I'm trying to work out the symmetry factor for a scattering inthe $\phi^4$ theory.
My initial and final states are:
$\left|i\right> = \left|p_1,p_2 \right>$
$\left|f\right> = \left|p_3,p_4 \right>$
The expression for the amplitude to first order in $\lambda$ is of the form:
$$\left<f\right|\frac{-i\lambda}{4!}\phi(x)\phi(x)\phi(x)\phi(x)\left|i\right>$$
So then I have (ommiting the factors of energy which cancels out in the end anyway and the factor of $\frac{-i\lambda}{4!}$):
$$\left<0\right|\int d^4x \left[a_{p_3} a_{p_4} \int\int\frac{d^3q_1d^3q_2}{(2\pi)^6}a_{q_1}^\dagger a_{q_2}^\dagger e^{ix(q_1 + q_2)}\left|0\right>\left<0\right|\int\int\frac{d^3q_3d^3q_2}{(2\pi)^6}a_{q_3} a_{q_4} e^{-ix(q_3 + q_4)} a_{p_1}^\dagger a_{p_2}^\dagger\right] \left|0\right>$$
Now, using the commutation relations I commute the annihilation operators past the creation operators on the RHS and the other way around on the LHS. I drop some delta functions and I get:
$$\left<0\right|\int d^4x\left[\int\int\frac{d^3q_1d^3q_2}{(2\pi)^6}(2\pi)^6\left( \delta(p_3-q_1)\delta(p_4-q_2) + \delta(p_4-q_2)\delta(p_3-q_1)\right) e^{ix(q_1 + q_2)}\left|0\right>\left<0\right|\int\int\frac{d^3q_3d^3q_2}{(2\pi)^6}(2\pi)^6 \left( \delta(p_1-q_3)\delta(p_1-q_4) + \delta(p_1-q_4)\delta(p_2-q_3)\right)e^{-ix(q_3 + q_4)} \right]\left|0\right>$$
Which then gives me:
$$ \int d^4x (2e^{ix(p_3 + p_4)}) (2e^{-ix(p_1 + p_2)}) = 4\delta(p_3+p_4-p_1-p_2)$$
Which is what I expect except for the factor of 4. Because now bringing back the factor of $\frac{-i\lambda}{4!}$:
$$4\delta(p_3+p_4-p_1-p_2)\frac{-i\lambda}{4!}$$
But I know from other sources that the factor of $4!$ should get canceled. I think that by using different fields to annihilate different particles we get a combinatoric factor of $4!$ as we can arrange the four fields in this many possible permutations. But then the final result is by a factor of $4$ off. Why do I get this extra factor next to my $\delta$?
Thanks!
PS. I couldn't find any info on how to break lines in long equations. Any tips for the future?
Best Answer
See this answer How to count and 'see' the symmetry factor of Feynman diagrams? for more, I do not see any symmetry factors of $O(\lambda)$ in that diagram. The main reason is, is that there are no internal lines that can give you a symmetry factor on the $O(\lambda)$.
Update 1: Checking your reference the $4!$ comes from the $4!$ possible contractions because this is a scattering of identical particles so you can contract the fields $4!$ different ways that are identical so what you actually doing is summing up all possible contractions, that is not the same as symmetry factor which occurs on internal lines.
Update 2 example: So for example if we were to label the $\phi$'s even though they are identical say $\phi_1\phi_2\phi_3\phi_4$, then you can contract $\phi_1 $ with $p_1$, $\phi_2$ with $p_2$, $\phi_3 $ with $p_3$, and $\phi_4$ with $p_4$. But you could also have contracted $\phi_1 $ with $p_2$, $\phi_2$ with $p_1$, $\phi_3$ with $p_3$, and $\phi_4$ with $p_4$ and so on ..., so since all these contractions are equal you can some over them and since there is $4!$ ways to contract you some over the $4!$ possibilities.