# The Hermiticity of the Laplacian (and other operators)

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

Alternatively: is the matrix representation of the Laplacian Hermitian?

i.e.

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

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.

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.
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}