# Mathematical proof on helicity of a massive fermion is not Lorentz invariant

What is the mathematical proof that the helicity of a massive spin-$1/2$ fermion is not Lorentz invariant?

Something is Lorentz invariant (e.g., $P_\mu P^\mu$) if it commutes with all the generators of the Lorentz group. Assuming this statement to be correct, the helicity operator, therefore, must not commute with all the generators $\frac{1}{2}\sigma_{\mu\nu}$ of Lorentz transformation. I should also be able to verify that in the limit the mass $m\to 0$, a Lorentz transformation doesn't change the helicity.

Is this line of reasoning sufficient to prove the failure of Lorentz invariance of the helicity operator?

• I've edited the question. Maybe the random downvoter reconsider the downvote or like to explain the reason? – SRS May 26 '18 at 13:07

Let $|\boldsymbol p,\sigma\rangle$ be a one-particle state, with $\boldsymbol p$ being its linear momentum, $$\boldsymbol P|\boldsymbol p,\sigma\rangle:=\boldsymbol p|\boldsymbol p,\sigma\rangle\tag{2.5.1}$$ with $\boldsymbol P$ the momentum operator, the generator of translations.

The label $\sigma$ is called the helicity of the particle. It is defined as the third component of its angular momentum in a special reference frame: $$J_z|\boldsymbol p^\star,\sigma\rangle:=\sigma|\boldsymbol p^\star,\sigma\rangle\tag{2.5.39}$$ with $\boldsymbol J$ the angular-momentum operator, the generator of rotations.

For a massive particle, the special reference frame is taken as the rest frame of the particle, $\boldsymbol p^\star=\boldsymbol 0$. For a massless one, it is taken as the reference frame such that $\boldsymbol p^\star=(0,0,\omega)$, with $\omega$ the energy of the particle.

Ref.1 works out the transformation laws of $|\boldsymbol p,\sigma\rangle$ under Lorentz transformations. In general, it is given by $$U(\Lambda)|\boldsymbol p,\sigma\rangle=\sum_{\sigma'} D_{\sigma\sigma'}(W(\Lambda,\boldsymbol p))|\boldsymbol p,\sigma\rangle\tag{2.5.23}$$ up to an inconsequential normalisation constant. Here, $W$ is a Wigner rotation and $D$ is a representation of the little group of $\boldsymbol p^\star$.

The key point is that the matrix $D$ for massless particle is diagonal, $$D_{\sigma\sigma'}(W)\mathrm e^{i\theta\sigma}\delta_{\sigma\sigma'}\tag{2.5.42}$$ and therefore the Lorentz transformation $U(\Lambda)$ doesn't change the value of $\sigma$. In the massive case, $D$ is not (in general) diagonal, and therefore Lorentz transformations do change the value of $\sigma$. In short, $\sigma$ is Lorentz invariant for massless particles, and Lorentz covariant for massive ones.

References.

1. Weinberg S. - Quantum theory of fields, Vol.1.
• Great! I have just one doubt. Unlike the operator $P_\mu P^\mu$, the operator $J_z$ is not Lorentz invariant. Shouldn't we also find the transformation of the operator under Lorentz transformation and act the transformed operator on the transformed state to draw the conclusion? @AccidentalFourierTransform – SRS May 29 '18 at 15:21
• @SRS The transformation of $J_z$ is given by Ref.1., eqs. 2.4.18-2.4.24. I'm not sure what you mean though: helicity is given, by definition, by the expression I wrote. If you were to use a different expression, you would calculate something different. – AccidentalFourierTransform May 30 '18 at 14:29

The usual way of seeing this is to note that a boost along the direction of travel of the particle with a speed greater than the particle's speed will flip the direction of the momentum. It does not affect the spin. The helicity, the component of spin along the direction of the particle momentum, therefore changes sign. If the particle is massless then its speed is $c$ and as a boost with $v>c$ is not possible the helicity cannot be changed.

You should be able to get away with e.g. performing a Lorentz transformation on a helicity eigenstate (e.g. a Weyl Spinor), and readily see it is no longer a helicity eigenstate.

I can try to write up some equations showing as much later, if you prefer and feel this line of thinking answers your question.

• That would be fine but I would also be able to check the limit $m\to 0$ in which case a Lorentz transformation doesn't change the helicity. – SRS May 25 '18 at 9:33
• I think I had this derivation exactly as part of a Relativistic QM course (so it can't be that hard :P) – rubenvb May 25 '18 at 10:24