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?
Best Answer
Well this is a quick disclaimer, I don't work in QTF, so perhaps you should crosscheck this with someone.
The operator $(\hat a_p^\dagger \hat a_p)$ is simply the number operator, $\hat N_p$, which counts the number of excitations in a field mode labelled with its propagation vector $\vec{p}$. So if you consider the commutator with the destruction operator (which removes one excitation from a field mode) it will be non-zero as long as they both act on the same mode:
$$ [\hat N_p, \hat a_{p'}] = \hat N_p \hat a_{p'} - \hat a_{p'} \hat N_p = (N_p-\delta_{p,p'}) \hat a_{p'} - N_p \hat a_{p'} = - \delta_{p,p'} \hat a_{p'} $$
Check out equations (4.47) and (4.48) in the material you provided.
Then I would say that the comutator with the Hamiltonian is:
$$ \int \frac{d^3p}{(2\pi)^3} \, (- \delta_{p,p'} \hat a_{p'}) = - \hat a_{p'} $$