# Getting rid of double delta function in Feynman rules

[1] A very simple example of feynman rule for scalar fields.

After computing the diagram i have got the following:

$$-i(2\pi)^4g^2\int d^4q \frac{i}{q^2 -m^2c^2}\delta^{(4)}(p_1 - p_3 -q) \delta^{(4)}(p_2 + q -p_4)$$

I'm a little confused about how the integral approached, it integrated over one delta function to get

$$-ig^2\frac{1}{(p_4 - p_2)^2 -m^2c^2}(2\pi)^4\delta^{(4)}(p_1+p_2 - p_3 - p_4)$$

Am i allowed to do that? I mean I have $q$ in both delta functions. Can I just integrate over one of it? It doesn't sound right. What I'm missing here?

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If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Apr 18 '13 at 13:39

That looks correct to me. Consider the basic property of the delta functions $$\int dx f(x) \delta(x-a) = f(a).$$ Nothing forbids $f(x)$ to be a composite function, for example $f(x) \equiv g(x)\delta(x-b)$, so $f(a) = g(a) \delta(a-b)$. Hence we get, $$\int dx f(x) \delta(x-a) \equiv \int dx \, g(x)\delta(x-b) \delta(x-a) = g(a)\delta(a-b).$$
@Aftnix: The factor of $\delta^{(4)}(p_{\mathrm{final}}-p_{\mathrm{initial}})$ is common to all scattering amplitudes, whether it be string theory or QED. It ensures momentum conservation. You can define new Feynman rules to get rid of the delta functions altogether, and just compute the part that changes based on the process which you need for cross sections, decay rates, etc. – JamalS Apr 9 '14 at 19:38