reilly said:
This indicates to me, as I've said before, that virtual states, and virtual particles are artifacts of perturbation theory, and they do not show up in exact solutions -- as in the central force problem described by the Schrodinger Eq.
These are (as often) very wise words.
What you essentially mean is that what we want to calculate is:
( A | B ), and when we do that, using a "basis expansion of unity", then:
<br />
<A | B > = <A|1|B> = <A| \sum_k |k><k| B> = \sum_k <A|k> <k|B> <br />
and we call then the |k> states, virtual states through which our initial state A goes to reach B, and that the choice of this basis |k> is totally arbitrary and just a way to help us organize our calculation. Often, one even doesn't use the FULL set of |k> but limits oneself to a subset (as in the cutoff of a series devellopment).
If you're smart enough to calculate directly (A|B) without needing such a (truncated) basis to help you, then of course you wouldn't have a clue what that other guy is talking about, as "virtual" particles or anything.
However, what I wanted to point out was the following. In all generality, you can never say what you'd like to calculate. For one experimenter, (A|B) is important, and he'll calculate (A|B), and then the probability of obtaining B starting from A, using the good old |(A|B)|^2 ;
but for his peer, it might be more important to measure (A|C). Now, it might very well be, that C is "further down the line" than B, and then we could think of having |B1>, |B2> ... |Bk> as a basis which contains the B cases (in the case the first experimenter is going to calculate several possible outcomes of his experiment, and not just a yes/no answer)
And now, what was the "end" result for the first experimenter, (namely the B states), are now the "virtual states" of the calculation of the second one:
<A|C> = \sum_k <A|B_k> <B_k|C>
Now, when working out
|<A|C>|^2 = \sum_k \sum_l \left((<A|B_k><B_l|A>)(<C|B_k><B_l|C>)\right) =<br />
\sum_k \sum_l \left((<B_l|A><A|B_k>)(<B_l|C><C|B_k>)\right) =<br />
\sum_k \sum_l A_{lk}C_{lk}
When the off-diagonal contributions in the matrix representation of A and C in the basis B can somehow be neglected, one sees that this sum reduces to:
\sum_k A_{kk} C_{kk}
In other words, to go from A to C one goes, in a classical way from A to B_k (with probability A_{kk}), and then from B_k to C with probability C_{kk}, and one sums over all these intermediate results. This is when I say that one could consider the B states as real.
However, if the off-diagonal terms in A_{kl} and C_{kl} are not neglegible in the sum above, then one needs to consider the B states as "virtual" because they cause quantum interference (in other words, neglecting them will give you an error on (A|C) calculated as if B_k were real states, using classical probability).
When the situation is such, that for all thinkable C, we can go through the last sum, then B is, for all practical purposes, a real state.