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Consider the lagrangian of the real scalar field given by $$\mathcal L = \frac{1}{2} (\partial \phi)^2 - \frac{1}{2} m^2 \phi^2 - \frac{\lambda}{4!} \phi^4$$

Disregarding snail contributions, the only diagram contributing to $ \langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle$ at one loop order is the so called dinosaur:

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To argue the symmetry factor $S$ of this diagram, I say that there are 4 choice for a $\phi_y$ field to be contracted with one of the final states and then 3 choices for another $\phi_y$ field to be contracted with the remaining final state. Same arguments for the $\phi_x$ fields and their contractions with the initial states. This leaves 2! permutations of the propagators between $x$ and $y$. Two vertices => have factor $(1/4!)^2$ and such a diagram would be generated at second order in the Dyson expansion => have factor $1/2$. Putting this all together I get $$S^{-1} = \frac{4 \cdot 3 \cdot 4 \cdot 3 \cdot 2!}{4! \cdot 4! \cdot 2} = \frac{1}{4}$$ I think the answer should be $1/2$ so can someone help in seeing where I lost a factor of $2$?

I could also evaluate $$\langle p_4 p_3 | T (\phi(y)^4 \phi(x)^4) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + (\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 :| p_1 p_2 \rangle + \dots $$ where dots indicate diagrams generated via this correlator that do not contribute at one loop. (I don't know the latex for the Wick contraction symbol so I just write contract). Is there a way to find out the symmetry factor from computing the term $(\text{contract}(\phi(x) \phi(y)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2: | p_1 p_2 \rangle?$

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Let's start with the external legs on the left. There are eight possible places for the first upper-left external leg to attach: it can attach to one of the four possible $\phi_x$ fields, or to one of the four possible $\phi_y$ fields. The lower-left external leg then only has three choices, since if the first leg attached to the $\phi_x$ field, this leg must also attach to a $\phi_x$ field, and similarly for $\phi_y$. So attaching these legs gives a factor of $2\times 4\times 3$.

Now, let's do the legs on the right. If the legs on the left attached to $\phi_x$, the legs on the right must attach to $\phi_y$, and vice-versa. So there are only four choices for the upper-right external leg, and three choices for the upper-left external leg. Thus, attaching these legs gives a factor of $4\times 3$.

Finally, let's attach the internal legs. The first leg has two places to attach, and the second only has one. So we get a factor of $2$.

Overall, the Dyson series gives us a $\frac{1}{2!}$, and the vertices give us a $\frac{1}{4!4!}$, so the symmetry factor is

$$ \frac{2\times 4 \times 3\times 4\times 3\times 2}{2!4!4!}=\frac{1}{2} $$

Your mistake was in neglecting the factor of two that comes about from permuting the role of $\phi_x$ and $\phi_y$.

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  • $\begingroup$ Thanks for your answer, but I did notice the factor of $2$ coming from attaching the internal legs, it's just I had an extra factor of $2$ in the denominator also (coming from the expansion of the exponential in the Dyson expansion) which cancelled this. Why don't you have this factor of 2 from the Dyson expansion in your answer? $\endgroup$ – CAF Dec 6 '16 at 22:13
  • $\begingroup$ @CAF Oh, my mistake. I thought your factor of 2 came from interchanging $\phi_x$ and $\phi_y$. In general, the factor of $n!$ from the Dyson expansion almost always cancels the factor of $n!$ coming from permuting the vertices, so I neglected both these factors. I will edit. $\endgroup$ – Jahan Claes Dec 6 '16 at 22:16
  • $\begingroup$ @CAF Edited, hopefully that helps more! $\endgroup$ – Jahan Claes Dec 6 '16 at 22:24
  • $\begingroup$ Thanks! Yes I think I nearly understood but then I tried to apply my example to the two loop sunset diagram in $\phi^4$ with two external legs. That has symmetry factor 1/12. I said there were 4 ways to attach an x field to the external field and 4 ways to attach a y field to the other external field. That leaves 3! ways to attach the propagators. My numerator is then $4 \cdot 4 \cdot 3!$ and denominator is $4! \cdot 4! \cdot 2$ This yields the right answer but here I didn't take into account the factor of 2 coming from permuting the role of x and y? +1 $\endgroup$ – CAF Dec 6 '16 at 22:41
  • $\begingroup$ @CAF According to Peskin and Schroder, the sunset diagram has a symmetry factor of 1/6. I don't know where you found 1/12 from, but I'm pretty sure that's not right. $\endgroup$ – Jahan Claes Dec 7 '16 at 3:06
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The diagram of interest (the $s$ channel dinosaur) is generated at second order in the Dyson expansion (alongside the $t$ and $u$ channel dinosaur diagrams and the snail $1 \text{PI}$ reducible diagrams) within the correlator $\langle p_4 p_3 | T( \mathcal L(y) \mathcal L(x)) | p_1 p_2 \rangle$. Using Wick's theorem, we can write this explicitly as $$\langle p_4 p_3 | T( \mathcal L(y) \mathcal L(x)) | p_1 p_2 \rangle = \langle p_4 p_3 | : \phi(y)^4 \phi(x)^4 : | p_1 p_2 \rangle + \dots + X(\text{C}(\phi (y) \phi(x)))^2 \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle + \dots $$ where dots indicate other diagrams not of interest, $\mathcal L(x) = -\frac{\lambda}{4} \phi(x)^4$ and $X$ is the number of ways the contraction $(\text{C}(\phi(y) \phi(x)))^2$ can be formed.

There are four possible $\phi(x)$ fields that can be contracted with four possible $\phi(y)$ fields. This means a single contraction shows up $4 \cdot 4$ times. There are then three $\phi(x)$ fields to contract with three $\phi(y)$ fields. This gives a factor $3 \cdot 3$. To avoid an overcount, we divide by $2!$. Therefore, $$X = \frac{4 \cdot 4 \cdot 3 \cdot 3}{2!}$$

Now, we may write the remaining correlator in which the fields are contracted with the in and out external states in terms of field creation and annihilation operators to yield $$ \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle = 4 \langle p_4 p_3 |(\phi(y)^- \phi(x)^- \phi(y)^+\phi(x)^+ )| p_1 p_2 \rangle.$$ where we explicitly normal ordered. The computation leads to $$ \langle p_4 p_3 | : \phi(y)^2 \phi(x)^2 : | p_1 p_2 \rangle = 4 \left(e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} + \begin{cases} p_1 \rightarrow -p_4 \\ p_2 \rightarrow p_1 \\ p_3 \rightarrow p_3 \\ p_4 \rightarrow -p_2 \end{cases} + \begin{cases} p_1 \rightarrow -p_4 \\ p_2 \rightarrow p_2 \\ p_3 \rightarrow p_3 \\ p_4 \rightarrow -p_1 \end{cases} + \left\{ y \leftrightarrow x \right\} \right) $$ where the permutations of the momenta within the braces give rise to the $t$ and $u$ channel dinosaur diagrams. Focusing only on the $s$ channel contribution and inserting back into the $\mathcal T$ matrix element we obtain the expression $$\mathcal A = \frac{(- \lambda)^2}{(4!)^2} \int \text{d}^4 y \,\,\text{d}^4 x \frac{\mathrm{i}^2}{2!} \left(\frac{4 \cdot 4 \cdot 3 \cdot 3}{2!}\right) (\mathrm{i} \Delta_F(y-x))^2 4(e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} + (y \leftrightarrow x)), $$ with $(C(\phi(y) \phi(x)))^2 \equiv (\mathrm{i} \Delta_F (y-x))^2$, the Feynman propagator. The $( y \leftrightarrow x)$ term yields the same contribution as that displayed, readily seen by relabelling the indices on the space time measures. Collecting all numerical factors in front of the integration we arrive at $$\mathcal A = \underbrace{\left(\frac{4 \cdot 4 \cdot 3 \cdot 3 \cdot 4 \cdot 2}{4!\cdot 4! \cdot 2! \cdot 2!}\right)}_{1/2} (-\mathrm{i}\lambda)^2 \int \text{d}^4 y \,\,\text{d}^4 x \,\,(\mathrm{i} \Delta_F(y-x))^2 e^{-i(p_1 +p_2)y} e^{i (p_3+p_4)x} $$ which is the coordinate space representation (easily recovered using Feynman's rules in position space) of the diagram in question with symmetry factor $S = 2$.

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