# Proving 3D Hamiltonian operator is Hermitian

A Hermitian operator $$H$$ is defined as $$\int f^*(Hg) d^3\vec{r} =\int (Hf)^*gd^3\vec{r}$$ where $$f$$, $$g$$ are 3D square integrable functions and the integrals are taken over all coordinates.

I am trying to prove that the Hamiltonian operator is Hermitian.

The Hamiltonian operator is given by $$H=\frac{-\hbar^2}{2m}\nabla^2+V(\vec{r}).$$ Proving that operator $$V$$ is Hermitian is trivial. How can I prove that the Laplacian operator $$\nabla^2$$ is also Hermitian?

I read that a hint is to integrate the identity $$f^*\nabla^2g-g\nabla^2f^*=\nabla\cdot(f^*\nabla g-g\nabla f^*).$$ How can the RHS of this identity be integrated to show that the Laplacian operator $$\nabla^2$$ is Hermitian, hence proving that the 3D Hamiltonian is Hermitian?

Integrating the RHS from your hint, using the divergence theorem (and spherical polar coordinates to better make the point): $$\int_V \mathrm{d}^3\mathbf{r}\, \nabla\cdot(f^*\nabla g-g\nabla f^*) = \oint_S r^2\mathrm{d}\Omega \, (f^*\nabla g-g\nabla f^*)\bigg\vert_{r=R},$$ where the last integral is a surface integral evaluated at the edge of the volume $$V$$, located at $$r=R$$.

For an integral over all of space, $$R\rightarrow \infty$$. But if the wavefunctions $$f$$ and $$g$$ are square-integrable (i.e. $$\int |f|^2\, \mathrm{d}^3\mathbf{r}= 1$$), then $$f(r\rightarrow \infty) = g(r\rightarrow \infty) = 0$$.

Hence, the RHS of the equality above is $$0$$. It is zero $$\forall V$$ and $$\forall S$$ so we can take the integrand to actually be zero:

$$\nabla\cdot(f^*\nabla g-g\nabla f^*) = 0.$$

Using the fact that

$$\nabla\cdot(f^*\nabla g-g\nabla f^*) = f^*\nabla^2g-g\nabla^2f^*,$$

then, we have that: $$f^*\nabla^2g-g\nabla^2f^* = 0 \quad \Rightarrow \quad f^*\nabla^2g = g\nabla^2f^*.$$

The last result is what you need to prove that $$f^*(Hg) = (Hf)^\ast g$$.