[Physics] Kinetic energy operator in second quantization formalism


If we want to express a quantum mechanical oeprator $ \hat{A}$ in second quantization formalism, it is

$$ \hat{A} = \sum_{\alpha, \beta} \langle \alpha | \hat{A}|\beta \rangle c^{\dagger}_{\alpha}c_{\beta} $$

So if we represent the kinetic operator $ \hat{T} = -\frac{\hbar^2}{2m}\nabla^2 $, it is written formally

$$ \hat{T} = -\frac{\hbar ^2}{2m}\int d\vec{r} ~\Psi_{r}^{\dagger} \nabla^2 \Psi_r $$

But here what does $\nabla^2 $ mean? It should be a number, because it is the matrix element $ \langle \alpha | \hat{A}|\beta \rangle $ in the first equation. But I can't guess what should it be.

In the derivation of kinetic operator in second quantization form, we calculate the matrix element between the two position space eigenfunctions, the delta function. So Actually $\nabla^2 $ means the diagonal component of the delta function, but what is it in terms of number? Just $ \infty $ or 0 like delta function?

Best Answer

In second quantization formalism the kinetic energy reads (as you have written) $$\hat{T} = -\frac{\hbar^2}{2m}\int d^3r\ \hat{\Psi}^{\dagger}(\vec{r}) \nabla^2 \hat{\Psi}(\vec{r})$$ If you expand field operator in orthonormal basis $$\hat{\Psi}(\vec{r}) = \sum\limits_{k}\hat{a}_{k}\phi_{k}(\vec{r}),$$ you can write $$\hat{T} = -\frac{\hbar^2}{2m}\sum\limits_{k,n}\hat{a}_{k}^{\dagger}\hat{a}_n\int d^3r\ \phi_{k}^{*}(\vec{r}) \nabla^2 \phi_{n}(\vec{r}).$$ Integral can now be easily calculated - this is just a number. This is a standard thing in condensed matter that brings you e.g. to the Bose-Hubbard model. The integral is usually called the hopping integral. In the context of QFT we also expand the fields in order to diagonalize the Hamiltonian and find elementary excitations (particles).

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