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?


1 Answer 1


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 $f^*(Hg) = (Hf)^\ast g$.


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