It seems to me different people are saying different things. Let me explain my confusion a little bit more.
Avodyne said:
Your spin operator is the correct one. After boosting, in general you get a superposition of states of definite spin along any given axis. You only retain a spin eigenspinor if you boost in the direction of the original spin axis.
I understand that I can retain the spinor direction if I boost it in the direction of the spin from the rest frame. What I don't understand is how to find out the direction of a spinor that's boosted in some other direction from rest frame. And as I said before, there is no way you can write the boosted spinor as any mixture of the eigenstates of that spin operator, so I don't think that spin operator is the correct one. Another possibility is proposed by Bill_K:
Bill_K said:
The idea that the spin "points in a particular direction" is not valid. By taking a superposition of states, you can quantize the spin along any axis you choose. As Avodyne says, the easiest axis to quantize along in this case is along the direction of motion. The spin states are then the helicity states.
But I can't make sense of a spin that doesn't point in a particular direction, especially since I started with a spin that points in a particular direction. I merely boosted the system to another frame, why should the spin now become ambiguous? I can reduce my original question to this:
if I have a gyroscope pointing in +z direction in the rest frame, what direction is it pointing in if I see it from another inertia frame that's moving in say -x direction with velocity v? I would think this is the same question for our spin, since the lorentz transformation should work exactly the same on a group theory level.
Since the direction of the helicity eigenstates are unambiguous, I attempted the following way to find out the direction of the boosted spinor. Please take a look and see whether this logic is right:
First, Helicity spinors moving with momentum \mathbf{P} can be obtained by first boosting a spin-up spinor in the +z direction (parallel to spin orientation) and then rotate to \mathbf{P} direction (see Page 195 in Sakurai's Advanced Quantum Mechanics). If the particle is moving in the xz plane, the helicity spinor can be written as the following in the standard representation:
<br />
\begin{align}<br />
u^{+}_{H}(\mathbf{P})=\sqrt{E+M}<br />
\begin{bmatrix}<br />
\cos\dfrac{\theta}{2}\\<br />
\sin\dfrac{\theta}{2}\\<br />
\dfrac{|\mathbf{P}|}{E+M}\cos\dfrac{\theta}{2}\\[6pt]<br />
\dfrac{|\mathbf{P}|}{E+M}\sin\dfrac{\theta}{2}<br />
\end{bmatrix},<br />
\quad<br />
u^{-}_{H}(\mathbf{P})=\sqrt{E+M}<br />
\begin{bmatrix}<br />
-\sin\dfrac{\theta}{2}\\<br />
\cos\dfrac{\theta}{2}\\<br />
\dfrac{|\mathbf{P}|}{E+M}\sin\dfrac{\theta}{2}\\[6pt]<br />
-\dfrac{|\mathbf{P}|}{E+M}\cos\dfrac{\theta}{2}<br />
\end{bmatrix},<br />
\end{align}<br />
where \theta is the angle between \mathbf{P} and +z direction.
Now we can easily write the Dirac spinor in standard representation as a combination of these two Helicity spinors:
\begin{align}
u^{+}_{D}(\mathbf{P})=\sqrt{E+M}
\begin{bmatrix}
1\\[8pt]
0\\[8pt]
\dfrac{P_{z}}{E+M}\\[8pt]
\dfrac{P_{x}}{E+M}
\end{bmatrix}
=\sqrt{E+M}
\begin{bmatrix}
1\\[8pt]
0\\[8pt]
\dfrac{|\mathbf{P}|\cos\theta}{E+M}\\[8pt]
\dfrac{|\mathbf{P}|\sin\theta}{E+M}
\end{bmatrix}
=\cos\dfrac{\theta}{2}u^{+}_{H}(\mathbf{P})-\sin\dfrac{\theta}{2}u^{-}_{H}(\mathbf{P}).
\end{align}
From this I can conclude that the above "spin-up" Dirac spinor is the "+" helicity spinor rotated by "-θ". And since the "+" helicity spinor is rotated from +z direction by "θ" angle, this Dirac spinor is again pointing in +z direction. So the above derivation demonstrates that Dirac spinor with "spin-up" always points in the +z direction. In other words, a "spin-up" spinor in the rest frame, when boosted to some arbitrary direction, remains pointing in the +z direction. One can prove similar results for spin down. This suggests that a single boost from rest frame doesn't change the spin orientation. And the gyroscope in a moving frame would still point in the same direction as in the rest frame.
In order for the above logic be consistent, the spin operator for a moving particle has to be modified to e^{-i\boldsymbol\eta\mathbf{K}}\Sigma e^{i\boldsymbol\eta\mathbf{K}}, if the spinor is boosted by e^{-i\boldsymbol\eta\mathbf{K}}, namely the operator has to be transformed in the opposite way in order to maintain the same spin eigenvalue. It's as if this is just a matter of change of basis. Speaking of which, this logic seems to suggest that for every different momentum in different direction, we need to change the basis for the spin operator in order to get a sensible answer.
Does this make sense? If my logic is correct, is there a simpler way to see why the spin orientation doesn't change direction under a single boost from the rest frame?
I haven't thought this through yet, but one "contradiction" to the above conclusion can be presented as follows: the commutation relation [J_{i}, K_{j}]=i\epsilon_{ijk}K_{k} says that boost and rotation don't commute unless \mathbf{J} and \mathbf{K} are collinear, which suggest that if I first rotate a spin that's originally pointing in +z by θ, and then boost it in \mathbf{P} direction, it would end up in a different state if I were to first boost it in P direction and then rotate the spinor by the same angle. Since the first set of operations gives me the helicity spinor, and the second one should differ, so when I first boost the spinor, it has to rotate by some amount so that later when I rotate by θ it doesn't end up in the same state. Is this right? I feel like something's wrong with this argument, but I can't put my finger on it yet. How does one resolve this contradiction?
And last but not least is DrDu's response:
DrDu said:
I would be careful with what is meant with spin vs. spinors.
The direction of spin is the direction of the Pauli-Lyubarsky vector and not of the spinor whose direction does not change under a single boost (but eventually under a combination of several boosts along different directions).
The claim that the spinor direction does not change seems to agree with my above derivation. But I don't know what you mean by the direction of spin vs spinors. What does it mean that spin has a different direction than spinor? What do they separately represent? More explicitly, if I say I have a "spin-up" Dirac spinor, what do you say is the spin direction and spinor direction according to your definition? Which one is the physically observed one? For example, if I put a stream of high velocity electron through a Stern-Gerlach apparatus (the momentum P is approximately in horizontal direction but spins are detected in vertical direction), do the ones deflected up end up in a state that has spin up or spinor up?
You need to review your understanding of angular momentum!
