= Solution
Direct multiplication of the <gamma matrices> in the <chiral representation> gives
$$
\boxed{\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3=
\begin{pmatrix}-I_2&0\\0&I_2\end{pmatrix}}.
$$
For example $\sigma^1\sigma^2\sigma^3=iI_2$ by the <Pauli matrix multiplication law>, and the block multiplications give the displayed signs. Every $\gamma^\mu$ is off-diagonal in this representation, whereas the <chirality matrix> has opposite diagonal blocks. Therefore $\gamma^5\gamma^\mu+\gamma^\mu\gamma^5=0$ for all $\mu$.
The left-handed <chirality> condition makes the plane-wave coefficient $\lambda(p)=\binom{\chi}{0}$. With $p\cdot x=Ex^0-\mathbf p\cdot\mathbf x$, the <Dirac equation> gives
$$
\not p\,\lambda(p)=0
\quad\Longleftrightarrow\quad
(EI_2+\boldsymbol\sigma\cdot\mathbf p)\chi=0.
$$
Since $(\boldsymbol\sigma\cdot\mathbf p)^2=|\mathbf p|^2I_2$, its <determinant> is $E^2-|\mathbf p|^2$. A nonzero <spinor> therefore exists exactly when
$$
\boxed{p^2=0}.
$$
For nonzero <momentum>, put $r=|\mathbf p|$ and write its direction as $(\sin\vartheta\cos\varphi,\sin\vartheta\sin\varphi,\cos\vartheta)$. Normalized <eigenvectors> of $\boldsymbol\sigma\cdot\widehat{\mathbf p}$ are
$$
\chi_-=
\begin{pmatrix}-e^{-i\varphi}\sin(\vartheta/2)\\\cos(\vartheta/2)\end{pmatrix},
\qquad
\chi_+=\begin{pmatrix}\cos(\vartheta/2)\\e^{i\varphi}\sin(\vartheta/2)\end{pmatrix}.
$$
The general nonzero solution is $\lambda(p)=C\binom{\chi_-}{0}$ for $E=r$ and $\lambda(p)=C\binom{\chi_+}{0}$ for $E=-r$, with arbitrary nonzero complex normalization $C$. At $p=0$ any constant two-component $\chi$ solves the equation, but the <momentum> direction and <helicity> are undefined. For the physical positive-energy branch this is the <left-handed massless plane-wave spinor>.
A rotation by angle $\alpha$ about $\widehat{\mathbf p}$ acts on the two-component <spinor> as
$$
S(R)=e^{-i\alpha\boldsymbol\sigma\cdot\widehat{\mathbf p}/2}.
$$
The plane-wave argument is unchanged because the rotation fixes $\mathbf p$, and $S(R)\chi_-=e^{i\alpha/2}\chi_-$. Thus the positive-energy solution has <helicity> $h=-1/2$, where a helicity-$h$ state has rotation phase $e^{-i\alpha h}$. The negative-energy coefficient has the opposite <spinor> phase.
In a quantized complex left-handed <Weyl field>, the positive-frequency term annihilates particles and the negative-frequency term creates antiparticles. Reinterpreting negative energy also reverses its spatial <momentum>; for the positive physical <momentum> the two <spinor> coefficients have the same left-chiral kernel. Because one multiplies an <annihilation operator> and the other a <creation operator>, the corresponding particle and antiparticle state phases are conjugate. The <helicity content of a quantized Weyl field> is consequently
$$
\boxed{h_{\rm particle}=-\tfrac12,\qquad h_{\rm antiparticle}=+\tfrac12}.
$$
There is one particle spin state at each nonzero <momentum>, rather than the two helicities supplied by a full massless <Dirac field>. The right-handed <Weyl field> interchanges these signs. This is the massless relation between <chirality> and <helicity>; it does not identify them for a massive field.
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