Given an alternate representation of the Dirac equation

\label{eqn:diracAlternate:20}
H =
\begin{bmatrix}
m c^2 + V_0 & c \hat{p} \\
c \hat{p} & – m c^2 + V_0
\end{bmatrix},

calculate the constant momentum solutions, the Heisenberg velocity operator $$\hat{v}$$, and find the form of the probability density current.

### Plane wave solutions

The action of the Hamiltonian on

\label{eqn:diracAlternate:40}
\psi =
e^{i k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}

is
\label{eqn:diracAlternate:60}
\begin{aligned}
H \psi
&=
\begin{bmatrix}
m c^2 + V_0 & c (-i \Hbar) i k \\
c (-i \Hbar) i k & – m c^2 + V_0
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
e^{i k x – i E t/\Hbar} \\
&=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}
\psi.
\end{aligned}

Writing

\label{eqn:diracAlternate:80}
H_k
=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}

the characteristic equation is

\label{eqn:diracAlternate:100}
0
=
(m c^2 + V_0 – \lambda)
(-m c^2 + V_0 – \lambda)
– (c \Hbar k)^2
=
\lr{ (\lambda – V_0)^2 – (m c^2)^2 } – (c \Hbar k)^2,

so

\label{eqn:diracAlternate:120}
\lambda = V_0 \pm \epsilon,

where
\label{eqn:diracAlternate:140}
\epsilon^2 = (m c^2)^2 + (c \Hbar k)^2.

We’ve got

\label{eqn:diracAlternate:160}
\begin{aligned}
H – ( V_0 + \epsilon )
&=
\begin{bmatrix}
m c^2 – \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 – \epsilon
\end{bmatrix} \\
H – ( V_0 – \epsilon )
&=
\begin{bmatrix}
m c^2 + \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

so the eigenkets are

\label{eqn:diracAlternate:180}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix}.
\end{aligned}

Up to an arbitrary phase for each, these are

\label{eqn:diracAlternate:200}
\begin{aligned}
\ket{V_0 + \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon – m c^2) }}
\begin{bmatrix}
c \Hbar k \\
\epsilon -m c^2
\end{bmatrix} \\
\ket{V_0 – \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon + m c^2) }}
\begin{bmatrix}
-c \Hbar k \\
\epsilon + m c^2
\end{bmatrix} \\
\end{aligned}

We can now write

\label{eqn:diracAlternate:220}
H_k =
E
\begin{bmatrix}
V_0 + \epsilon & 0 \\
0 & V_0 – \epsilon
\end{bmatrix}
E^{-1},

where
\label{eqn:diracAlternate:240}
\begin{aligned}
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
-\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k < 0. \end{aligned} Here the signs have been adjusted to ensure the transformation matrix has a unit determinant. Observe that there's redundancy in this matrix since $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon - m c^2 } } = \sqrt{ \epsilon + m c^2 }$$, and $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon + m c^2 } } = \sqrt{ \epsilon - m c^2 }$$, which allows the transformation matrix to be written in the form of a rotation matrix \label{eqn:diracAlternate:260} \begin{aligned} E &= \inv{\sqrt{2 \epsilon} } \begin{bmatrix} \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\ \frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } \end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } }
\end{bmatrix}, \qquad k < 0 \\ \end{aligned} With \label{eqn:diracAlternate:280} \begin{aligned} \cos\theta &= \frac{c \Hbar \Abs{k}}{ \sqrt{ 2 \epsilon( \epsilon - m c^2) } } = \frac{\sqrt{ \epsilon + m c^2} }{ \sqrt{ 2 \epsilon}}\\ \sin\theta &= \frac{c \Hbar k}{ \sqrt{ 2 \epsilon( \epsilon + m c^2) } } = \frac{\textrm{sgn}(k) \sqrt{ \epsilon - m c^2}}{ \sqrt{ 2 \epsilon } }, \end{aligned} the transformation matrix (and eigenkets) is $$\label{eqn:diracAlternate:300} \boxed{ E = \begin{bmatrix} \ket{V_0 + \epsilon} & \ket{V_0 - \epsilon} \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}. }$$ Observe that \ref{eqn:diracAlternate:280} can be simplified by using double angle formulas \label{eqn:diracAlternate:320} \begin{aligned} \cos(2 \theta) &= \frac{\lr{ \epsilon + m c^2} }{ 2 \epsilon } - \frac{\lr{ \epsilon - m c^2}}{ 2 \epsilon } \\ &= \frac{1}{ 2 \epsilon } \lr{ \epsilon + m c^2 - \epsilon + m c^2 } \\ &= \frac{m c^2 }{ \epsilon }, \end{aligned} and $$\label{eqn:diracAlternate:340} \sin(2\theta) = 2 \frac{1}{2 \epsilon} \textrm{sgn}(k ) \sqrt{ \epsilon^2 - (m c^2)^2 } = \frac{\Hbar k c}{\epsilon}.$$ This allows all the $$\theta$$ dependence on $$\Hbar k c$$ and $$m c^2$$ to be expressed as a ratio of momenta $$\label{eqn:diracAlternate:360} \boxed{ \tan(2\theta) = \frac{\Hbar k}{m c}. }$$

### Hyperbolic solutions

For a wave function of the form

\label{eqn:diracAlternate:380}
\psi =
e^{k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},

some of the work above can be recycled if we substitute $$k \rightarrow -i k$$, which yields unnormalized eigenfunctions

\label{eqn:diracAlternate:400}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

where

\label{eqn:diracAlternate:420}
\epsilon^2 = (m c^2)^2 – (c \Hbar k)^2.

The squared magnitude of these wavefunctions are

\label{eqn:diracAlternate:440}
\begin{aligned}
(c \Hbar k)^2 + (m c^2 \mp \epsilon)^2
&=
(c \Hbar k)^2 + (m c^2)^2 + \epsilon^2 \mp 2 m c^2 \epsilon \\
&=
(c \Hbar k)^2 + (m c^2)^2 + (m c^2)^2 \mp (c \Hbar k)^2 – 2 m c^2 \epsilon \\
&= 2 (m c^2)^2 \mp 2 m c^2 \epsilon \\
&= 2 m c^2 ( m c^2 \mp \epsilon ),
\end{aligned}

so, up to a constant phase for each, the normalized kets are

\label{eqn:diracAlternate:460}
\begin{aligned}
\ket{V_0+\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

After the $$k \rightarrow -i k$$ substitution, $$H_k$$ is not Hermitian, so these kets aren’t expected to be orthonormal, which is readily verified

\label{eqn:diracAlternate:480}
\begin{aligned}
\braket{V_0+\epsilon}{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
-i c \Hbar k &
m c^2 – \epsilon
\end{bmatrix}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix} \\
&=
\frac{ 2 ( c \Hbar k )^2 }{2 m c^2 \sqrt{(\Hbar k c)^2} } \\
&=
\textrm{sgn}(k)
\frac{
\Hbar k }{m c } .
\end{aligned}

### Heisenberg velocity operator

\label{eqn:diracAlternate:500}
\begin{aligned}
\hat{v}
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ H} \\
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ m c^2 \sigma_z + V_0 + c \hat{p} \sigma_x } \\
&= \frac{c \sigma_x}{i \Hbar} \antisymmetric{ \hat{x} }{ \hat{p} } \\
&= c \sigma_x.
\end{aligned}

### Probability current

Acting against a completely general wavefunction the Hamiltonian action $$H \psi$$ is

\label{eqn:diracAlternate:520}
\begin{aligned}
i \Hbar \PD{t}{\psi}
&= m c^2 \sigma_z \psi + V_0 \psi + c \hat{p} \sigma_x \psi \\
&= m c^2 \sigma_z \psi + V_0 \psi -i \Hbar c \sigma_x \PD{x}{\psi}.
\end{aligned}

Conversely, the conjugate $$(H \psi)^\dagger$$ is

\label{eqn:diracAlternate:540}
-i \Hbar \PD{t}{\psi^\dagger}
= m c^2 \psi^\dagger \sigma_z + V_0 \psi^\dagger +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x.

These give

\label{eqn:diracAlternate:560}
\begin{aligned}
i \Hbar \psi^\dagger \PD{t}{\psi}
&=
m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi -i \Hbar c \psi^\dagger \sigma_x \PD{x}{\psi} \\
-i \Hbar \PD{t}{\psi^\dagger} \psi
&= m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x \psi.
\end{aligned}

Taking differences
\label{eqn:diracAlternate:580}
\psi^\dagger \PD{t}{\psi} + \PD{t}{\psi^\dagger} \psi
=
– c \psi^\dagger \sigma_x \PD{x}{\psi} – c \PD{x}{\psi^\dagger} \sigma_x \psi,

or

\label{eqn:diracAlternate:600}
0
=
\PD{t}{}
\lr{
\psi^\dagger \psi
}
+
\PD{x}{}
\lr{
c \psi^\dagger \sigma_x \psi
}.

The probability current still has the usual form $$\rho = \psi^\dagger \psi = \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2$$, but the probability current with this representation of the Dirac Hamiltonian is

\label{eqn:diracAlternate:620}
\begin{aligned}
j
&= c \psi^\dagger \sigma_x \psi \\
&= c
\begin{bmatrix}
\psi_1^\conj &
\psi_2^\conj
\end{bmatrix}
\begin{bmatrix}
\psi_2 \\
\psi_1
\end{bmatrix} \\
&= c \lr{ \psi_1^\conj \psi_2 + \psi_2^\conj \psi_1 }.
\end{aligned}