# A simple model that exhibits emergent symmetry?

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?

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@user10001:Thanks for your corrections. – Kai Li Mar 22 '13 at 19:50
How about this one: $H=S_1^z S_2^z+\frac{1}{4}(S_1^z+S_2^z)$. The second term breaks the degeneracy of the excited states in your model. The operator $S_1^+S_2^++S_1^-S_2^-$ does not commute with the Hamiltonian, but the two ground states are its eigenstates. My logic is to construct a symmetry operator that commutes with the Hamiltonian only in the subspace of the ground states. – Tengen Mar 23 '13 at 6:58
I'm thinking about the one dimensional antiferromagnetic Haldane chain (with open boundary condition). The gound state is 4-fold degenerate only in thermodynamic limit. That means the energy difference between a state with total spin 0 and a state with total spin 1 becomes zero in thermodynamic limit. I'm wondering how to get it from symmetry point of view. – Tengen Mar 23 '13 at 7:10
@Tengen:The ground states in your model are the same as mine, I wonder whether there exist simple models whose ground states all have emergent symmetries. – Kai Li Mar 23 '13 at 9:36
My model is an example. The gound state subspace is spanned by $|\uparrow\downarrow\rangle$ and $|\downarrow\uparrow\rangle$. They all have the symmetry as I defined above. But only a one dimentional subspace spanned by $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ has spin rotation symmetry. – Tengen Mar 23 '13 at 11:01

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.
Thanks for your answer.But I can't understand your explanation clearly cause your language seems a little hard to me. Do you mean that the two dimer ground states of MG model both have the same symmetry $U(i,j)$ while the Hamiltonian $H$ doesn't have ? – Kai Li Apr 1 '13 at 18:56
Ok. How about the superposition of the two dimer states?For example, whether the ground state $\psi=\lambda_1\psi_1+\lambda_2\psi_2$ is still a eigenstate of $U(i,j)$ ? Thanks a lot. – Kai Li Apr 2 '13 at 13:35
Nope. There are $N$ $U(1)$ groups: $U(i,i+1), i=1,2,...,N$. $|\psi_1\rangle$ is invariant under the actions of half of there groups, and $|\psi_2\rangle$ the other half. – Tengen Apr 3 '13 at 4:48