Is the Laplacian operator, $\nabla^{2}$, a Hermitian operator?

Alternatively: is the matrix representation of the Laplacian Hermitian?


$$\langle \nabla^{2} x | y \rangle = \langle x | \nabla^{2} y \rangle$$

I believe that $\nabla^{2}$ is Hermitian (if it was not, then the Hamiltonian in the time-independent Schroedinger equation would not be Hermitian), but I do not know how one would demonstrate that this is the case.

More broadly, how would one determine whether a general operator is Hermitian? One could calculate every element in a matrix representation of the operator to see whether the matrix is equal to it's conjugate transpose, but this would neither efficient or general.

It is my understanding that Hermiticity is a property that does not depend on the matrix representation of the operator. I feel that there should be a general way to test the Hermiticity of an operator without evaluating matrix elements in a particular matrix representation.

Apologies if this question is poorly posed. I am not sure if I need to be more specific with the definitions of "Hermitian" and "Laplacian". Feel free to request clarification.


2 Answers 2


In general, one needs to write down the integrals for $\langle\phi|\Delta\psi\rangle$ and $\langle\Delta\phi|\psi\rangle$ and transform them into each other using integration by parts.

Hermiticity does not depend on the basis (matrix representation) used. But it depends on the boundary conditions imposed, as one needs to ensure that integration by parts does not generate nonhermitian boundary terms.

With the boundary conditions usually used in quantum mechanics (square integrability in $R^n$), it is a self-adjoint operator, and in particular Hermitian.



Hermitian means self-adjoint with respect to a conjugate-linear form. In this case, the form is $\langle\phi|\psi\rangle=\int\phi\psi^*$, where the integral is over ${\mathbb R}^3$. You know this because $p=\psi\psi^*$ gives the probability of finding a particle at $x$, so $\langle\psi|\psi\rangle=\int p=1$ for a single particle. That’s not meant to be a proof, just a way to rule out almost any other possible conjugate-linear form you might think of.

Self-adjoint means that $\langle\nabla^2\phi|\psi\rangle=\langle\phi|\nabla^2\psi\rangle$. In this case, $$\int(\nabla^2\phi)\psi^* =\int\phi(\nabla^2\psi)^* =\left(\int(\nabla^2\psi)\phi^*\right)^*.$$ So let’s do that. With liberal use of integration by parts and things vanishing at infinity when they are supposed to, $$\begin{align} \int(\nabla^2\phi)\psi^* &=\sum\int\frac{d^2\phi}{d x_i^2}\psi^* =-\sum\int\frac{d\phi}{d x_i}\frac{d\psi^*}{d x_i}\\ &=-\sum\left(\int\frac{d\psi}{d x_i}\frac{d\phi^*}{d x_i}\right)^* =\left(\int(\nabla^2\psi)\phi^*\right)^*. \end{align}$$


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