How to get QFT operator expectation values?

1. Sep 21, 2008

tim_lou

I am having some great difficulty getting intuition out of the standard quantization of the Klein-Gordon Lagrangian.

consider the H operator. In QM, the expectation values for H in any eigenstates |n> is just
<n|H|n>

now, in QFT, suppose I have a state |p> in the universe, what do I get if I measure the energy?

well, simply the eigenvalues under H, so E_p. But if I go ahead and try:
<p|H|p>, i get $$\delta (0) E_p (2\pi)^3 (2E_p)$$
so, how should I make sense of <p|H|p> ?

in general, suppose I have an operator, Q, corresponding to a measurement of some observables, how do I find the expectation values? specially when the states are not eigenstates of Q?

One more question, what does the state
$$\left|\psi \right> =a\left| 0 \right> + b\left | p \right>$$ mean? and how should it be normalized?
i.e. $$\left< \psi \left| \psi \right>$$ should be what?

Also, In the usual QM, we can roughly think of psi as a state who's probability of being in 0 is |a|^2 and probability of being in p is |b|^2. However, that is completely based on the fact that <0|0> = <p|p>=1, <0|p>=0. in QFT, <p|p> gives delta function at zero, so how to interpret psi?

Last edited: Sep 21, 2008
2. Sep 21, 2008

olgranpappy

The problem you are having doesn't seem to be with QFT, but rather with non-normalizable states... like eigenstates of the momentum operator. Even in single particle quantum mechanics if you try to take expectation values with these states you will end up with a dirac-delta evaluated at zero.

Maybe try to explain in a bit more detail exactly what you are trying to do.

3. Sep 22, 2008

tim_lou

Ah, I see. For some reason, the idea that those interpretations are the same as in the usual momentum eigenstates escape me. I see it clearly now - specially after a good night's sleep. I'll think for a bit and see if I have any further question though.