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

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, |y_I(t)> is defined from the state vector in the Schroedinger picture through |y_I(t)>=U₀^T(t,t₀)|y_S(t)>.

.

Here we used

, .

We now can write

.

Therefore

We can rewrite this differential equation in the form of an integral equation.

¯_________________________­

This integral equation can be solved by iteration.

The ket |y_I(t)> can therefore be expanded in a power series of the form

.

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.

.

The interaction picture is often used when H_0S is time-independent and W_S(t) is a small correction to H_0S.  Assume that the problem governed by H_0S is already solved, either exactly, or in some approximation.  Assume W_S(t)=0 for t<0.  Then |y_I(0)>=|y_S(0)>.  Neglecting higher order terms we have

,

with .

Let {|n>} be an orthonormal eigenbasis of H₀ and let the system be in the eigenstate |m> at t=0, i.e. |y_I(0)>=|m>.  We have H₀|m>=E_m|m>,

The probability P(E_k,t) of finding the system in the eigenstate |k> of H₀ at time t, i.e. the probability of measuring the eigenvalue E_k, is |<k|y_I(t)>|²..  (The predictions of quantum mechanics are independent of the representation.)

.

With <k|m>=0 we have

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

If W_S(t) is independent of time, i.e. if a constant small term is added to the Hamiltonian at t=0 then

.

Here and we have used

Problem:

Consider a one-dimensional, infinitely deep well.  Let V(x)=0, for 0<x<a, and V(x)=¥ everywhere else.  The eigenstates of   are  with eigenvalues

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 f₃(x), what is the probability of finding it in f₁(x) at time t?

The evolution of the mean value of an observable

In the Heisenberg picture we have an equation for the evolution of operators.

 .

The corresponding equation in the Schroedinger picture describes the evolution of the mean value of an observable.

.

If A does not explicitly depend on time and [A,H]=0, then the mean value <A> is constant.

Let . Then

.

These two equations express Ehrenfest’s theorem.

<R(t)> denotes the position of the center of the wave packet at time t.  Is the trajectory followed by the center of the wave packet the same as predicted by the laws of classical mechanics?

The classical force at the position <R(t)> of the center of the wave packet is  .

In general  , which is the average force over the whole wave packet.

Therefore the trajectory of the center of the wave packet is in general not the one predicted by classical mechanics.  However if the dimensions of the wave packet are much smaller than the dimensions over which V(r) or changes appreciably, we reach the classical limit.