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.

  • 4
    $\begingroup$ Why not calculate the commutator and find out? $\endgroup$
    – wsc
    Nov 17, 2013 at 21:11

2 Answers 2


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 $\sum_i\mathbf{S}_i^2$.

  • $\begingroup$ How is it possible that the system Hamiltonian commutes with each component of the total spin $\pmb{S}=\sum_i\pmb{S}_i$ ? I guess that $H$ commutes just with $S_z$. Isn't it? $\endgroup$
    – AndreaPaco
    Feb 6, 2020 at 16:49
  • 1
    $\begingroup$ @AndreaPaco If a Hamiltonian is symmetric under a Lie group, it must commute with all generators of that group. Since the three spin operators generate the SU(2) group, all the spin components must commute with the Hamiltonian. If $H$ only commuted with $S_z$, it would mean that the symmetry group is SO(2), i.e. rotations about z-axis, which is a subgroup of SO(3). [Note: SU(2) is the double cover of SO(3)] $\endgroup$
    – Arkya
    Mar 25, 2020 at 21:41

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.


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