# Amplitudes for ##\phi^4## theory

• A
Homework Helper
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## Summary:

I'm trying to work out theamplitudes for the various Feynman diagrams from the Dyson series expansion of the S-matrix. Sometimes I'm left with constant integrals over all space and other anomalies.

## Main Question or Discussion Point

This is QFT for the gifted amateur, chapter 19, which is generating the various Feynman diagrams and rules. Some calculations are given but I encounter various problems when trying to work them all out.

The starting point is that we want to calculate:
$$\langle q| \hat S | p \rangle = (2\pi)^3 \sqrt{2E_q}\sqrt{2E_p}\langle 0|\hat a_{\vec q} \hat S a_{\vec p}^{\dagger}| 0 \rangle$$
Where:
$$\hat S = T[1 + (\frac{-i\lambda}{4!})\int d^4x \ \hat \phi(x)^4 + (\frac{-i\lambda}{4!})^2(\frac 1 {2!})\int d^4xd^4y \ \hat \phi(x)^4 \hat \phi(y)^4 + \dots]$$
First, if we take the term in ##\lambda##, we have the term:
$$\langle 0|T[\hat a_{\vec q} \hat \phi(x)^4 a_{\vec p}^{\dagger}]| 0 \rangle$$
And, using Wick's theorem I get two non-zero terms coming out of this. The first is covered in the book:
$$12\langle 0|\ [\hat a_{\vec q} \hat \phi(x)] \ [\hat \phi(x) \hat \phi(x)] \ [\hat \phi(x) a_{\vec p}^{\dagger}]| 0 \rangle$$
Where I've used ##[ \ \ ]## to indicate a contraction.

But, I was also looking at the term:
$$3\langle 0|\ [\hat a_{\vec q} a_{\vec p}^{\dagger} ] \ [\hat \phi(x) \hat \phi(x)] \ [\hat \phi(x) \hat \phi(x)]| 0 \rangle$$
Is this term valid? In any case, it leads to an infinity:
$$\delta^4(q-p) \int d^4x \bigg ( \int \frac{d^4k}{(2\pi)^4} \big ( \frac{i}{k^2 - m^2 + i\epsilon} \big ) \bigg )^2$$

A similar thing happens for the ##\lambda^2## term. I have an extra term in the integral that does not correspond to any diagram:
$$[\hat a_{\vec q} a_{\vec p}^{\dagger} ] \ [\hat \phi(x) \hat \phi(y)]^4$$
I can see why the diagram would not make sense, but I can't see why that term vanishes from the integral.

Finally, similar terms crop up in trying to calculate the integrals for other diagrams. I get the right answer except I still have an integral of the form ##\int d^4 x## in front of everything. Mathematically, it all comes back to the same issue, as above.

Any help would be very welcome.

Thanks.

Last edited:
• etotheipi

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PS I think it might boil down to what to do with the Feynman propagator when ##x = y##. If this generates a delta function ##\delta^4(0)##, then that might sort things out?

vanhees71
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That's one of the most simple examples for divergent contributions to higher-order corrections depicted by socalled tadpole diagrams, i.e., a line in a Feynman diagram that connects one spacetime-point with a loop. These are divergent and have to be renormalized. In ##\phi^4## theory the most simple example is the one-loop contribution to the self-energy. It provides simply an infinite constant contribution to ##m^2## and has to be subtracted such that you get at one-loop order the physical mass of the particle.

For details, see Sect. 5.4.1 in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

• dextercioby and PeroK
These "anomolous" terms all go away when you calculate amplitudes from correlation functions via the LSZ theorem. Specifically, the expression
$$\delta^4(q-p) \int d^4x \bigg ( \int \frac{d^4k}{(2\pi)^4} \big ( \frac{i}{k^2 - m^2 + i\epsilon} \big ) \bigg )^2$$
represents a vacuum diagram which factors out and does not appear in scattering amplitudes. It corresponds to this Feynman diagram: See Peskin & Schroeder Chapter 4 for a discussion of vacuum diagrams and how they factor out in calculations, and Chapter 7 for how scattering amplitudes are extracted from correlation functions. If you continue with your current approach to calculating scattering amplitudes you'll also find strange divergences with external leg corrections, i.e. loops on external legs, which as you might expect are also fixed by the correlation function approach. The way to think about these problems is that your states ##|q\rangle## and ##|p\rangle## are not "good" states in the interacting theory and need to be modified to get rid of all the odd divergences (besides the genuine divergences in the QFT).

• vanhees71 and PeroK