Why can the spin of a relativistic particle not be orthogonal to its momentum?

I have read that the 3-momentum of a relativistic particle cannot be orthogonal to its spin 3-vector. When thinking about how the spin vector transforms when the particle approaches light speed, it seems clear that it cannot be orthogonal to the boost direction, but I'm wondering what is the exact mechanism that prevents it.

An example for clarification: Let's take an electron at rest in the lab frame, which has been put in a spin-up eigenstate ($z$-axis). Then, a potential is added, such that the electron gains some $x$-momentum.

Since the spin was pure $z$ axis and the momentum pure $x$, some relativistic process must change the spin vector to prevent it from being orthogonal to the momentum.

EDIT: I'm starting to think that this must be answered with Dirac spinors. The $x$ momentum must have some effect on the spin vector direction. But I cannot see the quantitative answer, so I'm still looking for an answer.

• Where did you read this? – Ryan Unger Jan 8 '16 at 23:39
• @0celo7 Some lecture notes I took some years ago. – Bass Jan 8 '16 at 23:45
• So this is only absolutely true for luminal (i.e. massless) particles (that's why a spin-1 photon has only two polarization states), but starting with an unpolarized population and boosting results is Lorentz focusing working to reduce the average transverse component. – dmckee Jan 8 '16 at 23:52
• @dmckee Not sure if that answers my question. I'm talking about a polarized particle. How can the Lorentz transformation reduce the transverse componente if there is only transverse component? This would mean that the magnitude of the spin vector decreases, wouldn't it? – Bass Jan 8 '16 at 23:59

I think you are just referring to a simple kinematical phenomenon. Consider a massive particle carrying a $3$-vector $\vec{S} \equiv (s_x,s_y,s_z)$ in its rest frame, which exists just because the particle is massive.

This vector may describe the spin, but not necessarily it. Now suppose that the particle travels with a speed $v$ along the direction $z$ if referring to the laboratory reference frame whose spatial $z$-axis is collinear with the one of the rest frame (up to a pair of internal rotations the decomposition theorem of Lorentz group guarantees that this is a standard situation).

The vector $\vec{S}$ defines a four vector $S^\mu$ whose components in the laboratory frame are $$\left(\frac{vs_3}{c\sqrt{1-\frac{v^2}{c^2}}},s_x,s_y, \frac{s_3}{\sqrt{1-\frac{v^2}{c^2}}}\right)\:.$$ You see that, among the last three spatial components, the component parallel to the momentum (in the direction $z$) $\frac{s_3}{\sqrt{1-\frac{v^2}{c^2}}}$ dominates with respect to the normal components $s_x,s_y$ as $v \sim c$.

In the limit $v=c$, if one assumes that something like the parallel component exists, the normal components cannot be defined at all. However the rest frame itself do not exist, so this case cannot be easily treated and another approach is more suitable, the one arising from the classification of unitary irreducible representations of Lorentz group. In a sense, however, the remnant of the spatial components of $S^\mu$ in the laboratory frame, in this limit case is the so called helicity, which could be viewed as always parallel (or antiparallel) to the $3$-momentum (see ACuriousMind's answer).

• Thank you! In your example, what happens if $s_3=0$ in the lab frame? In this case, the parallel component cannot dominate the other components in the $v\sim c$ limit. Furthermore, the time component $S^0$ vanishes regardless of $v$. In the $v\sim c$ limit, this would yield zero helicity, wouldn't it? (that's what I meant by spin orthogonal to momentum). – Bass Jan 9 '16 at 10:07
• Indeed the correct viewpoint is that orthogonal and parallel components decouple. But this concerns classical particles, for quantum ones one should study the representations of the Poincaré group and sees that, for $m =0$, in a sense, only the parallel component survives. – Valter Moretti Jan 9 '16 at 10:15
• But this isn't obvious if you consider $S^\mu$ as a 4-vector, as you did, is it? I mean, the components of your $S^\mu$ approach $(0,s_x,s_y,0)$ in the $v\sim c$ limit. So there must be some other transformation at work that makes $S^3$ non-zero. Would it be better to ask this again after I've studied Poincaré group representations (which I intend to)? – Bass Jan 9 '16 at 10:19
• I do not understand, as $v \to c$ you do not obtain $(0,s_x,s_y,0)$ but something like $(\infty, s_x,s_y, \infty)$. To get a sensible limit you should rescale the infinite components. However all that is cumbersome. The right procedure is to study the unitary strongly continuous irreducible representations of the Poincar\'e group and focus on the generators of the 3-rotations. If the particle is massive (the mass is a Casimir operator) these generators are naturally decomposed into an orbital and spinorial part. If $m=0$ the structure is more complicated and an observable enters the game. – Valter Moretti Jan 9 '16 at 10:59
• Sorry, I misunderstood your comment! Yes for $s_3=0$ you have $(0,s_x,s_y,0)$. The point is that, in the quantum formalism to assume that, for $v=c$, the Pauli-Lubanski vector is not parallel to the momentum (technically this is your claim), leads to pathological representations of Poincaré group (formally, particles with continuous spin, none obseverved them). – Valter Moretti Jan 9 '16 at 11:12

The question is probably about helicity, which is the proper quantum field theoretic notion for deducing the relation of the spin of a particle to its momentum.

Let us consider a Dirac spinor with a chosen realization of the gamma matrices as $\gamma^0 = \left(\begin{matrix}0 & \mathbf{1}_2 \\ \mathbf{1}_2 & 0 \end{matrix}\right)$ and $\gamma^i = \left(\begin{matrix} 0 & \sigma^i \\ - \sigma^i & 0 \end{matrix}\right)$ where $\sigma^i$ are the ordinary Pauli matrices. Then the helicity operator is defined as $$h = \frac{1}{2}\vec p \cdot\vec \sigma = \epsilon_{ijk}p^i S^{jk}$$ where $S^{jk} = \frac{\mathrm{i}}{4}[\gamma^i,\gamma^j]$ are the rotation generators (since generally, $S^{\mu\nu} = \frac{\mathrm{i}}{4}[\gamma^\mu,\gamma^\nu]$ are the Lorentz generators).

Obviously, helicity is a measure for whether or not spin and momentum are aligned. In particular, a zero eigenvalue means that spin and momentum are orthogonal.

Now, the space of Dirac spinors splits into left- and right-handed spinors, which are just the eigenspaces of $h$ with eigenvalue $-1/2$ and $+1/2$, respectively. There is no zero eigenvalue of $h$, so there are no relativistic free fermion states which have their spin orthogonal to their momentum.

This helicity is not really a "good" object to think about for two reasons:

1. If the fermion is massive, the left- and right-handed spinors do not decouple, i.e. the helical eigenspaces are not preserved by time evolution. Hence helicity is not conserved under time evolution for massive particles.

2. Helicity is not a Lorentz-invariant notion for massive particles. To each momentum of a massive particle, there is a boosted frame in which it points in the opposite direction, but the spin is unchanged under a Lorentz transformation. Hence, helicity may change its sign from frame to frame, and is thus not invariant.

• I do not think your answer is completely correct: if you were considering a spin $1$ particle you would find the $0$ eigenvalue of $h$. The point is the mass. As soon as $m \to 0$, the helicity becomes an independent operator (the spin ceases to exist) and only two eigenvalues remains. It does not depend on the fact one is dealing with fermions or bosons as the photon (boson, improperly speaking, spin $1$) shows. – Valter Moretti Jan 9 '16 at 9:25
• @ValterMoretti: You're right, my argument doesn't work for a massive bosonic object. I just guessed this might be what the question is about because we didn't get a more precise reference than "I have read that the 3-momentum of a relativistic particle cannot be orthogonal to its spin 3-vector." – ACuriousMind Jan 9 '16 at 13:25