I am currently working through the first chapter of Molecular Electronic-Structure Theory (Helgaker et al.). They are demonstrating that, in a complete basis set, commutation relations that hold for operators in their first quantization representation also hold for those operators in their second quantization representation. Using their notation, in first quantization: $$[r_{\alpha}^c, p_{\beta}^c] = i\delta_{\alpha\beta}^c$$ where \begin{align} r_{\alpha}^c&=\sum_{j=1}^N r_{\alpha}^c(j)\\ p_{\alpha}^c&=\sum_{j=1}^N p_{\alpha}^c(j)\\ \delta_{\alpha\beta}^c&=\sum_{j=1}^N \delta_{\alpha\beta}(j) \end{align} and the Greek letters index cartesian directions. The corresponding second quantization representation of $r_{\alpha}^c$, for instance, would be $$\hat{r}_{\alpha}=\sum_{PQ}[r_{\alpha}^c]_{PQ}\,a_{P}^{\dagger}a_{Q}$$ where $a_{P}^{\dagger}$, $a_{Q}$ are the creation/annihilation operators, respectively, and $$[r_{\alpha}^c]_{PQ}=\int\phi_{P}^*(x)\,r_{\alpha}^c\,\phi_{Q}(x)dx$$ for spin orbitals $\phi_P(x)$, $\phi_Q(x)$. My issue is near the last line of their derivation (1.5.31). In a complete basis set, they state that: \begin{align} [\hat{r}_{\alpha}, \hat{p}_{\beta}] &= \sum_{P,Q=1}^{\infty} [[r_{\alpha}^c, p_{\beta}^c]]_{PQ}\,a_{P}^{\dagger}a_{Q}\\ &= \sum_{P,Q=1}^{\infty} [i\delta_{\alpha\beta}^c]_{PQ}\,a_{P}^{\dagger}a_{Q}\\ &= i\delta_{\alpha\beta}\hat{N} \end{align} where $\hat{N} = \sum_Pa_{P}^{\dagger}a_{P}$ is the number operator. These last two lines are giving me trouble. I understand that a $\delta_{PQ}$ comes out of $[\delta_{\alpha\beta}^c]_{PQ}$ due to the orthonormality of the spin orbitals, which collapses the P,Q sum into $\hat{N}$. However, I do not understand how $\delta_{\alpha\beta}^c$ becomes $\delta_{\alpha\beta}$, as $\delta_{\alpha\beta}^c$ is not at all affected by the integral and should just be able to be pulled out entirely (this is what happens to get the $\delta_{PQ}$ in the first place!). I feel like I am missing something very basic.
Thanks!
