Propagator using Functional QFT

[jfy4]

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:
<br /> \langle \psi (x&#039;, t&#039;) | \psi (x,t) \rangle <br />
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
<br /> \langle \psi (x&#039;, t&#039;) | \psi (x,t) \rangle = \int \mathcal{D}\phi \, \langle \psi (x&#039;, t&#039;) |\phi \rangle \langle \phi | \psi (x,t) \rangle = \int \mathcal{D}\phi \, \psi^{&#039; *}[\phi] \, \psi [\phi]<br />
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.

 

[jfy4]

Wait, I think I got it. For a free scalar field the propogator would be like
<br /> \langle 0 | \varphi (x) \varphi(x&#039;) | 0 \rangle<br />
Then we put in a complete set of eigenstates
<br /> \int \mathcal{D}\phi \, \langle 0 | \varphi (x) | \phi \rangle \langle \phi | \varphi(x&#039;) | 0 \rangle = \int \mathcal{D}\phi \, \phi(x) \, \phi(x&#039;) \, \psi^{*}_{0}[\phi] \, \psi_{0}[\phi] <br />
Next would be to explicitly calculate what \psi_{0}[\phi] would be and then do the functional integral I believe.

 

[andrien]

seems you are somewhat confused with it.see here 3.5
http://www.nhn.ou.edu/~milton/p6433/chap3.pdf

 

[jfy4]

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
<br /> a(\vec{k}) = \int d^3 x \, e^{-i \vec{k}\cdot \vec{x}}(\omega(\vec{k})\varphi(x) + i\pi(x) )<br />
and solving for \Psi_{0}[\tilde{\phi}] using
<br /> a(\vec{k}&#039;)\Psi_{0}[\tilde{\phi}]=\omega(\vec{k})\tilde{\phi}(\vec{k}&#039;)\Psi_{0}[\tilde{\phi}]+\frac{\delta \Psi_{0}[\tilde{\phi}]}{\delta \tilde{\phi}(\vec{k}&#039;)}=0<br />
and I get
<br /> \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]<br />
Now I am in the process of solving the gaussian integral
<br /> \int\mathcal{D}\tilde{\phi} \, \tilde{\phi}(\vec{k})\tilde{\phi}(\vec{k}&#039;) \, \Psi_{0}^{*}[\tilde{\phi}] \, \Psi_{0}[\tilde{\phi}]<br />
but I can't seem to get it to work yet . . .

 

[andrien]

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

 

[xepma]

Wait, I think I got it. For a free scalar field the propogator would be like
<br /> \langle 0 | \varphi (x) \varphi(x&#039;) | 0 \rangle<br />
Then we put in a complete set of eigenstates
<br /> \int \mathcal{D}\phi \, \langle 0 | \varphi (x) | \phi \rangle \langle \phi | \varphi(x&#039;) | 0 \rangle = \int \mathcal{D}\phi \, \phi(x) \, \phi(x&#039;) \, \psi^{*}_{0}[\phi] \, \psi_{0}[\phi] <br />
Next would be to explicitly calculate what \psi_{0}[\phi] 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.