Does the non-zero orbital angular momentum (or z-component of angular momentum) of a stationary state of hydrogen atom imply motion of electron (or at least the probability density $|\Psi|^2$) around the nucleus?

I really don't know how to start this but:
the expectation value for velocity of a particle can be derived from $$\frac{d\langle x \rangle}{dt} = \int x \frac{\partial}{\partial t}|\Psi|^2dx \tag{1}$$ and applying some integration by parts and some substitution of the time-dependent schrodinger equation (as shown in Griffiths book of QM), we will have

\begin{equation}\frac{d\langle x \rangle}{dt} = -\frac{i\hbar}{m}\int \Psi^*\frac{\partial \Psi}{\partial x}dx \tag{2}\end{equation}

and the expectation value for momentum is just \begin{equation}\langle p_x\rangle = m\frac{d\langle x\rangle}{dt} = -i\hbar\int \Psi^*\frac{\partial \Psi}{\partial x}dx \tag{3}\end{equation} so that in applying the momentum operator $p_x = -i\hbar\partial/\partial x$, we are just getting values for the expectation value of momentum.
However, the operator for the z-component of angular momentum reads \begin{equation}L_z = -i\hbar(x\partial/\partial y - y\partial/\partial x) = xp_y - yp_x \tag{4}\end{equation} But for a stationary state of a hydrogen atom, $|\Psi|^2$ is not time-dependent, because \begin{equation}|\Psi|^2 = \Psi^*\Psi = \psi^* e^{iE_nt/\hbar}\psi e^{-iE_nt/\hbar} = \psi^* \psi \tag{5}\end{equation} and the tacked time dependence factor cancels out (This does not occur if the wave-function is a superposition of states, because the tacked time-dependence factor does not necessarily cancel out because of difference in energies). Which also means the "electron cloud (Probability density)" is not moving/changing, in contradiction to what others say that the motion of the electron cloud and not the electron itself is the reason for the presence of angular momentum - because not even the electron cloud is changing.

And it is also contradictory for me that the z-component of angular momentum should be present, since it involves momentum operators and if you look at equation 1 from which the momentum operator is derived, $\frac{\partial}{\partial t}|\Psi|^2 = 0$ if $|\Psi|^2$ is not time dependent, which means $\frac{d\langle x \rangle}{dt} = 0$, $\langle p_x\rangle = m\frac{d\langle x\rangle}{dt} = 0$, and if I am correct, applying $p_x = -i\hbar\partial/\partial x$ should yield $0$?

  • $\begingroup$ An energy eigenstate is stationary. However, if you apply a symmetry breaking external field like a magnetic field, then the situation changes and the energy levels of formerly degenerate states split up and then we get "motion" in the way that you mentioned: the superposition of the formerly degenerate states is not stationary any longer. Having said this... the detailed theory is a lot more complicated because of the necessary relativistic treatment of spin-orbital coupling and the addition of spin-spin coupling. $\endgroup$
    – CuriousOne
    Commented Jul 28, 2016 at 10:47
  • $\begingroup$ Even in classical mechanics, if you have a Liouville density of electrons in a circular orbit but with undefined phase then $⟨\mathbf r⟩$ is identically zero, but the electrons are obviously moving. So, how exactly are your results surprising, again? $\endgroup$ Commented Jul 28, 2016 at 10:52
  • 1
    $\begingroup$ @EmilioPisanty But this is a single electron, where we can describe its motion by change in expectation of position over time. We can never really say that it's moving or not if the change in expectation value is zero. $\endgroup$ Commented Jul 28, 2016 at 10:56

2 Answers 2


This is one of the mysteries of quantum mechanics. If you could measure the velocity of an electron, you would get a non-zero value. But what you can't do is use that velocity to predict where to find it next. The act of measurement disturbs the electron in an essential way.

One popular interpretation of quantum mechanics is a statistical one. This says that the wave function provides the probability density for finding the result of a measurement on an ensemble of identically prepared systems. That is, I start with an atom and measure the velocity of its electron. I get a value. I then prepare an identical atom and measure the velocity of its electron. I get a different value. This is very different from classical mechanics. It also might make a connection for you to @EmilioPisanty comment.

There is no way in which our traditional notion of "orbit" makes sense. There's no way our traditional notion of angular momentum applies. We notice that atoms behave as if they have angular momentum, and then go about building a mathematical structure that describes it. We find that the idea of electrons moving from here to there just has no place. Our description of nature does not include the idea that electrons in atoms move from here to there the way macroscopic objects do.


Aside from the fact that everything about the electron in an atom must be understood in the sense of quantum statistics, as already pointed out in the other answer and the comments, there is still a definite sense in which electrons do "circle around the nucleus" in (stationary) eigenstates of well-defined angular momentum.

Think of electron eigenfunctions as "standing waves" in the field of the nucleus. The angular part of a state of well-defined angular momentum, which is basically the spherical harmonic $Y_l^m(\theta, \phi) \sim e^{im\phi} P_l^m(\cos\theta)$, represents a (stationary) rotating traveling wave around the quantization axis, while the angular parts of the equivalent (and degenerate) atomic orbitals are superpositions of the traveling waves and therefore represent "standing waves".

For instance, for the $p_x$ and $p_y$ orbitals the rotating and counter-rotating traveling waves are given by $Y_1^1 \sim e^{i\phi} \sin\theta$, $Y_1^{-1} \sim e^{-i\phi} \sin\theta$, while the $p_x$ and $p_y$ orbitals themselves correspond to "standing modes" $Y_1^1 + Y_1^{-1} \sim \sin\theta \cos\phi$ and $Y_1^1 - Y_1^{-1} \sim \sin\theta \sin\phi$ respectively.

To see that we really deal with rotating traveling waves it is sufficient to consider the probability current $$ {\bf J}({\bf x}) = \frac{\hbar}{2m_ei} \left[ \Psi^*({\bf x}) \nabla\Psi({\bf x}) - \Psi({\bf x}) \nabla\Psi^*({\bf x}) \right] $$ Like the probability density, and unlike the average momentum or angular momentum, the current has the advantage that it is well-defined as a local quantity, at least wherever the wavefunction and its gradient are well-defined. For eigenfunctions of the form $$ \Psi_{nlm}({\bf r}) \sim e^{im\phi} R_{nl}(r )P_l^m(\cos\theta) $$ $$ R_{nl} = R^*_{nl}, \;\;P_l^m = (P_l^m)^* $$ it amounts to $$ {\bf J}_{nlm}(r, \theta, \phi) \sim \frac{\hbar}{2m_ei} \left[ e^{-im\phi} R_{nl}(r )P_l^m(\cos\theta) \left(e^{im\phi} P_l^m(\cos\theta)\frac{\partial R_{nl}(r )}{\partial r} {\bf \hat r}+ \frac{1}{r} e^{im\phi} R_{nl}(r )\frac{\partial P_l^m}{\partial \theta} {\bf \hat \theta} + \right.\\ \left. + \frac{1}{r\sin\theta} R_{nl}(r )P_l^m(\cos\theta) \frac{\partial e^{im\phi}} {\partial \phi}{\bf \hat \phi} \right) - \right.\\ \left. - e^{im\phi} R_{nl}(r )P_l^m(\cos\theta)\left[ e^{-im\phi} P_l^m(\cos\theta)\frac{\partial R_{nl}(r )}{\partial r}{\bf \hat r} + \frac{1}{r} e^{-im\phi} R_{nl}(r )\frac{\partial P_l^m}{\partial \theta}{\bf \hat \theta} + \right.\\ \left. + \frac{1}{r\sin\theta} R_{nl}(r )P_l^m(\cos\theta) \frac{\partial e^{-im\phi}} {\partial \phi } {\bf \hat \phi} \right) \right] $$ or $$ {\bf J}_{nlm}(r, \theta, \phi) \sim m \frac{\hbar}{2m_er\sin\theta} \left[ R_{nl}(r )P_l^m(\cos\theta) \right]^2 {\bf \hat \phi} = m \left( \frac{\hbar |\Psi({\bf x})|^2}{2m_er\sin\theta}\right) {\bf \hat \phi} $$ That is, the probability current of $\Psi$ lies along the ${\bf \hat \phi}$ unit vector as expected for a wave rotating, but stationary, around the quantization axis.

For the "standing waves" of the regular (real-valued) atomic orbitals it is also possible to draw a very neat analogy with the modes of a circular drum, see this Wikipedia section on Qualitative understanding of shapes (of atomic orbitals).

Caution: The current calculated above is sufficient to justify the point, but is incomplete. It does not take into account the electron's spin in the Coulomb field of the nucleus, but only the "kinematic" part due to $\Psi$ alone. For a complete form including spin see (Probability current of) Spin-s particle in electromagnetic field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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