I am trying to derive the free Hamiltonian of a particle, specifically the kinetic and potential energy, which under second quantization, takes the form of
$$ \hat{T} = -\frac{\hbar^{2}}{2m}\int d\textbf{r} \left(\hat{\psi}^{\dagger}(\textbf{r}) \nabla^{2} \hat{\psi}(\textbf{r}) \right) \tag{1} \\ \hat{U} = \int d\textbf{r}\left(U(\textbf{r})\hat{\psi}^{\dagger}(\textbf{r})\hat{\psi}(\textbf{r})\right) $$
So I start with the fact that a generic one-body operator takes the form of
$$ \hat{O}=\sum_{\alpha,\alpha^{'}}\langle\alpha|o|\alpha^{'}\rangle a_{\alpha}^{\dagger}a_{\alpha^{'}} = \sum_{\alpha,\alpha^{'}}o_{\alpha,\alpha^{'}} a_{\alpha}^{\dagger}a_{\alpha^{'}} \tag{2} $$ And I know that the transformation of basis ($\textbf{r}$ in this case) is given as
$$ a_{\textbf{r}}=\sum_{\alpha} \langle\textbf{r}|\alpha\rangle a_{\alpha} \\ \rightarrow \psi(\textbf{r})=\sum_{\alpha}\phi_{\alpha}(\textbf{r})a_{\alpha} \tag{3} $$ where $\psi(\textbf{r})$ is now a field operator satisfying the commutation relation $[\psi(\textbf{r}),\psi^{\dagger}(\textbf{r})]=\delta(r-r^{\prime})$. Using Eq. (2), the potential operator $\hat{U}$ is written as $$ \hat{U}=\sum_{\alpha,\alpha^{'}}\langle\alpha|U(\textbf{r})|\alpha^{'}\rangle a_{\alpha}^{\dagger}a_{\alpha^{'}}\\ =\sum_{\alpha,\alpha^{'}}\int\int dr dr^{'}\langle\alpha|\textbf{r}\rangle\langle\textbf{r}|U(\textbf{r})|\textbf{r}^{'}\rangle\langle\textbf{r}^{'}|\alpha^{'}\rangle a_{\alpha}^{\dagger}a_{\alpha^{'}} \\ =\sum_{\alpha,\alpha^{'}}\int\int dr dr^{'}\phi_{\alpha}^{*}(\textbf{r})\langle\textbf{r}|U(\textbf{r})|\textbf{r}^{'}\rangle\phi_{\alpha^{'}}(\textbf{r}^{'}) a_{\alpha}^{\dagger}a_{\alpha^{'}} \\ =\int\int dr dr^{'}\langle\textbf{r}|U(\textbf{r})|\textbf{r}^{'}\rangle\psi^{\dagger}(\textbf{r})\psi(\textbf{r}^{'}) $$ where I have used (two) completeness relation $\int|\textbf{r}\rangle\langle\textbf{r}|d\textbf{r} = \mathbb{I}$ and $\int|\textbf{r}^{'}\rangle\langle\textbf{r}^{'}|d\textbf{r}^{'} = \mathbb{I}$
So at this step, I know that the delta function from the commutation relation of $\psi(\textbf{r})$ is used because I would need it to kill off one integral. But I don't see why legitimately I can factor out $U(\textbf{r})$ from $\langle\textbf{r}|U(\textbf{r})|\textbf{r}^{'}\rangle$ so that $$ \int\int dr dr^{'}\langle\textbf{r}|U(\textbf{r})|\textbf{r}^{'}\rangle\psi^{\dagger}(\textbf{r})\psi(\textbf{r}^{'}) \\ =\int\int dr dr^{'}U(\textbf{r})\langle\textbf{r}|\textbf{r}^{'}\rangle\psi^{\dagger}(\textbf{r})\psi(\textbf{r}^{'}) \\ =\int\int dr dr^{'}U(\textbf{r})\delta(\textbf{r}-\textbf{r}^{'})\psi^{\dagger}(\textbf{r})\psi(\textbf{r}^{'})\\ =\int dr U(\textbf{r})\psi^{\dagger}(\textbf{r})\psi(\textbf{r}) $$ so that I get $\hat{U}$ from Eq.(1). Is the factoring out of $U(\textbf{r})$ permitted here? How should I evaluate it in the case of kinetic energy where I have to evaluate $\langle\textbf{r}|\nabla^{2}|\textbf{r}^{'}\rangle$?
Thanks
