Feynman Vertex Rule, Correlator, 2 different coupling constants

[binbagsss]

Homework Statement

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.

Homework 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 $<0|T(\phi(x)\phi(y))|0>$ T the time- correlator function.By Dyson’s formula I am to look at

$_0 < 0 | T(\phi_0(x)\phi_0(y)) \int^{\infty}_{\infty}e^\frac{-1}{h} dt H_{int} (t) |0>_0$ Where the underscore 0 denotes free-field theory. (h is h/2pi 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:

$_0 < 0 | T(\phi_0(x)\phi_0(y)) \int^{\infty}_{\infty}e^\frac{-i}{h} dt H_{int} (t) |0>_0$

Now let me look at the 2nd term from the exponential expansion. I have, dropping the factorial constants:$\int \int dz_1 dz_2 0_ < 0 | T(\phi_0(x)\phi_0(y)) ( \lambda \phi_0^4(z_1) + g \phi_0^4(z_1)) . ( \lambda \phi_0^4(z_2) + g \phi_0^4(z_2)) |0>_0$

= sole $\lambda^2$ term + sole $g^2$ term + $2\lambda g \int \int dz_1 dz_2 < 0 | T(\phi_0(x)\phi_0(y)) ( \lambda \phi_0^4(z_1) \lambda \phi_0^4(z_2) |0>_0$

Which, as far as I can see, I can apply Wicks theorem to get:

$\int \int dz_1 dz_2 G(x-y) = G(x-z_1)G(y-z_2)G(z_1-z_3)^3$

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.

 

[binbagsss]

Just to add apologies my action is probably wrong, to be honest I wanted an example which would yield a fully connected diagram associated to the cross-term, as I am after the general concept- i.e not that there is no rule associated with a 'cross-term vertex' just because such an interaction would yield only disconnected graphs which may be discarded. I initially had a phi^3 + phi^4 term, however at second order of the exponential expansion, and with a two field correlatator, this yielded only disconnected graphs.

 

[binbagsss]

Nevermind, I understand now, cross-terms are allowed I misinterpreted, however, when you expand out the quadratic order of the exponential term, since the internal vertices are integrated over you get $2 \lambda g$ so wouldn't the symmetry rules (or the vertex rule ) for such interactions need to be slightly modified to account for this factor of 2? many thanks