4 added 72 characters in body edited Sep 27 '13 at 13:59 Kai Li 1,9191414 silver badges2525 bronze badges For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $$H=\sum_{}\mathbf{S}_i\cdot\mathbf{S}_j$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}=\sum_{}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$$ 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. For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $$H=\sum_{}\mathbf{S}_i\cdot\mathbf{S}_j$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}=\sum_{}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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. For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $$H=\sum_{}\mathbf{S}_i\cdot\mathbf{S}_j$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}=\sum_{}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$$ 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. 3 added 179 characters in body; edited tags edited Sep 6 '13 at 10:47 Kai Li 1,9191414 silver badges2525 bronze badges For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $$H$$$$H=\sum_{}\mathbf{S}_i\cdot\mathbf{S}_j$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}$$$$H_{MF}=\sum_{}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$$ be the resulting mean-field Hamiltonian, andwhere $$(\chi_{ij},\eta_{ij})$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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. For example, consider a spin-1/2 Heisenberg Hamiltonian $$H$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}$$ be the resulting mean-field Hamiltonian, and $$\psi_{1,2}$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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. For example, consider a spin-1/2 AFM Heisenberg Hamiltonian $$H=\sum_{}\mathbf{S}_i\cdot\mathbf{S}_j$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}=\sum_{}(f^\dagger_i\chi_{ij}f_j+f^\dagger_i\eta_{ij}f_j^\dagger+H.c.)$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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 added 112 characters in body edited Sep 1 '13 at 6:08 Kai Li 1,9191414 silver badges2525 bronze badges For example, consider a spin-1/2 Heisenberg Hamiltonian $$H$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}$$ be the resulting mean-field Hamiltonian, and $$\psi_{1,2}$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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: $$\left \langle \phi_1 \mid H \mid \phi_1 \right \rangle>\left \langle \phi_2 \mid H \mid \phi_2 \right \rangle$$$$\frac{\left \langle \phi_1 \mid H \mid \phi_1 \right \rangle}{\left \langle \phi_1 \mid \phi_1 \right \rangle}>\frac{\left \langle \phi_2 \mid H \mid \phi_2 \right \rangle}{\left \langle \phi_2 \mid \phi_2 \right \rangle}$$ ? If it's true, then how to prove it rigorously ? Thanks a lot. For example, consider a spin-1/2 Heisenberg Hamiltonian $$H$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}$$ be the resulting mean-field Hamiltonian, and $$\psi_{1,2}$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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: $$\left \langle \phi_1 \mid H \mid \phi_1 \right \rangle>\left \langle \phi_2 \mid H \mid \phi_2 \right \rangle$$ ? If it's true, then how to prove it rigorously ? Thanks a lot. For example, consider a spin-1/2 Heisenberg Hamiltonian $$H$$, and we perform a Schwinger-fermion($$\mathbf{S}_i=\frac{1}{2}f^\dagger_i\mathbf{\sigma}f_i$$) mean-field study. Let $$H_{MF}$$ be the resulting mean-field Hamiltonian, and $$\psi_{1,2}$$ 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 $$P=\prod_i(2\hat{n}_i-\hat{n}_i^2)$$ 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. 1 asked Aug 31 '13 at 17:53 Kai Li 1,9191414 silver badges2525 bronze badges