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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?

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2 Answers 2

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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.

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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.

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