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?

  • 3
    $\begingroup$ presumably the operators depend somehow on $p$? $\endgroup$ Jan 14, 2022 at 17:10
  • $\begingroup$ @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... $\endgroup$
    – Geop
    Jan 14, 2022 at 17:12
  • $\begingroup$ @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 $\endgroup$
    – Geop
    Jan 14, 2022 at 17:13
  • $\begingroup$ 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? $\endgroup$ Jan 14, 2022 at 17:22
  • $\begingroup$ @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. $\endgroup$
    – Geop
    Jan 14, 2022 at 17:42

1 Answer 1


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:

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

  • $\begingroup$ That is perfect, I can't believe I didn't see this myself. Thank you so much for your help. $\endgroup$
    – Geop
    Jan 14, 2022 at 17:54
  • 1
    $\begingroup$ You’re welcome! Btw, there was initially a missing minus sign in my reply. Now fixed! $\endgroup$ Jan 14, 2022 at 18:04

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