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Probably I'm missing something trivial here. When calculating a commutator of Klein Gordon Hamiltonian with annihilation/creation operator it seems that the operators are inserted under the integral, for example: $$ [H, a_p^\dagger]=[\int \frac{d^3p}{(2*\pi)^3}\omega_pa_p^\dagger a_p, a_p^\dagger]=\int \frac{d^3p}{(2*\pi)^3}\omega_p[a_p^\dagger a_p, a_p^\dagger]. $$ Is there a justification for that or we just need to use another momentum variable for the annihilation/creation operator when inserting it under the integral, i.e $$ [H, a_p^\dagger]=\int \frac{d^3p}{(2*\pi)^3}\omega_p[a_p^\dagger a_p, a_q^\dagger]~? $$

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    $\begingroup$ you definitely should use another variable for the operators in the integral description of $H$ and the annihilation/creation operator with which you commute the Hamiltonian. Just take advantage of the fact that $p$ in the integral is a dummy variable and you can relabel it as whatever you like. $\endgroup$ – user245141 Oct 22 '19 at 10:01
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In the expression for the Hamiltonian, $\mathbf{p}$ is indeed a dummy variable. To stay consistent in notation, you should be renaming the dummy variable, not the variable subscripting $a^\dagger$ on the LHS. For example, rename $p\to p'$ in the integral:

$$[H,a^\dagger_\mathbf{p}] = \left[\int\frac{d^3p'}{(2\pi)^3}\omega_{\mathbf{p}'}a^\dagger_{\mathbf{p}'}a_{\mathbf{p}'}\,,\,a^\dagger_\mathbf{p}\right] = \int\frac{d^3p'}{(2\pi)^3}\omega_{\mathbf{p}'}[a^\dagger_{\mathbf{p}'}a_{\mathbf{p}'},a^\dagger_\mathbf{p}]\,.$$

As you can see, $\mathbf{p}'$ is integrated over. If you continue with the derivation here, you will end up with a $\delta^{(3)}(\mathbf{p}-\mathbf{p}')$ in your integral, collapsing it so that the final answer is in terms of $\mathbf{p}$ only.

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