Kinetic energy Hamiltonian in second quantized form and the delta function

Question

I'm trying to transform the kinetic energy Hamiltonian to momentum basis. Starting with: $$\hat{T} = -\frac{\hbar^2}{2m}\int d^3r\ \hat{\Psi}^{\dagger}(\vec{r}) \nabla^2 \hat{\Psi}(\vec{r})$$ I can expand the field operators in terms of momentum eigenstates, since the system is translationally invariant: $$\hat{\Psi}(\vec{r}) = \frac{1}{\sqrt{V}}\sum\limits_{\vec{k}}e^{i\vec{k}\cdot\vec{r}}\hat{a}_{\vec{k}}.$$ I get: \begin{align} \hat{T} = &-\frac{\hbar^2}{2m}\frac{1}{V}\sum\limits_{\vec{k}_1,\vec{k}_2}\hat{a}_{\vec{k}_1}^{\dagger}\hat{a}_{\vec{k}_2}\int d^3r\ e^{-i{\vec{k}_1}\cdot\vec{r}} \nabla^2 e^{i\vec{k_2}\cdot\vec{r}} \\= &\frac{\hbar^2}{2m}\frac{1}{V}\sum\limits_{\vec{k}_1,\vec{k}_2}\hat{a}_{\vec{k}_1}^{\dagger}\hat{a}_{\vec{k}_2}\vec{k}_2^2\int d^3r\ e^{-i({\vec{k}_1}-\vec{k}_2)\cdot\vec{r}} \\= &\frac{\hbar^2}{2m}\frac{1}{V}\sum\limits_{\vec{k}_1,\vec{k}_2}\hat{a}_{\vec{k}_1}^{\dagger}\hat{a}_{\vec{k}_2}\vec{k}_2^2 V\delta(\vec{k}_1 - \vec{k}_2). \end{align} Here I don't see how this expression should simplify to $$\sum\limits_{\vec{k}}\frac{\hbar^2\vec{k}^2 }{2m}\hat{a}_{\vec{k}}^{\dagger}\hat{a}_{\vec{k}}\, ,$$ given that the sum over $\vec{k}_1=\vec{k}_2$ gives $\delta (0)=+\infty$. Did I make a mistake somewhere? Could this be due to the delta function switching interpretation between a Dirac and a Kronecker delta?

Answer 1

Unless you are, for some unknown reason, insisting that your momentum spectra are discrete, those sums of $k_1$ and $k_2$ should really be integrations. This that the immediate consequence of resolving the difficulty with the Dirac delta that you are describing.

I will also note that $\Psi$ is generically a function of both space and time, even in the canonical formalism. So you should really be expanding by Fourier modes in time as well. All of this would be worked out in detail towards the beginning of most standard references on quantum field theory, such as Peskin and Schroeder.