Propagator using Functional QFT

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jfy4
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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.
 
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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.
 
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 . . .
 
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it does not seem that it will work.Are you sure with it?
 
jfy4 said:
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.

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.