Factorising a 4D Dirac delta function in a rest frame 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? 
 A: 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
A: 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 $\vec{p} = \vec{0}$)
