 Solution

Let ${\chi }_{\pm }=(1,\pm 1{)}^T/\sqrt{2}$. The eigenvalues of ${\Sigma }_1$ are $\pm 1/2$, each with multiplicity two, and orthonormal bases for their eigenspaces areA general linear combination of the two positive-energy solutions has upper spinor $\xi =a{\chi }^1+b{\chi }^2$ and lower spinor $(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. An x-spin eigenvector needs $\xi \propto {\chi }_{\pm }$ and the lower spinor to have the same ${\sigma }_1$ eigenvalue. The Fixed-axis spin obstruction for positive-energy Dirac spinors isIts norm is $2\sqrt{p_y^2+p_z^2}$, so it vanishes exactly when $p_y=p_z=0$. For example with $\mathbf{p}=p\hat{z}$, the upper ${\chi }_{+}$ produces lower ${\chi }_{-}$, excluding a common eigenvector for every $p\ne 0$. In the nonrelativistic limit, the lower components are $O(p/m)$, and $({\psi }_1\pm {\psi }_2)/\sqrt{2}$ approach x-spin eigenvectors. At rest they are exact.

The prompt's unrestricted claim of impossibility at relativistic energies needs qualification: for momentum parallel to x these combinations remain exact eigenvectors at any energy. For a beam transverse to x, such as the conventional z-directed beam, the intended impossibility holds, and the lower components are no longer negligible. A transported relativistic spin observable differs from this fixed matrix; helicity is the conserved projection along momentum for a free particle.