# Electron spin

by jobsism
Tags: electron, spin
P: 3,015
From Spin (physics) Wikipedia article:

 Spin is a type of angular momentum, where angular momentum is defined in the modern way as the "generator of rotations" (see Noether's theorem).[1][2] This modern definition of angular momentum is not the same as the historical classical mechanics definition, L = r × p. (The historical definition, which does not include spin, is more specifically called "orbital angular momentum".)
If you go to the section Spin and rotations, it says:

 Mathematically speaking, these matrices (referring to n dimensional unitary transformation matrices) furnish a unitary projective representation of the rotation group SO(3). Each such representation corresponds to a representation of the covering group of SO(3), which is SU(2). There is one n-dimensional irreducible representation of SU(2) for each dimension, though this representation is n-dimensional real for odd n and n-dimensional complex for even n (hence of real dimension 2n).
P: 1,732
 Quote by Dickfore You didn't read the upper part of my post carefully
I think you did not read the first sentence of mine carefully. :-)
I was more interested in whether SR is necessary to account for
half-integer 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 non-relativistic QM.
 P: 3,015 what is a basis element of a lie algebra?
 Sci Advisor 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.
P: 3,015
 Quote by dextercioby 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.
Which one is it?
 Sci Advisor HW Helper P: 11,866 Depends on the algebra, of course. For u(1), it's simply 'x', a real number.
P: 584
 Quote by G01 No. It is not correct to think of the electrons intrinsic angular momentum as resulting from the electron physically rotating about an axis. In fact this problem is usually solved during a standard junior level quantum mechanics course. You can treat the electron as a spinning spherical shell of radius $r_c=\frac{e^2}{4\pi\epsilon_omc^2}$, the "classical electron radius." Now, calculate how fast the electron must be spinning in order to have angular momentum equal to $\hbar/2$. You'll see that the tangential speed of a point on the electrons equator would have to spinning faster than the speed of light for everything to work out and give the correct angular momentum. Thus, the intrinsic angular momentum of an electron cannot be due to the electron rotating about an axis.
I have read this argument many times in several books, but somehow failed to understand it. Maybe you could explain it? My problem is I consider a point particle with a mass equal to that of electron and having a circular orbit with a radius of the order of the Compton wavelength (or even classical electron radius). If the (tangential) velocity of this particle is very close to the speed of light, its momentum can be arbitrarily high, so the angular momentum can also be arbitrarily high. I don't quite see how this reasoning is faulty. Do you?
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 Quote by jobsism What exactly is electron spin?The thing is, I had read in some general books on QM that spin of an electron is a non-classical degree of freedom,i.e.,an intrinsic angular momentum it possesses, and that it is NOT to be confused with a physically spinning electron. But in my physics textbook, it says that an electron physically spins about it's axis,due to which it has a magnetic dipole moment. So which is it?
According to modern physics, it does not spin round an axis. It would need to spin twice the angle it began with just to return to it's original restarting point - this is because we say particles have no structure - they are dimensionless.

Spin is simply an angular momentum. It is an intrinsic property of all matter, and there have been no psuedoscalar fields which predict zero-spin particles ever been detected.
 Sci Advisor Thanks P: 2,152 Pions are (pseudo-)scalar particles with spin 0. However, they are not elementary but the bound state of a quark-antiquark pair (coming in three flavors, the charged pions and the neutral pion). By definition, the spin of an elementary massive particle $s \in \{0,1/2,1,3/2,\ldots \}$ 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 $h=\pm s$ for each momentum state (of course for $s=0$ there's only one helicity state with helicity 0).
P: 189
 Quote by akhmeteli I have read this argument many times in several books, but somehow failed to understand it. Maybe you could explain it? My problem is I consider a point particle with a mass equal to that of electron and having a circular orbit with a radius of the order of the Compton wavelength (or even classical electron radius). If the (tangential) velocity of this particle is very close to the speed of light, its momentum can be arbitrarily high, so the angular momentum can also be arbitrarily high. I don't quite see how this reasoning is faulty. Do you?

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.
P: 584
 Quote by ytuab In your spin state, the spnning speed is almost light speed, and the relativistic mass is too heavy to move it.
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 )
I would think you can only add speeds like you do when the speeds are not relativistic, which is not the case.
 P: 2,258 the speed of the electron in ground state of the bohr model of hydrogen is ac
P: 584
 Quote by granpa the speed of the electron in ground state of the bohr model of hydrogen is ac
I think you'll agree that there are better theories than the Bohr model. For example, for the Dirac equation, eigenvalues of velocity projections are +-c (see, e.g., Dirac's book "Principles...", the chapter where he discusses Zitterbewegung.)
 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...