# How to commute a Hamiltonian (integral form) with its operators?

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?

• presumably the operators depend somehow on $p$? Jan 14, 2022 at 17:10
• @ZeroTheHero All I know about the operator is that it depends on p, as it is represented as $\hat{a} (p)$, but I do not know the equation that shows this dependence. How can this help me solve the commutation, I don't quite see it...
– Geop
Jan 14, 2022 at 17:12
• @CosmasZachos , I noticed the error in my writing, I meant to write the commutation relation, and not that it commutes, my apologies, I will correct it straight away
– Geop
Jan 14, 2022 at 17:13
• So you understand that (4.38) of your first reference amounts to an infinity of completely decoupled oscillators of the type described in WP, but in a peculiar normalization? What, specifically, is your problem? Jan 14, 2022 at 17:22
• @CosmasZachos I didn't know that (4.38) meant that the oscillators are decoupled. But then if they are so, does that mean that $\hat{H} = \bar{h} \omega (a a^\dagger - \frac{1}{2} )$ and that $a = \frac{1}{2} (q + ip)$ as shown in WP? How is that so? And what do you mean "peculiar normalisation"? Thank you so much for your help.
– Geop
Jan 14, 2022 at 17:42

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:

$$\int \frac{d^3p}{(2\pi)^3} \, (- \delta_{p,p'} \hat a_{p'}) = - \hat a_{p'}$$

• That is perfect, I can't believe I didn't see this myself. Thank you so much for your help.
– Geop
Jan 14, 2022 at 17:54
• You’re welcome! Btw, there was initially a missing minus sign in my reply. Now fixed! Jan 14, 2022 at 18:04