# 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?

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.
• Ok so where does the integral $\int{d^4 x}$ come from? – mathripper Apr 16 at 11:47
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).