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

I start with a single particle Hamiltonian, like:

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

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).


1 Answer 1


$\newcommand{\bra}[1]{\langle #1 \vert}$ $\newcommand{\ket}[1]{\vert #1 \rangle}$ In the position basis $\hat{p}^2$ acts as $$ -\nabla^2 \psi(x) = \bra{x}\hat{p}^2\ket{\psi}$$ From \begin{align} \bra{\psi}\hat{p}^2\ket{\psi}&=\int\mathrm{dr}\,\bra{\psi}\ket{r}\bra{r}\hat{p}^2\ket{\psi}\\ &=\int\mathrm{dr}\,\psi(r)^*(-\nabla_r^2\psi(r))\\ &=\int\mathrm{drdr'}\,\psi(r')^*(-\delta(r-r')\nabla_r^2)\psi(r)\\ &=\int\mathrm{drdr'}\,\bra{\psi}\ket{r'}\bra{r'}\hat{p}^2\ket{r}\bra{r}\ket{\psi} \end{align} it follows by comparison that $$\bra{r'}\hat{p}^2\ket{r} =-\delta(r-r')\nabla_r^2$$ The $\delta$ simply tells you, that $\hat{p}^2$ is diagonal in the position basis. This argument certainly lacks mathematical rigor. One has to accept the fact that the position eigenstates can be used in this fashion. Or learn some functional analysis.

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.

  • $\begingroup$ Thanks! A bit off-topic but I wonder if there is something important in that term being diagonal in a certain basis. Is there something special in the connection between position and momentum such that the coefficients are diagonal? $\endgroup$
    – Ralph
    Apr 1, 2015 at 17:51
  • $\begingroup$ @Ralph See the edit for my take at it. $\endgroup$
    – Nephente
    Apr 1, 2015 at 18:18

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