Consider a charge conjugation operator which acts on the Dirac field($\psi$) as
$$\psi_{C} \equiv \mathcal{C}\psi\mathcal{C}^{-1} = C\gamma_{0}^{T}\psi^{*}$$ Just as we can operate the parity operator on the Lagrangian, and we say that a theory has a symmetry if $$\mathcal{P}\mathcal{L}(t,x^{i})\mathcal{P}^{-1} = \mathcal{L}(t,-x^{i})$$

Suppose we operate $\mathcal{C}$ on the Dirac Lagrangian what should we get? $$\mathcal{C}\mathcal{L}_{Dirac}(x^{\mu})\mathcal{C}^{-1} = \mathcal{L}^{*}_{Dirac}(x^{\mu}) ?$$ in analogy to the transformation of the scalar field $\phi$ under charge conjugation.

On a same note one can ask what equation should $\psi_{C}$ satisfy? Should it satisfy the conjugated Dirac equation as $$(i\gamma^{\mu}\partial_{\mu} + m)\psi_{C} = 0 ?$$ If so can someone give me the physical interpretation for it.

I am asking this question as I want to explicitly use $\psi_{C}$ and check whether it keeps the Dirac Lagrangian invariant. I have done a calculation by substituting $\psi_{C}$ in the Dirac equation and have found it is not satisfying as shown below.

$$(i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} = i(\gamma_{\mu}C\gamma_{0}^{T})\partial^{\mu}\psi^{*} - (C\gamma_{0}^{T})m\psi^{*}$$ We will use $C^{-1}\gamma_{\mu}C = - \gamma_{\mu}^{T}$ and $\{\gamma_{\mu}^{T},\gamma_{\nu}^{T}\} = 2g_{\mu\nu}$.
Consider \begin{align} \gamma_{\mu}C\gamma_{0}^{T} &= CC^{-1}\gamma_{\mu}C\gamma_{0}^{T} \\ &= -C\gamma_{\mu}^{T}\gamma_{0}^{T} \\ &= C\gamma_{0}^{T}\gamma_{\mu}^{T} \end{align} Hence substituting back we will get
\begin{align} (i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} &= C\gamma_{0}^{T}(i\gamma_{\mu}^{T}\partial^{\mu}\psi^* - m\psi^*) \\ &= C\gamma_{0}^{T}[(i\gamma_{\mu}^{T}\partial^{\mu}\psi^* - m\psi^*)^{T}]^{T} \\ &= C\gamma_{0}^{T}(i\psi^{\dagger}\gamma_{\mu}\partial^{\mu} - m\psi^{\dagger})^{T} \\ &= C\gamma_{0}^{T}[(i\bar{\psi}\gamma_{0}\gamma_{\mu}\partial^{\mu} - m\bar{\psi}\gamma_{0})]^{T} \\ \end{align} Now \begin{align} \gamma_{0}\gamma_{\mu}\partial^{\mu} &= (\gamma_{0}\partial^{t} + \gamma_{i}\partial^{i})\gamma_{0} \end{align} If we substitute back we will get \begin{align} (i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} &= C\gamma_{0}^{T}[\{i\bar{\psi}(\gamma_{0}\partial^{t} + \gamma_{i}\partial^{i}) - m\bar{\psi}\}\gamma_{0}]^T \\ &\neq 0 \end{align}


2 Answers 2


The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation.

Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry transformations that arises about your statement regarding the parity ($\mathbf{x}\to-\mathbf{x}$). Your equation \begin{align} \mathcal{P}\mathcal{L}(t,x^{i})\mathcal{P}^{-1} = \mathcal{L}(t,-x^{i}) \end{align} is correct. But this, in and of itself, doesn't "say that a theory has a symmetry." In fact, it places a restriction on the theory -- that the theory must be an even function of the position vector. For example, for a real scalar field, $\phi(t,\mathbf{x})$, the kinetic energy operator in the Lagrangian density, $\partial_\mu \partial^\mu \phi(x)$ is invariant under parity. In fact, it is only the action ($S=\int d^4x \mathcal{L}$) that need be invariant under the symmetry transformation. (Often this reduces to the invariance of the Lagrangian density.) So each term of the Lagrangian (density) need not be invariant (though this is often the case).

Returning to the Dirac equation -- the full statement, in words, of invariance of the theory of the interaction of electrons with light (QED) under the discrete transformations ($\mathcal{P},\mathcal{C}$, & $\mathcal{T}$) is that the theory is invariant under each of them separately or in any combination. (QED is less "interesting" than the electroweak theory in this regard since the electroweak theory appears to violate all three of these separately -- but perhaps not all simultaneously.)

We have to remember that the invariance under $\mathcal{C}$ requires the transformation not only of the wave function, $\psi_{C} \equiv \mathcal{C}\psi\mathcal{C}^{-1} = C\gamma_{0}^{T}\psi^{*}$ but also the charge, $q\to -q$. In the case of the free Dirac equation/Lagrangian considered above, the charge does not feature, so it's not directly relevant to the present discussion but it's important to keep in mind.

Now the direct answers to your questions. (I won't do the algebra since, if you can carry out the calculations for the Dirac equation itself, then the transformation of the Lagrangian should be straightforward.)

"Suppose we operate C on the Dirac Lagrangian what should we get?" The corrected relation is:

\begin{align}\mathcal{C}\mathcal{L}_{Dirac}(x^{\mu})\mathcal{C}^{-1} = \mathcal{L}_{Dirac}(x^{\mu})\end{align}

(Incidentally, your relation turns out to be all right since the Lagrangian (density) must be a Hermitian, scalar operator, so $\mathcal{L}^* = \mathcal{L}^\dagger = \mathcal{L}$. EDIT: Thanks to Omkar for pointing out that this is wrong. $\mathcal{L}^*\ne\mathcal{L}$.)

On a same note one can ask what equation should ψC satisfy? Should it satisfy the conjugated Dirac equation as \begin{align} > (i\gamma^{\mu}\partial_{\mu} + m)\psi_{C} = 0? \end{align}

If you're using the "West Coast" metric $(+1,-1,-1,-1)$ then the equation that $\psi_C$ should satisfy is the one above with $m\to -m$. That is, the free Dirac equation is the same for $\psi$ and $\psi_C$. This is because the masses of the particle and anti-particle are identical, as first divined by Dirac. (If you're using the metric of the opposite sign, then you're equation is correct.)

  • $\begingroup$ Thank You for your reply. I agree with your argument regarding the Lagrangian. And even your argument regarding the form of the equation sounds correct and something we should expect given that the Lagrangian is same. However I did a calculation and I found otherwise. I will put up the calculation in the question, maybe I am making some error. $\endgroup$
    – Omkar
    Commented Oct 22, 2012 at 18:50
  • $\begingroup$ another point is that even though $\mathcal{L}$ is hermitian, I think it is not real, as the term $\bar{\psi}\psi$ is not real. Hence you can't say $\mathcal{L}^* = \mathcal{L}$ $\endgroup$
    – Omkar
    Commented Oct 22, 2012 at 20:26
  • $\begingroup$ I see an error in the above. You should have: $\gamma_{0} \gamma_{\mu}\partial^{\mu} = (\gamma_{0}\partial^{t} - \gamma_{i}\partial^{i})\gamma_{0}$. I wouldn't do it this way though. I get the correct result by starting with the Dirac equation for $\psi$, taking the complex conjugate and showing that $C \propto \gamma^2$ (in what's sometimes called the Dirac representation of the $\gamma$ matrices). And you're right that $\mathcal{L}^*\ne\mathcal{L}$. Sorry for that. $\endgroup$
    – MarkWayne
    Commented Oct 22, 2012 at 21:38
  • $\begingroup$ Sorry again -- $C$ should be $\propto \gamma^2\gamma^0$. (I'm using Bjorken & Drell conventions.) $\endgroup$
    – MarkWayne
    Commented Oct 22, 2012 at 21:49
  • 1
    $\begingroup$ Omkar, this is a common mistake you've made in the above line for $\gamma_\mu\partial^\mu$. Consider $a_\mu b^\mu = a_0 b^0 + a_i b^i$ -- there's no minus sign since you haven't lowered the index yet. The correct relation is: $\gamma_\mu\partial^\mu = \gamma_0 \partial_0 - \sum_{i=1}^3 \gamma_i \partial_i$. $\endgroup$
    – MarkWayne
    Commented Oct 23, 2012 at 16:18

It depends on the definition of charge conjugation operator. Your definition $$ \psi_c = C\gamma_0^T\psi^\star. $$ changes the sign of the mass term in Dirac equation (or equivalently the sign of kinetic term if we multiply the equation by a minus sign). If we add an extra $\gamma^5$ (or $i\gamma^5$, which amounts to a global phase transformation.) to your definition of charge conjugation, $$ \psi_c = \gamma^5C\gamma_0^T\psi^\star, $$ voila! the charge conjugation would get you back at the original Dirac equation by virtue of $$ \gamma^5\gamma^\mu = -\gamma^\mu\gamma^5. $$


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