Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.10.

In deriving the adiabatic theorem, Griffiths (in his section 10.1) shows that the solution to the time-dependent Schrödinger equation can be written as

where the form an orthonormal set of functions that are eigenfunctions of the Hamiltonian at a particular instant of time, and is the dynamic phase. The coefficients are the usual weighting factors, and they depend only on time.

Later in the derivation, he arrives at a differential equation for the :

In the adiabatic approximation, this equation has the approximate solution

where is the geometric phase. In particular, if the system starts in a definite eigenstate then so

with the result that the overall solution becomes

that is, the system stays in the state over time, although its phase can change.

We can extend the adiabatic approximation recursively by using the first approximation 5 to generate the next approximation. We can do this by inserting 5 into 2 and then solving the resulting differential equation. The sum in 2 is reduced to a single term where , the eigenstate in which the system starts at .

This correction to the basic adiabatic approximation now has the ability to predict transitions from the initial state to other states where . We can apply this to the forced oscillator, where we found that in the adiabatic approximation

Here, is the forcing term, and the adiabatic approximation is obtained by assuming that changes very slowly, or to be precise:

To work out the correction, we need to find in 8. We can do this using the raising and lowering operators for the harmonic oscillator. In particular, the momentum operator can be written in terms of them as

Also, recall that the effects of are

How does this help us? We need to find the derivative , so we get, defining :

where the last line follows because and doesn’t depend on .

Now the momentum operator is

so our derivative is

Using the orthonormality of the we have

with all other matrix elements equal to zero.

Returning to 8 we can work out the phase terms from 10 and 11.

Therefore, since ,

[These answers aren’t the same as those given in Griffiths’s question (although the square moduli are the same) but I can’t see anything wrong with my derivation. Comments welcome.]

Note that

so 8 predicts that , thus the sum of the square moduli of the s is greater than 1. However, these values for the s are correct only to first order in . To get the second order corrections, we’d need to insert 28 and 29 back into 2 and integrate again to get new values for the s, which would give for . The process can be continued as long as we like, giving an adiabatic series.