# Relativistic Fermi Golden Rule?

On online slide notes, it is mentioned that:

Fermi Golden Rule: $$P_{if}=\frac{2\pi}{\hbar}|M_{if}|^2\rho_f$$ where $\rho_f$ is density of final sates --number of quantum states per unit volume - states in a cubical box wih side $l$ wih periodic boundary conditions - $\overrightarrow{p}=2\pi \overrightarrow{n}/l$

$$dp(p)=\frac{nbofstates}{l^3}=\frac{d^3p}{(2\pi )^3}$$

Then he wrote: $$\frac{d^3p}{2E(2\pi )^3}=\theta(p^0)\delta(p^2) \frac{d^4p}{(2\pi)^3}$$

and said conventional to make this relativistic.

I did not understand how can this be conventional? That is I did not understand also how did this derive from the Non-Relativistic equation above.

• More on relativistic FGR: physics.stackexchange.com/q/12034/2451 – Qmechanic May 16 '15 at 11:31
• Yes, I saw that one but it is the $\theta(p^0)$ that worries me and the conventional part. @Qmechanic – Beyond-formulas May 16 '15 at 11:33
• It's "conventional" because there would be others ways to write down a Lorentz invariant measure, but this is the one used. For why this is a Lorentz invariant measure, see joshphysics' answer to another question. – ACuriousMind May 16 '15 at 11:55
• Which online slide notes? – Qmechanic Aug 4 '15 at 12:30