quantum field theory - Charge conjugation in Dirac equation

According to Dirac equation we can write,

$$\begin{equation} \left(i\gamma^\mu( \partial_\mu +ie A_\mu)- m \right)\psi(x,t) = 0 \end{equation}$$ We seek an equation where $e\rightarrow -e$ and which relates to the new wave functions to $\psi(x,t)$ . Now taking the complex conjugate of this equation we get

$$\begin{equation} \left[-i(\gamma^\mu)^* \partial_\mu -e(\gamma^\mu)^* A_\mu - m \right] \psi^*(x,t) = 0 \end{equation}$$ If we can identify a matrix U such that $$\begin{equation} \tilde{U} (\gamma^\mu)^* ( \tilde{U} )^{-1} = -\gamma^\mu \end{equation}$$ where $1 =U^{-1} U$.

I want to know that, why and how did we do the last two equation. More precisely, I want to know more details and significance of the last two equations.

Answer

The Dirac equation for a particle with charge $e$ is

$$\left[\gamma^\mu (i\partial_\mu - e A_\mu) - m \right] \psi = 0$$ We want to know if we can construct a spinor $\psi^c$ with the opposite charge from $\psi$. This would obey the equation $$\left[\gamma^\mu (i\partial_\mu + e A_\mu) - m \right] \psi^c = 0$$ If you know about gauge transformations $$\psi \rightarrow \exp\left( i e \phi\right) \psi$$ (together with the compensating transformation for $A_\mu$, which we don't need here), this suggests that complex conjugation is the thing to do: $$\psi^\star \rightarrow \exp\left( i (-e) \phi\right) \psi^\star$$ So it looks like $\psi^\star$ has the opposite charge. Let's take the complex conjugate of the Dirac equation: $$\left[-\gamma^{\mu\star} (i\partial_\mu + e A_\mu) - m \right] \psi^\star = 0$$ Unfortunately this isn't what we want. But remember that spinors and $\gamma$ matrices are only defined up to a change of basis $\psi \rightarrow S \psi$ and $\gamma^\mu \rightarrow S \gamma^\mu S^{-1}$. Possibly we can find a change of basis that brings the Dirac equation into the form we want. Introduce an invertible matrix $C$ by multiplying on the left and inserting $1 = C^{-1}C$ (note that $C$ is the more common notation for your $\tilde{U}$): $$\begin{array}{lcl} 0 &= & C \left[-\gamma^{\mu\star} (i\partial_\mu + e A_\mu) - m \right] C^{-1} C\psi^\star \\ &= & \left[-C\gamma^{\mu\star}C^{-1} (i\partial_\mu + e A_\mu) - m \right] C\psi^\star \end{array}$$

Note that if we can find a $C$ which obeys $-C\gamma^{\mu\star}C^{-1} = \gamma^\mu$ then $C\psi^\star$ makes a perfectly good candidate for $\psi^c$! It turns out that one can indeed construct $C$ satisfying the condition and define charge conjugation as

$$\psi \rightarrow \psi^c = C\psi^\star$$

You can see this more explicitly in terms of two component spinors in the Weyl basis:

$$\psi = \left( \begin{matrix} \chi_\alpha \\ \eta^{\dagger}_{\dot{\alpha}} \end{matrix} \right)$$ (the notation follows the tome on the subject). The charge conjugate spinor in this representation is $$\psi^c = \left( \begin{matrix} \eta_\alpha \\ \chi^{\dagger}_{\dot{\alpha}} \end{matrix} \right)$$ So charge conjugation is $$\eta \leftrightarrow \chi$$ This representation explicitly brings out the two oppositely charged components of the Dirac spinor, $\eta$ and $\chi$, and shows that charge conjugation acts by swapping them.

To recap: we want to define a charge conjugation operation so that given a $\psi$ with some electric charge $e$, we can get a $\psi^c$ with charge $-e$. Complex conjugating the Dirac equation gets us there, but the resulting spinor $\psi^\star$ is in a different spinor basis so the Dirac equation is not in standard form. We introduce a change of basis $C$ to get the Dirac equation back in standard form. The necessary conditions for this to work are that $C$ is invertible (otherwise it wouldn't be a change of basis and bad things would happen) and $-C\gamma^{\mu\star}C^{-1} = \gamma^\mu$.