# Entanglement entropy of one Majorana fermion. $(1/2)log2$?

Physics Asked by Geralt on November 13, 2020

I am trying to see that the contribution to the entanglement entropy (EE) of one Majorana fermion is $$frac{1}{2}log2$$.

I’m working with the covariance matrix $$Gamma$$ of the Majorana fermions. This matrix is skew-symmetric, so that the eigenvalues come in pairs $$pmlambda$$, with $$0leqlambdaleq1$$.
$$begin{equation} Gamma=begin{bmatrix} begin{matrix}0 & lambda_1 \ -lambda_1 & 0end{matrix} & 0 & cdots & 0 \ 0 & begin{matrix}0 & lambda_2 \ -lambda_2 & 0end{matrix} & & 0 \ vdots & & ddots & vdots \ 0 & 0 & cdots & begin{matrix}0 & lambda_r\ -lambda_r & 0end{matrix} end{bmatrix} end{equation}$$

Now, in order to compute the EE of one subsystem A, one takes $$Gamma_A$$, the restriction of $$Gamma$$ to $$A$$. By construction the new matrix $$Gamma_A$$ is also skew-symmetric.
The EE is given by
$$begin{equation} S(A)=-sum_{k=1}^{d_{A/2}} frac{1+lambda_k}{2}log{frac{1+lambda_k}{2}}+ frac{1-lambda_k}{2}log{frac{1-lambda_k}{2}} end{equation}$$

This formula comes from realizing that the density matrix associated $$rho_A$$ decomposes into $$m$$ independent 2-state (i.e. complex fermions) systems. This formula can be found, for instance here:

So suppose that my subsystem A involves and odd number of Majorana fermions. As a consequence, one eigenvalue will be 0. In fact we can think in the simplest one subsystem: just one Majorana.
If I use the formula above $$S(lambda=0)=log(2)$$ which is unphysical as this is the entropy that corresponds to a complex fermion. So my guess is that this formula is only valid for subsystems that involves an even number (2D) of Majorana fermions, where (D) complex fermions can be accomodated. If I am true, how can I prove that the EE of just one Majorana is $$1/2log(2)$$?