Quantum Mechanics for Engineers 

© Leon van Dommelen 

D.38 Timedependent perturbation theory
The equations to be solved are
To simplify the use of perturbation theory, it is convenient to use a
trick that gets rid of half the terms in these equations. The trick
is to define new coefficients and by

(D.22) 
The new coefficients and are physically just
as good as and . For one, the probabilities are
given by the square magnitudes of the coefficients, and the square
magnitudes of and are exactly the same as
those of and . That is because the exponentials have
magnitude one. Also, the initial conditions are unchanged, assuming
that you choose the integration constants so that the integrals are
initially zero.
The evolution equations for and are

(D.23) 
with .
Effectively, the two energy expectation values have been turned into
zero. However, the matrix element is now timedependent, if it was
not already. To check the above evolution equations, just plug in the
definition of the coefficients.
It will from now on be assumed that the original Hamiltonian
coefficients are independent of time. That makes the difference in
expectation energies constant too.
Now the formal way to perform timedependent perturbation theory is to
assume that the matrix element is small. Write as
where is a scale factor. Then
you can find the behavior of the solution in the limiting process
by expanding the solution in powers of
. The definition of the scale factor
is not important. You might identify it with a small physical
parameter in the matrix element. But in fact you can take
the same as and as an additional mathematical
parameter with no meaning for the physical problem. In that approach,
disappears when you take it to be 1 in the final answer.
But because the problem here is so trivial, there is really no need
for a formal timedependent perturbation expansion. In particular, by
assumption the system stays close to state , so the
coefficient must remain small. Then the evolution
equations above show that will hardly change. That allows
it to be treated as a constant in the evolution equation for
. That then allows to be found by simple
integration. The integration constant follows from the condition that
is zero at the initial time. That then gives the result cited
in the text.
It may be noted that for the analysis to be valid,
must be small. That ensures that is correspondingly small
according to its evolution equation. And then the change in
from its original value is small of order
according to its evolution equation. So the
assumption that it is about constant in the equation for
is verified. The error will be of order .
To be sure, this does not verify that this error in decays
to zero when tends to infinity. But it does, as can
be seen from the exact solution,
By splitting it up into ranges no larger than
and no larger than 1, you can see
that the error is never larger than order for
no larger than 1. And it is of order
outside that range.
Finally, consider the case that the state cannot just transition to
one state but to a large number of them, each with its
own coefficient . In that case, the individual
contributions of all these states add up to change .
And must definitely stay approximately constant for the
above analysis to be valid. Fortunately, if you plug the approximate
expressions for the into the evolution equation for
, you can see that stays approximately
constant as long as the sum of all the transition probabilities does.
So as long as there is little probability of any transition at
time , timedependent perturbation theory should be OK.