Electrons orbiting the nucleus: angular momentum

[PeroK]

It is nothing to do with classical or quantum intuition. It is a simple physical question: does an electron move around a nucleus? I do not mean in the sense of an electron orbiting in the sense of the Earth orbiting the sun. I simply mean: does the electron move around a nucleus at all?

Define "move".

 

[barnflakes]

Define "move".

If you take the rest frame of the nucleus, does the electron have non-zero velocity (or better, momentum)?

 

[mfb]

If you take the rest frame of the nucleus, does the electron have non-zero velocity (or better, momentum)?

The electron does not have a property "velocity" or "momentum". At least not as a number. There is a momentum operator that transforms one wave function into a different one, so asking if it is zero does not help much (what is zero?).

 

[DrDu]

It is nothing to do with classical or quantum intuition. It is a simple physical question: does an electron move around a nucleus? I do not mean in the sense of an electron orbiting in the sense of the Earth orbiting the sun. I simply mean: does the electron move around a nucleus at all?

I think it is helpful to look at correlation functions here. Let's say we ask whether the electron moves around the nucleus and the z-axis passing through it. Then we can calculate the correlation function to find the electron at $\vec{r}(t)=t \vec{\Omega}_z\times \vec{r}(0)$ if it was at $\vec{r}(0)$ at t=0. $\vec{\Omega}_z$ is angular velocity. If this function is non vanishing for some value of $\Omega_z$, then the electron apparently is rotating the nucleus. Note that it is not necessary to measure neither of the positions at any time, only their correlation. Compare this to a radar trap: There, to measure the velocity of a car, you also don't need to know its position at any time. It is sufficient to measure the shift in the frequency of the radar wave.

 

[vanhees71]

As is written in the above quoted paper in great detail and in almost any textbook on quantum theory, the electron and proton are in an complicated entangled state when the atom is in an energy eigenstate. The solution is simple product state in relative and center-momentum (equal to center-mass in non-relativistic physics) coordinates. The center-momentum coordinates describe the atom as a hole, and it's of course free and thus it's in a (generalized) momentum eigenstate when it is in an energy eigenstate. The center-momentum part is that of a single quasiparticle with the reduced mass as mass in the Coulomb potential, and it must be in the corresponding energy eigenstate of the center-mass motion. Since it's a central-potential problem that are the usual hydrogen wave function which are common eigenstates of energy, $\vec{L}^2$ and $L_z$. The angular momentum in the restframe of the atom by definition refers to the total spin of the atom. So there doesn't need to move anything for that spin to make sense. In the ground state it has no spin at all, because the relative-wave function is an s-state, for p-states the atom has spin 1, for d-states spin 2, and so on. In a stationary state nothing moves by definition!

It's a different question, if you ask about the electron's or the proton's motion. For the electron in the above paper the reduced density matrix is calculated, which answers all your questions conerning the electron, but it only tells you about statistical properties of the observables you seem to look at (angular momentum and what you call "motion", which I'd associate with momentum), as usual in quantum theory if the state is not an eigenstate of the observables you look at.

 

[PeroK]

If you take the rest frame of the nucleus, does the electron have non-zero velocity (or better, momentum)?

Let's take the ground state $\psi_{100}$. The expected value of momentum is 0 (by symmetry), so you need to look at the expected value of the Kinetic Energy (or momentum squared). The expected value of the KE is $-E_1$ where $E_1$ is the total energy of the ground state (hence the expected value of the potential is $2E_1$).

If you measure it, chances are it is moving if the definition of moving is non-zero KE.

But, the expected value of the total AM squared is 0, hence a measurement of $L^2$ will always give 0. So, it isn't orbiting the nucleus, it's moving radially!

Other states with non-zero azimuthal quantum number do, of course, have non-zero $L^2$.

This is where I found a long discussion on Physics Stack Exchange on this very question:

http://physics.stackexchange.com/qu...ly-radial-motion-in-the-hydrogen-ground-state

There are lots of explanations here, some of which are over my head. But, the conclusion I came to is that you can only push your classical notions of what motion is so far.

The other thing I noticed is that in another state $\psi_{210}$, say, the orbital angular momentum does not account for all of the expected KE. There must still be a component of "radial motion".

Perhaps one answer to your question is:

An electron in a state $\psi_{n00}$ is oscillating radially and an electron in any other state is orbiting the nucleus.

But, this goes against one of the central principles of QM that you can only know what you measure. To define an orbital motion you need to fill in the gaps between measurements and, of course, one measurement of position throws the system out of its stationary state. How would you ever determine that an electron was orbiting the nucleus in an orbit of a given shape?

You simply cannot watch an electron go round and round (or oscillate back and forwards ghosting through the nucleus) the way you can watch a planet or a pendulum. It's a different world down there!

 

[Nugatory]

If you take the rest frame of the nucleus, does the electron have non-zero velocity (or better, momentum)?

The expectation value of a measurement of the square of the electron's momentum is non-zero. Does this mean that it is "moving"?

If the electron were a classical object, you would be justified in concluding from this that the electron was continuously changing its position, and hence that it was moving (in the sense that the word "moving" is generally understood). However, it is not a classical particle, so this conclusion is not justified - it requires an additional assumption, valid for classical objects but not for electrons, that the particle has a position even in the absence of a position measurement.

There are some fairly serious difficulties in defining "the rest frame of the nucleus", and I suspect that that notion seems natural to you only because you're making similarly unjustified classical assumptions about the nucleus. The standard undergraduate treatment of the hydrogen atom is done in what appears to be "the rest frame of the nucleus", but that's not what's really going on. That treatment is not a solution for a quantum system consisting of nucleus and electron (for that, you need something like the paper that @vanhees71 linked), it's the solution to the much simpler problem of a single particle in an inverse-square field, which happens to be a pretty good approximation of the actual problem. You should bear this in mind when interpreting that passage in Griffiths that you quoted above - Griffiths is an introductory textbook, and contains its fair share of lies to children simplifcations adopted for pedagogical convenience.

 

[DrClaude]

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