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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.

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Your first equation is missing $eA_{\mu}$. –  Vladimir Kalitvianski Jan 4 '13 at 13:04
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And the second equation is missing a $\star$ on the $\psi$. –  Michael Brown Jan 4 '13 at 13:06

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up vote 13 down vote accepted

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 $-C\gamma^{\mu\star}C^{-1} = \gamma^\mu$.

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How can Gauge transformations be written like this? :$$ \psi \rightarrow \exp\left( i e \phi\right) \psi $$ –  Unlimited Dreamer Jan 4 '13 at 16:13
    
@Forhad_jnu What that means is that you replace $\psi$ with a phase rotated version of itself. Here $\phi=\phi(x)$ is the spacetime dependent phase. This is the standard U(1) gauge transformation of electrodynamics. I haven't written the compensating change you need to make to $A_\mu$ to make the action invariant. Some people don't put the explicit factor of $e$ in the exponent, opting instead to absorb it into the normalisation of the photon field. –  Michael Brown Jan 7 '13 at 6:56
    
Why do not use $C e C^{-1} = -e $?, As I understand, charge conjugation operator has to be change sign of e –  user55944 Aug 7 at 16:32
    
Wow +1 for the link/reference! –  Love Learning Oct 17 at 22:12

The key here is that the gamma matrices are given by their commutation relationships and those do not determine a unique representation for the matrices.

If you start from the Dirac equation

$$\gamma^\mu (i\partial_\mu - e A_\mu) \Psi = m \Psi$$

and make the following generic transformation $\Psi = U \Psi'$ with $U$ a constant matrix with inverse $UU^{-1}=1$ the equation becomes

$$\gamma^\mu U (i\partial_\mu - e A_\mu) \Psi' = m U \Psi'$$

multiplying by the inverse matrix

$$U^{-1} \gamma^\mu U (i\partial_\mu - e A_\mu) \Psi' = m \Psi'$$

this is equivalent to the original equation if $\gamma^\mu{'} = U^{-1} \gamma^\mu U$. This relation guarantee that the new matrices satisfy the same commutation relationships than the original.

Regarding the specific case of charge conjugation I think the most fundamental approach uses the CPT theorem. In this case parity is trivial therefore it remains charge conjugation (C) and time reversal (T). The Dirac equation is invariant if both the sign of the charge and time are reversed. This is the basis for Stuckelberg Feynman interpretation of antiparticles as particles travelling backward in time.

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