Quantization of Angular Momentum Under Boosts

[michael879]

Can anyone clear this up? The only way for $|\vec{J}|^2$ to be a Lorentz scalar is if $|\vec{N}|^2 = E^2|\vec{x}|^2+t^2|\vec{p}|^2-2Et\vec{x}\cdot\vec{p}$ is also a scalar. However, it is incredibly easy to show that this quantity is NOT a scalar.

 

[strangerep]

Can anyone clear this up? The only way for $|\vec{J}|^2$ to be a Lorentz scalar is if $|\vec{N}|^2 = E^2|\vec{x}|^2+t^2|\vec{p}|^2-2Et\vec{x}\cdot\vec{p}$ is also a scalar. However, it is incredibly easy to show that this quantity is NOT a scalar.

So? What exactly are you trying to "clear up"?

 

[michael879]

So? What exactly are you trying to "clear up"?

Everybody on this thread has claimed that total quantum angular momentum is a Lorentz scalar, but I pretty clearly showed that it can't be, and I really don't understand why anybody would think it was? I mean yes, the total angular momentum of a system in it's rest frame is a "scalar" in the sense that everybody agrees on it's value. But you've fixed a frame here, so in no way is this a true scalar...

 

[WannabeNewton]

Nobody said $|\vec{S}|^2$ is a Lorentz scalar. What was stated, and what is absolutely trivial to see, is that $S_{\mu}S^{\mu}$ is a Lorentz scalar. The equality $S_{\mu}S^{\mu} = |\vec{S}|^2$ clearly only holds in the instantaneous rest frame of $u^{\mu}$, the 4-velocity with respect to which $S^{\mu} = \epsilon^{\mu\nu\alpha\beta}u_{\nu}S_{\alpha\beta}$ is defined. It's as simple as that.

 

[michael879]

Nobody said $|\vec{S}|^2$ is a Lorentz scalar. What was stated, and what is absolutely trivial to see, is that $S_{\mu}S^{\mu}$ is a Lorentz scalar. The equality $S_{\mu}S^{\mu} = |\vec{S}|^2$ clearly only holds in the instantaneous rest frame of $u^{\mu}$, the 4-velocity with respect to which $S^{\mu} = \epsilon^{\mu\nu\alpha\beta}u_{\nu}S_{\alpha\beta}$ is defined. It's as simple as that.

Then why bring up $S^\mu$ at all? My question was whether $\vec{S}$ is frame dependent or not. That is: If I make a measurement of an electron's spin will it always be $\hbar/2$ or will it be dependent on the velocity of the electron and the orientation of the measurement?

If you take $|S_i|=\hbar/2$ only in the rest frame of a particle, and say that the angular momentum will deviate under measurements in another frame, there is still a problem! What about the photon, which has no rest frame?

 

[samalkhaiat]

Then why bring up $S^\mu$ at all? My question was whether $\vec{S}$ is frame dependent or not.

$$\vec{ S } = \frac{1}{2} \vec{ \sigma }$$

Are Pauli's matrices frame dependent?

 

[michael879]

$$\vec{ S } = \frac{1}{2} \vec{ \sigma }$$

Are Pauli's matrices frame dependent?

The pauli matrices are used in non-relativistic QM, so I suspect that they would be invariant under boosts. HOWEVER, they are NOT invariant under rotations, so I would also guess that in a relativistic quantum theory the spin operators would also vary under boosts. Also $S^\mu$ is a 4-(pseudo?)vector, that transforms appropriately under Lorentz transformations. This would naively make the 3-vector component $\vec{S}$ frame dependent.

Also, I think it has already been established in this thread that the spin vector IS frame dependent, which is why I rephrased the question to whether or not the MAGNITUDE of the spin vector is also frame dependent. In classical relativistic physics, it is easy enough to show that the magnitude of an angular momentum can change under boosts (depending on their relative orientation). I would be shocked if quantum spin did not obey this same relation, as it would appear to violate Lorentz invariance...

 

[samalkhaiat]

The pauli matrices are used in non-relativistic QM, so I suspect that they would be invariant under boosts.

Your frist sentence DOES NOT implies your second sentence. What kind of reasoning is this?

HOWEVER, they are NOT invariant under rotations,

Can you prove this to me? Is the NUMBER 1 invariant under rotation?

Also $S^\mu$ is a 4-(pseudo?)vector, that transforms appropriately under Lorentz transformations.

This is the only correct statement I read in this post.

This would naively make the 3-vector component $\vec{S}$ frame dependent.

I think you need to formulate a meaningful question, otherwise you will get meaningless answer.

Also, I think it has already been established in this thread that the spin vector IS frame dependent,

Has it, well I don't usually read long threads, but I can tell you if that is true it will be the greatest DISCOVERY of this century, because one can simply choose a frame and see a fermion rurns into boson.
I would suggest you read about the representation of Lorentz group, and understand the difference between massive and massless eigenvalues of its Casmir operator. I wrote (some time ago) a long post explaining these stuff, I will link to it if I can find it.

 

[michael879]

Your frist sentence DOES NOT implies your second sentence. What kind of reasoning is this?

I said I "suspect", I never offered any proof. We were talking about boosts in a non-relativistic theory, so its pretty irrelevant what the answer is...

Can you prove this to me? Is the NUMBER 1 invariant under rotation?

The pauli matrices, when used in the spin operator, transform like a 3-vector. σ_z transforms into σ_x under a 90° rotation around the y axis. So yes, they are very frame dependent...

Has it, well I don't usually read long threads, but I can tell you if that is true it will be the greatest DISCOVERY of this century, because one can simply choose a frame and see a fermion rurns into boson.
I would suggest you read about the representation of Lorentz group, and understand the difference between massive and massless eigenvalues of its Casmir operator. I wrote (some time ago) a long post explaining these stuff, I will link to it if I can find it.

Maybe you should read a thread before you comment on it... My question has changed quite a bit throughout this thread. And no, the frame dependence of spin is not a great discovery and is actually pretty boring. I was referring to the rotation of the spin vector when changing reference frames, NOT a change in the magnitude.

My question here is really simple, and still hasn't been answered.. Let me just try to make it as specific as possible:

I have a massive system in a state with some angular momentum, let's say $\hbar$. If I were to measure its total angular momentum (assuming this is possible, I don't really see how one would), would I always measure $\sqrt{2}\hbar$? From the responses in this thread the answer is yes, which led to the question: why? In classical relativistic mechanics the total angular momentum of any system is not Lorentz invariant, and can be altered by boosting the system. How can total angular momentum go from being the magnitude of a 3-vector, to being a Lorentz scalar in the transition from classical to quantum mechanics?? Surely all classical systems can be described as very large quantum systems, so shouldn't all angular momenta be lorentz scalars??

 

[samalkhaiat]

The pauli matrices, when used in the spin operator, transform like a 3-vector

.
No, they DO NOT transform much in the same way that ordinary numbers do not transform. This seems to become a common misunderstanding now days. So, let me explain to you how the spin vector transforms under an $SO(3)$ rotation. The spin vector of one particle state is given by
$$S_{ i } = \int d^{ 3 } x \ \psi^{ \dagger } ( x ) \sigma_{ i } \psi ( x ) \ , \ \ (1)$$
where $\psi ( x )$ is the spinor “wave function”. Being a 3-vector, it has to obey the same (infinitesimal) transformation law of the coordinates:
$$\bar{ x }_{ i } = x_{ i } + \omega_{ i j } x_{ j } \ ,$$
where
$$\omega_{ 1 2 } = \theta_{ 3 } , \ \omega_{ 2 3 } = \theta_{ 1 } , \ \omega_{ 3 1 } = \theta_{ 2 } ,$$
are the infinitesimal rotation angles about the coordinates z, x and y. So, the spin vector must transform according to
$$\bar{ S }_{ i } = S_{ i } + \omega_{ i j } S_{ j } \ . \ \ \ (2)$$
Now comes the important point: under space rotation, the transformation of the spin vector rests entirely upon the spinors (NOT THE SIGMA’S). That is
$$\bar{ S }_{ i } = \int d^{ 3 } x \bar{ \psi }^{ \dagger } \sigma_{ i } \bar{ \psi } \ . \ \ (3)$$
So, eq(2) has to be achieved by the infinitesimal transformations
$$\bar{ \psi } = ( 1 + \eta ) \psi , \ \ \ \bar{ \psi }^{ \dagger } = \psi^{ \dagger } ( 1 + \eta^{ \dagger } ) \ . \ \ (4)$$
To determine this transformation law (i.e. to find $\eta$), first we observe that
$$\bar{ \psi }^{ \dagger } \bar{ \psi } = \psi^{ \dagger } ( 1 + \eta^{ \dagger } ) ( 1 + \eta ) \psi \equiv \psi^{ \dagger } \psi \ ,$$
is a scalar only if
$$\eta^{ \dagger } = - \eta \ . \ \ \ (5)$$

Putting (5) and (4) in (3), we find
$$\bar{ S }_{ i } = S_{ i } + \int d^{ 3 } x \ \psi^{ \dagger } ( \sigma_{ i } \eta - \eta \sigma_{ i } ) \psi \ . \ \ (6)$$
Now, comparing (6) with (2) leads to
$$[ \sigma_{ i } , \eta ] = \omega_{ i j } \sigma_{ j } \ .$$
This system of equations can be solved unambiguously by
$$\eta = \frac{ i }{ 2 } ( \omega_{ 1 2 } \sigma_{ 3 } + \omega_{ 2 3 } \sigma_{ 1 } + \omega_{ 3 1 } \sigma_{ 2 } ) = \frac{ i }{ 2 } \vec{ \theta } \cdot \vec{ \sigma } \ .$$
We recognise (the exponential of) this (I hope) as the correct $SO(3) = SU(2) / Z_{ 2 }$ unitary transformation of spinors. So, you must keep this in mind, coordinate transformations do not change the matrices which generate the unitary transformations in Hilbert space. So, when you make coordinate transformations, Dirac and Pauli matrices must be treated like ordinary numbers.

Maybe you should read a thread before you comment on it...

Yeah thanks, maybe next time :)

My question here is really simple, and still hasn't been answered.. Let me just try to make it as specific as possible:

I have a massive system in a state with some angular momentum, let's say $\hbar$. If I were to measure its total angular momentum (assuming this is possible, I don't really see how one would), would I always measure $\sqrt{2}\hbar$? From the responses in this thread the answer is yes, which led to the question: why? In classical relativistic mechanics the total angular momentum of any system is not Lorentz invariant, and can be altered by boosting the system. How can total angular momentum go from being the magnitude of a 3-vector, to being a Lorentz scalar in the transition from classical to quantum mechanics?? Surely all classical systems can be described as very large quantum systems, so shouldn't all angular momenta be lorentz scalars??

The magnitude of spin angular momentum IS the eigen-value of the INVARIANT Casmir operator. In the non-relativistic ($SO(3)$) case, this $SO(3)$-INVARIANT is given by $s ( s + 1 )$, and in the (massive) relativistic ($SO(3 , 1)$) case, this $SO(1,3)$-INVARIANT is given by $- m^{ 2 } s ( s + 1 )$.

 

[strangerep]

Sam,

Since michael879 seems ungrateful for your efforts, I'd like to thank you in public for persevering patiently in this thread (which I and other SAs had essentially given up on), and the (obviously-nontrivial) effort involved in your post #45 above.

Michael879,

FYI, Sam is one of the most knowledgeable people I've ever met on the Internet. Therefore, it would be in your interests to soften your attitude. Part of the reason why you hadn't got a satisfying answer previously lies in the ambiguity of your posts. E.g., when talking about "vectors" in the context of both the non-relativistic and relativistic cases, ambiguities can sometimes be minimized by saying "3-vector" or "4-vector" explicitly. Not doing so can lead to misunderstanding and talking at crossed purposes.

 

[samalkhaiat]

Sam,

Since michael879 seems ungrateful for your efforts, I'd like to thank you in public for persevering patiently in this thread (which I and other SAs had essentially given up on), and the (obviously-nontrivial) effort involved in your post #45 above.

Thank you. It is a pleasure to be here helping others understand the mathematical tools necessary for comprehending the physical world and appreciating its beauty. I wish I have more time to spare, and contribute more in here. It is fun to be here even though it is not very challenging.

Students always ask: “How can you solve commutator equations?” So, for the sake of completeness, I will now solve the following equations from post #45
$$[ \sigma_{ i } , \eta ] = \omega_{ i j } \sigma_{ j } \ . \ \ \ (1)$$
Recall that $\sigma_{ i }$ together with 2 by 2 identity matrix $I$ form a complete set for expanding any 2 by 2 matrix. So, we may write (repeated indices are summed over)
$$\eta = c I + c_{ j } \sigma_{ j } \ .$$
Now, equation (5) in post# 45 leads to the following sequence of implications
$$eq(5) \ \Rightarrow \ \eta \in su(2) \ \Rightarrow \ \mbox{ Tr } \eta = 0 \ \Rightarrow \ c = 0 \ .$$
Therefore
$$\eta = c_{ j } \ \sigma_{ j } \ . \ \ \ (2)$$
Our task now is finding the constant numbers $c_{ i }$. From (1) and (2), we get
$$\omega_{ i k } \ \sigma_{ k } = c_{ j } \ [ \sigma_{ i } , \sigma_{ j } ] \ . \ \ (3)$$
Using the Lie algebra of $SU(2)$
$$[ \sigma_{ i } , \sigma_{ j } ] = 2 i \ \epsilon_{ i j k } \ \sigma_{ k } \ ,$$
in the RHS of equation (3), we find
$$\omega_{ i k } \ \sigma_{ k } = 2 i \ c_{ j } \ \epsilon_{ i j k } \ \sigma_{ k } \ . \ \ (4)$$
Anti-commuting both sides of (4) with $\sigma_{ n }$, and using the Clifford algebra $\{ \sigma_{ n } , \sigma_{ k } \} = 2 \delta_{ n k }$, we find
$$\epsilon_{ i j n } \ c_{ j } = - \frac{ i }{ 2 } \ \omega_{ i n } \ . \ \ \ (5)$$
Contracting both sides of (5) with $\epsilon_{ l m n }$, and using the identity
$$\epsilon_{ l m n } \ \epsilon_{ i j n } = \delta_{ l i } \ \delta_{ m j } - \delta_{ l j } \ \delta_{ m i } \ ,$$
we find
$$\delta_{ l i } \ c_{ m } - \delta_{ m i } \ c_{ l } = \frac{ - i }{ 2 } \omega_{ i n } \ \epsilon_{ l m n } \ . \ \ \ (6)$$
Contracting (6) with $\delta_{ i m }$, we find
$$c_{ l } - 3 \ c_{ l } = \frac{ - i }{ 2 } \ \omega_{ i n } \ \epsilon_{ l i n } \ .$$
Thus
$$c_{ j } = \frac{ i }{ 4 } \ \epsilon_{ j i n } \ \omega_{ i n } \ . \ \ \ (7)$$
Substituting (7) in (2), we get our final solution
$$\eta = \frac{ i }{ 4 } \ \epsilon_{ j i n } \ \omega_{ i n } \ \sigma_{ j } = \frac{ i }{ 2 } ( \omega_{ 1 2 } \ \sigma_{ 3 } + \omega_{ 2 3 } \ \sigma_{ 1 } + \omega_{ 3 1 } \ \sigma_{ 2 } ) \ ,$$
or
$$\eta = \frac{ i }{ 2 } \theta_{ i } \ \sigma_{ i } \ \ .$$

Sam

 

[michael879]

Ok I'm just coming off as a jackass here and my question hasn't been answered at all, so I'm giving up.

One last thing before I leave this thread:

.
Now comes the important point: under space rotation, the transformation of the spin vector rests entirely upon the spinors (NOT THE SIGMA’S). That is
$$\bar{ S }_{ i } = \int d^{ 3 } x \bar{ \psi }^{ \dagger } \sigma_{ i } \bar{ \psi } \ . \ \ (3)$$

I would argue this is totally just a matter of semantics. I said something like "when you rotate your frame of reference, apply a rotation to all operators", and you're telling me "no, that's wrong you need to rotate all the wave functions instead". Just look at your equation (1), if you rotate the spinors you end up with: spinor * R * sigma * R * spinor. If instead you rotate sigma you also end up with: spinor * R * sigma * R * spinor. In the end the math is identical so I really don't see why this even needed to be brought up in this discussion...

 

[strangerep]

Since michael879 is "leaving this thread", maybe I shouldn't bother further. But for the benefit of other readers...

Just look at your equation (1), if you rotate the spinors you end up with: spinor * R * sigma * R * spinor. If instead you rotate sigma you also end up with: spinor * R * sigma * R * spinor. In the end the math is identical [...].

Wrong.

One must also apply the rotation to the "i" index on both sides. The R's are 2x2 matrices, whereas a rotation on the "i" index involves a 3x3 matrix which I'll denote by D. The transformation of the sigmas is then more like:
$$\sigma'_i ~=~ D_i^{~j} R^\dagger \sigma_j R$$The effect of D cancels the effect of the two R's, leaving each sigma unchanged, i.e., $\sigma'_i = \sigma_i$.

A similar thing happens in relativistic case with the Dirac equation and gamma matrices. Each gamma matrix remains unchanged under a Lorentz transformation. (For more detail, see any textbook on relativistic quantum mechanics.)

 

[michael879]

Since michael879 is "leaving this thread", maybe I shouldn't bother further. But for the benefit of other readers...

Wrong.

One must also apply the rotation to the "i" index on both sides. The R's are 2x2 matrices, whereas a rotation on the "i" index involves a 3x3 matrix which I'll denote by D. The transformation of the sigmas is then more like:
$$\sigma'_i ~=~ D_i^{~j} R^\dagger \sigma_j R$$The effect of D cancels the effect of the two R's, leaving each sigma unchanged, i.e., $\sigma'_i = \sigma_i$.

A similar thing happens in relativistic case with the Dirac equation and gamma matrices. Each gamma matrix remains unchanged under a Lorentz transformation. (For more detail, see any textbook on relativistic quantum mechanics.)

Then remove the D SO(3) rotation?? You're applying the rotation AND its inverse, so of course you end up at the same pauli matrix. This seems like a completely trivial issue of passive vs. active transformations.. Any observable will have the form $<\psi|O|\psi'>$ which under a global rotation becomes $<\psi|R^{-1}OR|\psi'>$ regardless of whether you apply the rotation R to the wavefunctions $\psi$ or the operators O!

Here's an easy example: Rotate your reference frame by 90° around the z axis. The SU(2) rotation matrix is
$$\dfrac{1}{\sqrt{2}}\left(1-i\sigma_z\right)$$
if you apply this rotation to the three pauli matrics you get:
$$\sigma_z\rightarrow\sigma_z$$
$$\sigma_x\rightarrow-\sigma_y$$
$$\sigma_y\rightarrow\sigma_x$$
which is exactly the same as applying the SO(3) rotation to the coordinate axes (i.e. D "applied" to the pauli indices in your notation), or applying the SU(2) rotation to the wave function.

What you're doing is applying all 3 methods, pointing out that two of them cancel out and claiming the third one is "correct". Passive transformations are the inverses of their respective active transformations though, so I could just as easily claim that D (the passive rotation of the coordinate system) cancels out the rotation of the wave function and rotating the pauli matrices is "correct" (the alternate active rotation).

I am just baffled how this thread became about something so irrelevant..

 

[strangerep]

Then remove the D SO(3) rotation?? You're applying the rotation AND its inverse, so of course you end up at the same pauli matrix. This seems like a completely trivial issue of passive vs. active transformations..

You miss the point. The pauli matrices are objects of composite type: each has 1 vector index, and 2 spinor indices. The correct representations of any given physical rotation must be applied to the respective indices simultaneously. The net effect is that, for a single physical rotation, the Pauli matrices remain unchanged.

I am just baffled how this thread became about something so irrelevant..

Well, when understood properly, the point is not irrelevant. Without understanding it correctly, one cannot understand non-relativistic QM with spin.

In any case, it's your thread. You asked questions, then tended to contradict the answers by making incorrect or ill-informed statements -- which others then felt obliged to correct, and so it continued. A number of people have tried to help you in this thread. Those people have helped lots of others here on PF. The common factor in this thread is you, not them.

Which textbook(s) are you working from anyway?