For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $H=\sum_{<ij>}\mathbf{S}_i\cdot\mathbf{S}_j$, and we perform a Schwinger-fermion($\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$) mean-field study.

Let $H_{MF}=\sum_{<ij>}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$ be the resulting mean-field Hamiltonian, where $(\chi_{ij},\eta_{ij})$ is the mean-field ansatz. And let $\psi_{1,2}$ represent two exact eigenstates of $H_{MF}$ with energies $E_{1,2}(E_1>E_2)$, say $H_{MF}\psi_{1,2}=E_{1,2}\psi_{1,2}$.

Now we can construct the physical spin states $\phi_{1,2}$ by applying the projective operator $P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$(Note that $P\neq \prod _i(1-\hat{n}_{i\uparrow}\hat{n}_{i\downarrow})$) to $\psi_{1,2}$(where $\hat{n}_i=f^\dagger_{i\uparrow}f_{i\uparrow}+f^\dagger_{i\downarrow}f_{i\downarrow}$), say $\phi_{1,2}=P\psi_{1,2}$, and generally we don't expect that $\phi_{1,2}$ are the exact eigenstates of the original spin Hamiltonian $H$.

My question is: $\frac{\left \langle \phi_1 \mid H \mid \phi_1 \right \rangle}{\left \langle \phi_1 \mid \phi_1 \right \rangle}>\frac{\left \langle \phi_2 \mid H \mid \phi_2 \right \rangle}{\left \langle \phi_2 \mid \phi_2 \right \rangle} $ ? If it's true, then how to prove it rigorously ? Thanks a lot.

  • 2
    $\begingroup$ Too much quantities are not defined precisely, and no reference is given, so you had better to completely rewrite your question, or to ask a new question. $\endgroup$ – Trimok Sep 1 '13 at 14:14
  • $\begingroup$ @ Trimok, I added some details to my post. $\endgroup$ – Kai Li Sep 6 '13 at 10:49

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