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 $1$.)

  • $\begingroup$ 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. $\endgroup$
    – kaylimekay
    Dec 14, 2020 at 12:01
  • $\begingroup$ @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? $\endgroup$
    – Ma Joad
    Dec 14, 2020 at 12:34
  • $\begingroup$ Yes, and this precisely cancels the factors in your denominator, so the overall coefficient is 1. $\endgroup$
    – kaylimekay
    Dec 14, 2020 at 12:50

1 Answer 1


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.

fyi it is quite cumbersome and even passé to do this calculation working out the contractions explicitly. That is what Feynman diagrams are for.

  • $\begingroup$ Is $10$ the correct coefficient? (That's what I am asking.) $\endgroup$
    – Ma Joad
    Dec 14, 2020 at 11:26
  • $\begingroup$ What do you get when you draw the appropriate Feynman diagram(s)? $\endgroup$ Dec 14, 2020 at 11:29
  • $\begingroup$ 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. $\endgroup$
    – Ma Joad
    Dec 14, 2020 at 11:32
  • 1
    $\begingroup$ Feynman diagrams are quick, but direct calculations helps us to find out the rules of Feynman diagrams, which is the point. $\endgroup$
    – Ma Joad
    Dec 14, 2020 at 11:33
  • $\begingroup$ Think again about the $6!$. $\endgroup$ Dec 14, 2020 at 11:40

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