D.9 Momentum operators are Hermitian

To check that the linear momentum operators are Hermitian, assume that $\Psi_1$ and $\Psi_2$ are any two proper, reasonably behaved, wave functions. By definition:

\begin{displaymath}
\langle\Psi_1\vert{\widehat p}_x\Psi_2\rangle =
\int_{x=...
...* \frac{\hbar}{{\rm i}} \Psi_{2,x} {\,\rm d}x{\rm d}y{\rm d}z
\end{displaymath}

Here the subscript $x$ indicates differentiation with respect to $x$. This can be rewritten as
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int_{x=-\infty}^\infty\int_...
...\frac{\hbar}{{\rm i}} \Psi_2
\right] {\,\rm d}x{\rm d}y{\rm d}z
$\hfill(1)}$
as can be checked by simply differentiating out the product in the first term.

Now the first term in the integral can be integrated with respect to $x$ and is then seen to produce zero. The reason is that $\Psi_1$ and $\Psi_2$ must become zero at large distances, otherwise their square integral cannot be zero. That leaves only the second term. And that equals

\begin{displaymath}
\langle {\widehat p}_x\Psi_1\vert\Psi_2\rangle =
\int_{x...
...si_1}{\partial x}\right)^*
\Psi_2{\,\rm d}x{\rm d}y{\rm d}z
\end{displaymath}

(Recall that the complex conjugate of ${\rm i}$ is $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$, hence the minus sign.)

For the mathematically picky, it is maybe a good idea to examine the claim that the first term in the integral in (1) integrates to zero a bit more closely. The integral is definitely zero if the system is not in infinite space, but in a periodic box. The reason is that in that case the lower and upper limits of integration are equal and drop out against each other. To be rigorous in infinite space, you will at first need to limit the region of integration to a distance no more than some large number $R$ away from the origin. (It will be assumed that the initial inner product is well defined, in the sense that the integral has a finite limit in the limit $R\to\infty$. In that case, the integral can be approximated to arbitrary accuracy by just taking $R$ large enough.) Using the divergence theorem, the first integral is

\begin{displaymath}
\int_{S} \left(\Psi_1^* \frac{\hbar}{{\rm i}} \Psi_2\right) n_x {\,\rm d}S
\end{displaymath}

where $S$ is the surface of the sphere $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R$ and $n_x$ the $x$-​component of the unit vector ${\hat\imath}_r$ normal to the surface. Now since the absolute square integrals of the wave functions are finite, for large enough $R$ their square integrals outside $R$ become arbitrarily small. The Cauchy Schwartz inequality then says that the above integral integrated with respect to $r$ must become vanishingly small. And that is not possible unless the integral itself becomes vanishingly small at almost all locations. So you can define a sequence for $R$ where the inner product with the linear momentum swapped over approaches the original inner product. In particular, the two inner products are equal to the degree that they are well defined in the first place.