If I have $$ \hat{H} = \int d^3r \hat{\psi}^\dagger(\textbf{r}) H_1(\textbf{r})\hat{\psi}(\textbf{r})$$

How does commutation work with the $\hat{\psi}(\textbf{r})$ with the $H_1(\textbf{r})$? I know the commutator relationship

$$ \hat{\psi}(\textbf{r})\hat{\psi}^\dagger(\textbf{r'}) \pm \hat{\psi}^\dagger(\textbf{r'})\hat{\psi}(\textbf{r}) = \delta(r-r')$$

But I don't know what the commutator is for $H_1$ and $\psi$. I was under the impression that $H_1$ is a scalar number, so I can commute it however, but my notes say to sandwich it between creation and annihilation operators, so I'm not sure how to change the order of terms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.