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?