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I am currently studying particle physics and recently reached the part of particle decay. Here we converted the Fermi's Golden Rule:

$$\Gamma_{fi} = \frac{2\pi}{\hbar}|T_{fi}|^2 \rho(E_i) $$

to its relativistic counterpart:

$$\Gamma_{fi} = \frac{(2\pi)^4\hbar^2c^3}{2E_a} \int |\mathcal {M}_{fi}|^2 \delta(E_a - E_1 - E_2) \delta^3(\vec{p}_a -\vec{p}_1 - \vec{p}_2) \frac{d^3\vec{p}_1}{(2\pi\hbar)^32E_1}\frac{d^3\vec{p}_2}{(2\pi\hbar)^32E_2} $$

for the decay process $ a \rightarrow 1 + 2$ with a perturbed hamiltonian $ H = H_0 + H'$.

It's obvious to check wether or not this relativistic correct by checking if its Lorentz invariant or not. However I can't seem to understand why $\mathcal M$ is Lorentz invariant. I know that $\mathcal M$ is given by:

$$\mathcal M = \langle\psi'_a|H'|\psi'_1\psi'_2\rangle$$

with $\psi'$ represent the relativistically normalised wave function given as:

$$ \psi' = \sqrt{\frac{2E}{c}}\psi$$

but what about $H'$? I know the Hamiltonian changes when we assume relativity, but is this accounted for in this case?

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In my view, now that you want to get the relativistic decay rate, then you would use relativistic quantum field theory instead of nonrelativistic quantum mechanics. Then the relativistic invariant transition matrix $\mathcal{M}$ is calculated by Feynman diagrams, under which the Lorentz indices are fully contracted. For detail description, you may refer to the Chapter 4 and 5 of $Quantum\ Field\ Theory$ by Peskin.

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