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One elementary detector

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

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 :


Arkadiusz Jadczyk
Thu Feb 22 09:58:31 MET 1996