# Klein-Gordon Hamiltonian commutator with annihilation (creation) operator

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 $$[H, a_p^\dagger]=\int \frac{d^3p}{(2*\pi)^3}\omega_p[a_p^\dagger a_p, a_q^\dagger]~?$$

• 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. – user245141 Oct 22 '19 at 10:01

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.