# Symmetries in the Hubbard model

I would like to understand both in an intuive and in a mathematical way the meaning of the sentence "The Hubbard Hamiltonian has an SU(2) symmetry". What are the symmetry transformations that leave the Hubbard Hamiltonian unchanged?

Moreover, I would like to understand how is it possible to generalize the Hubbard model to the case of SU(N) symmetry. In particular, what does N stand for? What are the conserved quantities of the SU(2) Hubbard model and of the SU(N) Hubbard model?

The Hubbard model is described by nearest-neighbor hopping $$H_t = -t\sum [c^{\dagger}_{\sigma,i}c_{\sigma,j}+\rm{h.c.} ]$$, which preserves the spin-orientation, plus an on-site Coulomb replusions $$H_U = U \sum n_{i,\uparrow}n_{i,\downarrow}$$. Both of these terms are invariant to a global rotation of the spins. The first term will still describe each flavor of spins hopping from site to site, and the other will still be $$U$$ if there are two electrons on the site and zero otherwise. Therefore, a global rotation of the spins, which is described by $$SU(2)$$, leaves the Hamiltonian invariant. This helps us as we can characterize the states with total spin and total $$S_z$$, which are conserved.

In order to expand to an $$SU(N)$$ symmetry, we need to characterize the particles with higher degrees of freedom, such that their symmetry is not spin-rotations of $$SU(2)$$ but a general $$SU(N)$$.

Edit after question in comments:

For $$SU(N)$$ it's not enough to have larger spins. An example of how to generate $$SU(N)$$ might be along the following lines - instead of just one type of electrons let's say we have $$N$$-types of electrons, each coming from a different source. For example, in the 1d Hubbard model, we can proximitize $$N$$ wires of spinless electrons. So now instead of labeling each by the spin index we label it by the index $$\nu=1,\ldots,N$$ that tells us at which wire the electron is. We can now define "rotations" in the basis of these wires, which simply signify basis change. These are rotations in $$N$$-dimensional space, that will be characterized by the symmetry group $$SU(N)$$. Depending on the type of Hamiltonian and which terms we allow, it might be symmetric under this rotations (that is - it will not care in which basis we write our electrons). We usually call this extra degree of freedom "flavor".

• Let's define $\mathcal{S}^\alpha = \sum_j S^{\alpha}_j$ as the total spin operator in the direction $\alpha$. Then $e^{-i\theta^{\alpha}\mathcal{S}^{\alpha}}H_{t,U}e^{i\theta^{\alpha}\mathcal{S}^{\alpha}} = H_{t,U}$. This is the invariacne of the Hamiltonian. For $SU(N)$ you need to define other degrees of freedom, that rotate not with the $3$ angular momentum operators but let's say the $8$ generators of $SU(3)$. The argument then follows identically.
– user245141
Commented Feb 7, 2020 at 11:02
• No, this is not correct. Spin $1$ particles also belong to $SU(2)$ symmetry group, as are any general spin particles. It's not the number of degrees of freedom but rather how they transform under rotations. All spins are always defined by just 3 operators - $S^x, S^y$ and $S^z$. These 3 operators maintain the Lie-algebra of $SU(2)$ $[S^\alpha, S^\beta] = i\epsilon_{\alpha,\beta,\gamma} S^{\gamma}$, and any transformation in spin-basis is done by them, regardless of the magnitude of spin. For $SU(N)$ you need something different, such that you will have other "rotations".
– user245141
Commented Feb 7, 2020 at 11:46
• I will try to expand my reply
– user245141
Commented Feb 7, 2020 at 11:52
• For $SU(2)$ you just keep the spin label of $\uparrow, \downarrow$. You can look at it as two wires of spinless fermions, with wire index replacing the spin projection index. They are equivalent.
– user245141
Commented Feb 7, 2020 at 14:14
• oh yeah you're right. What I described is characterized by the symmetry group $SU(2)\times SU(N)$.
– user245141
Commented Feb 7, 2020 at 14:26

Let me give a completely mathematical description, since I think the accepted answer is more physically motivated.

Let $$U\in SU(2)$$. Then its second-quantization $$W_U$$ on the $$n$$-particle Fock space would be just $$U\otimes\cdots\otimes U$$. In particular, if $$U=e^{-iS}$$ for some $$S\in i \text{su}(2)$$ (its Lie algebra), then $$W_U = \exp(-ic^*S c)$$ where $$c^* S c$$ is the Jordan map of $$S$$, i.e. short-hand for $$c^* Sc=\sum_\varphi c^*(S\varphi) c(\varphi) = S\otimes I\otimes \cdots \otimes I+\cdots+I\otimes\cdots\otimes I\otimes S$$ where the summation is over any orthonormal basis $$\varphi$$ of the single-particle Hilbert space $$\mathscr{H}$$. It should be noted that $$W_U c^*(\varphi)W^*_U= c^*(U\varphi)$$ Also notice that if $$n=n_\uparrow +n_\downarrow$$, then $$n^2 =n+2n_\uparrow n_\downarrow$$ Also notice that $$n=c^* c$$ and that $$W_U nW_U^*=\sum_sc^*(Us)c(Us)$$ SInce $$U$$ is unitary, we see that $$Us$$ is just another orthonormal basis. Since the I already claimed (and can be easily proven) that the shorthand $$c^*c$$ is independent of orthonormal basis, we see that $$W_U n W_U^*=n$$ Hence, $$W_U n_\uparrow n_\downarrow W_U^* = n_\uparrow n_\downarrow$$ Therefore, the interaction term is $$SU(2)$$-invariant. You could do something similar with the hopping term. Also, I ignored the lattice sites in my derivation, but the same proof with slight modifications hold in the general case.

From this proof, one thing becomes clear. If you want to generalize this interaction term to $$SU(N)$$, we would need an interaction term that looks like $$n^2-n$$ (or some other function of $$n=c^* c$$).

• +1 I find this answer very useful. There is something that bothers me a little. Suppose I had spin-$3/2$ particles. I would feel then that $n = n_{3/2} + n_{1/2} + n_{-1/2} +n_{-3/2}$ would be invariant under both $SU(2)$ and $SU(4)$. Is it that for $k$ degrees of freedom, $n$ is invariant under $SU(k)$, and that a spin-$3/2$ model which was $SU(2)$ but not $SU(4)$ symmetric would require a more complicated interaction term (more complicated than a polynomial in $n$) to break $SU(4)$? Commented Jun 7, 2021 at 3:36
• Any $s$ spin is just an irreducible representation of $SU(2)$, so $n$ is always $SU(2)$ invariant (in the sense of representations). I haven't really thought about $SU(N)$ tbh, but my guess would be to look at the irreps of $SU(N)$, which are usually labelled using weights and the Weyl group. Commented Jun 8, 2021 at 6:56
• I agree. My comment was more on that it seems that one will automatically get $SU(N)$ using a polynomial in $n$ for a spin $(N-1)/2$ particle, which would require more complicated interactions to break $SU(N)$ if it was not desired (and preserve the subgroup of $SU(2)$.) Commented Jun 8, 2021 at 7:00