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Let's have two forms of Majorana equation.

First form (standart or spinor representations of gamma-matrices).

$$ i\gamma^{\mu} \partial_{\mu}\Psi - m\Psi = 0, \quad \Psi = \Psi_{c} = \hat {C} \bar {\Psi}^{T} = \begin{pmatrix} \Psi_{a} \\ \bar {\Psi}^{\dot {a}}\end{pmatrix}, \quad \hat {C} = diag (\varepsilon_{\alpha \beta}, \varepsilon^{\dot {\alpha} \dot {\beta}}) = diag (-i\sigma_{y}, i\sigma_{y}). $$ I only say that the spinor is equal to its charge-conjugated, so the corresponding particle doesn't have an electric charge.

The second form (Majorana representation of gamma-matrices).

$$ i\tilde {\gamma}^{\mu} \partial_{\mu}\Psi - m\Psi = 0, $$ where all coefficients are real, so we can take $\Psi$ as the real function (for the Majorana fermions we must take $\Psi$ as the real function).

So, the question: how these forms are connected? First, of course, I must get the unitary transformation $\Psi ' = \hat {U} \Psi$, which leads to $\tilde {\gamma}_{\mu} = \hat {U}^{+}\gamma_{\mu} \hat {U}$. But what to do with the definition of charge conjugation? Do I need to transform $C$-matrix? If I think correctly, will transformation $\hat {U}^{+} \hat {C} \hat {U}$ lead to the new charge conjugation operation, which consists only complex conjugation of the spinor?

An edit.

I decided to check my assumptions about charge conjugating. It is determined as $$ \Psi^{c} = \hat {C} \gamma_{0}^{T} \Psi^{*}, $$ where $\hat {C}$ refers to the charge conjugation operator.

I started from spinor basis: $$ \gamma_{0} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \hat {C} = \begin{pmatrix} -i\sigma_{y} & 0 \\ 0 & i\sigma_{y}\end{pmatrix}. $$ Standart (or Dirac) basis: $$ U_{spinor\to standart} = U_{1} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix} \Rightarrow $$ $$ \gamma_{0}^{Dirac} = U_{1}^{+}\gamma_{0}U_{1} =\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}, \quad \hat {C}^{Dirac} = -\begin{pmatrix} 0 & i\sigma_{y} \\ i\sigma_{y} & 0\end{pmatrix}. $$ Finally, Majorana basis: $$ U_{standart \to Majorana} = U_{2} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & \sigma_{y} \\ \sigma_{y} & -1\end{pmatrix} \Rightarrow $$ $$ \gamma_{0}^{Majorana} = U_{2}^{+}\gamma_{0}^{Dirac}U_{2} =\begin{pmatrix} 0 & \sigma_{y} \\ \sigma_{y} & 0\end{pmatrix}, \quad \hat {C}^{Majorana} = \begin{pmatrix} -i & 0 \\ 0 & i\end{pmatrix}. $$ So $$ \Psi^{c} = \begin{pmatrix} 0 & i\sigma_{y} \\ -i\sigma_{y} & 0\end{pmatrix}\Psi^{*} \neq \Psi^{*}. $$

It is very strange, because Majorana fermion is real in Majorana representation, so charge conjugation must be equal to complex conjugation.

Where is the mistake?

One more edit.

The answer is found.

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1 Answer 1

up vote 1 down vote accepted

The transformation rule for $\hat{C}$ or that of $\hat{C}\gamma_{0}^{T}$ is different from the usual one.

The whole charge conjugation operation is given by $VK$, where $V\equiv\hat{C}\gamma_{0}^{T}$ is unitary and $K$ is complex conjugation. (i.e., $VK$ is antiunitary.) Then, under a basis transformation, $$ VK \ \rightarrow \ U^{\dagger}VK U = U^{\dagger}VU^{\ast} K. $$ Hence $V$ transforms as $V \ \rightarrow \ U^{\dagger}VU^{\ast}$. My guess is that this is the piece that was missing in your derivation.

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Yes, you're right, thank you! –  Andrew McAddams Feb 6 at 21:37

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