The Interaction Picture

In many treatments in current journals you find authors using the interaction picture.  Assume the Hamiltonian of an arbitrary system is H0(t), and the corresponding evolution operator is U0(t,t0).  We have
(iħ∂/∂t)U0(t,t0) = H0(t)U0(t,t0),   U0(t0,t0) = I.
Now assume that the system is perturbed in such a way that the Hamiltonian becomes
H(t) = H0(t) + W(t).  For such a system the state vector in the interaction picture, |ΨI(t)> is defined from the state vector in the Schroedinger picture through
I(t)> = U0†(t,t0)|ΨS(t)>.

How does |ΨI(t)> evolve?
iħ∂|ΨI(t)>/∂t = iħ∂/∂t(U0†(t,t0)|ΨS(t)>)
=  iħ[∂/∂t(U0†(t,t0)]|ΨS(t)>) + iħU0†(t,t0)[∂/∂t(|ΨS(t)>]
=  -U0†(t,t0)H0s(t)|ΨS(t)>) + U0†(t,t0)Hs(t)|ΨS(t)>.
Here we used
iħ∂U0(t,t0)/∂t = H0(t)U0(t,t0),  -iħ∂U0†(t,t0)/∂t = U0†(t,t0)H0(t).

We now can write
iħ∂|ΨI(t)>/∂t = -U0†H0S(t)U0U0†S(t)> + U0†Hs(t)U0U0†S(t)>
= -H0I(t)ΨI(t)> + (H0I(t) + WI(t))|ΨI(t)> = WI(t))|ΨI(t)>.
Therefore
iħ∂|ΨI(t)>/∂t = WI(t))|ΨI(t)>.

We can rewrite this differential equation in the form of an integral equation.
d|ΨI(t)> = (iħ)-1WI(t))|ΨI(t)>dt.
t0t d|ΨI(t)>  = (iħ)-1t0t WI(t')|ΨI(t')>dt'.
I(t)> = |ΨI(t0)> + (iħ)-1t0t WI(t')|ΨI(t')>dt'
   ŻŻ_______________________________ŻŻ
This integral equation can be solved by iteration.
The ket |ΨI(t)> can be expanded in a power series of the form
I(t)> = {I + (iħ)-1t0t dt'WI(t') + (iħ)-2t0t dt'WI(t') ∫t0t' dt''WI(t'') + ... }|Ψ0(t)>.
The interaction picture assigns part of the time dependence to the state vectors, and part to the operators.  Following the method used to derive the equation for the evolution of the operators in the Heisenberg picture, we can derive the equation for the evolution of operators in the interaction picture.
dAI/dt = (∂AS/∂t)I+  (iħ)-1[AI,H0I].

When do we use the interaction picture?
The interaction picture is often used when H0S is time-independent and WS(t) is a small correction to H0S
Assume that the problem governed by H0S is already solved, either exactly, or in some approximation. 
Assume WS(t) = 0 for t < 0.  Then |ΨI(0)> = |ΨS(0)>.  Neglecting higher order terms we have
I(t)> = |ΨI(t0)> + (iħ)-1t0t WI(t')|ΨI(0>dt'
with WI(t) = exp(iH0t/ħ) WS(t) exp(-iH0t/ħ).

Let {|n>} be an orthonormal eigenbasis of H0 and let the system be in the eigenstate |m> at t = 0, i.e. |ΨI(0)> = |m>.  We have H0|m> = Em|m>,
The probability P(Ek,t) of finding the system in the eigenstate |k> of H0 at time t, i.e. the probability of measuring the eigenvalue Ek, is |<k|ΨI(t)>|2..  (The predictions of quantum mechanics are independent of the representation.)
<k|ΨI(t)> = <k|m> + (iħ)-10t <k|WI(t')|m>dt'
= <k|m> + (iħ)-10t exp(i(Ek - Em)t/ħ) <k|WS(t')|m>dt'.
With <k|m> = 0 we have
P(Ek,t) = (1/ħ2)|∫0t exp(i(Ek - Em)t/ħ) <k|WS(t')|m>dt'|2.

This is the result of first order time dependent perturbation theory.  P(Ek,t) is the transition probability.

If WS(t) is independent of time, i.e. if a constant small term is added to the Hamiltonian at t = 0 then
P(Ek,t) = (1/ħ2)|<k|WS|m>|2| (exp(i(Ek - Em)t/ħ - 1)/(i(Ek - Em)/ħ)|2

= (1/ħ2)|<k|WS|m>|2| (exp(-iEmt/ħ - exp(-iEkt/ħ)/(i(Ek - Em)/ħ)|2
= (|<k|WS|m>|2/(Ek - Em)2) (2 sin(ωkmt/2))2
= (|<k|WS|m>|2/(Ek - Em)2) 4 sin2kmt/2).
Here ωkm = (Ek - Em)/ħ and we have used |e - e| = 2sin((θ - φ)/2).

Problem:

Consider a one-dimensional, infinitely deep well.  Let U(x) = 0, for 0 < x < a, and U(x) = ∞ everywhere else.  The eigenstates of H0 = p2/(2m) + U(x) are Φn(x) = (2/a)1/2sin(nπx/a) with eigenvalues En = n2π2ħ2/(2ma2).
Assume that at t = 0 you put a coin on the bottom of the well.
W(x) = W0, a/4 < x < 3a/4 for t > 0, W(x) = 0 everywhere else, H = H0 + W.
If at t = 0 the system is in the state Φ3(x), what is the probability of finding it in Φ1(x) at time t?

Solution: