#### Classical Example

Under the charge conjugation (C for short), $e \to -e$, $\mathbf E \to - \mathbf E$, but the position vector remains invariant $\mathbf r(t) \to \mathbf r(t)$.

In the theories particle and antiparticle do not mix (e.g. Schroedinger's Equation, Paul's Equation), charge symmetry is almost trivial.

#### Dirac Theory

Dirac equation, \[(i\partial_\mu \gamma^\mu - eA_\mu\gamma^\mu + m)\psi(x) = 0

\] describes the relativistic motion of electrons as well as positrons. In general, the solution $\psi(x)$ contains both the electron state and the positron state. Here we take $g_{\mu\nu} = \mathrm{diag}\{-,+,+,+\}$.

In free particle theory where $A = 0$, there are four independent plane wave solutions: two positive energy solutions describe electrons $u_s(p) e^{+ip\cdot x}$ and two negative energy solutions describe the positrons $v_s(p) e^{-ip\cdot x}$, $E_{\mathbf p} = \sqrt{\mathbf p^2+m^2}$. The negative energy is a known problem. The standard interpretation is the Hole Theory.

A general solution is the superposition of the two pieces: $\psi(x) \equiv \psi^+(x) + \psi^-(x)$ where \[

\psi^+(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} b_s(\mathbf p) u_s(p) e^{+ip\cdot x}, \\

\psi^-(x) = \sum_{s=\pm}\int \frac{\mathrm{d}^3p}{(2\pi)^32E_{\mathbf p}} d^*_s(\mathbf p) v_s(p) e^{-ip\cdot x}. \] $b, d^*$ are scalar wave packet functions.

As a relativistic wave function approach, the charge conjugation would be implemented as a "unitary" spinor matrix: $C \bar C \triangleq C (\beta C^\dagger \beta) \equiv 1$. This matrix should transform a plane wave electron to a plane wave positron or vice versa, i.e. $C u_s(p)e^{+ip\cdot x} \sim v_s(p) e^{-ip\cdot x}$. We can immediately see that this is not possible if the charge conjugation spares the exponential function. We conclude that in Dirac theory, the charge conjugation must be implemented as an

**anti-unitary spinor operator**. Thus, we require \[

C \left( u_s(p) e^{+ip\cdot x}\right) =\eta_c v^*_s(p) e^{-ip\cdot x} \\

C \left( v_s(p) e^{-ip\cdot x} \right) = \xi_c u^*_s(p) e^{+ip\cdot x} \] where $|\eta_c| = |\xi_c| = 1$. For simplicity, we'll take these constant phases to unity. It is convenient to define the unitary part of $C$ by a new matrix $\mathcal C$ such that $C \psi(x) \triangleq \mathcal C \beta \psi^*(x) \equiv \mathcal C \bar\psi^T(x)$ (Here adding a $\beta$ is just a convention.). It is easy to see, $\mathcal C$ is unitary $\mathcal{C} \bar{\mathcal{C}}= 1$. The theory must be invariant under the charge conjugation, or $\mathcal L \overset{C}{\to} \mathcal L$, which implies $\mathcal C \gamma_\mu = - \gamma_\mu^T \mathcal C$. Applying to the $u,v$ spinors, $\mathcal C \bar u^T_s(p) = v_s(p), \mathcal C \bar v^T_s(p) = u_s(p)$.

Heuristically, the transformation can be viewed as a two-step procedure:

- swap the particle species, by exchanging $u_s(p)$ and $v_s(p)$;
- reverse the time, by conjugating the exponential factor.

*satisfies*the following Dirac equation: \[

(i\partial_\mu \gamma^\mu {\color \red +} eA_\mu\gamma^\mu + m)\psi^c(x) = 0

\] Note that the sign is needed to preserve the charge symmetry. Because in a full charge conjugation, $\psi(x) \to \psi^c(x), A^\mu \to -A^\mu$.

#### Quantum field theory

In quantum field theory, charge conjugation is implemented as a**unitary operator**, which we shall call $C$ (not be confused with the anti-unitary spinor operator). The definition is simple: $C^{-1} b_s(p) C = d_s(p)$, $C^{-1} d_s(p) C = b_s(p)$, where $b_s(p), d_s(p)$ are the electron and positron annihilation operators, respectively. It can be shown (after some algebra) that for the quantum fields, $C^{-1} \psi(x) C = \psi^c(x) \equiv \mathcal C \bar\psi^ T(x)$. Superficially, this resembles the Dirac equation result introduced in last section, although we should point out that here $\psi(x)$ is a

*field operator*instead of a

*relativistic wave function*.