This is a quick and simple question. I'm studynig about a charge conjugation tranformation over a complex scalar field, $\psi\left(x\right)$,

$$ \psi\left(x\right)\rightarrow C\psi\left(x\right)C^{-1}=\eta_{c}\psi^{\dagger}\left(x\right), $$

where I'm told $C$ is an unitary operator and $\eta_{c}$ is a phase factor. I should prove the Klein-Gordon Lagrangian density is invariant under this transformation. To do that I need to derive how $\psi^{\dagger}\left(x\right)$ transforms but I'm confused about what $\eta_{c}$ really is and how it's acted upon by $C$. Is it a scalar or an operator? Can you please check if this is correct: $$ C\psi^{\dagger}\left(x\right)C^{-1}\overset{?}{=}\eta_{c}^{-1}\psi\left(x\right) $$

$$ C\eta_{c}\psi^{\dagger}\left(x\right)C^{-1}\overset{?}{=}\eta_{c}C\psi^{\dagger}\left(x\right)C^{-1}=\psi\left(x\right) $$ I guessed the first expression so that the second would take me back to the original $\psi\left(x\right)$, but I don't really understand why $\psi^{\dagger}\left(x\right)$ would transform like that. If $\eta_{c}$ is a scalar wouldn't this be

$$ C\psi^{\dagger}\left(x\right)C^{-1}\overset{?}{=}\eta_{c}^{*}\psi\left(x\right)? $$

I know, this is pretty basic but I'm now confused.


1 Answer 1


You can just take the Hermitian conjugate of the transformation law and use the unitarity of $C$:

$$ \begin{align} C^{-1 \dagger} \psi^\dagger(x) C^\dagger &= \eta_c^* \psi(x) \\ \implies C \psi^\dagger(x) C^{-1} &= \eta_c^* \psi(x)\,.\end{align}$$

Incidentally, this is the same as the transformation law you postulated iff $\eta_c$ is a unit-modulus complex number – check this! Having $|\eta_c| = 1$ guarantees that conjugating twice returns the original field, which is a sensible thing to demand from a charge-conjugation operator.

  • $\begingroup$ Thanks. So just to confirm, am I free to commute $\eta_{c}$ and $\eta_{c}^{*}$ with $C$, $C^{-1}$, $\psi$ and $\psi^{\dagger}$? $\endgroup$
    – johani
    Jul 3, 2018 at 12:59
  • $\begingroup$ @johani Yes. $\eta_c$ is (by assumption) nothing more than a complex number. $\endgroup$
    – gj255
    Jul 3, 2018 at 13:17

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