# Electron speeds in orbitals

1. Apr 4, 2009

### Quantom

Do electrons in higher energy orbitals move faster than ones in lower energy orbitals?

2. Apr 4, 2009

### f95toli

No, because electrons do note "move" around the nucleus.
The idea of the atom being composed of electrons "orbiting" the nucleus just happened to be one of the first models of the atom; it is not correct meaning the concept of "speed" does not apply in this case.

If you want you can think of the the electron wavefunction being "smeared out" around the nucleus, the orbitals are then the positions where you are most likely to find an electron.

3. Apr 4, 2009

### feynmann

How can electrons have momentum if they do not move in atoms?

4. Apr 4, 2009

### Hootenanny

Staff Emeritus
The quantum mechanical definition of momentum does not require "movement" since as f95toli said, the concept of "moment" is not well defined for particles on the quantum scale. Instead, the momentum of a particle is defined in terms of an operator on the wave function of the particle.

5. Apr 4, 2009

### feynmann

If the momentum operator operate on wavefunction, you will get the eigenvalue of momemtum operator, which is the value of momentum itself.
The reason the Schrödinger Equation works is that the speed of electron in hydrogen atom is much slower than the speed of the light. So the electron definitely moves in the atom, Otherwise, we would not need the Dirac's equation for relativistic quantum mechanics

Last edited: Apr 4, 2009
6. Apr 4, 2009

### Quantom

well doesn't the wave function propagate or oscillate at a certain rate or am i wrong about that? And if electrons don't actually move in the atom, they cannot be relativistic?

7. Apr 5, 2009

### Hootenanny

Staff Emeritus
Yes.
No. The eigenvalues of an operator are not observables. The momentum of a system is given by the expectation value of the momentum operator, not it's eigenvalues. Assuming that the wave function, $\psi$ is normalised, the expectation value of any operator $\hat{A}$ is given by,

$$\langle\hat{A}\rangle = \langle\psi | \hat{A} | \psi\rangle$$

So in effect, the expectation value of an observable is obtained by summing each eigenvalue multiplied by its corresponding probability.
As I sad above, "movement" is not well defined on the quantum scale. Classically, one could say a particle moves if at time x it is at position y and at time x' it is at y'. However, in quantum mechanics, one never knows precisely where the particle is. The best you can do is say that at time t, the particle has a certain probability of being at position x.

This is precisely the reason why you will very rarely hear a physicist saying that the electrons move slowly, instead they would say that the electron energies are non-relativistic.

An important point to realise is that the wave function does not correspond to any physically meaningful/observable quantities. When we say a wave function, it doesn't mean that the particle follows the path defined by the wave function, or that the particle oscillates like the wave function.

The wave function is merely a mathematical tool for describing a system of particles.

8. Apr 5, 2009

### clem

You can measure an electron speed in a bound state by
v_rms=\sqrt<p^2/m^2>. In hydrogen, this gives v~1/n.

9. Apr 5, 2009

### alxm

f95toli, you're wrong. Electrons don't 'move' in the classical sense. But saying they don't have a velocity because they don't move classically is misleading. The exact nature of their motion isn't relevant to the question. They have momentum and kinetic energy, and that that kinetic energy is different for different orbitals.

Which is unsurprising since kinetic energy is the largest contributor to the overall orbital energy.
Also, 'orbitals' are not orbits, or |psi|^2, orbitals are the wave function(s).

And yes, electrons can reach relativistic velocities. The electrons in highest energy (core) orbitals of heavy atoms have a significant relativistic momentum.

Last edited: Apr 5, 2009
10. Apr 5, 2009

### sokrates

The concept of movement DOES make sense in the quantum scale. Obviously, you've never heard of quantum mechanical current flow operators, and needless to say current = movement. (See Non-Equilibrium Green's Function methodology). NEGF, Quantum transport, etc...

You don't need to know the exact momentum and displacement of a particle to see whether it is moving or not. If the probability of finding an electron at a particular position changes with time then we conclude that the electron is moving. Because after sufficient time, the probability of finding the electron at that initial position will be zero from which we conclude that the electron is now elsewhere, i.e, it is moving.

You might have never heard a physicist talk about a moving electron, but I assume it's because elementary QM texts (or courses) almost never deal with non-equilibrium problems.

I hear (from world-wide known physicists), and use, the concept of a moving electron in a quantum mechanical sense everyday.

I understand that it's difficult to visualize all the subtle aspects of QM, but saying 'movement does not make sense in quantum scale' is at the very least deeply flawed.

Last edited: Apr 5, 2009
11. Apr 5, 2009

### f95toli

Indeed, but note that Quantom was asking if electrons in higher orbitals move faster than electrons in lower energy states, i.e. the way I understood the question he was thinking of the classical concept of speed; something along the way of a "planetary system".

12. Apr 5, 2009

### f95toli

No one is saying that "movement doesn't make sense"; what we are saying is that the classical concept of speed doesn't make sense for an electron in an atom.
This is obviously not true in the general case; an obvious example would be tunnelling experiments where electrons are "shuttled" between islands by an applied RF voltage. This is still a quantum mechanical system but is is designed in such a way that the electrons are localized to the island most of the time meaning you can calculate their average speed if you want (people are trying to use these systems to make current standards; the current is proportional to the frequency of the applied voltage).
Hence, whether or not one can talk about "speed" in the classical sense depends on the system.

13. Apr 5, 2009

### sokrates

I am quoting what Hootenanny said just a few posts before:
The velocity might make sense in an atom (or a molecule) that is connected to two reservoirs. (the hallmark of nanotechnology)
In fact, there are such systems and the velocities are defined as escape rates.

So even in an atom, you cannot explicitly "ban" the concepts of speed, and movement.

Last edited: Apr 5, 2009
14. Apr 6, 2009

### Hootenanny

Staff Emeritus
I have indeed encountered current operators and am aware of quantum transport and similar phenomenon. Although I am indeed a student, I have not studied elementary quantum mechanics texts for some time.

I will concede that my statement regarding the movement was poorly formed. My meaning was that movement is not defined in the same sense quantum mechanically as it is classically. I thought that this was clear from the following sentence "Classically, one could say a particle moves if at time x it is at position y and at time x' it is at y'. However, in quantum mechanics, one never knows precisely where the particle is. The best you can do is say that at time t, the particle has a certain probability of being at position x". I was using the term "not well defined" in the same sense as the phrase often used in elementary texts: "the position of a particle is not well defined". Obviously, this statement doesn't mean that the concept of position doesn't exists.

I would also like to point out that I never said that "the concept of movement doesn't make sense".

15. Apr 6, 2009

### f95toli

This is slightly OT but escape rates has nothing as such to do the velocity. The escape rate of a system is the average rate (in units of events/s) at which it escapes from one local energy minimum to another, usually either via thermal excitation (over the energy barrier) via tunnelling ("through" the barrier). In the thermal case this rate is given by the famous Arrhenius formula which is simply the attempt rate multiplied by a Boltzmann factor (+a numerical pre-factor which is of the order 1).
Note that the escape is from one state , to another; it doesn't have to include "motion" of any kind.
(I spent about two years measuring the phase escape rate of Josephson junctions when I was a PhD student).

16. Apr 6, 2009

### sokrates

Ok, I see the point. I think I was more concentrated on the 'definition of movement'.
And indeed, in an isolated hydrogen atom it's difficult to talk about a movement of this kind because even the statistical probabilities are not changing... Ok, electron has a momentum but you cannot infer a velocity because momentum is exactly given.

I take back my hasty remarks and apologize. But you weren't crystal clear either : )

17. Apr 6, 2009

### sokrates

Hmm. Good point. I always imagined the escape rates as velocities. My reasoning is twofold:
1) Unit-wise it makes sense!
2) Eventhough it is a level-to-level transition, the levels in real space are spatially apart. So there must be some kind of movement involved? (And maybe that's why we really talk about escaping... How can you escape without moving?)

Please let me know what you think... This is something I want to clarify.
Thanks for the remarks

18. Apr 6, 2009

### Staff: Mentor

There is no generally accepted answer to questions like this. Different people have different answers, depending on which interpretation of QM they subscribe to. There's no way to decide the answer by experiment (at least not yet), so people argue about it on the basis of personal philosophical or metaphysical preferences.

19. Apr 6, 2009

### feynmann

Isn't this so-called Copenhagen interpretation? Why is the Copenhagen interpretation being taken for granted here? In Bohm's version of quantum mechanics, the wave function does correspond to any physically meaningful/observable quantities. It will guide the particle how to move.

20. Apr 7, 2009

### alxm

No, it's not the Copenhagen or any 'interpretation'. The wave function does not correspond to anything which is directly observable, and in that sense, it lacks any physical meaningl.

Any interpretation of the physical 'meaning' of the wave function is therefore precisely that: An interpretation. You cannot distinguish experimentally between interpretations (or they wouldn't be called interpretations, they'd be scientific theories). Saying "the wave function has meaning under the Bohm interpretation" is itself a rather meaningless position.