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Spinorbit interaction 
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#1
Mar1914, 02:00 AM

PF Gold
P: 574

Fron wiki:
spin–orbit interaction causes shifts in an electron's atomic energy levels due to electromagnetic interaction between the electron's spin and the magnetic field generated by the electron's orbit around the nucleus. This is detectable as a splitting of spectral lines. The splitting is given by [itex]\mu * B[/itex] (mu_{e}* B_{orbit}) I have a few questions: 1 does the angular momentum of the electron L_{e} play any role in this interaction? (L_{e  Bo}) 2 how is experimentally determined the value of L(e) = 1/2 h_{bar}? 3 is there a magnetic field of the electron B_{e}? if yes , what is its value; if not , how come? 4 if there is /or were a B_{e}, how would this interact with the magnetic field of the orbit B_{o}? (B_{e}  B_{o)} Your help is greatly appreciated 


#2
Mar1914, 02:07 AM

Sci Advisor
P: 3,593

Yes, the magnetic field responsible for the splitting is due to the orbital moment of the electron. However, the correct calculation of the
splitting is somewhat more complicated as special relativistic effects come into play, namely so calle Thomas precession. 


#3
Mar1914, 02:12 AM

PF Gold
P: 574

Thanks, DrDu.
Can anyone answer some questions individually, please? In question 4 I meant:  The orbiting (around the nucleus) generates an angular momentum L(o) = [itex]\hbar[/itex], a magnetic field B(o) = ?, and a magnetic moment [itex]\mu[/itex](o)= [itex]\frac{1}{2}\hbar[/itex], is that right?,  The spinning (around itself) generates a magnetic moment [itex]\mu[/itex](e) = [itex] \frac{1}{2}\hbar[/itex] an angular momentum L(e) = [itex] \frac{1}{2}\hbar[/itex] , is this right too? , now is there a B(e) ? 


#4
Mar1914, 10:11 AM

Thanks
P: 1,948

Spinorbit interaction
The magnetic moment isn't [itex]\mu[/itex](e) = [itex] \frac{1}{2}\hbar[/itex]. That doesn't even have correct units. The electron's magnetic dipole moment is given by [tex]\boldsymbol{\mu} = g \frac{e}{2m_e} \mathbf{L}[/tex]
(BTW the Angular momentum isn't [itex]L[/itex](e) = [itex] \frac{1}{2}\hbar[/itex] either. it is [itex]L[/itex](e) = [itex] \frac{\sqrt{3}}{2}\hbar[/itex]. The zcomponent of the angular momentum is [itex]L_z[/itex](e) = [itex] \pm \frac{1}{2}\hbar[/itex]) You should look up magnetic dipole on an introductory book to find out the shape of a dipole's magnetic field. 


#5
Mar2014, 08:12 AM

PF Gold
P: 574

 why is it necessary to say that L_{z}(e) is 1/2 the angular momentum of the orbit (L_{o}=[itex]\hbar[/itex]) and then multiply 1/2 by ≈2 (g_{e})? do you know why it wasn't reckoned from the beginning that it is about the same?  when you talk about the zcomponent of L, are you referring to QM theory? In the classical model the plane of the orbit determines z, but in QM, I gather, there is no orbit, and the spin of the electron is fictitious, so there is no spinning, what is then axis z? if it is arbitrary, then the axis of mu is also arbitrary? If I may recap L_{z}(o)= 1.054*10_{34} J*s, L_{z}(e) = ≈ L(o), L(e) = ≈√3L(o) = 1.825*10^{34} J*s, [itex]\mu[/itex] (γ*L) =1.76*10^{11}*1.8*10^{34} =3.2*10^{23} J/T  is this all correct? because here they say:"the splitting of each of them is about 0.016 nm, corresponding to an energy difference of about 0.000045 eV. This corresponds to an internal magnetic field on the electron of about 0.4 Tesla." Now 0.000045eV= 7.2*10^24 J and in order to get 0.4T, [itex]\mu[/itex] should be 1.8*10^{23}J/T. why so? lastly,  the value of L_{o} is different in the various orbits and [itex]\hbar[/itex] is multiplied by n, is the value of L_{z} (e) always the same = ≈[itex]\hbar[/itex]?  to get the energy of the split of the fine structure we multiplied [itex]\mu[/itex] * B, can you tell me in brief what is the magnetic field in QM model, how do you calculate its strength (from the wave equation?) do you get 0.4 T in QM? Thanks again 


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