How could I solve the following commutation $[\hat{H} , \hat{a}] $ when there seems to be no cancellations in its expansion? In this:
$$\hat{H} = \int \frac{d^3 p}{(2\pi)^3} a^\dagger a \tag{1}$$
I thought that perhaps by expanding this:
$$[\hat{H} , \hat{a}] = \int \frac{d^3 p}{(2\pi)^3} (a^\dagger a a - aa^\dagger a) \tag{2} $$
The Hamiltonian above was calculated from the "The Schrodinger equation from a Lagrangian density", as shown in the link.
On wikipedia there is simple demonstration on the commutation relations of a Hamiltonian operator with its operators , but I don't this works in this case because my Hamiltonian is so different from the one on wiki. Is this so?