# 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?

• Hint: you took the expectation of a vector and somehow got a scalar! This is incorrect!
– user12029
Jul 7, 2017 at 22:24
• @NeuroFuzzy True. I forgot to include the $\mathbf e_r$. But this doesn't help much, does it? Jul 7, 2017 at 22:25
• It does. Go back to the definition of the inner product in position space, and note that the integral over a spherical shell of a radial vector field is zero!
– user12029
Jul 7, 2017 at 22:37
• Another way to see that the mean momentum is zero is to note that, because of how Fourier transforms work, this wavefunction remains spherically symmetric in momentum space.
– J.G.
Jul 7, 2017 at 22:42
• @NeuroFuzzy The $\mathbf e_r$ of the operator is just a notation. I could as easily calculated $\hat p_r = -i\hbar\frac{\partial}{\partial r}$ alone, yielding the same thing. Jul 7, 2017 at 22:43

(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.
• Hi. Well. If $\hat p_r$ has no eigenvalue, how can we say that expected momentum is the zero (vector)? Jul 8, 2017 at 23:16
• $\hat p_r$ has eigenvalues, but the physics is also in the eigenvectors. The eigenvectors are not normalizable, hence not physical, meaning $\hat p_r$ is not an observable, i.e. there is no physical quantity associated to the radial momentum. This is a bit counterintuitive as there is a well-defined notion of radial momentum in classical physics. This notion does not carry over to QM. Jul 8, 2017 at 23:51