# How do we show that for massless fermions, Helcity and Chirality align?

The Helicity operator of a representation of the Lorentz group is given by $$h = \varepsilon_{ijk}S^{jk}\frac{P^i}{|P|}$$ where $$S^{\mu\nu}$$ are the generators of the Lorentz group.

In the $$(\frac{1}{2},0)\oplus(0,\frac{1}{2})$$ rep, for a massless Dirac spinor with momentum purely in the $$+z$$ direction, the helicity operator becomes:

$$h=\frac{1}{2}\left(\begin{array}{ll} \sigma_z & 0 \\ 0 & \sigma_z \end{array}\right)$$ Which has splits our space of spinors into two eigen-subspaces: $$\psi = \psi_+ + \psi_- \quad \rightarrow \quad \psi_+ = \left(\begin{array}{l} a \\ 0 \\ b \\ 0 \\ \end{array}\right), \quad \psi_- = \left(\begin{array}{l} 0 \\ c \\ 0 \\ d \\ \end{array}\right)$$

We also have the Chirality operator $$\gamma^5$$, which when working in the Chiral basis, is also diagonal: $$\gamma^5=\left(\begin{array}{ll} -I_2 & 0 \\ 0 & I_2 \end{array}\right)$$ Which again splits our space of spinors into $$2$$ eigen-subspaces $$\psi = \psi_L + \psi_R \quad \rightarrow \quad \psi_L = \left(\begin{array}{l} a \\ c \\ 0 \\ 0 \\ \end{array}\right), \quad \psi_R = \left(\begin{array}{l} 0 \\ 0 \\ b \\ d \\ \end{array}\right)$$ Clearly if we only allow cases when $$c = b = 0$$ then these operators are the same up on this subspace (up to a constant), but that doesn't seem to necessarily be the case.

What am I missing that allows us to say that for massless fermions, Helicity and Chirality are the same?

You need an input from dynamics. The Weyl hamiltonan for a right-handed (positive chirality) fermion is $$H= {\boldsymbol \sigma}\cdot {\bf p}$$ so with $${\bf p}=(0,0,p)$$ positive energy eigenstates have their spin a +1 eigenstate of $$\sigma_z$$, i.e their spin is
parallel to the momentum (positive helicity). Negative energy correspond to antiparticles and have negative helicity.

Mike Stone stirred my thought process and made me realize that physical spinors must obey the (massless) Dirac equation. For the specific set-up at hand this leads to the conditions that: $$\begin{split} \sigma_z\>\psi_L &= I_2 \>\psi_L \\ \sigma_z\>\psi_R &= -I_2 \>\psi_R \end{split}$$ Where I'm abusing the notation to make $$\psi_{L/R}$$ represent the $$2$$-component Weyl spinors. This forces the very condition $$c=b=0$$ in order for the eigen-spaces of the operators on physical spinors to coincide.