The Bethe Ansatz

Counting spinon statesc.h.e.s.c

• Two-spinon states
  

  Adding up the \(S=0\) and \(S=1\) two-spinon states, we obtain

  \begin{equation*} \frac{N(N-2)}{8} + 3 \times \frac{N(N+2)}{8} = \frac{N(N+1)}{2} = \left( \begin{array}{c} N+1 \\ 2 \end{array} \right) \end{equation*}

  which is the correct total number of two-spinon states.

• Four-spinon states
  

  Adding the numbers of states in all the \(S=0, 1\) and \(2\) sectors gives (note: as expected, there are 2 \(S=0\) representations, 3 \(S=1\) ones, and one \(S=2\))

  \begin{align*} \frac{N (N-2) (N-4) (N-6)}{384} + \frac{(N+2)N(N-2)(N-4)}{384} + 3\times \frac{(N+2) N (N-2) (N-4)}{128} + \nonumber \\ + 5\times \frac{(N+4)(N+2) N (N-2)}{384} = \left( \begin{array}{c} N+1 \\ 4 \end{array} \right) \hspace{5cm} \end{align*}

  which is the expected total number of four-spinon states.