Jul 25, 2014

Charge Symmetry

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

When the E&M field interaction is included, it can be shown that the charge conjugated wave function 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.