Charge Conjugate Waves

Assume that, in addition to rotation, boost and parity symmetry, the Dirac equation also has a symmetry under charge conjugation. We wish to write the Dirac equation in a way that makes the symmetry between electron and positron clear. Start from the Dirac equation and include the coupling to the EM field with the substitution that \bgroup\color{black}$\vec{p}\rightarrow\left(\vec{p}+{e\over c}\vec{A}\right)$\egroup.

\begin{eqnarray*}
{\partial\over\partial x_\mu}\gamma_\mu\psi+{mc\over\hbar}\psi...
...\over\hbar c}A_\mu\right)\gamma_\mu\psi+{mc\over\hbar}\psi=0 \\
\end{eqnarray*}


The strategy is to try to write the charge conjugate of this equation then show that it is equivalent to the Dirac equation with the right choice of charge conjugation operator for \bgroup\color{black}$\psi$\egroup. First of all, the sign of \bgroup\color{black}$eA_\mu$\egroup is expected to change in the charge conjugate equation. (Assume the equation, including the constant \bgroup\color{black}$e$\egroup is the same but the sign of the EM field \bgroup\color{black}$A_\mu$\egroup changes.) Second assume, for now, that the Dirac spinor is transformed to its charge conjugate by the operation

\begin{displaymath}\bgroup\color{black}\psi^C=S_C\psi^* \egroup\end{displaymath}

where we are motivated by complex scalar field experience. \bgroup\color{black}$S_C$\egroup is a 4 by 4 matrix. The charge conjugate equation then is

\begin{eqnarray*}
\left({\partial\over\partial x_\mu}-{ie\over\hbar c}A_\mu\right)\gamma_\mu S_C\psi^*+{mc\over\hbar}S_C\psi^*=0. \\
\end{eqnarray*}


Take the complex conjugate carefully remembering that \bgroup\color{black}$x_4$\egroup and \bgroup\color{black}$A_4$\egroup will change signs.

\begin{eqnarray*}
\left({\partial\over\partial x_i}+{ie\over\hbar c}A_i\right)\g...
...r c}A_4\right)\gamma_4^*S_C^*\psi
+{mc\over\hbar}S_C^*\psi=0 \\
\end{eqnarray*}


Multiply from the left by \bgroup\color{black}$S_C^{*-1}$\egroup.

\begin{eqnarray*}
\left({\partial\over\partial x_i}+{ie\over\hbar c}A_i\right)S_...
...A_4\right)S_C^{*-1}\gamma_4^*S_C^*\psi
+{mc\over\hbar}\psi=0 \\
\end{eqnarray*}


Compare this to the original Dirac equation,

\begin{eqnarray*}
\left({\partial\over\partial x_\mu}+{ie\over\hbar c}A_\mu\righ...
...{ie\over\hbar c}A_4\right)\gamma_4\psi
+{mc\over\hbar}\psi=0 \\
\end{eqnarray*}


The two equations will be the same if the matrix $S_C$ satisfies the conditions.

\begin{eqnarray*}
S_C^{*-1}\gamma_i^*S_C^*=\gamma_i \\
S_C^{*-1}\gamma_4^*S_C^*=-\gamma_4. \\
\end{eqnarray*}


Recalling the \bgroup\color{black}$\gamma$\egroup matrices in our representation,

\bgroup\color{black}$\gamma_1=\pmatrix{0 & 0 & 0 & -i \cr 0 & 0 & -i & 0 \cr 0 & i & 0 & 0 \cr i & 0 & 0 & 0 \cr}$\egroup \bgroup\color{black}$\gamma_2=\pmatrix{0 & 0 & 0 & -1 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr -1 & 0 & 0 & 0 \cr}$\egroup \bgroup\color{black}$\gamma_3=\pmatrix{0 & 0 & -i & 0 \cr 0 & 0 & 0 & i \cr i & 0 & 0 & 0 \cr 0 & -i & 0 & 0 \cr}$\egroup \bgroup\color{black}$\gamma_4=\pmatrix{1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0 \cr 0 & 0 & 0 & -1 \cr}$\egroup
note that \bgroup\color{black}$\gamma_1$\egroup and \bgroup\color{black}$\gamma_3$\egroup are completely imaginary and will change sign upon complex conjugation, while \bgroup\color{black}$\gamma_2$\egroup and \bgroup\color{black}$\gamma_4$\egroup are completely real and will not. The solution in our representation (only) is

\begin{displaymath}\bgroup\color{black} S_C^*=S_C^{*-1}=S_C=S_C^{-1}=\gamma_2. \egroup\end{displaymath}

It anti-commutes with \bgroup\color{black}$\gamma_1$\egroup and \bgroup\color{black}$\gamma_3$\egroup producing a minus sign to cancel the one from complex conjugation. It commutes with \bgroup\color{black}$\gamma_2$\egroup giving the right + sign. It anti-commutes with \bgroup\color{black}$\gamma_4$\egroup giving the right - sign.

The charge conjugate of the Dirac spinor is given by.

\bgroup\color{black}$\displaystyle \psi'= \gamma_2\psi^*$\egroup
Of course a second charge conjugation operation takes the state back to the original \bgroup\color{black}$\psi$\egroup.

Applying this to the plane wave solutions gives.

\begin{eqnarray*}
\psi^{(1)}_{\vec{p}}=\sqrt{mc^2\over\vert E\vert V} u^{(1)}_{...
...{-\vec{p}} e^{i(-\vec{p}\cdot\vec{x}-\vert E\vert t)/\hbar} \\
\end{eqnarray*}


The charge conjugate of an electron state is the ``negative energy'' electron state, the absence of which would produce a positron of the same energy, momentum, spin, and velocity as the electron.That is, the conjugate is the hole needed to make a positron with the same properties as the electron except that it has opposite charge.

Let us take one more look at a plane wave solution to the Dirac equation, for example \bgroup\color{black}$\psi^{(1)}_{\vec{p}}$\egroup and its charge conjugate, from the point of view that a positron is an electron moving backward in time. Discard the idea of the ``negative energy'' sea. Assume that we have found a new solution to the field equations that moves backward in time rather than forward.

\begin{eqnarray*}
\psi^{(1)}_{\vec{p}}=\sqrt{mc^2\over\vert E\vert V} u^{(1)}_{...
... V} v^{(1)}_{\vec{p}} e^{i(-\vec{p}\cdot\vec{x}+Et)/\hbar} \\
\end{eqnarray*}


The charge conjugate of the electron solution is an electron with the same charge \bgroup\color{black}$-e$\egroup, opposite momentum \bgroup\color{black}$-\vec{p}$\egroup, and spin opposite to the original state. It satisfies the equation with the signs of the EM fields reversed and, because the sign of the \bgroup\color{black}$Et$\egroup term in the exponential is reversed, it behaves as a positive energy solution moving backward in time, with the right momentum and spin.

Our opinion of the ``negative energy'' solutions has been biased by living in a world of matter. We know about matter waves oscillating as \bgroup\color{black}$e^{i(\vec{p}\cdot\vec{x}-Et)/\hbar}$\egroup. There is a symmetric set of solutions for the same particles moving ``backward in time'' oscillating as \bgroup\color{black}$e^{i(-\vec{p}\cdot\vec{x}+Et)/\hbar}$\egroup. These solutions behave like antiparticles moving forward in time. Consider the following diagram (which contributes to Thomson scattering) from two points of view. From one point of view, an electron starts out at \bgroup\color{black}$t_1$\egroup, lets say in the state \bgroup\color{black}$\psi^{(1)}_{\vec{p}}$\egroup. At time \bgroup\color{black}$t_3$\egroup, the electron interacts with the field and makes a transition to the state \bgroup\color{black}$\psi^{(4)}_{\vec{p''}}$\egroup which travels backward in time to \bgroup\color{black}$t_2$\egroup where it again interacts and makes a transition to \bgroup\color{black}$\psi^{(1)}_{\vec{p'}}$\egroup. From the other point of view, the electron starts out at \bgroup\color{black}$t_1$\egroup, then, at time \bgroup\color{black}$t^2$\egroup, the field causes the creation of an electron positron pair both of which propagate forward in time. At time \bgroup\color{black}$t_3$\egroup, the positron and initial electron annihilate interacting with the field. The electron produced at \bgroup\color{black}$t_2$\egroup propagates on into the future.

\epsfig{file=figs/positron.eps,height=3in}
No reference to the ``negative energy'' sea is needed. No change in the ``negative energy'' solutions is needed although it will be helpful to relabel them with the properties of the positron rather than the properties of the electron moving backward in time.

The charge conjugation operation is similar to parity. A parity operation changes the system to a symmetric one that also satisfies the equations of motion but is different from the original system. Both parity and charge conjugation are good symmetries of the Dirac equation and of the electromagnetic interaction.The charge conjugate solution is that of an electron going backward in time that can also be treated as a positron going forward in time.

Jim Branson 2013-04-22