# How can an electron in a steady-state have a non-zero speed?

1. Nov 10, 2013

### Mandragonia

In quantum mechanics one applies the time-independent Schroedinger Equation to a system, in order to find steady-state solutions for the wave function of a particle. One of the best-known examples is the Hydrogen atom. Here the procedure gives one the electron orbitals for the various quantum numbers. All the properties of the electron in an orbital can be computed from the corresponding wave function.

For example, the kinetic energy of the electron can be computed for each orbital. It has a positive, clearly non-zero, expectation value. One can also calculate the mean speed of the electron in the Hydrogen atom. It is found to be 1/137th of the speed of light.

Now, as far as I can see it, there arises a sort of paradox.

[1] The electron's orbitals are derived from the time-independent Schroedinger equation. Therefore the probability functions (absolute value of Psi squared) are also time-independent. This has led physicists to conclude that the electron in the orbital is stationary, not involved in any motion. They speak out against the popularized idea that the electron is rapidly moving in a cloud around the nucleus. They think this popular view is in conflict with the true (stationary) character of the orbital.

[2] On the other hand, as I pointed out above, from the mathematical form of the orbital one can compute different properties of the electron. Including its dynamical properties, such as the kinetic energy. This way one finds that the electron has a mean speed of 1/137th of the speed of light. From this fact one must draw the conclusion that the electron is always moving.

I would like to hear your expert opinions on this issue! And so I come to the following question.

How do physicists reconcile:
[1] the stationary (time-independent) nature of an electron's probability function in an orbital, with
[2] the fact that the non-zero value for the mean speed of the electron implies that it is perpetually in motion?

2. Nov 10, 2013

### atyy

The stationary probability distribution is not the reason that the picture of an electron classically orbiting the proton is discouraged. The picture of classical trajectories (having at each time a definite position) is discouraged because the electron does not have at each moment a particular position before the measurement, but is in a superposition of many positions at each moment.

3. Nov 11, 2013

### hilbert2

If an electron is in a stationary state, its velocity vector has zero expectation value, which means that the electron is equally likely to be moving to any direction. The speed of the electron does not need to have zero value. In fact the only wave function that corresponds to zero speed is the constant function $\psi(x,y,z)=C$, and obviously the electron would not be in any way localized in such a state, as required by the uncertainty principle.

4. Nov 11, 2013

### Mandragonia

Thank you very much. I could not agree more !!!

Perhaps I can sum it up it as follows:
* An electron in an orbital is perpetually in motion. [Since a non-moving electron would violate the uncertainty principle.]
* The mean electron speed as derived from the average kinetic energy is physically real.
* The orbital is a solution of the time-independent Schroedinger equation. It obtains the property of time-independence from the fact that the wave function is a simultaneous superposition of wave packets at all the positions allowed. By construction, the superposition is done in such a way that all effects of motion on the probability function cancel.

Is this correct?

5. Nov 11, 2013

### kith

That's not a conclusion from the formalism but an interpretative statement. The mainstream Copenhagen interpretation is different because it asserts that motion is simply indefinite until a momentum measurement is performed. There's also the de Broglie-Bohm interpretation which says that the electron is standing still (since the HUP is a statistical statement, this isn't in contradiction with it).

6. Nov 12, 2013

### Mandragonia

The average speed of the electron in the 1s orbital in the Hydrogen atom is 2000 km/s. This corresponds roughly to 6.5 * 10^15 revolutions per second around the nucleus. Now you may call this a semi-classical interpretative statement. I think these numbers are pretty convincing though. And of course a dynamic picture connects much better to the classical model of an electron moving in an orbit (as seen for high quantum numbers) than a stationary picture.

7. Nov 12, 2013

### ZapperZ

Staff Emeritus
... and since you like classical model better, then explain a consequence of the classical model that is missing here - where is the continual cyclotron-type EM radiation from the orbiting electron?

Zz.

8. Nov 12, 2013

### Naty1

Of what are you then 'convinced'??

edit: oops, I see Zapper got to an issue immediately.

9. Nov 12, 2013

### Mandragonia

No thanks! This is beyond the scope of this thread.
It is a well-known question though; you can find the explanation in every QM handbook.

Last edited: Nov 12, 2013
10. Nov 12, 2013

### ZapperZ

Staff Emeritus
That makes no sense in the context of CLASSICAL E&M, which you espouses. For example, I put a charge particle in a circular path, it WILL radiate. Classical E&M tells you so, and it does this without having to change state! You cannot get away from that if you chose to adopt that classical scenario. The classical pictures produce NO GROUND STATE. It doesn't prevent the electron from crashing in into the nucleus!

What you have done is pick and choose what works and you chose to ignore what doesn't. If appears that you were more comfortable with the classical picture, and yet, you seem to have zero issues by making ad-hoc elimination of the consequence of such a picture.

Fortunately, physics has no luxury to be as wishy-washy as that.

Zz.

11. Nov 12, 2013

### Naty1

12. Nov 12, 2013

### Jilang

OK, it cannot move in a classical sense, so how does it have angular momentum?

13. Nov 12, 2013

### ZapperZ

Staff Emeritus
The same way we say elementary particles have spin but there's nothing spinning.

I have no idea if you've derived at how the angular momentum quantum number came about from the spherical harmonic solution to the wavefunction. If you haven't, have a look at it. It has more to do with the symmetry of the orbital wavefunction.

Zz.

14. Nov 12, 2013

### Mandragonia

May I remind you that it was YOUR OWN DECISION to change the subject of this thread to "a classical perpective on cyclotron-type EM radiation as emitted (or not) by an electron in a QM system". It is obscure what this subject could possibly mean, and you provided no help whatsoever. Yet you challenged me to come up with an answer (why?), which you then started to pick apart "because classical and QM arguments were mixed up". Well, what did you expect? Perhaps your question was rhetorical, merely an amusing attempt to provoke an incorrect response.

15. Nov 13, 2013

### Naty1

Did that [incorrect] conclusion come from this thread?

As you already know, an electron DOES 'move' in a classical sense, but it's angular momentum arises from quantum mechanical origins in a manner which cannot be completely explained clasically.

The first two responses in this thread explain in general terms, and Zapper mentions harmonic solutions/symmetry:
Post #3 here
Spin and "intrinsic angular momentum"

16. Nov 13, 2013

### ZapperZ

Staff Emeritus
Please note that you said that you accepted the classical picture MORE than the quantum mechanical explanation. You wanted the atom to have the same planetary orbital, classical picture. Remember this?

This is EQUIVALENT to having an electron moving around in circle, just like a cyclotron! And in such a classical picture, classical E&M requires that such a motion will RADIATE EM field!

Your explanation on why it doesn't makes no sense, because you then dipped into quantum mechanics and argued about having a ground state! There is no such thing if you buy into the classical picture. Besides, even if the electron is in a ground state only, it should still radiate based on classical E&M. The electron is undergoing an acceleration when it is in such a motion!

That is why I said you picked and chose from those two different ideas, and what you came up with is something that is completely self-contradictory.

Zz.

17. Nov 13, 2013

### Mandragonia

No, that is incorrect. I have never said such a thing. I don't even know what the contemporary "quantum mechanical explanation" is of the apparent contradiction between movement and non-movement of the electron in an orbital. That is why I asked the experts on this forum for their opinion on this matter! Personally I prefer a picture in which the electron is in motion. But that is all.

Absolutely not!!! I merely referred to statements in QM handbooks, that if you go to higher orbitals (say n=1000), then the probability density becomes radially more pronounced around a central value, and furthermore it becomes more confined to a plane. Therefore, the authors claim, the quantum orbits start to resemble classical orbits; or rather the elliptical orbits of the Bohr model.

The only reason I referred to this is because in the latter the electron is assumed to be really moving. Of course one must always keep in mind that these models are incorrect in certain ways. For that very reason I am NOT an advocate of these models.

18. Nov 13, 2013

### Jilang

Yes, my Eisberg & Resnick book covered that. I get that in the ground state which is spherically symmetric the angular momentum is zero. But with the other shaped orbitals is there a net charge flow around the atom?

19. Nov 13, 2013

### ZapperZ

Staff Emeritus
There may be a net electric dipole moment, yes, but I wouldn't call it a charge flow because of all the discussion we already had in this thread. Again, it has more to do with the symmetry of the orbital, i.e. the shape, rather than actually anything having a movement.

Zz.

20. Nov 13, 2013

### Bill_K

From the Schrodinger Equation can be derived a continuity equation ∂ρ/∂t + ∇·J = 0 where ρ = ψ*ψ is the probability density and J = i(ψ*∇ψ - ψ∇ψ*) is the probability current density. For a real solution, J vanishes identically.

However for solutions with definite angular momentum, ψ is typically not real. ψ ~ eimφ, and J is nonzero. There is a steady flow of probability current about the z-axis. If ψ is the wavefunction of an electron, this also represents a flow of electric current, producing a magnetic dipole moment.

Last edited: Nov 13, 2013
21. Nov 13, 2013

### Naty1

22. Nov 13, 2013

### Jilang

Oh dear, now I am really confused. Who to trust, Mentor or Sci advisor, or are you saying the same thing? What is the distinction between a real solution and solutions with a definite angular momentum. Perhaps not real until it's measured, is that right?

23. Nov 13, 2013

### Bill_K

By real I mean nothing philosophical, just ψ = ψ*.

24. Nov 13, 2013

### Jilang

I get it! T=0, it is measured.

25. Nov 13, 2013

### Naty1

I love it: for an imaginary electron state, ψ, there is a 'real' current probability.
Who knew??