# Why is the relativistic transition matrix M Lorentz invariant

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?

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.