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:


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 $\frac{-i\lambda}{4!}$:


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 $\delta$?


PS. I couldn't find any info on how to break lines in long equations. Any tips for the future?

  • $\begingroup$ Draw the Feynman diagrams to find the symmetries $\endgroup$ Apr 20, 2017 at 23:07
  • $\begingroup$ Well that's what I'm not sure about. To $O(\lambda)$ the diagrams are just crosses, i.e. 4 legs connected to one vertex. If I number vertices from left bottom corner going clockwise, is (1,2,3,4) the same or different as (1,2,4,3), i.e. if I swap the out-going legs, does that contribute a new diagram? What about (1,2,3,4) and (4,3,2,1), i.e. mirrored diagrams where in and out going legs were swapped? If the former are distinct and latter aren't then I get 6 distinct diagrams which is what I want. Is the rule: if you can rotate/reflect diagram onto any other one, then it doesn't contribute? $\endgroup$
    – Piotr
    Apr 21, 2017 at 9:30
  • $\begingroup$ you can only have symmetry factor for internal lines, external points are labeled with given positions so can't swap them arbitrarily. By the way the way you have written the amplitude are you concatenating the interaction Lagrangian with the final and initial states ? $\endgroup$ Apr 21, 2017 at 14:31
  • $\begingroup$ I think so; I thought I can get the symmetry factor from considering which field is contracted with which particle in the final/initial state, as any of the fields can do the job of creating/annihilating any of the particles, correct? In a diagram that is just a cross I don't really have any internal lines. All external legs are connected to the same one point, aren't they? Looking at link I only have the very first type of diagrams and according to previous comment the distinct diagrams are link. Is this right? $\endgroup$
    – Piotr
    Apr 21, 2017 at 16:22
  • 1
    $\begingroup$ @gautampk yeah but I need to be able to do it the messy way, too. I expand the fields into creation/annihilation operators but throw away some of them as they immediately annihilate the $\left | 0 \right>\left<0\right|$ I inserted in the middle. So all is left is one creation/annihilation operator per field. So the operators with q-subscripts come from the fields. $\endgroup$
    – Piotr
    Apr 22, 2017 at 12:54

1 Answer 1


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.

  • $\begingroup$ Yes, this is confusing. My main source is link, section 3.5.3. It makes intuitive sense to me and that's where I take the $4!$ diagrams from. But then I already have that factor of 4 when I calculate the amplitude of a 'single diagram' that I then wish to multiply by $4!$, the number of diagrams. Do you see what I mean? $\endgroup$
    – Piotr
    Apr 21, 2017 at 19:47
  • $\begingroup$ @Piotr I have expanded on my answer see my explanation. $\endgroup$ Apr 21, 2017 at 20:12
  • 2
    $\begingroup$ Yeah, I get that. But what I think I've done above is I calculated the amplitude for one diagram, whichever of the $4!$ contractions it was. Then I was going to multiply it by the number of diagrams which is, as you said, $4!$. But during my calculation of the single-diagram amplitude I already picked up a factor of 4, if I times that by an additional $4!$ I won't get the amplitude my source (and other sources) claim, I'll get 4 times that. So my question is: where did that 4 come from, since I think I did my single-diagram calculation correctly. Thanks for your help! $\endgroup$
    – Piotr
    Apr 21, 2017 at 20:45

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