
#19
Jun2011, 03:33 AM

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#20
Jun2111, 03:13 AM

Sci Advisor
P: 1,732

I was more interested in whether SR is necessary to account for halfinteger spin. Afaict, it's not  but you seemed to be saying the contrary in your earlier post. So I'll tidy up my sloppiness a little and restate... All we need are the so(3) (or su(2)) commutation relations, and the requirement that the usual basis elements of the so(3) (or su(2)) Lie algebra are represented as Hermitian operators on a Hilbert space (and also quadratic products of elements thereof, such as J^2, etc). That's why I referred to derivations in those textbooks on nonrelativistic QM. 



#21
Jun2111, 08:54 AM

P: 3,015

what is a basis element of a lie algebra?




#22
Jun2111, 09:14 AM

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HW Helper
P: 11,866

I'm surprised you ask. A Lie algebra is a vector space with one additional internal mapping, the Lie product. As a vector space, it has at least a basis with at least one vector as its element.




#23
Jun2111, 09:52 AM

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#24
Jun2111, 12:48 PM

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HW Helper
P: 11,866

Depends on the algebra, of course. For u(1), it's simply 'x', a real number.




#25
Jun2111, 12:50 PM

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#26
Jun2111, 01:17 PM

P: 113

Spin is simply an angular momentum. It is an intrinsic property of all matter, and there have been no psuedoscalar fields which predict zerospin particles ever been detected. 



#27
Jun2111, 01:43 PM

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Thanks
P: 2,152

Pions are (pseudo)scalar particles with spin 0. However, they are not elementary but the bound state of a quarkantiquark pair (coming in three flavors, the charged pions and the neutral pion).
By definition, the spin of an elementary massive particle [itex]s \in \{0,1/2,1,3/2,\ldots \}[/itex] denotes the representation of the rotation group of particle's rest state (momentum 0). I.e., the momentum eigenspace for momentum eigenvalue 0 is (2s+1)dimensional. For massless particles the situation is slightly more complicated, and for each s there are only two helicity states with [itex]h=\pm s[/itex] for each momentum state (of course for [itex]s=0[/itex] there's only one helicity state with helicity 0). 



#28
Jun2111, 06:16 PM

P: 189

In your spin state, the spnning speed is almost light speed, and the relativistic mass is too heavy to move it.<br> But the electron has orbital motion (kinetic energy) and orbital angular momentum, too. If you consider the orbital kinetic energy, its speed exceeds the speed of light. ( v = c + a ) So the spinning electron must stop. (This is the reason why its mass is very big.) Of course, you can not imagine the free spinning electron, which is flying in the air. And you can not accelerate the free electron any more by V, because it exceeds the light speed. (Electron spin 1/2 always exists.) Originally the correct energy of Schrodinger equation can be gotten from the electron's rest mass (or reduced mass). If you use the heavy mass , the electron is always stopping near the nucleus, which gives different energy result. 



#29
Jun2111, 06:30 PM

P: 584





#30
Jun2111, 08:12 PM

P: 2,258

the speed of the electron in ground state of the bohr model of hydrogen is ac




#31
Jun2111, 09:11 PM

P: 584





#32
Jun2511, 01:43 PM

P: 109

Thanks a lot, guys! :D I'm quite the beginner in QM. So,I still don't completely understand all of it, but I think I'm on the way...




#33
Jun2511, 06:15 PM

Sci Advisor
P: 1,867

But you can't model electron spin as if it were spinning on its axis; it's not SO(3). Spinors aren't vectors. Or in layman's terms: A rotating object doesn't reverse its direction if you rotate it 360 degrees  and an electron does. That fact is much more critical to electron behavior than the actual "rotation" itself (spinorbit coupling being a relatively small effect in many cases). Since without it, you have no Pauli principle! 


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