Fermion Determinant When we calculate fermion determinant for either Majorana or Weyl spinors, why do we get an extra factor of half as the coefficient of the determinant?
 A: If you mean by $1/2$ the power in the determinant, it is due to the fact, that for Dirac fermion you have two types of fermions, let call them $\eta, \bar{\eta}$, and the quadratic term in the action has following the following form:
$$
\sum_i \bar{\psi}_i M_{i j} \psi_j
$$
The integration over $N$ pairs of grassmanian variables $\bar{\eta}_i, \eta_i$ by rules of grassmanian integration gives:
$$
\sum_{\sigma} (-1)^{\sigma} \prod_{i = 1}^{N} M_{i ,\sigma(i)}
$$
Where the sum is over all permutations, and this sum gives a determinant by its definition. For the case of Weyl or Majorana fermion, you have only one species of fermion, so for a skew-symmetric matrix, the nonzero answer will emerge, when one takes $N/2$ pairs from the exponent $e^{\sum_i \bar{\psi}_i M_{i j} \psi_j}$ (here we $N$ has to be even, and $M_{i j}$ -skew-symmetric, the contraction of symmetric pair of indices with antisymmetric produces zero). The resulting expression will be Pfaffian:
$$
\text{pf} (M) = \frac{1}{2^{N/2} (N/2)!} \sum_{\sigma} (-1)^{\sigma} \prod_{i = 1}^{N /2} M_{\sigma(2i), \sigma(2i+1)}
$$
For a skew-symmetric matrices of size $2N$, there holds following relation:
$$
\det(M) = \text{pf} (M)^2
$$
From there originates the power of $1/2$.
A: I suggest  you read my  Gamma matrices, Majorana fermions, and discrete symmetries in Minkowski and Euclidean signature that discusses these issues. I'd appreciate any feedback...
