Dirac free particle with $x$-momentum For a free particle with momentum $\mathbf{p}=p\mathbf{x}$, the Dirac Hamiltonian is
\begin{equation}
H=\alpha_xp+\beta m = \begin{pmatrix}
m & 0 & 0 &p\\
0 & m & p & 0\\
0 & p & -m & 0\\
p & 0 & 0 & -m
\end{pmatrix} \ ,
\end{equation}
of which one eigenstate is
\begin{equation}
u_1 = \begin{pmatrix} 1 \\ 0 \\0\\ \frac{p}{E_p+m}\end{pmatrix} \ .
\end{equation}
This should also be an eigenstate of the helicity $\hat{h} = \mathbf{\Sigma}\cdot\mathbf{\hat p}=  \begin{pmatrix} \sigma_x & 0 \\ 0 & \sigma_x \end{pmatrix}$. However it is pretty clear to see that
\begin{equation}
u_1^\dagger\hat hu_1=\begin{pmatrix}
1 & 0 & 0 & \frac{p}{E_p+m}
\end{pmatrix}\begin{pmatrix}
0 & 1 & 0 & 0\\
1 & 0&0&0\\
0 &0&0&1\\
0&0&1&0
\end{pmatrix}
\begin{pmatrix}
1 \\0 \\ 0 \\ \frac{p}{E_p+m}
\end{pmatrix}=0 \ ,
\end{equation}
which should not be the case. I think I'm getting confused with something notational/simple here; please let me know.
 A: In the single-particle Dirac theory, the operators $\vec{p}$ and $h=\vec{\Sigma}\cdot\vec{p}$ both commute with the Hamiltonian.  That means that it is possible to find eigenstates of $H=\vec{\alpha}\cdot\vec{p}+\beta m$ that are also simultaneous eigenstates of $\vec{p}$ and $h=\vec{\Sigma}\cdot\vec{p}$.  However, it does not mean that an arbitrary eigenstate of the three-momentum $\vec{p}$ will also be an eigenstate of $h=\vec{\Sigma}\cdot\vec{p}$, which is what you seem to be aiming for in your example.
The simultaneous eigenstates of $h=\Sigma_{x}$ when $\vec{p}=p\hat{x}$ are (in the Dirac representation used in the question)
$$u=\left[ \begin{array}{c}
u_{>} \\ u_{<} \end{array}\right]=\left[ \begin{array}{c}
1 \\ \pm 1 \\ \mp\frac{p}{E_{p}+m} \\ \frac{p}{E_{p}+m}\end{array}\right].$$
Note that both the large ($u_{>}$) and small ($u_{<}$) components of the Dirac four-spinor are separately two-spinor eigenvectors of $\sigma_{x}$. What you have tried to do with the $u$ in the question is take a a spinor in which $u_{>}$ is an up eigenstate of $\sigma_{z}$, and what you found was exactly what was to be expected—that the energy and momentum eigenstate with this $u_{>}$ was not an eigenstate of $\Sigma_{x}$ (or $\Sigma_{z}$), unless $p=0$.
