1. Dec 8, 2006

Let be the Kelin-gordon equation (m=0) with a potential so:

$$(-\frac{\partial ^{2}}{\partial t^{2}}+V(x) )\Phi=0$$

my question is if you consider the wave function above as an operator..is the K-G operator of the form:

$$<0|T(\Phi(x)\Phi(x')|0>$$ T=time ordered

I think that in both cases..we use the same wave function but once is an scalar (or an spinor for electrons) and the other is an escalar...:shy: :shy:

2. Dec 9, 2006

### dextercioby

1.I don't know who Kelin was. Maybe you could supply some reference.

2. Your equation, misses a laplacian.

3. You depicted the Feynman Green function, which is a Green function for the operator written with a Laplacian.

All of course, if you mean "Klein-Gordon"

Daniel.

3. Dec 9, 2006

I apologize "DSextercioby"... i missed the keyboard.. yes i was referring Klein-Gordon equation with rest mass m=0 so:

$$(-\frac{\partial ^{2}}{\partial t^{2}}+\nabla +V(x))\Phi=0$$

then if you define the Green function by $$G(x,x')=<0|T(\Phi(x)\Phi(x'))0>$$

then my question were if the "Phi" wave function defined in both G and K-G equation is the same ,but in one case is an operator and in the other is an scalar with T=time ordered product.

- By the way i looked at the paper by Scwinger ..taking the Dirac equation with Electromagnetism:

$$(i\gamma_{\mu}\partial _{\mu}-eA_{\mu}+m)\Psi =0$$

he got the Green function (i don't know how he did it..:grumpy: ), he got the functional equation:

$$\partial _{\mu}-eA_{\mu}+m+\frac{\delta}{\delta J_{\mu}}G(x,x')=\delta(x-x')$$

4. Dec 9, 2006

### dextercioby

In the field eqn, the $\varphi (x)$ is not a wavefunction, it is a classical field.

In the VEV of the time-ordered product, it is an operator acting on a Fock space. It still keeps the scalar behavior wrt restricted Poincare' transformations.

As for the second part of your post, please supply the reference to Schwinger's paper.

Daniel.

5. Dec 9, 2006