My question is about a step in this paper: PhysRevB.65.165113 (X.G. Wen) or arxiv page 6.

Or alternatively: PhysRevB.90.174417 or arxiv page 3.

All papers on spin liquids and the projective symmetry group derive the mean field Hamiltonian in the same way.

Background: The Hilbert space $(\mathbb C^2)^{\otimes d}$ of $d$ spin-$1/2$ particles on a lattice is embedded in a larger fermionic Fock space with creation operators $f_{i\uparrow}^\dagger, f_{i\downarrow}^\dagger$ for each site $i$. The Heisenberg model Hamiltonian \begin{align} H = \sum_{ij} J_{ij} \, \vec S_i \cdot \vec S_j \end{align} is written in terms of the fermionic operators and a mean field decoupling is performed yielding \begin{align} H_{MF} = \sum_{ij} (\Psi_i^\dagger U_{ij} \Psi_j + \text{h.c.}) + \text{const.} , \qquad \text{where }\Psi_i = \begin{bmatrix}f_{i\uparrow} &f_{i\downarrow}^\dagger \end{bmatrix}^t \end{align} This is the part that I understand. But then an additional Lagrange multiplier term is added to the Hamiltonian. The Fock space generated by $f_{i\alpha}$ is bigger than $(\mathbb C^2)^{\otimes d}$. Only those states are "physical", i.e. correspond to states in $(\mathbb C^2)^{\otimes d}$, which satisfy a one-particle per site constraint: \begin{align} \langle \psi|f_{i\uparrow}^\dagger f_{i\uparrow} + f_{i\downarrow}^\dagger f_{i\downarrow} |\psi \rangle = 1 \end{align} Wen says in his paper that such a constraint can be forced by adding the (site dependent?) Lagrange multiplier term \begin{align} + \sum_{i} a_3 (f_{i\uparrow}^\dagger f_{i\uparrow} + f_{i\downarrow}^\dagger f_{i\downarrow} -1) + \left[(a_1 + i a_2) f_{i\uparrow}f_{i\downarrow} + \text{h.c.}\right] \end{align} to the Hamiltonian. $a_\mu$ are real numbers. The second paper furthermore tells that $a_\mu$ can be obtained by the condition \begin{align} \partial E_g/ \partial a_\mu = 0 \end{align} where $E_g$ is the ground state energy of the mean field Hamiltonian.

Remark: All the Fock space, fermion, spin stuff, etc. is probably not needed to answer the question. $f_{i\alpha}$ are just any linear operators on the Hilbert space satisfying canonical anticommutaion relations. The Hilbert space is finite dimensional.


How can an additional term enforce a constraint on the eigenstates of the Hamiltonian? How do Lagrange multiplier work in second quantized quantum mechanics (mathematically)?

I see that the first term corresponds to the condition that each site is single occupied, and that the other two terms correspond to the conditions that no site is not and that no site is doubly occupied.

They look quite similar as Lagrange multiplier terms in multivariable calculus. I understand how Lagrange multipliers work there. But I don't understand how they work in quantum mechanics. I don't need a complete explanation, just a hint where to start searching. Any help is highly appreciated.

  • $\begingroup$ The claim in the second paper about $\partial E_g/\partial a_\mu=0$ actually cites the first paper, which mentions this under equation (26). In general $E_g$ might not be differentiable - I don't know if this matters here. $\endgroup$ – Dap Apr 6 '19 at 11:16

The ground state energy is a classical minimization problem, over the unit sphere. Consider the general problem $H=H_a=V+aC$ with $H,C$ self-adjoint operators on a finite dimensional Hilbert space. The problem is:

  • minimize $E_g=\langle\psi| H| \psi\rangle=\langle\psi| V| \psi\rangle+a\langle\psi| C| \psi\rangle$
  • such that $\langle\psi|\psi\rangle=1$

A ground state $|\psi\rangle$ for $H_{a+\epsilon}$ is close to a ground state for $H_a.$ This leads to an approximation like $\langle \psi | H_{a+\epsilon} |\psi\rangle = E_g + \epsilon \langle \psi |C|\psi\rangle + O(\epsilon^2).$ So $\partial E_g/\partial a=\langle \psi| C|\psi\rangle.$


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