# Transverse Spin

1. May 19, 2014

### michael879

Is the spin of an electron still $\hbar$/2 in the direction transverse to its momentum? Classically angular momentum transforms non-trivially under boosts, so I was wondering if this applies to quantum mechanics too. A more general question would be: Is angular momentum quantized into units of $\hbar$ under general boosts?

2. May 19, 2014

### Jilang

In the Stern Gerlach experiment the spin always seems to be measured transverse to the momentum of the particle. I would ask whether it is also quantised along the direction of travel.

3. May 19, 2014

### michael879

I'm more asking about theoretical predictions than experimental confirmations. Helicity doesn't change at all under parallel boosts

4. May 19, 2014

### Staff: Mentor

The component of an individual electron's spin angular momentum along any axis is always measured to be either $+\hbar/2$ or $-\hbar/2$. Before measurement, the probabilities of these two values depend on how the electron was prepared and on which axis you make the measurement along.

5. May 19, 2014

### michael879

But doesn't that violate relativity? For a classical system, two people won't measure the same angular momentum in general. For an electron you could use the argument that it's a point particle, but for an extended system where angular momentum is still quantized I don't see how to reconcile it with relativity

6. May 19, 2014

### Einj

Spin is an intrinsic property of the particle and therefore it doesn't change with boosts. We can measure its value along any direction we want and the result of a measure for that direction is going to be $\pm\hbar/2$.

In Quantum Mechanics one defines J=L+S, where S is the spin of the particle and L is the orbital angular momentum.

7. May 19, 2014

### DrDu

Spin is defined as angular momentum in the rest frame. Hence, it does change under boosts.
A more formal treatment involves the Pauli-Lubanski vector:
http://en.wikipedia.org/wiki/Pauli–Lubanski_pseudovector

8. May 19, 2014

### michael879

So are you saying we WOULD measure an electron's angular momentum to be something other than hbar/2 in general?? As I said above, it seems necessary to obey relativity, but everybody else seems to disagree

9. May 19, 2014

### Einj

In Quantum Mechanics you can't really separate L and S, you always consider J=L+S. S is intrinsic, L is not.

10. May 19, 2014

### michael879

I'm not separating them here either. I'm just asking about angular momentum quantization in boosted reference frames, spin or orbital

11. May 19, 2014

### Staff: Mentor

In any inertial reference frame, the measured component of an electron's spin along any axis is always either $+\hbar/2$ or $-\hbar/2$. What changes when you boost from one frame to another is the probabilities of these two outcomes.

The magnitude of an electron's spin is always $\sqrt{s(s+1)}\hbar = \frac{\sqrt{3}}{2}\hbar$. It is a Lorentz scalar.

12. May 19, 2014

### michael879

Right, I can see that now from the link you gave. However, classically total angular momentum isn't a Lorentz scalar is it??

13. May 19, 2014

### WannabeNewton

In a relativistic setting the angular momentum can be written in terms of a 4-vector $S^{\mu}$ so $S^{\mu}S_{\mu}$ is certainly a Lorentz scalar. However don't confuse "Lorentz scalar" with "invariance of total angular momentum measurements across all Lorentz frames". $S_{\mu}S^{\mu}$ is the total angular momentum as measured in the frame of the background observer whose 4-velocity is used explicitly in the definition of $S^{\mu}$: $S^{\mu} = \epsilon^{\mu\nu\gamma \delta}u_{\nu}S_{\gamma\delta}$.

14. May 19, 2014

### Bill_K

Also known as Wμ, the Pauli-Lubanski pseudovector.

15. May 19, 2014

### michael879

In 4-d space angular momentum is the spatial components of a rank-2 antisymmetric tensor... And regardless, we're talking about the length of a 3-vector here, not a 4-vector or tensor. Also, I'm not sure what you mean by "don't confuse Lorentz scalar with invariance of total angular momentum measurements across all Lorentz frames". A Lorentz scalar is defined as being invariant across all Lorentz frames..

I really don't see how the total angular momentum could be a Lorentz scalar, it's the cross-product of two 3-vectors and it mixes with N under Lorentz transformations (http://en.wikipedia.org/wiki/Relati...ntertwine_of_L_and_N:_Lorentz_transformations)

16. May 19, 2014

### WannabeNewton

So? What does that have to do with anything that was stated? $S_{\mu\nu}$ is exactly what's used to define $S^{\mu}$ as is clear from the definition I wrote above. Angular momentum doesn't have a unique mathematical form-I can describe it as a 2-form or a pseudo-vector.

A Lorentz scalar has the same value in all frames but that doesn't mean it corresponds to (in this case) the angular momentum measurement actually made by the observer in a given frame. Clearly $S^{\mu}S_{\mu}$ corresponds only to the angular momentum measurement made by the observer with respect to which the 3-volume element is defined from the 4-volume element. It's a very simple and basic distinction.

Bill's link explains all this perfectly.

17. May 19, 2014

### Bill_K

Well, maybe if you took a look at the reference I gave, you'd understand. Yes, Wμ is a 4-vector, but it's orthogonal to the momentum 4-vector Pμ, and therefore is a 3-vector in the rest frame of the particle. And yes, it's built from a rank-2 antisymmetric tensor.

18. May 19, 2014

### michael879

I already read that link when DrDu posted it, my confusion is how that translates to the classical realm, where L2 is not Lorentz invariant.

*edit* also, I don't see how that wikipedia answers my question at all. All the relevant information in it basically just condenses to what jtbell pointed out, that the eigenvalues of L2 are Lorentz scalars. It's an interesting page but it doesn't really explain anything..

Last edited: May 19, 2014
19. May 19, 2014

### Bill_K

Somewhere in this thread did we switch the topic from S to L? It started out being about S.

The Wμ vector specifically relates to spin, being the angular momentum vector in the rest frame of the particle. That's the quantity whose magnitude is ħ/2.

The orbital angular momentum is Lμν = x pν]. Which one are you asking about?

20. May 19, 2014

### michael879

Ah sorry, I'm asking about both (or either). And I get that both J2 and S2 are constant Lorentz scalars. What I can't wrap my head around is where this comes from?? Because angular momentum is not a scalar classically!

*edit* I may have explained this badly, but this question is really just about quantized angular momentum (any kind) under Lorentz boosts

21. May 19, 2014

### strangerep

1) What do you mean by "classically"? Do you mean classical-relativistic or classical-nonrelativistic?

2) When you say "angular momentum", do you mean orbital, intrinsic or total? And do you mean the angular momentum (pseudo-)vector or its magnitude?

Again, are you asking about orbital, intrinsic or total angular momentum in the quantum setting? And are you asking about quantization of its magnitude, or quantization of its projection along a particular axis?

22. May 19, 2014

### michael879

Well my question has evolved a bit since the OP, but what I'm asking about now is:
1) classical-relativistic
2) I mean the magnitude of any of the angular momenta vectors
3) I'm only asking about the overall magnitude now, because it has been established that the spin vector does rotate under boosts

23. May 19, 2014

### strangerep

Then,.... provided one understands that the Pauli--Lubanski (pseudo)vector is the appropriate generalization of angular momentum in a relativistic context, its magnitude $W^2 := W_\mu W^\mu$ is indeed a Lorentz scalar.

(One also needs to know that, although the orbital and intrinsic parts of angular momentum remain distinct in a nonrelativistic context, they mix together under general Lorentz transformations. Hence one prefers to talk only about total angular momentum in the relativistic context.)

24. May 20, 2014

### DrDu

When you consider a bound system, e.g. a hydrogen atom, $J^2$ or, if you ignore S, $L^2$ is in fact the spin of the compound system. Hence it is relativistically invariant.

25. May 20, 2014

### michael879

Right, I understand that the relativistic quantum operator W is a Lorentz 4-vector who's magnitude shows that the total angular momentum is a Lorentz scalar. What I don't understand is how this works for a classical system, which should be identical to the quantum predictions with some limits applied.

If you treat the Pauli--Lubanski pseudovector as a classical 4-vector (rather than a quantum operator) you find that $W^\mu W_\mu = 0$, so I'm not really clear how meaningful this vector is outside of quantum mechanics (I understand that it describes spin so this isn't surprising at all).

I'm not really sure what the confusion is at this point, you guys keep giving me purely quantum answers to my question on how to relate quantum/classical predictions. From what I've gathered the total angular momentum operator $\vec{J}^2$ is Lorentz invariant, but the corresponding classical variable $\vec{J}^2$ is definitely not a Lorentz invariant. Even if we completely ignored spin this discrepancy would remain, and I've never seen such a huge difference between the classical and quantum realms before... Basically this is saying that if I have a spinning ball, I will measure it's total angular momentum differently at different velocities. However, if I shrink this ball down to quantum scales it will suddenly have an invariant total angular momentum?? Even at the classical level, all angular momentum should be quantized (even though its too small to detect), so that if quantum mechanics predicts that angular momentum is Lorentz invariant we should also be able to measure that on macroscopic systems...

The only possible explanation I have is that I screwed up somewhere and angular momentum is in fact Lorentz invariant classically. However I'm pretty confident this isn't the case..