Electron spin

Dickfore

You didn't read the upper part of my post carefully where I said that the Lie algebra of so(3) is isomorphic to su(2), thus making the corresponding Lie groups locally isomporphic. However, globally SU(2) is a double cover of SO(3) and SU(2) is the only simply connected group.

According to the theorems for the relations of Lie groups and Lie algebras, the correspondence between the representations of the Lie algebra and the Lie group from it is generated via the exponential map is only true for the simply connected groups. Thus, the representations of the lie algebra given above is a representation for SU(2).

Dickfore

From Spin (physics) Wikipedia article:

If you go to the section Spin and rotations, it says:

strangerep

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.

Dickfore

what is a basis element of a lie algebra?

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.

Dickfore

Which one is it?

dextercioby

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

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?

Goldstone1

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.

vanhees71

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 [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).

ytuab

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.

akhmeteli

I would think you can only add speeds like you do when the speeds are not relativistic, which is not the case.

granpa

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

akhmeteli

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.)

jobsism

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...

alxm

Good quotes; I feel this is seldom emphasized enough. The problem when describing spin as 'intrinsic angular momentum' is that it gives the impression that it can be described as classical (or orbital) angular momentum - as if the electron is 'spinning on its own axis', even though it's not. The oft-cited rationale that it can't be doing that "because it'd spin faster than the speed of light" easily furthers that misconception. -What if you're working non-relativistically? That's usually the case when describing electrons in atoms/molecules. If that's the case, there's no reason to care about exceeding the speed of light. Can't you model electron spin that way anyway?

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 (spin-orbit coupling being a relatively small effect in many cases). Since without it, you have no Pauli principle!