5
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ Hint: you took the expectation of a vector and somehow got a scalar! This is incorrect! $\endgroup$ – user12029 Jul 7 '17 at 22:24
  • $\begingroup$ @NeuroFuzzy True. I forgot to include the $\mathbf e_r$. But this doesn't help much, does it? $\endgroup$ – Physicist137 Jul 7 '17 at 22:25
  • $\begingroup$ 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! $\endgroup$ – user12029 Jul 7 '17 at 22:37
  • $\begingroup$ 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. $\endgroup$ – J.G. Jul 7 '17 at 22:42
  • 1
    $\begingroup$ @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. $\endgroup$ – Physicist137 Jul 7 '17 at 22:43
2
$\begingroup$

(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.

$\endgroup$
4
  • $\begingroup$ Hi. Well. If $\hat p_r$ has no eigenvalue, how can we say that expected momentum is the zero (vector)? $\endgroup$ – Physicist137 Jul 8 '17 at 23:16
  • $\begingroup$ $\hat p_r$ 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. $\endgroup$ – ZeroTheHero Jul 8 '17 at 23:51
  • $\begingroup$ I dont know if it is too late to comment, but the radial part of the momentum of a particle has such an issue. Dirac was aware of this and he has mentioned it in his book. R. Penrose has mentioned it in his book, Road to Reality in the Section: Quantum Hamiltonians.Also check this arxiv.org/pdf/0704.0373.pdf $\endgroup$ – Chetan Waghela Feb 28 '20 at 7:27
  • $\begingroup$ @ChetanWaghela Thanks for the reference. FYI I know I read this point on radial momentum in Messiah and probably elsewhere. $\endgroup$ – ZeroTheHero 19 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.