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Propagator using Functional QFT

  1. Dec 29, 2012 #1
    Hi,

    I am trying to write down the propagator for a scalar field theory, but I want to try and get it in the functional representation. My plan is to compute the following:
    [tex]
    \langle \psi (x', t') | \psi (x,t) \rangle
    [/tex]
    which gives the amplitude to go from x' to x. Now I guess I have to interpret this state as the ground state of the scalar field, since next I want to drop in a complete set of states
    [tex]
    \langle \psi (x', t') | \psi (x,t) \rangle = \int \mathcal{D}\phi \, \langle \psi (x', t') |\phi \rangle \langle \phi | \psi (x,t) \rangle = \int \mathcal{D}\phi \, \psi^{' *}[\phi] \, \psi [\phi]
    [/tex]
    Is this procedure correct so far? Can I assume the wave functional is the ground state of the field theory in order to continue? Thanks.
     
  2. jcsd
  3. Dec 29, 2012 #2
    Wait, I think I got it. For a free scalar field the propogator would be like
    [tex]
    \langle 0 | \varphi (x) \varphi(x') | 0 \rangle
    [/tex]
    Then we put in a complete set of eigenstates
    [tex]
    \int \mathcal{D}\phi \, \langle 0 | \varphi (x) | \phi \rangle \langle \phi | \varphi(x') | 0 \rangle = \int \mathcal{D}\phi \, \phi(x) \, \phi(x') \, \psi^{*}_{0}[\phi] \, \psi_{0}[\phi]
    [/tex]
    Next would be to explicitly calculate what [itex]\psi_{0}[\phi] [/itex] would be and then do the functional integral I believe.
     
  4. Dec 31, 2012 #3
  5. Dec 31, 2012 #4
    Ah, the generating functional approach, I am trying to stay away from that right now. I want to compute the two point correlation function for a free scalar field theory using "wave functional" representation. I'm still trying to make it work though . . .

    currently, I am working with the lowering operator
    [tex]
    a(\vec{k}) = \int d^3 x \, e^{-i \vec{k}\cdot \vec{x}}(\omega(\vec{k})\varphi(x) + i\pi(x) )
    [/tex]
    and solving for [itex]\Psi_{0}[\tilde{\phi}][/itex] using
    [tex]
    a(\vec{k}')\Psi_{0}[\tilde{\phi}]=\omega(\vec{k})\tilde{\phi}(\vec{k}')\Psi_{0}[\tilde{\phi}]+\frac{\delta \Psi_{0}[\tilde{\phi}]}{\delta \tilde{\phi}(\vec{k}')}=0
    [/tex]
    and I get
    [tex]
    \Psi_{0}[\tilde{\phi}] = N \exp \left[-\frac{1}{2} \int d^3k \, \tilde{\phi}(\vec{k})\omega(\vec{k}) \tilde{\phi} ( \vec{k} ) \right]
    [/tex]
    Now I am in the process of solving the gaussian integral
    [tex]
    \int\mathcal{D}\tilde{\phi} \, \tilde{\phi}(\vec{k})\tilde{\phi}(\vec{k}') \, \Psi_{0}^{*}[\tilde{\phi}] \, \Psi_{0}[\tilde{\phi}]
    [/tex]
    but I can't seem to get it to work yet . . .
     
    Last edited: Dec 31, 2012
  6. Jan 1, 2013 #5
    it does not seem that it will work.Are you sure with it?
     
  7. Jan 2, 2013 #6
    You have to be careful here. You're probably using a coherent-state basis, which is a basis corresponding to eigenvalues of the field operators. This basis is overcomplete, so you need a compensating factor for this. Look at for instance Altland and Simons.
     
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