- The problem statement, all variables and given/known data
Vertex Feynamnn rule for computing the time correlator of fields under an action such as, for example,
Say $S_{int} [\phi] =\int d^4 x \lambda \frac{\phi^4(x)}{4!} + g \frac{\phi^4(x)}{4!} $, $\lambda$ and $g$ the coupling constants.
- Relevant equations
see below
- The attempt at a solution
The solution is the usual a vertex rule for $\lambda$ like interactions with and a separate vertex rule for $g $ like interactions I don’t get why there is no vertex rule for a combined $\lambda g $ like interaction- please see my working below:
$S_{int} [\phi] =\int d^4 x \lambda \frac{\phi^4(x)}{4!} + g \frac{\phi^4(x)}{4!} $, $\lambda$ and $g$ the coupling constants. (apologies signs are probably wrong here , please ignore)
Say I am evaluating $\langle 0|T(\phi(x)\phi(y))|0\rangle $ T the time- correlator function.
By Dyson’s formula I am to look at
$ _0 \langle 0 | T(\phi_0(x)\phi_0(y)) \int^{\infty}_{\infty}e^\frac{-1}{h} dt \ H_{int} (t) |0\rangle _0 $ Where the underscore 0 denotes free-field theory. ($h$ is $h/2\pi$ apologies).
(only need to look at this rather than normalising it also and computed the correlators of the normalisation by noting that ‘bubble’ diagrams will cancel).
Expanding out the exponential I am looking at:
$ _0 \langle 0 | T(\phi_0(x)\phi_0(y)) \int^{\infty}_{\infty}e^\frac{-i}{h} dt H_{int} (t) |0\rangle _0 $
Now let me look at the 2nd term from the exponential expansion. I have, dropping the factorial constants:
$\iint dz_1 dz_2 0_ \langle 0 | T(\phi_0(x)\phi_0(y)) ( \lambda \phi_0^4(z_1) + g \phi_0^4(z_1)) . ( \lambda \phi_0^4(z_2) + g \phi_0^4(z_2)) |0\rangle _0 $
= sole $\lambda^2 $ term + sole $g^2$ term + $2\lambda g \iint dz_1 dz_2 \langle 0 | T(\phi_0(x)\phi_0(y)) ( \lambda \phi_0^4(z_1) \lambda \phi_0^4(z_2) |0\rangle _0 $
Which, as far as I can see, I can apply Wicks theorem to get:
$\iint dz_1 dz_2 G(x-y) = G(x-z_1)G(y-z_2)G(z_1-z_3)^3$
Where $G(x-y) $ is the propagator.
For the case where there are two different fields, a field for each interaction term, it is clear that there would be no cross terms since you can’t contract fields of a different type. However in this case I don’t understand why there is no vertex rule for cross $\lambda g$ terms.