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How to get QFT operator expectation values?

  1. Sep 21, 2008 #1
    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 [tex]\delta (0) E_p (2\pi)^3 (2E_p)[/tex]
    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
    [tex]\left|\psi \right> =a\left| 0 \right> + b\left | p \right>[/tex] mean? and how should it be normalized?
    i.e. [tex]\left< \psi \left| \psi \right>[/tex] 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. jcsd
  3. Sep 21, 2008 #2

    olgranpappy

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    Homework Helper

    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.
     
  4. Sep 22, 2008 #3
    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.
     
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