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?

  • $\begingroup$ What is your notion of "second quantization formalism" (for me it's just an archaic name for QFT, which doesn't make much sense here)? Also, note that the position space "eigenfunctions" aren't proper functions, but distributions, so the $\nabla^2$ can't act on them to produce a function, all you might get is another distribution, so you can't expect to be able to talk about the "matrix elements" here. $\endgroup$
    – ACuriousMind
    Commented Dec 9, 2015 at 14:21
  • $\begingroup$ @ACuriousMind Isn't the name second quantion formalim general? I mean the formalism using creation and annihilation operators in condensed matter physics. In many books, they just represnet the kinetic operator like above expression. Then do you mean taht above expression is just wrong because we can't talk about the matrix elements between delta function? Then how can we express the kinetic operator in terms of creation and annihilation operators? $\endgroup$
    – user42298
    Commented Dec 9, 2015 at 14:32

1 Answer 1


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).

  • $\begingroup$ In the first equation (kinetic operator in terms of field operator), is the $\nabla^2$ operator or number? If it is operator, then how does it operate on $ \hat{\Psi(\vec{r})}$ right to the $\nabla^2$? $\endgroup$
    – user42298
    Commented Dec 9, 2015 at 15:09
  • $\begingroup$ In the derivation of the second quantized Hamiltonian e.g. Schwabl, you firstly introduced creation and annihilation operators $\hat{a}_k$ that transfer you between $N$ particle and $N\pm 1$ particle states. In this case, there was no $\nabla$ just an operator. Then you go into position space and both integral over space variables and nabla appear acting on functions $\phi_{k}$. $\endgroup$
    – WoofDoggy
    Commented Dec 9, 2015 at 15:38
  • $\begingroup$ Why would it make sense to move the creation/annihilation operators past or around the $\nabla^2$ sign? $\endgroup$ Commented Jun 11, 2017 at 3:02

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