V|2)]1/2/2
c1(t) = exp{- iω21t/2} [cos βt + iω21 sin βt /2β]
c2(t) = -i|V| exp{(iω21/2)t} sin β t/β