# Is there a direct magnetic interaction between two electrons?

1. Jun 3, 2011

### jbb

Spin orbit coupling accounts for the relative motion of a magnetic moment interacting with an electric field.

Is there a direct coupling of the magnetic moments of, say, the electrons in an atom as well? Does a magnetic field emanate from spin 1/2 particles in general?

2. Jun 3, 2011

### kof9595995

I believe there is, just like hyperfine structure involves spin-spin interaction of electron and nucleus, but I have not seen any calculations on spin-spin interaction between eletrons in textbooks.

3. Jun 3, 2011

### clem

Any particles with magnetic moments, including electrons have a spin.spin interaction energy. It could occur for electrons in atoms, and is important for quarks in baryons. Hyperfine splitting in atoms is affected by the spin.spin interaction of the nucleus and an orbiting electron.

4. Jun 3, 2011

5. Jun 3, 2011

### SpectraCat

Typically, for electrons in atoms, we talk about coupling of angular momentum (e.g. spin orbit coupling), rather than magnetic interaction, between electrons. However the effects are not really separable .. part of the magnetic moment of the electrons in atoms arises from their intrinsic angular momentum in the first place.

For multi-electron atoms, you can use angular momentum coupling rules to come up with term symbols for the electronic states. These term symbols use angular momentum coupling rules to associate angular momentum quantum numbers for the total angular momentum, orbital angular momentum, and spin angular momentum of the overall electronic wavefunction. The total angular momentum (quantum number J), is always a good quantum number (neglecting the nuclear hyperfine interaction), but the orbital angular momentum quantum number (L) and spin angular momentum quantum number (S), are only approximately good (at best). By the way, a "good" quantum number is one corresponding to an operator that commutes with the Hamiltonian ... an "approximately good" quantum number corresponds to an operator that does not strictly commute with the Hamiltonian, but where the coupling is small compared to the energy level spacing. If the spin-orbit interaction is particularly large, it doesn't even make sense to talk about overall orbital angular momentum (L) and overall spin angular momentum (S) any more. Rather, it is more appropriate to couple the orbital (l) and spin (s) angular momenta to obtain a total angular momentum (j) for each individual electron, and then treat the coupling of these angular momenta to obtain the overall total angular momentum (J). This scheme is called j-j coupling.

Anyway, the point of all of this is that the magnetic dipole of an atom can be determined from the total angular momentum quantum number, taking into account the angular momentum coupling according to the Lande g-factor.

The wikipedia pages on this stuff are pretty good. If you are interested, you might want to look at the ones on the electron magnetic dipole moment, Term symbols, and angular momentum coupling to start.

6. Jun 5, 2011

not if you could get to absalute zero and stop the spin of the electron at atomic level all magnetic feilds would not exist.

7. Jun 5, 2011

### haael

As a side question: doesn't it depend on frame of reference? For two electrons in rest, there is only electrostatic repell between them. But if we change our frame so that they start moving, they will produce electric current and thus magnetic attraction.

Someone here said some time ago, that photons and gauge bosons in general are not frame-independent, but obey gauge symmetries. So I can believe that there are photons in one frame and not in the other, but I still have problems understanding it.

8. Jun 5, 2011

### SpectraCat

I don't think this is completely correct .. the electrons will still have a magnetic interaction through their spins, even when at rest. A two electron state will always have a total spin of 0 (antiparallel spins) or 1 (parallel spins), and these two spin cases will be non-degenerate, as far as I know.

Last edited: Jun 5, 2011
9. Jun 5, 2011

### haael

That's not the point. Let's consider 2 charged 0-spin particles then. When at rest, they repell each other. When moving (or rather, when we change the viewpoint to a moving frame), their trajectories get curved in each other's magnetic field, so attractive force appears. If we switch to a frame moving fast enough, the attraction will balace with the repell, or even overcome it.

So we can have attracting or repelling particles depending on the frame we choose. I completly miss something here.

10. Jun 5, 2011

### SpectraCat

That is not possible either ... all charged particles have non-zero spin as far as I know.

I am sorry that I don't know the answer to your question .. I am not familiar enough with relativistic effects in quantum mechanics to give a sensible answer. My guess is that the spin-spin interaction may be an important component to the answer however, since the spins are a manifestation of the full relativistic treatment of quantum fields, if I recall correctly.

11. Jun 5, 2011

### clem

No, the electron magnetic moment and its spin do not depend on temperature.
Nothing can stop it spin.

12. Jun 5, 2011

### clem

Charged pions have zero spin.

13. Jun 5, 2011

### SpectraCat

You are right .. I wasn't thinking of mesons when I wrote that (or even of bosons at all). I suppose the charged rho mesons would also be examples of charged particles with zero spin.

Has anyone studied the interactions of charged, spin-zero mesons? Do they even exist long enough for such studies to be undertaken? I guess there would be no Pauli-principle for such particles (since they are bosons). In fact, I guess in principle you could even have a BEC of charged mesons, if they hung around long enough to be cooled and trapped.

Ok .. I will shut up now .. the water is way deeper than my head and there may be sharks coming

14. Jun 5, 2011

### clem

You are missing something. The force between two spin zero particles of equal charge does vary with frame, but it is always repulsive.