= Solution
For spin-up along the first axis, take the normalized $+1$ eigenspinor of $\sigma_1$, namely $\xi_1=(1,1)^T/\sqrt2$. At rest this gives
$$
\boxed{u_{1,+}(\mathbf0)=\sqrt{\frac m2}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}.}
$$
For momentum $(E,0,0,E)$, using the square roots displayed in part (i) with the same label gives
$$
\boxed{u_{1,\mathrm{label}}(E\widehat{\mathbf e}_3)=\sqrt E\begin{pmatrix}0\\1\\1\\0\end{pmatrix}.}
$$
This is the <massless limit of a transverse canonical-spin label>: it is the equal superposition of the negative-helicity left-chiral mode and the positive-helicity right-chiral mode. It solves the massless <Dirac equation> and has norm squared $2E$.
A qualification is essential if “spin along the first axis” is interpreted literally at zero mass. The fixed transverse rotation generator is $\Sigma_1/2$, with $\Sigma_1=\operatorname{diag}(\sigma_1,\sigma_1)$, and the massless Hamiltonian at this momentum is $H=E\gamma^0\gamma^3$. They anticommute. If $Hu=Eu$ and $\Sigma_1u=u$ both held, then $H\Sigma_1u=-\Sigma_1Hu=-Eu$, whereas $H\Sigma_1u=Hu=Eu$, forcing $u=0$. Thus there is no nonzero positive-energy massless solution with that literal transverse-spin eigenvalue. The displayed nonzero spinor is the transported rest-spin label requested by the massive-to-massless construction, rather than a transverse-spin eigenstate.
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