= Solution
Let $\chi_\pm=(1,\pm1)^T/\sqrt2$. The <eigenvalues> of $\Sigma_1$ are $\pm1/2$, each with multiplicity two, and orthonormal bases for their <eigenspaces> are
$$
v_{\pm,1}=\frac1{\sqrt2}(1,\pm1,0,0)^T,\qquad
v_{\pm,2}=\frac1{\sqrt2}(0,0,1,\pm1)^T.
$$
A general linear combination of the two positive-energy solutions has upper spinor $\xi=a\chi^1+b\chi^2$ and lower spinor $(\boldsymbol\sigma\cdot\mathbf p)\xi/(E+m)$. 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> is
$$
[\sigma_1,\boldsymbol\sigma\cdot\mathbf p]\chi_\pm=2i(p_y\sigma_3-p_z\sigma_2)\chi_\pm.
$$
Its 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 $p\ne0$. In the <nonrelativistic limit>, the lower components are $O(p/m)$, and $(\psi_1\pm\psi_2)/\sqrt2$ 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.
Back to article page
