Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is generating functional and vacuum-to-vacuum boundary conditions in QFT?

  1. Jul 21, 2011 #1
    Hello everyone :) I'm reading the book QFT - L. H. Ryder, and I don't understand clearly what are the generating functional Z[J] and vacuum-to-vacuum boundary conditions? Help me, please >"<
     
  2. jcsd
  3. Jul 21, 2011 #2
    I don't know that book, but if it is an introduction to QFT, it should definitely explain those terms. The generating functional is given by

    [itex]Z \left[ J \right]=<0|0>[/itex],

    which can be expressed in terms of a path integral. It is called "generating" functional because you can apply functional derivatives with respect to J on it in order to gain vacuum expectation values for operators. What they mean by vacuum-to-vacuum boundary conditions could depend on the context, but it is probably the normalization

    [itex]Z \left[ 0 \right]=0[/itex].
     
  4. Jul 21, 2011 #3
    Thank you very much! ^^

    I had carefully read the book again. The vacuum-to-vacuum boundary conditions turned out to be [itex] \psi(t_i) = \psi_i [/itex] and [itex] \psi(t_f) = \psi_f [/itex].

    :D And, are the operators you talk about above the field operators?
    [itex] \dfrac{\delta Z[J]}{\delta J(t_1)\ldots \delta J(t_n)} = i^n \bra 0 \lvert T(q(t_1)\ldots q(t_n)) \rvert 0 \ket [/itex]
     
  5. Jul 22, 2011 #4
    Ah, I see.

    Yes, that's exactly what I meant!
     
  6. Jul 22, 2011 #5
    Oh, yeah, it's now clearer for me ^^ Thank you!
     
  7. Jul 22, 2011 #6
    It's meant to be
    [itex]Z \left[ 0 \right]=1[/itex],
    sorry!
     
  8. Jul 23, 2011 #7
    Does it mean vacuum is still vacuum if there is no source ?
     
  9. Jul 23, 2011 #8
    I guess you could put it like that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is generating functional and vacuum-to-vacuum boundary conditions in QFT?
  1. Vacuum in Qft (Replies: 7)

Loading...