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Where does the magnetic field in spinorbit coupling come from? 
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#1
Dec2412, 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? 


#2
Dec2412, 03:44 PM

Mentor
P: 11,928

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.



#3
Dec2412, 04:19 PM

Sci Advisor
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P: 4,160

The standard derivation of spinorbit 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 oneline 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 twocomponent 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 spinorbit term from that. 


#4
Dec2512, 01:14 AM

P: 1,020

Where does the magnetic field in spinorbit 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')/xx' d^{3}x',apart from some constant.Now one can evaluate B by B=∇×A. 


#5
Dec2512, 01:35 PM

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PF Gold
P: 2,015

The Dirac equation (in the rest frame of the proton) gives spin orbit coupling for the electron.



#6
Dec2612, 02:57 AM

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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. 


#7
Dec2612, 04:32 AM

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#8
Dec2612, 09:08 AM

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PF Gold
P: 2,015

'paradoxes' caused by too much talk and too little calculation. 


#9
Dec2612, 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



#10
Dec2612, 12:10 PM

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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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