Let be given by Eq. (3), and let

Then satisfies the Schrödinger equation

This differential equation, together with initial data , is easily seen to be equivalent to the following integral equation:

By taking the Laplace transform and by the convolution theorem we get the Laplace transformed equation:

Let us consider the case of a maximally sharp measurement. In this case
we would take , where **|a>** is some Hilbert space vector.
It is not assumed
to be normalized; in fact its norm stands for the strength of the coupling
(notice that **<a|a>** must have physical dimension ).
Taking look at the formula (4)
we see that now and so we need to know
rather than the full propagator . Multiplying Eq. (11)
from the left by **<a|** and from the right by we obtain:

where is the Laplace transform of :

Thu Feb 22 09:58:31 MET 1996