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In my Quantum Field Theory course, I am having some trouble with the derivation of the Feynman rules from a Lagrangian. While I am confident I can get it under control by the time the exams come around, in the mean time I'm stuck on a homework problem that gives the following Lagrangian density for a scalar field $\phi$:

$$ \mathcal{L} = \frac{1}{2} \, \partial _\mu \phi \, \partial ^\mu \phi - \frac{m^2}{2} \phi ^2 - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4 $$

As I understand it, the first two terms are simply the Klein-Gordon Lagrangian, so we're left with an interaction Lagrangian of

$$ \mathcal{L} _I = - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4 $$

I am then asked to give the interaction vertices for this Lagrangian, which I think are just all possible vertices with three or four arrowless lines connected to them (?).

The second part of the question is: 'Determine the Feynman rules for the vertices (warning: the combinatorics will be non-trivial)' and this is where I have a problem. I really do not understand how to go about doing this.

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You have two interaction vertices in this theory, a local three point and four point interaction with coupling constants given by $\lambda_3$ and $\lambda_4$ respectively. A fully fledged calculation in all gory detail using wick's theorem will allow you to see that for every interaction vertex, given the standard inclusion of $3!$ and $4!$ of the terms within the interaction lagrangian, we have a factor of the form $-\mathrm{i}\lambda$. The feynman rule associated with an interaction vertex is precisely this factor, with $-\mathrm{i}\lambda_3$ for a $3$-point interaction and $-\mathrm{i}\lambda_4$ for a $4$-point one.

Let's do the aforementioned calculation in all gory detail restricting to the $\phi^4$ vertex:

Wick's theorem allows us to extract the correct coordinate space representation of the process with the result that the net numerical factor in front gives rise to the symmetry factor. We will see here that the $1/4!$ present in the interaction lagrangian is exactly cancelled by the cardinality of the symmetry group of the four point vertex (this makes it possible to say that at tree level, we have no non trivial symmetry factors). This vertex is produced (amongst other diagrams) in $$\langle p_3 p_4 | T( \phi(x)^4) |p_1 p_2 \rangle.$$ The term of interest is the one with no contractions, that is $\langle p_3 p_4 | : \phi(x)^4 :|p_1 p_2 \rangle$. Computing this should generate terms of the form $e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that it is equal to $e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots$ In the Dyson expansion we have therefore $$ \mathcal A = -\frac{i\lambda}{4!} \int \text{d}^4 x ( e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots) .$$

Using the result derived in my answer to the question here Symmetry factor via Wick's theorem, the result for $$\langle p_3 p_4 | : \phi(x)^2 \phi(y)^2 : | p_1 p_2 \rangle = 4(e^{-i(p_1+p_2)y} e^{i(p_3+p_4)x} + e^{i(p_4-p_1)y} e^{i(p_3-p_2)x} + e^{i(p_4-p_2)y} e^{i(p_3-p_1)x} + (x \leftrightarrow y)).$$ Specialising this to the case of $x=y$ as we have here, the above reduces to $4! \, e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that indeed the factor of $1/4!$ is cancelled.

The combinatoric argument without all this tedious calculation is that in the correlator we can contract four $\phi(x)$ fields with $p_1$, leaving three for $p_2$, two for $p_3$ and one for $p_4$. This gives $4!$ permutations of the prongs, again cancelling the $1/4!$.

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  • $\begingroup$ +1. I'll just add that I find it easier to see how the vertex factors come about by trying to calculate an n-point function like $\langle 0|T(\phi(z_1)\phi(z_2)\phi(z_3)\phi(z_4))|0\rangle$ rather than an S-matrix element, since S-matrix elements have the additional complication of contracting $\phi$ with kets $|p\rangle$ $\endgroup$ – Jahan Claes Dec 10 '16 at 20:17

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