# Feynman rules for this real scalar field in 2d

1. Jan 6, 2015

### gu1t4r5

1. The problem statement, all variables and given/known data
Consider the following real scalar field in two dimensions:

$S = \int d^2 x ( \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2 - g \phi^3)$

What are the Feynman rules for calculating $< \Omega | T(\phi_1 ... \phi_n ) | \Omega >$
2. Relevant equations

For a phi-4 theory in 4d:

Each propagator contributes a Feynman propagator $D_F (x-y)$
Each vertex z (4 lines to a point) contributes $\frac{-i g}{4!} \int d^4 z$
3. The attempt at a solution

I just wanted to check my understanding is okay. Adapting the rules for a phi-4 theory (this is phi-3 theory, yes?):

Each propagator contributes a Feynman propagator $D_F (x-y)$ (same as before)
Each vertex (now only 3 lines to a point because it is a phi-3 theory) contributes $\frac{-i g}{3!} \int d^3 z$ (because phi-3 not phi-4) or $\frac{-i g}{2!} \int d^2 z$ (because 2d not 4d)

In fourier space, one would integrate over momentum as $\int \frac{d^3 p}{(2 \pi)^3}$

Is this correct for adapting a phi-4 theory to phi-3?
Does going from 4d to 2d change anything here I'm missing?

Thanks

2. Jan 8, 2015

### Couchyam

Hi gu1t4r5,
You are correct that each propagator contributes a factor of $D_F(x-y)$. The remaining rules and combinatorial factors can be derived most easily using the path integral formalism and Gaussian integrals. If you prefer to work directly with particle creation and annihilation operators, you can equivalently use Wick's theorem as presented on page 89 of Peskin and Schroeder, or in the middle of page 261 in Weinberg vol. 1. As for the integrals over vertices, recall that all spacial data comes from the Lagrangian density; if you expand the interaction part of the Lagrangian density, you find that each vertex is an integral over a fixed dimension (2). The combinatorial factors associated with vertices depend on how you define your diagrams. Conventionally, a diagram is a sum of all terms in the perturbative expansion that have the same form when viewed as a product of (infinite dimensional) tensors or equivalently, terms whose graphs are essentially the same. You may want to check the factors multiplying the vertex integrals with how the coupling constant appears in the Lagrangian. For instance, it is conventional in $\phi^4$ theory for the interaction density to be proportional to $\frac{g}{4!}\phi^4$, because the factor of $1/4!$ cancels some combinatorial factors (namely, factors of 4! [typically] for each vertex) that you get when you crank out the tedious algebra of perturbation theory directly. You either include the $1/4!$ factor in your Feynman rules and do combinatorics later, or you anticipate cancelling the $1/4!$ and instead figure out what the so-called `symmetry factor' for each diagram is. Hope this helps!