# Feynman diagram for double-bubble vacuum graph in $\phi^4$ theory

I was trying to do an exercise from the book "QFT for the Gifted Amateur" by Tom Lancaster. It involves computing the momentum space amplitudes of some Feynman diagrams. I was trying to compute the amplitude of the following Feynman diagram:

Now to compute these diagrams in momentum space the book mentioned the following rules:

So following these (p.185) rules I came up with that the amplitude associated with this diagram should be:

$$(2\pi)^4 \delta^4(p-q) \frac{(-i\lambda)}{\text{symmetry factor}} \int{\frac{d^4q~ d^4p}{(2\pi)^8} \frac{1}{q^2 - m^2 + i \epsilon}\frac{1}{p^2 - m^2 + i \epsilon}}$$

However, this turned out be incorrect as the solution mentioned that the answer should be (solutions):

$$\frac{(-i\lambda)}{8} \int{\frac{d^4q ~d^4p}{(2\pi)^8} \frac{1}{q^2 - m^2 + i \epsilon}\frac{1}{p^2 - m^2 + i \epsilon}} \int{d^4 x}$$

So we have no delta function and some strange integration all over space. Can someone explain where I went wrong with my calculation and what is the purpose of the second integral in the solutions?

## 2 Answers

Regarding the lack of an overall energy-momentum conserving delta function and the integral over all space (which contributes an infinite factor), you can see where this comes from when first considering the position-space diagram (I won't do this in full detail with all the factors, but you can go through it yourself): we have the usual integral over $$\int d^4 x$$ along with the propagator $$\Delta (x-x)^2$$. But when we go to momentum-space, e.g. $$\Delta (x-y) = \int d^4p \, e^{-i p (x-y)} \frac{i}{p^2-m^2+ i \epsilon} \ \ ,$$ for the propagator $$\Delta(x-x)^2$$ we end up with the integral over $$\int d^4x$$ left over. For non-bubble diagrams (with external legs), one usually integrates over $$d^4x$$ to get an overall momentum-conserving delta function, but here that isn't possible (the exponential terms vanish). Hence the solution given in the answers linked.

If you fill in the details you should be able to obtain the same answer.

• Ok so where does the integral $\int{d^4 x}$ come from? – mathripper Apr 16 at 11:47
• @mathripper updated, this should answer the question fully now – Eletie Apr 16 at 12:49

The integral $$\int dx$$ is the volume $$VT$$ of space-time, $$V$$ being to the volume of the system and $$T$$ the time that one considers it for. The bubble diagram contributes to the vacuum-vacuum amplitude $$\langle 0, {\rm out}|0 {\rm, in}\rangle =e^{-iW}= e^{-iTV{\mathcal E}}$$ where $$\mathcal E$$ is the ground state (vacuum) energy density. Obviously $$V$$ and $$T$$ are as large as you care to make them, so it is $${\mathcal E}$$ that has physical significance. The quantity $$W$$ is the sum of all connected vacuum diagrams (no incoming or outgoing particles).