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

Question

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.

Answer 1

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).

Answer 2

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.