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Consider the Majorana fermions expressed mathematically in terms of the creation and annihilation operators of second quantization, the ordinary fermionic annihilation and creation operators $\alpha$ and $\alpha^{\dagger }$ can be written in terms of two Majorana operators $\gamma_{1}$ and $\gamma_{2}$ by: \begin{align} \alpha &= \frac{\gamma_1+i\gamma_2}{2}\\ \alpha^{\dagger} &= \frac{\gamma_1-i\gamma_2}{2} \end{align}

Squaring, one obtains \begin{align} \alpha^2+(\alpha^\dagger)^2 &= \frac{1}{4}\left((\gamma_1+i \gamma_2)^2+(\gamma_1-i \gamma_2)^2\right)\\ &= \frac{1}{4}\left( 2 \gamma_1^2-2\gamma_2^2+i\{\gamma_1,\gamma_2\}-i\{\gamma_1,\gamma_2\}\right)\\ &= \frac{\gamma_1^2-\gamma_2^2}{2} \tag{1} \end{align} Now, noting \begin{align} \gamma_1 &= \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{pmatrix}\\ \gamma_2 &= \begin{pmatrix} 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0\\ -i & 0 & 0 & 0 \end{pmatrix} \end{align} Inserting into (1) I obtain $$\alpha^2+(\alpha^{\dagger})^2 = - \mathbb{I}_4 \tag{2}$$

Now I presume that the negative sign on the right of (2) is wrong, can anyone spot my mistake?

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Hint: verify $\gamma_1^2=\gamma_2^2=-\Bbb I_4$. (You may wish to also check $\{\gamma_1,\,\gamma_2\}=\Bbb O_4$, so $\alpha^2=\alpha^{\dagger2}=0$.)

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  • $\begingroup$ Hello, thanks you for this, I have verified the identities you have in your response and can therefore see I initially made an error, and indeed, $\alpha^2+(\alpha^{\dagger})^2=0$ but the literature states that the square of these operators should equal the identity $\mathbb{I}_4$ so what am I missing? $\endgroup$
    – user104617
    Commented Aug 4, 2021 at 9:40
  • $\begingroup$ @SoboKevSpace What literature have you found claiming $\alpha^2$ is the identity? It obviously vanishes because $\alpha^2|0\rangle=\alpha0=0,\,\alpha^2|1\rangle=\alpha|0\rangle=0$. $\endgroup$
    – J.G.
    Commented Aug 4, 2021 at 9:43
  • $\begingroup$ It was in a PDF on spinor formalism that I'll try to find the link for; the claim is that the sum of the squares of the two Majorana operators above is the identity. I follow your ket logic above so I believe you above the PDF $\endgroup$
    – user104617
    Commented Aug 4, 2021 at 9:48

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