- #1
Petar Mali
- 290
- 0
[tex]\hat{S}^+_i=\sqrt{2S}(\hat{a}_i-\frac{1}{2S}\hat{a}^+_i\hat{a}_i\hat{a}_i)[/tex]
[tex]\hat{S}^-_i=\sqrt{2S}\hat{a}^+_i, \quad
\hat{S}^z_i=S-\hat{a}^+_i\hat{a}_i[/tex]
Why is in solid state physics often convenient to use this representation? It is obvious that
[tex](\hat{S}^-_i)^{\dagger}\neq \hat{S}^+_i[/tex]
And Hamiltonian of Heisenberg model is hermitian!
[tex]\hat{S}^-_i=\sqrt{2S}\hat{a}^+_i, \quad
\hat{S}^z_i=S-\hat{a}^+_i\hat{a}_i[/tex]
Why is in solid state physics often convenient to use this representation? It is obvious that
[tex](\hat{S}^-_i)^{\dagger}\neq \hat{S}^+_i[/tex]
And Hamiltonian of Heisenberg model is hermitian!