Fermion Operators

At this point, we can hypothesize that the operators that create fermion states do not commute. In fact, if we assume that the operators creating fermion states anti-commute (as do the Pauli matrices), then we can show that fermion states are antisymmetric under interchange. Assume \bgroup\color{black}$b_r^\dagger$\egroup and \bgroup\color{black}$b_r$\egroup are the creation and annihilation operators for fermions and that they anti-commute.
\bgroup\color{black}$\displaystyle \{b_r^\dagger,b_{r'}^\dagger\}=0 $\egroup
The states are then antisymmetric under interchange of pairs of fermions.

\begin{displaymath}\bgroup\color{black} b_r^\dagger b_{r'}^\dagger \vert\rangle = -b_{r'}^\dagger b_{r}^\dagger \vert\rangle \egroup\end{displaymath}

Its not hard to show that the occupation number for fermion states is either zero or one.

Jim Branson 2013-04-22