2
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Why not calculate the commutator and find out? $\endgroup$ – wsc Nov 17 '13 at 21:11
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 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 at 21:41
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.