# Product of deltas in kinetic second quantization hamiltonian

I am trying to derive the result for a kinetic hamiltonian in second quantization in term of the fields, that is: $\hat{H} = \int - \Psi^\dagger (r) \frac{\hbar^2\hat{\nabla}^2}{2m} \Psi(r)$

$$\hat{h} = -\frac{\hbar^2\hat{\nabla}_1^2}{2m}$$

and I want to obtain an analogous formula in second quantization, given a basis $|k\rangle$, applying the recipe:

$$\hat{H} = \sum_{k,l} \langle k | \hat{h} | l \rangle a^\dagger_k a_l$$

However when the basis is $|r\rangle$, I encounter some (formal) problems. What is $\langle s | \hat{h} | r \rangle$? Ignoring the constants, I know that $\nabla^2 | r \rangle$=$\nabla^2\delta(x-r)$, so the "matrix element" $\langle s | \hat{h} | r \rangle$ should be something along $$\int dx\ \delta(x-s) \nabla^2\delta(x-r)$$ I understand the $\delta$ as a linear functional, so mathematically I can't really define this integral. I don't know how to solve that, or to show that, combined with the field $\Psi(r)$ it represents the operator $\Psi(r) \rightarrow\nabla^2 \Psi(r)$. I'd really appreciate an explanation of this passage, intuitive or rigorous (better both).

Addendum: The meaning of $\hat{p}$ being diagonal in the pos. basis is locality. The integral displayed above does not mix wavefunctions at different points. That should be expected from a derivative operation. It only cares about an infinitesimal neighborhood of a point. The expectation value for some other non-diagonal operator $A$ would be $$\langle A \rangle = \int\mathrm{drdr'}\psi^*(r')A(r,r')\psi(r)$$ "mixing" the wavefunction at different point in space.