Let $M$ be the mass matrix for fermions $\psi_+$ and for $\psi_-$ (separately). It is obtained by $\not{D}\not{D^+}= -\partial_t^2+ M^2$

Then $M^2=r^2 \mathbb{Id_{16}}- \not{v}$, Now, the $16*16$ matrix $\not{v}$ has a zero trace, and it square is $\vec v^2 Id_{16}$, so the only possibility is that the matrix $\not{v}$ has 8 eigenvalues $v$, and 8 eigenvalues $-v$ (here $v$ means $\sqrt{\vec v^2}$). So the matrix $M^2$ has 8 eigenvalues $r^2+v$ and 8 eigenvalues $r^2-v$. This is true for $\psi_+$ and for $\psi_-$, while $\psi_3$ is obviously massless.

[EDIT]

The gamma matrices of $SO(9)$ are real, so $\not{B}$ is hermitian. $\partial_t$ is antihermitian (because $i\partial_t$ is hermitian), so starting with $\not{D}=\partial_t-\not{B}$, it is easy to see that $\not{D}^{\dagger}=-\partial_t-\not{B}$

If you neglect order 3 terms in the Lagrangian ($\psi^2Y, \psi^2A$), and apply Lagrange equation on $\psi_+$, you get $\not{D}\psi_-=0$. And, because $\psi_+ = (\psi_-)^*$, and $\not{B}$ is real, you have also $\not{D}\psi_+=0$

The mass matrix apply separately to $\psi_+$ and $\psi_-$, simply because $\psi_+ = (\psi_-)^*$, and the mass matrix is real.

This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Trimok