I was following along Mark Thomson's Modern Particle Physics, and stumble upe the derivation of d$n$ of Fermi golden rule on page 62:

"... For the decay of a particle to a final state consisting of $N$ particles, there are $N-1$ independent momenta in the final state. Thus, the numer of independent states for an N-particle final state is \begin{equation} dn=\prod^{N-1}_{i=1} dn_i = \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \tag{A}. \end{equation} This can be expressend in a more democratic form including the momentum space volume for the $N$th particle $d^3\text{p}_N$ and using delta-function to impose momentum conservation \begin{equation} dn=\prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \delta^3\left(\textbf{p}_a-\sum^N_{i=1}\textbf{p}_i\right)d^3\textbf{p}_N \tag{B}, \end{equation} where $\textbf{p}_a$ is the momentum of the decaying particle.... (and so on) "

Based on the explanation between the steps, I don't understand why the three-dimensional dirac-delta function appear out of nowhere. I do know that physical significances imposing momentum conservation, but why does it have to take the form of dirac-delta? is there an intermediate steps that I'm missing here?


After giving some thoughts and times, I finally came to an answer to my own question. I hope this might help other who get the same problem. This come to some sloppy writing and definition of dirac-delta:

\begin{equation} \int d^nx \delta^n(x-a) = 1\\ \end{equation} similiarly, \begin{equation} \int d^3\textbf{p}_N \;\delta^3\left(\textbf{p}_a-\sum^N_{i=1} \textbf{p}_i\right) = 1 \end{equation} Therefore, from (A) \begin{align} dn&=\prod^{N-1}_{i=1} dn_i, \\ &= \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3}, \\ &= \prod^{N-1}_{i=1} \frac{d^3\textbf{p}_i}{(2\pi)^3} \int \delta^3\left(\textbf{p}_a-\sum^N_{i=1}\textbf{p}_i\right)d^3\textbf{p}_N. \end{align} can be obtained (B) with slight correction to the book.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.