For example, consider a spin-1/2 AFM Heisenberg Hamiltonian H=∑<ij>Si⋅Sj, and we perform a Schwinger-fermion(Si=12f†iσfi) mean-field study.
Let HMF=∑<ij>(f†iχijfj+f†iηijf†j+H.c.) be the resulting mean-field Hamiltonian, where (χij,ηij) is the mean-field ansatz. And let ψ1,2 represent two exact eigenstates of HMF with energies E1,2(E1>E2), say HMFψ1,2=E1,2ψ1,2.
Now we can construct the physical spin states ϕ1,2 by applying the projective operator P=∏i(2ˆni−ˆn2i)(Note that P≠∏i(1−ˆni↑ˆni↓)) to ψ1,2(where ˆni=f†i↑fi↑+f†i↓fi↓), say ϕ1,2=Pψ1,2, and generally we don't expect that ϕ1,2 are the exact eigenstates of the original spin Hamiltonian H.
My question is: ⟨ϕ1∣H∣ϕ1⟩⟨ϕ1∣ϕ1⟩>⟨ϕ2∣H∣ϕ2⟩⟨ϕ2∣ϕ2⟩ ? If it's true, then how to prove it rigorously ? Thanks a lot.
This post imported from StackExchange Physics at 2014-03-09 08:43 (UCT), posted by SE-user K-boy