next up previous contents
Next: Spin correlations in the Up: Information and correlation in Previous: Symmetric spin Hamiltonian

 

Energy eigenvectors of the symmetric spin Hamiltonian

It is necessary to diagonalize this 2N dimensional Hamiltonian in order to find the distribution of states at thermodynamic equilibrium. The 2N operators tex2html_wrap_inline15147, tex2html_wrap_inline15149, and tex2html_wrap_inline15151 form a complete set of operators commuting with H and each other. The N total spin eigenvalues tex2html_wrap_inline15157, along with the N-1 successively larger susbsystem s values tex2html_wrap_inline15163 tex2html_wrap_inline15165 may be taken. Taking the total tex2html_wrap_inline15167 value m rounds out the set of 2N eigenvalues needed. The energy eigenvalue of any eigenstate depends only on its s value and is E=s(s+1)-N(1/2)(1/2+1)=s(s+1)-3N/4 for spin 1/2 particles. Since spin correlations are of interest, the basis where the tex2html_wrap_inline15179 are easily found must be used to express the energy eigenstates. Thus step one is to generate a complete list of energy eigenvectors tex2html_wrap_inline15181 with the energy eigenvalues described above. The second step is to make use of Clebsch-Gordon coefficients to transform the energy eigenbasis just generated to the tex2html_wrap_inline15183 basis. The description of how this is done is given in appendix 13.



David Wolf
Tue Mar 25 08:11:49 CST 1997