In the Quantum electrodynamics book (look at the problem) its authors Lifshitz and Berestetskei claim that operator of charge conjugation $\hat {C} = -\alpha_{2}$ in Majorana basis transforms as $\hat {C}^{M} = \alpha_{2}$. Here (they started from the standart (Dirac) representation of the gamma-matrices) $$ \alpha_{2} = \begin{pmatrix} 0 & \sigma_{y}\\ \sigma_{y} & 0\end{pmatrix}, \quad \beta = \begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}, $$ $$ \hat {C}^{M} = \hat {U}^{+} \hat {C} \hat {U}, \quad \hat {U} = \frac{1}{\sqrt{2}}(\alpha_{2} + \beta ) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & \sigma_{y}\\ \sigma_{y} & -1\end{pmatrix} = \hat {U}^{+}. $$ I tried to get their result, but I only got $$ \hat {C}^{M} = \frac{1}{2}\begin{pmatrix} 1 & \sigma_{y}\\ \sigma_{y} & -1\end{pmatrix}\begin{pmatrix} 0 & \sigma_{y}\\ \sigma_{y} & 0\end{pmatrix}\begin{pmatrix} 1 & \sigma_{y}\\ \sigma_{y} & -1\end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}, $$ because $\sigma_{y}^{2} = 1$.

Where is the mistake?

An edit.

I found the mistake. I used the wrong definition of transformation of charge conjugation operator under unitary spinor transformation. The correct one is $\hat {C}^{'} = \hat {U}\hat {C}\hat {U}^{T}$.

  • $\begingroup$ Where did you get the transformation matrix? Is this a well known result to switch between representations of the Gamma matrices? $\endgroup$ – JeffDror Feb 6 '14 at 10:51
  • $\begingroup$ @JeffDror : this matrix is given in the text of the problem linked above and connects Dirac and Majorana basis (en.wikipedia.org/wiki/Gamma_matrices#Other_representations ). The matrix can be derived from considerations of imaginarity of all of gamma-matrices in Majorana representation. I think that this is the well-known result for switching representations, since I saw it in some modern articles, like here, mdpi.com/2073-8994/2/4/1776 . $\endgroup$ – Andrew McAddams Feb 6 '14 at 11:12
  • $\begingroup$ It may be better to answer your own question if you feel it's solved, and accept the answer. This will make it more useful for those having similar problem and looking for a solution. $\endgroup$ – Ruslan Feb 28 '14 at 8:06

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