Momentum of a hydrogen system is imaginary. Why?

Question

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?

Answer 1

(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.