= Solution
For positive frequency take $p^0=E=\sqrt{|\mathbf p|^2+m^2}>0$, with $m\geq0$. Lowering the index in $p_\mu\sigma^\mu$ is important: put
$$
A=p\cdot\sigma=E I_2-\mathbf p\cdot\boldsymbol\sigma,\qquad
B=p\cdot\bar\sigma=E I_2+\mathbf p\cdot\boldsymbol\sigma.
$$
These are commuting <Hermitian matrices> with $AB=(E^2-|\mathbf p|^2)I_2=m^2I_2$. Their <eigenvalues> are $E\pm|\mathbf p|$, so they are positive definite for $m>0$ and <positive semidefinite matrices> for $m=0$. <Simultaneous diagonalization> defines their positive <square roots of a matrix> and gives $\sqrt A\sqrt B=mI_2$.
The plane-wave ansatz reduces the <Dirac equation> to $(\not p-m)u=0$, where
$$
\not p=\begin{pmatrix}0&A\\B&0\end{pmatrix}.
$$
For $u_L=\sqrt A\xi$ and $u_R=\sqrt B\xi$, commutativity gives $Au_R=m u_L$ and $Bu_L=m u_R$. Therefore
$$
\boxed{\psi_+(x)=\begin{pmatrix}\sqrt A\xi\\\sqrt B\xi\end{pmatrix}e^{-ip\cdot x}.}
$$
This is the <Hermitian square-root construction of Dirac plane waves>.
For negative frequency retain the future-directed momentum label $p$ and use $\psi_-(x)=v(\mathbf p)e^{+ip\cdot x}$. The equation is now $(\not p+m)v=0$. Its general solution is
$$
\boxed{v(\mathbf p)=\begin{pmatrix}\sqrt A\zeta\\-\sqrt B\zeta\end{pmatrix},\qquad\psi_-(x)=v(\mathbf p)e^{+ip\cdot x},}
$$
with arbitrary two-spinor $\zeta$. The minus between its two blocks gives $Av_R=-m v_L$ and $Bv_L=-m v_R$. For $m>0$ these parametrizations span the two-dimensional positive- and negative-frequency solution spaces. At nonzero null momentum the square roots have complementary rank-one supports and still give the full two-dimensional kernels. If $\xi^\dagger\xi=\zeta^\dagger\zeta=1$, both spinors have norm squared $2E$; the positive one also has $\bar uu=2m$.
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