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In a previous question Emergent symmetries I asked, Prof.Luboš Motl said that emergent symmetries are never exact. But I wonder whether the following example is an counterexample that has exact emergent spin rotational symmetry.

Just consider the simplest Ising model for two spin-1/2 system $H=\sigma_1^z\sigma_2^z$, it has two ground states, one of them is spin-singlet $|\uparrow\downarrow> -|\downarrow\uparrow> $ which possesses spin rotational symmetry, while the original Hamiltonian explicitly breaks it.

And I want to know if anyone knows some simple examples that all of the ground states have the emergent symmetry while the Hamiltonian doesn't have?

By the way,I remember that Prof.Xiao-gang Wen has said, a key difference between "topological degeneracy" and "ordinary degeneracy" is that the topological degeneracy is generally approximate while the ordinary degeneracy is exact. If the emergent symmetries are generally approximate, whether are there some connections between the topological degeneracy and emergent symmetries?

Comments: The two ground states of the above Ising example are degenerate. I wonder that whether an emergent symmetry could happen for a nondegenerate eigenstate ? For example, if an eigenstate of a Hamiltonian is non-degenerate, then this eigenstate must preserve all the symmetries of the Hamiltonian, and is there any possibility that this eigenstate has additional symmetry which is absence in the Hamiltonian ? Does someone know any example of this kind?

Thanks in advance.

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Kai Li
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2 Answers2

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I think the simplest example is very closely related to your suggestion of the two-site Ising model. Instead, consider the two-site XX-chain: $$ H = \sigma_1^x \sigma_2^x + \sigma_1^y \sigma_2^y. $$ Clearly the Hamiltonian has $U(1)$ symmetry (generated by $\sigma^z_1 + \sigma^z_2$) but it does NOT have full $SU(2)$ symmetry. However, its (unique!) ground state is the spin singlet $$ |\psi_\textrm{gs}\rangle = \frac{1}{\sqrt{2}} \left( |\uparrow_1 \downarrow_2\rangle -|\downarrow_1 \uparrow_2\rangle \right). $$ Hence, its unique ground state has an emergent $SU(2)$ symmetry.

Ruben Verresen
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The simplest model is the spin-1/2 chain with Majumdar–Ghosh interaction: $$H=\sum_i P_{3/2}(i-1,i,i+1),$$ where $P_{3/2}(i,j,k)$ is the projection operator that projects a state onto the subspace with total spin-3/2 on sites $i,j,k$. The ground states are two dimer states (see the figure on wikipedia Majumdar–Ghosh model): $$|\psi_1\rangle=\prod_i|\mathrm{singlet}\rangle_{2i,2i+1},$$ $$|\psi_2\rangle=\prod_i|\mathrm{singlet}\rangle_{2i-1,2i}.$$

If we define the symmetry transformation $U(i,j)=\exp(ia_{ij}P_0(i,j))$ where $P_0(i,j)$ is the singlet projection operator, then $$U(2i,2i+1)|\psi_1\rangle=\exp(i a_{2i,2i+1})|\psi_1\rangle,$$ $$U(2i-1,2i)|\psi_2\rangle=\exp(i a_{2i-1,2i})|\psi_2\rangle,$$ for any $i$. In other words, $|\psi_1\rangle$ supports a one dimensional representation of the group $U(2i,2i+1)$ (any $i$) which is not a symmetry of the original Hamiltonian. Similar for $|\psi_2\rangle$. It is exactly those emergent symmetries that make this model soluble.

More sophisticated examples can be found here: 0207106.

Tengen
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