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I'm working through a QFT problem and at one stage in the solutions we have this step:

$$\delta^{(4)}(p - q_1 - q_2) = \delta(E_1 +E_2 - M)\delta^{(3)}(\bf{q_1} - \bf{q_2}).$$

We are working in the rest frame of a meson with mass $M$ and the process is a decay to a nucleon anti-nucleon pair.

I cannot quite see why we are allowed to split the delta function this way. Can anyone break this down further for me?

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Always $$ \delta^4 (k) = \delta^1(k_0) \delta^1(k_1) \delta^1(k_2) \delta^1(k_3) $$ If the momenta in your question are on-shell, then $\vec p=0$ because of the frame chosen,and $p^0=E_{p}=M$ , $q_j^0=E_j$, for the "on-shellness". Putting everything together you get your equality

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  • $\begingroup$ This almost makes sense to me but just one thing: in the $\delta^3$ part why is it not $\delta^3(\bf{-q_1 - q_2})$? since the four momentum is $p - q_1 - q_2$? $\endgroup$ – Gaugegroup1996 May 13 at 21:36
  • $\begingroup$ @Gaugegroup1996 It should indeed be like that, it's probably a typo. $\endgroup$ – Javier May 14 at 1:09
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    $\begingroup$ Hint: what's the difference between $\delta(p)$ and $\delta(-p)$ ? @Gaugegroup1996 $\endgroup$ – tbt May 14 at 15:23
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We can quickly show this using a limit definition of the dirac delta, that is \begin{eqnarray*} \delta(x_1,\dots,x_n) &=& \lim_{\epsilon \to 0^+}\frac{1}{\epsilon^n}e^{-\pi (x_1^2+\dots + x_n^2) / \epsilon^2} \\ &=& \lim_{\epsilon \to 0^+}\frac{1}{\epsilon^n}e^{-\pi x_1^2 / \epsilon^2}\times\dots \times e^{-\pi x_n^2 / \epsilon^2} \\ &=& \delta(x_1)\times \dots \times \delta(x_n) \end{eqnarray*} All you need to do is recognize that $$ p - q_1 - q_2 = (p^0-q^0_1-q^0_2, \vec{p}-\vec{q}_1- \vec{q}_2)$$ You can put this straight into the definition above and get the required pieces (for example if $p$ represents the meson four momentum, clearly it only has one component which is $p^0 = M$ and $\vec{p} = \vec{0}$)

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