About SU(2) gauge symmetry of the large U limit of the Hubbard model

Question

I have been studying about the SU(2) symmetry in Heisenberg Hamiltonian with a paper 'SU(2) gauge symmetry of the large U limit of the Hubbard model' written by Ian Affleck et al(Phys. Rev. B 38, 745 – Published 1 July 1988). In the paper, they represent the spin in terms of fermionic operators with constraint that number of particle at each lattice is 1.

$$S_{x}=\frac{1}{2}c^{\alpha\dagger}_{x}\sigma_{\alpha}^{\beta}c_{x\beta}, \\constraint: c^{\dagger\alpha}c_{\alpha}=1$$

And they re-express the Hamiltonian using matrix $\Psi_{\alpha\beta}\equiv\left(\begin{array}{cc}c_{1}&c_{2}\\c_{2}^{\dagger}&-c_{1}^{\dagger}\end{array}\right)$(where number denotes the spin up and down) to show the SU(2) symmetry of the Heisenberg Hamiltonian explicitly. Also the constraint can be re-expressed as follows: $$\frac{1}{2}tr\Psi^{\dagger}\sigma^{z}\Psi=\frac{1}{2}(c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-c_{1}c_{1}^{\dagger}-c_{2}c_{2}^{\dagger})=c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-1=0$$ At this time $c^{\dagger}, c$ are operators.

And they write the lagrangian of this Hamiltonian. $$L=\frac{1}{2}\sum_{x}tr\Psi^{\dagger}_{x}(id/dt+A_{0x})\Psi_{x}-H$$ where $A_{0}=\frac{1}{2}\mathbf{\sigma}\cdot\mathbf{A_{0}}$. Here components of $\mathbf{A_{0}}$ are Lagrangian multipliers. This lagrangian has time dependent gauge symmetry. Here I have a problem. As far as i know $c^{\dagger}, c$ in Lagrangian are not operators anymore but grassmann variables. Therefore constraint of $(\mathbf{A_{0}})_{z}$ in the Largrangian becomes
$$\frac{1}{2}tr\Psi^{\dagger}\sigma^{z}\Psi=\frac{1}{2}(c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}-c_{1}c_{1}^{\dagger}-c_{2}c_{2}^{\dagger})=c^{\dagger}_{1}c_{1}+c_{2}^{\dagger}c_{2}=0$$ This is different from the original constraint condition!! I am really confused about this. Can anyone help me to solve this problem?

This post imported from StackExchange Physics at 2014-08-25 11:29 (UCT), posted by SE-user Iksu Jang