Operate on (5.7) with $\gamma^\nu \partial_v$ and show that each of the four components $\psi_i$ satisfies the Klein-Gordon equation
$$
\left(\square^2+m^2\right) \psi_i=0 .
$$
For a free particle, we can therefore seek four-momentum eigensolutions of Dirac's equation of the form
$$
\psi=u(\mathbf{p}) e^{-i p \cdot x},
$$
where $u$ is a four-component spinor independent of $x$. Substituting in (5.7), we have
$$
\left(\gamma^\mu p_\mu-m\right) u(\mathbf{p})=0
$$
or, using the abbreviated notation $\mathcal{A} \equiv \gamma^\mu A_\mu$ for any four-vector $A_\mu$,
$$
(p-m) u=0 \text {. }
$$
Since we are seeking energy eigenvectors, it is easier to use the original form, (5.1),
$$
H u=(\boldsymbol{\alpha} \cdot \mathbf{p}+\beta m) u=E u .
$$
There are four independent solutions of this equation, two with $E>0$ and two with $E<0$. This is particularly easy to see in the Dirac-Pauli representation of $\alpha$ and $\beta$. First, take the particle at rest, $\mathbf{p}=0$. Using (5.4), we have
$$
H u=\beta m u=\left(\begin{array}{cc}
m I & 0 \\
0 & -m I
\end{array}\right) u
$$
with eigenvalues $E=m, m,-m,-m$, and eigenvectors
$$
\left(\begin{array}{l}
1 \\
0 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
1 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
1 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
0 \\
1
\end{array}\right) .
$$
As we have just mentioned, the electron, charge $-e$, is regarded as the particle. The first two solutions therefore describe an $E>0$ electron. The $E<0$ particle solutions are to be interpreted, as before, as describing an $E>0$ antiparticle (positron) (see Section 3.6).
For $\mathbf{p} \neq 0,(5.21)$ becomes, using (5.4),
$$
H u=\left(\begin{array}{cc}
m & \boldsymbol{\sigma} \cdot \mathbf{p} \\
\boldsymbol{\sigma} \cdot \mathbf{p} & -m
\end{array}\right)\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right)=E\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right),
$$
where $u$ has been divided into two two-component spinors, $u_A$ and $u_B$. This EXERCISE 5.2 Operate on (5.7) with $\gamma^\nu \partial_v$ and show that each of the four components $\psi_i$ satisfies the Klein-Gordon equation
$$
\left(\square^2+m^2\right) \psi_i=0 .
$$
For a free particle, we can therefore seek four-momentum eigensolutions of Dirac's equation of the form
$$
\psi=u(\mathbf{p}) e^{-i p \cdot x},
$$
where $u$ is a four-component spinor independent of $x$. Substituting in (5.7), we have
$$
\left(\gamma^\mu p_\mu-m\right) u(\mathbf{p})=0
$$
or, using the abbreviated notation $\mathcal{A} \equiv \gamma^\mu A_\mu$ for any four-vector $A_\mu$,
$$
(p-m) u=0 \text {. }
$$
Since we are seeking energy eigenvectors, it is easier to use the original form, (5.1),
$$
H u=(\boldsymbol{\alpha} \cdot \mathbf{p}+\beta m) u=E u .
$$
There are four independent solutions of this equation, two with $E>0$ and two with $E<0$. This is particularly easy to see in the Dirac-Pauli representation of $\alpha$ and $\beta$. First, take the particle at rest, $\mathbf{p}=0$. Using (5.4), we have
$$
H u=\beta m u=\left(\begin{array}{cc}
m I & 0 \\
0 & -m I
\end{array}\right) u
$$
with eigenvalues $E=m, m,-m,-m$, and eigenvectors
$$
\left(\begin{array}{l}
1 \\
0 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
1 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
1 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
0 \\
1
\end{array}\right) .
$$
As we have just mentioned, the electron, charge $-e$, is regarded as the particle. The first two solutions therefore describe an $E>0$ electron. The $E<0$ particle solutions are to be interpreted, as before, as describing an $E>0$ antiparticle (positron) (see Section 3.6).
For $\mathbf{p} \neq 0,(5.21)$ becomes, using (5.4),
$$
H u=\left(\begin{array}{cc}
m & \boldsymbol{\sigma} \cdot \mathbf{p} \\
\boldsymbol{\sigma} \cdot \mathbf{p} & -m
\end{array}\right)\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right)=E\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right),
$$
where $u$ has been divided into two two-component spinors, $u_A$ and $u_B$. This call this quantum number the helicity of the state. The possible eigenvalues $\lambda$ of the helicity operator $\frac{1}{2} \boldsymbol{\sigma} \cdot \hat{\mathbf{p}}$ are
$$
\lambda=\left\{\begin{array}{l}
+\frac{1}{2} \text { positive helicity, } \\
-\frac{1}{2} \text { negative helicity. }
\end{array}\right.
$$
We see that no other component of $\boldsymbol{\sigma}$ has eigenvalues which are good quantum numbers.
With the above choice (5.25), of the spinors $\chi^{(s)}$, it is appropriate to choose $\mathbf{p}$ along the $z$ axis, $\mathbf{p}=(0,0, p)$. Then
$$
\frac{1}{2} \mathbf{\sigma} \cdot \hat{\mathbf{p}} \chi^{(s)}=\frac{1}{2} \sigma_3 \chi^{(s)}=\lambda \chi^{(s)}
$$
with $\lambda= \pm \frac{1}{2}$ corresponding to $s=1,2$, respectively.