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Does the Hubbard Hamiltonian $$H=-t\sum_{\langle ij\rangle \sigma}c_{i\sigma}^{\dagger}c_{j\sigma}+h.c.+U\sum_{i}n_{i\uparrow}n_{i\downarrow}$$ commute with $\sum_{i}\mathbf{S}_i^2$? where $\mathbf{S}$ is the spin angular momentum.

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    $\begingroup$ Why not calculate the commutator and find out? $\endgroup$ – wsc Nov 17 '13 at 21:11
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It is known that the Hubbard model possesses the global $SU(2)$ spin-rotation symmetry, which means that the Hamiltonian commutes with the total spin $\sum_i\mathbf{S}_i$(where $\mathbf{S}_i=\frac{1}{2}c_i^\dagger \mathbf{\sigma}c_i$), which is the generators of the global $SU(2)$ spin-rotation group, and it does not commute with $\sum_i\mathbf{S}_i^2$.

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