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Where does the magnetic field in spin-orbit coupling come from?

by alemsalem
Tags: coupling, field, magnetic, spinorbit
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alemsalem
#1
Dec24-12, 02:59 PM
P: 159
the proton is stationary, and we're assuming there is no magnetic field in the rest frame of the proton. I know if we move to the rest frame of the electron there will be a magnetic field.
but shouldn't we be able to analyze it in the rest frame of the proton?
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mfb
#2
Dec24-12, 03:44 PM
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It easier in the frame of the electron. Even in the frame of the proton, the easiest description would be "the electric field, if we would transfer it to the electron frame, has a magnetic component". In the proton rest frame, you have to consider electron spin, electron velocity and electric field at the same time.
Bill_K
#3
Dec24-12, 04:19 PM
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The standard derivation of spin-orbit coupling is total horseradish. For a classical particle one can talk about its orbit, its centripetal acceleration, and the relativistic Thomas precession that will act on the spin. None of these classical concepts apply to the quantum mechanical Dirac wavefunction for a bound state. Quantum mechanics can only be done in a nonaccelerating frame.

Unfortunately, most books satisfy themselves with a one-line handwaving derivation, and even more unfortunately, most students fall for it. A legitimate calculation can be found, for example, in the old but excellent QM book by Schiff. His approach involves taking the nonrelativistic limit, in which the Dirac spinor can be replaced by a two-component Pauli spinor, assuming a central potential, and simplifying the resulting Hamiltonian.

One can also write down the exact solution for the bound states of a Dirac particle in a Coulomb potential and extract the spin-orbit term from that.

andrien
#4
Dec25-12, 01:14 AM
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Where does the magnetic field in spin-orbit coupling come from?

I don't remember it but a proton can be described by a localized current.and can be described by current j=∇M,where M=mδ(x),m contains contribution from magnetic moment of proton.
this corresponds to vector potential
A=∫j(x')/|x-x'| d3x',apart from some constant.Now one can evaluate B by
B=∇A.
Meir Achuz
#5
Dec25-12, 01:35 PM
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PF Gold
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The Dirac equation (in the rest frame of the proton) gives spin orbit coupling for the electron.
DrDu
#6
Dec26-12, 02:57 AM
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Quote Quote by Bill_K View Post
The standard derivation of spin-orbit coupling is total horseradish.
SO coupling is not an intrinsically QM effect. Hence the explanation with Thomas precession is classically correct and physically illuminating, I don't see why it is horseradish.
The standard derivation in QM is via the Dirac equation, but it is somehow a "shut up and calculate" approach and the significance of the terms arising is usually done via comparison with the classically limiting expressions.
Bill_K
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Dec26-12, 04:32 AM
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SO coupling is not an intrinsically QM effect. Hence the explanation with Thomas precession is classically correct and physically illuminating, I don't see why it is horseradish.
Because the atom is an intrinsically QM system, and IMO it is downright harmful to pretend otherwise. In fact most of the confusion in students' minds about QM comes from their too-persistent image of little BBs in little well-defined orbits. Angular momentum exists classically too, but using a classical description of it on the atomic level in terms of little vectors, is not "illumination", just misleading.
The standard derivation in QM is via the Dirac equation, but it is somehow a "shut up and calculate" approach
This is not a ten-page calculation, more like ten lines. Even a small amount of calculation is sometimes looked down on, but that is the way the universe runs, not on intuition.
the significance of the terms arising is usually done via comparison with the classically limiting expressions.
When I see a term rxpσ in the Hamiltonian I don't need to look at the classical limit to tell me that it represents spin-orbit coupling.
Meir Achuz
#8
Dec26-12, 09:08 AM
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Quote Quote by DrDu View Post
SO coupling IS an intrinsically QM effect.
(My slight correction)

the significance of the terms arising is usually done via comparison with the classically limiting expressions.
Using classical 'intuition' to explain a quantum effect is the cause of most of the
'paradoxes' caused by too much talk and too little calculation.
alemsalem
#9
Dec26-12, 09:48 AM
P: 159
I guess its good to see both ways of doing it, with keeping in mind that its actually QM not a classical effect
BruceW
#10
Dec26-12, 12:10 PM
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Quote Quote by Bill_K View Post
Unfortunately, most books satisfy themselves with a one-line handwaving derivation, and even more unfortunately, most students fall for it. A legitimate calculation can be found, for example, in the old but excellent QM book by Schiff. His approach involves taking the nonrelativistic limit, in which the Dirac spinor can be replaced by a two-component Pauli spinor, assuming a central potential, and simplifying the resulting Hamiltonian.
Yeah, It is always annoying when a semi-classical explanation is half-heartedly used. I prefer it when the professor just says something like "there is a good explanation, but we don't have time to go into it now".

Also, although the explanation using spinors might not take many lines, I think that for students who are just starting to learn quantum mechanics, it might be a bit too much all at once to go into spinors and other stuff.


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