Momentum of a hydrogen system is imaginary. Why? The hydrogen ground eigenstate:
$$
\psi(r) = \frac{1}{\sqrt{\pi r_0^3}}\exp\left(\frac{-r}{r_0}\right)
$$
Notice how nice:
$$
\frac{\partial \psi}{\partial r}(r) = \frac{1}{\sqrt{\pi r_0^3}}\frac{-1}{r_0}\exp\left(\frac{-r}{r_0}\right) = \frac{-1}{r_0}\psi(r)
$$
The momentum operator:
$$
\mathbf{\hat p}\psi = -i\hbar\nabla\psi = -i\hbar\left[\frac{\partial\psi}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial\psi}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial\psi}{\partial\phi}\right]
$$
Given $\psi$ depends only on $r$, we can find its eigenvalue:
$$
\mathbf{\hat p}\psi = 
-i\hbar\frac{\partial \psi}{\partial r}(r)\mathbf e_r = 
\frac{i\hbar}{r_0}\psi(r)\mathbf e_r
$$
How is that possible an hermitian operator give an imaginary eigenvalue? The momentum operator is hermitian, right? And finally the grand result:
$$
\langle\mathbf{\hat p}\rangle = 
\langle\psi|\mathbf{\hat p}|\psi\rangle = 
\bigg\langle \psi \bigg|\frac{i\hbar}{r_0}\mathbf e_r\bigg| \psi \bigg\rangle =
\frac{i\hbar}{r_0}\mathbf e_r
$$
So, the average momentum of the hydrogen atom is an imaginary amount. Needless to say, this is precisely what our intuition should expect from this system. What am I doing wrong?
 A: (With the hope there is no typesetting errors) there are several excellent points in your question.
First recall that $\hat
p_x\mapsto -i\hbar{\partial\over \partial x}\, , \hat x\mapsto x\,
,$ with
$$
[\hat x,\hat p\,]=i\hbar\, .
$$
Let
$$
\hat p_r \mapsto  -i\hbar\left({\partial\over \partial r}+{1\over
r}\right)\, ,\qquad \hat r\mapsto  r\, , \tag{1}$$
and $f(r)$ be any function of $r$.
One easily shows that 
$\hat p_r$ and $\hat r$ have the right commutation relation and thus 
established that (1) is a putative $\hat p_r$.  
From the radial part of the Schrodinger equation 
$$
-{\hbar^2\over {2m r^2}}{d\over dr}\left(r^2{d\over
dr}\right)R(r)+(V(r)-E)R(r)-{\hbar^2\over 2m}{\ell(\ell+1)\over
r^2}R(r)=0\, .
$$
one shows that 
$$
{-{\hbar^2\over 2m r^2}}{d\over dr}\left(r^2{d\over dr}\right)R(r)
$$
is nothing but $\left(\hat p_r\right)^2 R(r)$.  This justifies the
additional $1/r$ factor in (1). 
To establish the conditions under which $\hat p_r$ is hermitian, note that
$$
0=\langle{R}\vert {\hat p_rR}\rangle-
\langle{R}\vert {\hat p_rR}\rangle^*
$$
where $R(r)$ is a square integrable function,
leads to the restriction
$$
\lim_{r\to 0}\,r\,R(r)=0\, , \tag{2}
$$
showing that $r\,R(r)$ must go to zero at the origin, as
shown in the usual study of the radial solutions.
Although $\hat p_r$ is hermitian, no observable is associated
with this operator.  To show this, note that, for any $\omega$, the
solution to $\hat p_r\,f(r)=\omega\,f(r)$ is, to within a constant,
$$
f(r)\propto \frac{e^{i\omega r/\hbar}}{r}\, ,
$$
which never satisfies the condition of Eqn.(2).  The
eigenvalue problem for $\hat p_r$ has no physically valid solution.  
