Quantization of Angular Momentum Under Boosts

[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?

 

[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.

 

[michael879]

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.

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

 

[jtbell]

Is the spin of an electron still $\hbar$/2 in the direction transverse to its momentum?

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.

 

[michael879]

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.

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

 

[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.

 

[DrDu]

Classically angular momentum transforms non-trivially under boosts, so I was wondering if this applies to quantum mechanics too.

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

 

[michael879]

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

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

 

[Einj]

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

 

[michael879]

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

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

 

[jtbell]

So are you saying we WOULD measure an electron's angular momentum to be something other than hbar/2 in general??

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.

 

[michael879]

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

 

[WannabeNewton]

However, classically total angular momentum isn't a Lorentz scalar is it??

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 $S^{\mu}$: $S^{\mu} = \epsilon^{\mu\nu\gamma \delta}u_{\nu}S_{\gamma\delta}$.

 

[Bill_K]

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 $S^{\mu}$: $S^{\mu} = \epsilon^{\mu\nu\gamma \delta}u_{\nu}S_{\gamma\delta}$.

Also known as W_μ, the Pauli-Lubanski pseudovector.

 

[michael879]

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 $S^{\mu}$: $S^{\mu} = \epsilon^{\mu\nu\gamma \delta}u_{\nu}S_{\gamma\delta}$.

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)

 

[WannabeNewton]

In 4-d space angular momentum is the spatial components of a rank-2 antisymmetric tensor...

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.

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..

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.

 

[Bill_K]

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)

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

 

[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..

 

[Bill_K]

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..

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

The W_μ 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 L_μν = x_[μ p_ν]. Which one are you asking about?

 

[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

 

[strangerep]

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!

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?

[...] this question is really just about quantized angular momentum (any kind) under Lorentz boosts

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?

 

[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

 

[strangerep]

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

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.)

 

[DrDu]

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.

.

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.

 

[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..

 

[DrDu]

If you treat the Pauli--Lubanski pseudovector as a classical 4-vector (rather than a quantum operator) you find that $W^\mu W_\mu = 0$,

I don't believe this!
I also urge you to calculate W for a hydrogen atom (ignoring the spin of the electron), even a classical one like in Bohrs model.
I read once a nice article treating exactly with your confusion and I think it was even written by E. Schroedinger himself. The conclusion was that what we call "angular momentum" in QM is in many casesr rather the spin of the system.

 

[michael879]

hmm well I'll admit it wasn't the most careful calculation but I'm pretty sure all the terms ended up canceling it out... I'll double check that

But either way, this isn't a spin phenomenon that I'm confused about! Any quantum angular momentum transforms the same as spin under boosts, and their magnitude is always a Lorentz scalar as long as they commute with the Hamiltonian right? On the other hand, in classical relativity angular momentum isn't remotely Lorentz invariant! Where do these two concepts diverge??

Also, if you found the title of that article I'd be interested to read it

 

[michael879]

I don't believe this!

Ok well here is my derivation, I did a much more drawn out one before but it's actually much simpler if you take the symmetries into account:

$$W^2 = \dfrac{1}{4}\epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu\lambda\delta\kappa}J^{\nu\rho}J_{\lambda\delta}p^\sigma p_\kappa \\ J^{\mu\nu} = x^\mu p^\nu - x^\nu p ^\mu \\ \epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu\lambda\delta\kappa} = \delta^{\lambda\delta\kappa}_{\nu\rho\sigma}$$
Where the last equation uses the generalized kronecker delta function, which is fully antisymmetric on the upper (lower) indices. Now just plugging that in,
$$W^2 = \dfrac{1}{4}\delta^{\lambda\delta\kappa}_{\nu\rho\sigma}(x^\nu p^\rho - x^\rho p^\nu)(x^\lambda p^\delta - x^\delta p^\lambda)p^\sigma p_\kappa$$
Now you could go through each of the 4x6=24 terms here and show that they all cancel out (which I did earlier), or you could just note that every term is of the form
$$\delta^{\lambda\delta\kappa}_{\nu\rho\sigma}x^a x_b p^c p^d p_e p_f$$
where a,b,c,d,e, and f are the different greek indices. Since these are not operators, I've been free to commute x's and p's which is why you end up with 4 p terms two with lower indices, two with upper indices. No matter what those indices are, you're going to end up with a symmetric pp term in conjunction with the antisymmetric delta function, giving you a result of W²=0. Of course none of this works in the quantum case because most of these terms wouldn't commute

*edit* This isn't really surprising, given that W describes relativistic spin (a purely quantum effect), is it?

**edit** Well I'm dumb, the entire operator is trivially 0.
$$W^\mu = \dfrac{1}{4}\epsilon^{\mu\nu\rho\sigma}J_{\nu\rho}p_\sigma \\ =\dfrac{1}{4}\epsilon^{\mu\nu\rho\sigma}(x_\nu p_\rho p_\sigma - x_\rho p_\nu p_\sigma) = 0$$

 

[strangerep]

By focussing only on orbital angular momentum in that way you get a misleading answer. Intrinsic angular momentum in a classical--relativistic context is a bit tricky. See below.

But either way, this isn't a spin phenomenon that I'm confused about! Any quantum angular momentum transforms the same as spin under boosts, and their magnitude is always a Lorentz scalar as long as they commute with the Hamiltonian right? On the other hand, in classical relativity angular momentum isn't remotely Lorentz invariant! Where do these two concepts diverge??

Can you access a copy of the Misner, Thorne & Wheeler "Gravitation" textbook? If so, it might help to study Box 5.6 on pp157-159 which covers angular momentum in flat spacetime (i.e., classical--special--relativistically).

 

[michael879]

Can you access a copy of Misner, Thorne & Wheeler "Gravitation" textbook? If so, it might help to study Box 5.6 on pp157-159 which covers angular momentum in flat spacetime (i.e., classical--special--relativistically).

Not that I'm aware of, I think I have a pretty solid grasp on special relativity though... Am I missing something?

 

[strangerep]

Not that I'm aware of, I think I have a pretty solid grasp on special relativity though... Am I missing something?

I suspect so, but let's see: What is the definition of angular momentum and total angular momentum in the classical--relativistic context?

 

[DrDu]

Ok well here is my derivation, I did a much more drawn out one before but it's actually much simpler if you take the symmetries into account:

**edit** Well I'm dumb, the entire operator is trivially 0.
$$W^\mu = \dfrac{1}{4}\epsilon^{\mu\nu\rho\sigma}J_{\nu\rho}p_\sigma \\ =\dfrac{1}{4}\epsilon^{\mu\nu\rho\sigma}(x_\nu p_\rho p_\sigma - x_\rho p_\nu p_\sigma) = 0$$

Ah, I see the error. You are considering a single particle. Both for a classical and a quantum particle, the orbital angular momentum does not contribute to spin.
If you use a compound particle, like a hydrogen atom, the momentum operators P and the p appearing in L aren't the same, as the first one refers to the center of mass while the second one to the momenta of the individual particles. In a hydrogen atom, you can assume P to equal the momentum of the nucleus, which has only one non-vanishing component P_0 in the rest frame, while the relevant part of J can be expressed in terms of L.

 

[michael879]

I suspect so, but let's see: What is the definition of angular momentum and total angular momentum in the classical--relativistic context?

$$J^{\mu\nu} = x^\mu p^\nu - x^\nu p^\mu \\ J_i = \epsilon_{ijk}J^{jk} \\ |\vec{J}|^2 = J_i J^i = \epsilon_{ijk}\epsilon^{ilm}J^{jk}J_{lm} = 2J^{jk}J_{jk}$$

Under Lorentz transformations $J^{\mu\nu}$ transforms like a 2-tensor, but $|\vec{J}|^2$ is the magnitude of a 3-vector, which transforms non-trivially and is not a scalar...

 

[michael879]

Ah, I see the error. You are considering a single particle. Both for a classical and a quantum particle, the orbital angular momentum does not contribute to spin.
If you use a compound particle, like a hydrogen atom, the momentum operators P and the p appearing in L aren't the same, as the first one refers to the center of mass while the second one to the momenta of the individual particles. In a hydrogen atom, you can assume P to equal the momentum of the nucleus, which has only one non-vanishing component P_0 in the rest frame, while the relevant part of J can be expressed in terms of L.

Ah your right, I have been considering a single particle for this entire discussion. I don't think that clears up my question but it does explain why I found W=0 :P

 

[michael879]

O well crap, I see the point you guys have been trying to make now...

$$P=(M,0,0,0) \\ W^2 = \dfrac{M^2}{4}\delta^{\lambda\delta 0}_{\nu\rho 0}J^{\nu\rho}_{\lambda\delta} = \dfrac{M^2}{4}\delta^{ij}_{kl}J^{kl}J_{ij} = \dfrac{M^2}{2}|\vec{J}|^2$$

So if W is a 4-vector then J must be a Lorentz scalar... That's bizarre because when I apply a Lorentz boost to the elements of the 3-vector J the magnitude I get is definitely not a scalar...

*edit* For example, $J^{\mu\nu}$ is clearly a Lorentz tensor, so $J^2$ should be a Lorentz scalar. Given that $J^{\mu\nu} = x^\mu p^\nu - x^\nu p^\mu$, $\vec{N}=E\vec{x}-t\vec{p}$, and $J^{ij}J_{ij} = 2|\vec{J}|^2$, you find that $J^2 = 2|\vec{J}|^2 - 2|\vec{N}|^2$. Now we know that the left hand side of this equation is a scalar, so the right hand side must be too. The N vector can be arbitrarily varied through Lorentz transformations though ($|\vec{N}|^2=E^2 |\vec{x}|^2 +t^2 |\vec{p}|^2 - 2Et\vec{x}\cdot\vec{p}$), so $|\vec{J}|^2$ can't be a scalar!

**edit** yea my logic in the first part of this post is flawed, because I fixed a reference frame! So the equation should really read:
$$W^2 = \dfrac{M^2}{2}|\vec{J}_0|^2$$
Where $\vec{J}_0$ is the angular momentum in the rest frame of P. This doesn't mean that the magnitude of the angular momentum is fixed!