# 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
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.