To check that the linear momentum operators are Hermitian, assume that
and are any two proper, reasonably behaved, wave
functions. By definition:
Now the first term in the integral can be integrated with respect to
and is then seen to produce zero. The reason is that and
must become zero at large distances, otherwise their square
integral cannot be zero. That leaves only the second term. And that
equals
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 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 . In that
case, the integral can be approximated to arbitrary accuracy by just
taking large enough.) Using the divergence theorem, the first
integral is