 Solution

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.