= Solution
Use signature $(+,-,-,-)$ and the free <Dirac equation> $(i\gamma^\mu\partial_\mu-m)\psi=0$. For positive $p^0=E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$, substitution of the two frequency signs gives
$$
\psi=u_s(p)e^{-ip\cdot x}:\quad(\not p-m)u_s(p)=0,
\qquad
\psi=v_s(p)e^{ip\cdot x}:\quad(\not p+m)v_s(p)=0.
$$
Multiplying by the opposite factor shows $p^2=m^2$. Each equation has two independent solutions for a massive field.
In the <Dirac basis>, $\gamma^0=\operatorname{diag}(I,-I)$ and $\gamma^i=\begin{pmatrix}0&\sigma_i\\-\sigma_i&0\end{pmatrix}$. For orthonormal two-component <spin> labels $\xi_s,\eta_s$, convenient solutions are
$$
\boxed{u_s(p)=\sqrt{E_{\mathbf p}+m}
\begin{pmatrix}\xi_s\\
\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E_{\mathbf p}+m}\xi_s
\end{pmatrix},\qquad
v_s(p)=\sqrt{E_{\mathbf p}+m}
\begin{pmatrix}
\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E_{\mathbf p}+m}\eta_s\\
\eta_s
\end{pmatrix}.}
$$
The two-block equations verify these expressions using $(\boldsymbol\sigma\cdot\mathbf p)^2=\mathbf p^2I$. At rest the positive-frequency solutions have only upper components and the negative-frequency solutions only lower components. Their normalizations are $u_s^\dagger u_r=v_s^\dagger v_r=2E_{\mathbf p}\delta_{sr}$ and $\bar u_su_r=2m\delta_{sr}$, $\bar v_sv_r=-2m\delta_{sr}$. Their completeness relations are
$$
\sum_su_s(p)\bar u_s(p)=\not p+m,\qquad
\sum_sv_s(p)\bar v_s(p)=\not p-m.
$$
The field expansion combines <fermionic annihilation operators> multiplying the positive-frequency waves with <antiparticle> <fermionic creation operators> multiplying the negative-frequency waves. The latter frequency sign is not the <energy> of a physical negative-energy <antiparticle> state.
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