Product of Majorana operators after a orthogonal transformation

Question

The question I want to ask is the following:

There are $N$ Majorana fermion modes: $\gamma_1, \gamma_2, \dots, \gamma_N$, and they satisfy the anti-commutation relation: $\{ \gamma_i, \gamma_j \} = 2\delta_{i,j}$. There can be some orthogonal transformation to these Majoranas: \begin{equation} \gamma_k \rightarrow \sum_{l} O_{k,m} \gamma_m \end{equation} here $O$ is an orthogonal matrix: $O^T O = I$. What I want to know is what happens to the product of all these Majoranas under this transformation, i.e., \begin{equation} \gamma_1 \gamma_2 \dots \gamma_N \rightarrow \sum_{l_1,\dots,l_N} O_{1,l_1} O_{2,l_2}\dots O_{N,l_N} \gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} = ? \end{equation} I wondering whether the right-hand side of the above equation can be further simplified. More specifically, I wonder if it is equal to: $\det(O) \, \gamma_1 \gamma_2 \dots \gamma_N$?

Answer 1

Yes! We can work it out for a general matrix $M$ that preserves the commutation relations of the Majorana operators (which are the orthogonal matrices). $$\gamma_k \mapsto \sum_l M_{kl} \gamma_l $$

Applying this transformation to the expression \begin{align} \gamma_1 \gamma_2 \dots \gamma_N \mapsto \sum_{l_1 l_2\dots l_N} M_{1,l_1} M_{2,l_2} \dots M_{N,l_N} \gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} \end{align} Let us rewrite the l.h.s. Using the commutation relations, we know that $$ \gamma_1 \gamma_2 \dots \gamma_N = \frac{1}{N!} \sum_{j_1j_2\dots j_N} \epsilon^{j_1j_2\dots j_N} \gamma_{j_1} \gamma_{j_2} \dots \gamma_{j_N}, $$ where $ \epsilon^{j_1j_2\dots j_N} $ is the totally antisymmetric Levi Civita tensor that calculates the parity of the permutation $j_\alpha$. $$ \epsilon^{j_1j_2\dots j_N} = \begin{cases}1 \text{ for even permutations} \\ -1 \text{ for odd permutations} \\ 0 \text{ if } j_\alpha = j_\beta \text{ for any } \alpha \neq\beta\end{cases} $$

Making this replacement \begin{align} \gamma_1 \gamma_2 \dots \gamma_N &= \frac{1}{N!} \sum_{j_1j_2\dots j_N} \epsilon^{j_1j_2\dots j_N} \gamma_{j_1} \gamma_{j_2} \dots \gamma_{j_N}\\ &\mapsto \frac{1}{N!} \sum_{j_1j_2\dots j_N} \epsilon^{j_1j_2\dots j_N} \sum_{l_1 l_2\dots l_N} M_{j_1,l_1} M_{j_2,l_2} \dots M_{j_N,l_N} \gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} \\ &=\frac{1}{N!} \sum_{l_1 l_2\dots l_N} \gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} \sum_{j_1j_2\dots j_N} \epsilon^{j_1j_2\dots j_N} M_{j_1,l_1} M_{j_2,l_2} \dots M_{j_N,l_N} \end{align} Let us focus on the second sum in that last expression. Due to the totally antisymmetric property of the Levi-Civita tensor, we can see that the sum is zero if two different $l_\alpha$ have the same value. We can also see that the sum is antisymmetric under a swap of two $l_\alpha$. Therefore, we have \begin{align} &=\frac{1}{N!} \sum_{l_1 l_2\dots l_N} \gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} \sum_{j_1j_2\dots j_N} \epsilon^{j_1j_2\dots j_N} \epsilon^{l_1 l_2\dots l_N}M_{j_1,1} M_{j_2,2} \dots M_{j_N,N} \\ &= \frac{1}{N!} \sum_{l_1 l_2\dots l_N} \epsilon^{l_1 l_2\dots l_N}\gamma_{l_1} \gamma_{l_2} \dots \gamma_{l_N} \text{Det}[M] \\ &= \text{Det}[M] \gamma_1 \gamma_2 \dots \gamma_N \end{align} where we use the formula for determinant in terms of the Levi-Civita tensor, and the earlier derived expression for $\gamma_1\gamma_2\dots\gamma_N$ respectively.