## Spinor solutions with alternate $$\gamma^0$$ representation.

January 2, 2019 phy2403 , ,

This follows an interesting derivation of the $$u, v$$ spinors [2], adding some details.

In class (QFT I) and [3] we used a non-diagonal $$\gamma^0$$ representation
\label{eqn:spinorSolutions:20}
\gamma^0 =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},

whereas in [2] a diagonal representation is used
\label{eqn:spinorSolutions:40}
\gamma^0 =
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}.

This representation makes it particularly simple to determine the form of the $$u, v$$ spinors. We seek solutions of the Dirac equation
\label{eqn:spinorSolutions:60}
\begin{aligned}
0 &= \lr{ i \gamma^\mu \partial_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= \lr{ i \gamma^\mu \partial_\mu – m } v(p) e^{i p \cdot x},
\end{aligned}

or
\label{eqn:spinorSolutions:80}
\begin{aligned}
0 &= \lr{ \gamma^\mu p_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= -\lr{ \gamma^\mu p_\mu + m } v(p) e^{i p \cdot x}.
\end{aligned}

In the rest frame where $$\gamma^\mu p_\mu = E \gamma^0$$, where $$E = m = \omega_\Bp$$, these take the particularly simple form
\label{eqn:spinorSolutions:100}
\begin{aligned}
0 &= \lr{ \gamma^0 – 1 } u(E, \Bzero) \\
0 &= \lr{ \gamma^0 + 1 } v(E, \Bzero).
\end{aligned}

This is a nice relation, as we can determine a portion of the structure of the rest frame $$u, v$$ that is independent of the Dirac matrix representation
\label{eqn:spinorSolutions:120}
\begin{aligned}
u(E, \Bzero) &= (\gamma^0 + 1) \psi \\
v(E, \Bzero) &= (\gamma^0 – 1) \psi
\end{aligned}

Similarly, and more generally, we have
\label{eqn:spinorSolutions:140}
\begin{aligned}
u(p) &= (\gamma^\mu p_\mu + m) \psi \\
v(p) &= (\gamma^\mu p_\mu – m) \psi
\end{aligned}

also independent of the representation of $$\gamma^\mu$$. Looking forward to non-matrix representations of the Dirac equation ([1]) note that we have not yet imposed a spinorial structure on the solution
\label{eqn:spinorSolutions:260}
\psi
=
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix},

where $$\phi, \chi$$ are two component matrices.

The particular choice of the diagonal representation \ref{eqn:spinorSolutions:40} for $$\gamma^0$$ makes it simple to determine additional structure for $$u, v$$. Consider the rest frame first, where
\label{eqn:spinorSolutions:160}
\begin{aligned}
\gamma^0 – 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}

\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix} \\
\gamma^0 + 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix},
\end{aligned}

so we have
\label{eqn:spinorSolutions:280}
\begin{aligned}
u(E, \Bzero) &=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix} \\
v(E, \Bzero) &=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix}
\end{aligned}

Therefore a basis for the spinors $$u$$ (in the rest frame), is
\label{eqn:spinorSolutions:180}
u(E, \Bzero) \in \setlr{
\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}
},

and a basis for the rest frame spinors $$v$$ is
\label{eqn:spinorSolutions:200}
v(E, \Bzero) \in \setlr{
\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}
}.

Using the two spinor bases $$\zeta^a, \eta^a$$ notation from class, we can write these
\label{eqn:spinorSolutions:220}
\begin{aligned}
u^a(E, \Bzero) &=
\begin{bmatrix}
\zeta^a \\
0
\end{bmatrix},
v^a(E, \Bzero) &=
\begin{bmatrix}
0 \\
\eta^a \\
\end{bmatrix}.
\end{aligned}

For the non-rest frame solutions, [2] opts not to boost, as in [3], but to use the geometry of $$\gamma^\mu p_\mu \pm m$$. With their diagonal representation of $$\gamma^0$$ those are
\label{eqn:spinorSolutions:240}
\begin{aligned}
\gamma^\mu p_\mu – m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}

m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix} \\
\gamma^\mu p_\mu + m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}
+
m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix} \\
\end{aligned}

Let’s assume that the arbitrary momentum solutions \ref{eqn:spinorSolutions:140} are each proportional to the rest frame solutions
\label{eqn:spinorSolutions:300}
\begin{aligned}
u^a(p) &= (\gamma^\mu p_\mu + m) u^a(E, \Bzero) \\
v^a(p) &= (\gamma^\mu p_\mu – m) u^a(E, \Bzero).
\end{aligned}

Plugging in \ref{eqn:spinorSolutions:240} gives
\label{eqn:spinorSolutions:320}
\begin{aligned}
u^a(p) &=
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
v^a(p) &=
\begin{bmatrix}
(\Bsigma \cdot \Bp) \eta^a \\
(E + m) \eta^a
\end{bmatrix},
\end{aligned}

where an overall sign on $$v^a(p)$$ has been dropped. Let’s check the assumption that the rest frame and general solutions are so simply related
\label{eqn:spinorSolutions:340}
\begin{aligned}
\lr{ \gamma^\mu p_\mu – m } u^a(p)
&=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix}
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
&=
\begin{bmatrix}
(E^2 – m^2 – \Bp^2) \zeta^a \\
0
\end{bmatrix} \\
&= 0,
\end{aligned}

and
\label{eqn:spinorSolutions:360}
\begin{aligned}
\lr{ \gamma^\mu p_\mu + m } v^a(p)
&=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix}
\begin{bmatrix}
(\Bsigma \cdot \Bp ) \eta^a \\
(E + m) \eta^a \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 \\
\Bp^2 + m^2 – E^2
\end{bmatrix} \\
&= 0.
\end{aligned}

Everything works out nicely. The form of the solution for this representation of $$\gamma^0$$ is much simpler than the Chiral solution that we found in class. We end up with an explicit split of energy and spatial momentum components in the spinor solutions, instead of factors involving $$p \cdot \sigma$$ and $$p \cdot \overline{\sigma}$$, which are arguably nicer from a Lorentz invariance point of view.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Claude Itzykson and Jean-Bernard Zuber. Quantum field theory. McGraw-Hill, 1980.

[3] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## PHY2403H Quantum Field Theory. Lecture 22: Dirac sea, charges, angular momentum, spin, U(1) symmetries, electrons and positrons. Taught by Prof. Erich Poppitz

This post is a synopsis of the material from the second last lecture of QFT I. I missed that class, but worked from notes kindly provided by Emily Tyhurst, and Stefan Divic, filling in enough details that it made sense to me.

[Click here for an unabrided PDF of my full notes on this day’s lecture material.]

Topics covered include

• The Hamiltonian action on single particle states showed that the Hamiltonian was an energy eigenoperator
\label{eqn:qftLecture22:140}
H \ket{\Bp, r}
=
\omega_\Bp \ket{\Bp, r}.
• The conserved Noether current and charge for spatial translations, the momentum operator, was found to be
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which could be written in creation and anhillation operator form as
\label{eqn:momentumDirac:261}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

Single particle states were found to be the eigenvectors of this operator, with momentum eigenvalues
\label{eqn:momentumDirac:262}
\BP a_\Bq^{s\dagger} \ket{0} = \Bq (a_\Bq^{s\dagger} \ket{0}).
• The conserved Noether current and charge for a rotation was found. That charge is
\label{eqn:qftLecture22:920}
\BJ = \int d^3 x \Psi^\dagger(x) \lr{ \underbrace{\Bx \cross (-i \spacegrad)}_{\text{orbital angular momentum}} + \inv{2} \underbrace{\mathbf{1} \otimes \Bsigma}_{\text{spin angular momentum}} } \Psi,

where
\label{eqn:qftLecture22:260}
\mathbf{1} \otimes \Bsigma =
\begin{bmatrix}
\Bsigma & 0 \\
0 & \Bsigma
\end{bmatrix},

which has distinct orbital and spin angular momentum components. Unlike NRQM, we see both types of angular momentum as components of a single operator. It is argued in [3] that for a particle at rest the single particle state is an eigenvector of this operator, with eigenvalues $$\pm 1/2$$ — the Fermion spin eigenvalues!
• We examined two $$U(1)$$ global symmetries. The Noether charge for the “vector” $$U(1)$$ symmetry is
\label{eqn:qftLecture22:380}
Q
=
\int \frac{d^3 q}{(2\pi)^3} \sum_{s = 1}^2
\lr{
a_\Bp^{s \dagger} a_\Bp^s

b_\Bp^{s \dagger}
b_\Bp^s
},

This charge operator characterizes the $$a, b$$ operators. $$a$$ particles have charge $$+1$$, and $$b$$ particles have charge $$-1$$, or vice-versa depending on convention. We call $$a$$ the operator for the electron, and $$b$$ the operator for the positron.
• CPT (Charge-Parity-TimeReversal) symmetries were also mentioned, but not covered in class. We were pointed to [2], [3], [4] to start studying that topic.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Dr. Michael Luke. Quantum Field Theory., 2011. URL https://www.physics.utoronto.ca/~luke/PHY2403F/References_files/lecturenotes.pdf. [Online; accessed 05-Dec-2018].

[3] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

[4] Dr. David Tong. Quantum Field Theory. URL http://www.damtp.cam.ac.uk/user/tong/qft.html.

## Explicit form of the square root of p . sigma.

December 10, 2018 phy2403 , , , ,

With the help of Mathematica, a fairly compact form was found for the root of $$p \cdot \sigma$$
\label{eqn:DiracUVmatricesExplicit:121}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.

A bit of examination shows that we can do much better. The leading scalar term can be simplified by squaring it
\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{ \sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} } }^2
&=
\omega_\Bp- \Norm{\Bp} + \omega_\Bp+ \Norm{\Bp} + 2 \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
&=
2 \omega_\Bp + 2 m,
\end{aligned}

where the on-shell value of the energy $$\omega_\Bp^2 = m^2 + \Bp^2$$ has been inserted. Using that again in the matrix, we have
\label{eqn:squarerootpsigma:160}
\begin{aligned}
\sqrt{ p \cdot \sigma }
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\begin{bmatrix}
\omega_\Bp- p^3 + m & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + m
\end{bmatrix} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
-p^2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
-p^3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \sigma^1
-p^2 \sigma^2
-p^3 \sigma^3
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0 – \Bsigma \cdot \Bp
}.
\end{aligned}

We’ve now found a nice algebraic form for these matrix roots
\label{eqn:squarerootpsigma:180}
\boxed{
\begin{aligned}
\sqrt{p \cdot \sigma} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \sigma } \\
\sqrt{p \cdot \overline{\sigma}} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \overline{\sigma}}.
\end{aligned}}

As a check, let’s square one of these explicitly
\label{eqn:squarerootpsigma:101}
\begin{aligned}
\lr{ \sqrt{p \cdot \sigma} }^2
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (p \cdot \sigma)^2 + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (\omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + \Bp^2) + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + 2 m (\omega_\Bp – \Bsigma \cdot \Bp) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp \lr{ \omega_\Bp + m } – (2 \omega_\Bp + 2 m) \Bsigma \cdot \Bp } \\
&=
\omega_\Bp – \Bsigma \cdot \Bp \\
&=
p \cdot \sigma,
\end{aligned}

which validates the result.

## Momentum operator for the Dirac field?

In the borrowed notes I have for last Monday’s lecture (which I missed) I see the momentum operator defined by
\label{eqn:momentumDirac:20}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

There’s a “use Noether’s theorem” comment associated with this. For the scalar field, using Noether’s theorem, we identified the conserved charge of a spacetime translation as the momentum operator
\label{eqn:momentumDirac:40}
P^i = \int d^3 x T^{0i} = – \int d^3 x \pi(x) \spacegrad \phi(x),

and if we plugged in the creation and anhillation operator representation of $$\pi, \phi$$, out comes
\label{eqn:momentumDirac:60}
\BP =
\inv{2} \int \frac{d^3 q}{(2\pi)^3} \Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger},

(plus $$e^{\pm 2 i \omega_\Bp t}$$ terms that we can argue away.)

It wasn’t clear to me how this worked with the Dirac field, but it turns out that this does follow systematically as expected. For a spacetime translation
\label{eqn:momentumDirac:80}
x^\mu \rightarrow x^\mu + a^\mu,

we find
\label{eqn:momentumDirac:100}
\delta \Psi = -a^\mu \partial_\mu \Psi,

so for the Dirac Lagrangian, we have
\label{eqn:momentumDirac:120}
\begin{aligned}
\delta \LL
&= \delta \lr{ \overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \Psi } \\
&=
(\delta \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \delta \Psi \\
&=
(-a^\sigma \partial_\sigma \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } (-a^\sigma \partial_\sigma \Psi ) \\
&=
-a^\sigma \partial_\sigma \LL \\
&=
\partial_\sigma (-a^\sigma \LL),
\end{aligned}

i.e. $$J^\mu = -a^\mu \LL$$.
To plugging this into the Noether current calculating machine, we have
\label{eqn:momentumDirac:160}
\begin{aligned}
\PD{(\partial_\mu \Psi)}{\LL}
&=
\PD{(\partial_\mu \Psi)}{} \lr{ \overline{\Psi} i \gamma^\sigma \partial_\sigma \Psi – m \overline{\Psi} \Psi } \\
&=
\overline{\Psi} i \gamma^\mu,
\end{aligned}

and
\label{eqn:momentumDirac:180}
\PD{(\partial_\mu \overline{\Psi})}{\LL} = 0,

so
\label{eqn:momentumDirac:140}
\begin{aligned}
j^\mu
&=
(\delta \overline{\Psi}) \PD{(\partial_\mu \overline{\Psi})}{\LL}
+
\PD{(\partial_\mu \Psi)}{\LL} (\delta \Psi)
– a^\mu \LL \\
&=
\overline{\Psi} i \gamma^\mu (-a^\sigma \partial_\sigma \Psi)
– a^\sigma {\delta^{\mu}}_{\sigma} \LL \\
&=
– a^\sigma
\lr{
\overline{\Psi} i \gamma^\mu \partial_\sigma \Psi
+ {\delta^{\mu}}_{\sigma} \LL
} \\
&=
-a_\nu
\lr{
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL
}.
\end{aligned}

We can now define an energy-momentum tensor
\label{eqn:momentumDirac:200}
T^{\mu\nu}
=
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL.

A couple things are of notable in this tensor. One is that it is not symmetric, and there’s doesn’t appear to be any hope
of making it so. For example, the space+time components are way different
\label{eqn:momentumDirac:220}
\begin{aligned}
T^{0k} &= \overline{\Psi} i \gamma^0 \partial^k \Psi \\
T^{k0} &= \overline{\Psi} i \gamma^k \partial^0 \Psi,
\end{aligned}

so if we want a momentum like creature, we have to use $$T^{0k}$$, not $$T^{k0}$$. The charge associated with that current is
\label{eqn:momentumDirac:240}
\begin{aligned}
Q^k
&=
\int d^3 x
\overline{\Psi} i \gamma^0 \partial^k \Psi \\
&=
\int d^3 x
\Psi^\dagger (-i \partial_k) \Psi,
\end{aligned}

or translating from component to vector form
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which is the how the momentum operator is first stated in [2]. Here the vector notation doesn’t have any specific representation, but it is interesting to observe how this is directly related to the massless Dirac Lagrangian

\label{eqn:momentumDirac:280}
\begin{aligned}
\LL(m = 0)
&=
\overline{\Psi} i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i (\partial_0 + \gamma_0 \gamma^k \partial_k) \Psi \\
&=
\Psi^\dagger i (\partial_0 – \gamma_0 \gamma_k \partial_k ) \Psi,
\end{aligned}

but since $$\gamma_0 \gamma_k$$ is a $$4 \times 4$$ representation of the Pauli matrix $$\sigma_k$$ Lagrangian itself breaks down into
\label{eqn:momentumDirac:300}
\LL(m = 0)
=
\Psi^\dagger i \partial_0 \Psi
+
\Bsigma \cdot \lr{ \Psi^\dagger (-i\spacegrad) \Psi },

components, and lo and behold, out pops the momentum operator density! There is ambiguity as to what order of products $$\gamma_0 \gamma_k$$, or $$\gamma_k \gamma_0$$ to pick to represent the Pauli basis ([1] uses $$\gamma_k \gamma_0$$), but we also have sign ambiguity in assembling a Noether charge from the conserved current, so I don’t think that matters. Some part of this should be expected this since the Dirac equation in momentum space is just $$\gamma \cdot p – m = 0$$, so there is an intimate connection with the operator portion and momentum.

The last detail to fill in is going from \ref{eqn:momentumDirac:260} to \ref{eqn:momentumDirac:20} using the $$a, b$$ representation of the field. That’s an algebraically messy looking job that I don’t feel like trying at the moment.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## Alternate Dirac equation representation

November 27, 2015 phy1520 , ,

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}