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=-...
...\Psi_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.