A commutation problem in Hubbard model 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.
 A: 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 $\sum_i\mathbf{S}_i^2$. 
A: A direct calculation was suggested in the comments, so I'm posting this for the sake of completeness.
Using the Pauli matrix identity $\sigma_{\alpha\beta}\cdot\sigma_{\gamma\delta}=2\delta_{\alpha\delta}\delta_{\beta\gamma}$ and the representation of the spin operators $\mathbf{S}_i=c_{i\alpha}^\dagger \sigma_{\alpha\beta}c_{i\beta}$, it is straightforward to check the following identity:
$$\mathbf{S}_i\cdot \mathbf{S}_j=-\frac{1}{2}\sum_{\alpha,\beta}c_{i\alpha}^\dagger c_{j\beta}^\dagger c_{i\beta} c_{j\alpha}-\frac{n_i n_j}{4}$$
In particular, we can check $$\mathbf{S}^2=\sum_i\mathbf{S}_i^2=\frac{1}{2}\sum_i(n_i-\frac{3}{2}n_i^2)=\frac{N}{2}-\frac{3}{4}\sum_in_i^2$$ The first term is proportional to the total number operator, which commutes with the Hubbard Hamiltonian since it is known to be a particle number-conserving Hamiltonian. So we only need to check the commutator with the second term. The hopping term is the potentially non-commuting term, because $n_i, n_{i\uparrow},n_{i\downarrow}$ are all simultaneously diagonalizable operators.
\begin{align}
\left[\sum_m n_m^2,\sum_{<ij>,\sigma}c_{i\sigma}^\dagger c_{j\sigma}\right] &= \sum_{\substack{m,\alpha,\beta\\<ij>,\sigma}} \left[n_{m\alpha}n_{m\beta},c_{i\sigma}^\dagger c_{j\sigma}\right]\\
&=\sum_{\substack{m,\alpha,\beta\\<ij>,\sigma}}\left(n_{m\alpha}\left[n_{m\beta},c_{i\sigma}^\dagger c_{j\sigma}\right]+\left[n_{m\alpha},c_{i\sigma}^\dagger c_{j\sigma}\right]n_{m\beta}\right)
\end{align}
Let's calculate one of these commutators,
\begin{align}
\left[n_{m\beta},c_{i\sigma}^\dagger c_{j\sigma}\right] &= c_{i\sigma}^\dagger\left[n_{m\beta}, c_{j\sigma}\right]+\left[n_{m\beta},c_{i\sigma}^\dagger \right]c_{j\sigma}\\
&=c_{i\sigma}^\dagger(-\delta_{jm}\delta_{\beta\sigma}c_{j\sigma})+(\delta_{im}\delta_{\beta\sigma}c_{i\sigma}^\dagger)c_{j\sigma}\\
&=(\delta_{im}-\delta_{jm})\delta_{\beta\sigma}c_{i\sigma}^\dagger c_{j\sigma}
\end{align}
Putting this back in,
\begin{align}
\left[\sum_m n_m^2,\sum_{<ij>,\sigma}c_{i\sigma}^\dagger c_{j\sigma}\right] &= \sum_{\substack{m,\alpha,\beta\\<ij>,\sigma}}\left[n_{m\alpha} (\delta_{im}-\delta_{jm})\delta_{\beta\sigma}c_{i\sigma}^\dagger c_{j\sigma} 
+ (\delta_{im}-\delta_{jm})\delta_{\alpha\sigma}c_{i\sigma}^\dagger c_{j\sigma} n_{m\beta}\right]\\
&=\sum_{<ij>,\sigma} \left[(n_i-n_j) c_{i\sigma}^\dagger c_{j\sigma} 
+ c_{i\sigma}^\dagger c_{j\sigma} (n_i-n_j)\right]
\end{align}
Now we can use the commutator $\left[n_{m\beta},c_{i\sigma}^\dagger c_{j\sigma}\right]$ derived above once more to take all the number operators to the right, leaving behind
$$\sum_{<ij>,\sigma}2 c_{i\sigma}^\dagger c_{j\sigma}(1+n_i-n_j) $$ which generically does not vanish. Thus it is clear that $\sum_i \mathbf{S}_i^2$ doesn't commute with the Hubbard Hamiltonian.
