# Interaction of three particles in $\phi^4$ theory

I am trying to compute, without using Feynman diagrams, the scattering amplitude in $$\phi^4$$ theory with three incoming and three outgoing particles. (See this question for an outline of $$\phi^4$$ theory.)

Let's restrict out attention to the case that three $$\phi$$-particles interact to give a single new $$\phi$$ particle, which then react to give a new set of three $$\phi$$-particles. (So the Feynman diagram is a tree graph.)

Initial and final states (with details of normalization omitted; they are not related to my question): $$|i\rangle = \sqrt{8\omega_{\vec p_1}\omega_{\ldots}} a^{\dagger}_{\vec p_1}a^{\dagger}_{\vec p_2}a^{\dagger}_{\vec p_3} |0\rangle\\ |f\rangle = \sqrt\ldots a^{\dagger}_{\vec p_1'}a^{\dagger}_{\vec p_2'}a^{\dagger}_{\vec p_3'} |0\rangle\\$$

I am sure the answer is a constant times $$(-i\lambda)^2 \frac{i}{(p_1+p_2+p_3)^2-m^2},$$ where $$p_1,p_2,p_3$$ are the $$4$$-momenta of initial particles. But I am not sure about whether I have got the correct constant in the front.

Here is how I get the constant.

I first use wick's theorem and consider contractions of the product $$\phi(x)^4\phi(y)^4$$ any of the $$\phi(x)$$ can be contracted with any of the $$\phi(y)$$, giving $$4 \times 4 =16$$ possible contractions. (We do not need to consider other contractions, including ones with more than four $$\phi$$'s contracted, since they are not part of the amplitude we are going to find - we are going to annihilate 3 particles and create three new ones.)

Next, of the remaining six $$\phi$$'s that are not contracted, any one of them can be used to annilate or create any of the particles $$\vec p_j,\vec p_j'$$. This gives $$6!$$ permutations.

The second-order term in Dyson series has the coefficient $$\frac{(-i)}{2}$$. The interaction term in $$\phi^4$$ has coefficient $$\frac{\lambda}{4!}$$. Putting all these numbers together, I end up with the coefficient $$\frac{16 \times 6!}{2\times 4!\times 4!}=10,$$ which seems quite large.

Is the answer $$10(-i\lambda)^2 \frac{i}{(p_1+p_2+p_3)^2-m^2}$$ correct?

Is the number $$10$$ correct?

Apparently, symmetric factors are not something to worry about, because we are calculating from first principles. (And anyway, the Feynman diagram I mentioned has symmetric factor $$1$$.)

• These 6! contractions that you are talking about do not all correspond to the same diagram. Some of them have your 3 initial particles all going to one vertex and the 3 final particles to the other. Then your propagator is in the s-channel as you have written. But other contractions have 1 or 2 initial & final particles at the same vertex. Then your propagator is in the t-channel. You need to identify these separately. Dec 14, 2020 at 12:01
• @kaylimekay Thank you for the hint. I now work out that $6!$ should be replaced by $2\times 3!\times 3!$, where $3!$ comes from permuting the initial and final 3 particles, and $2$ comes from the ordering - we can either first create $3$ new particles and then annihilate the old ones, or do annihilate first before creating. Is that right? Dec 14, 2020 at 12:34
• Yes, and this precisely cancels the factors in your denominator, so the overall coefficient is 1. Dec 14, 2020 at 12:50

Assuming you take the interaction term to be $$\lambda/4!$$ then, from the Feynman rules, the vertex for the interaction is indeed $$-i\lambda$$.The lowest order diagram is indeed three particles coming in, one (intermediate) propagator and three particles coming out. So you have indeed $$(-i\lambda)^2$$ times the propagator.
• Is $10$ the correct coefficient? (That's what I am asking.) Dec 14, 2020 at 11:26
• It doesn't seem to be $10$, but I do want to know where it goes wrong. Could you explain it a little bit? Thanks. Dec 14, 2020 at 11:32
• Think again about the $6!$. Dec 14, 2020 at 11:40