A simple model that exhibits emergent symmetry?

Question

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?

Thanks in advance.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

Comments

@Tengen: Ok, I see, you really give us an example of what I want, although its physical meaning is temporarily unclear. Thank you all the same.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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You can construct them easily as many as you like. As I said before, all you need is to construct a symmetry operator that commutes with the Hamiltonian only in the subspace of the ground states.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

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@Tengen: Yeah, that's right

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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@Tengen: For the Haldane chain you mentioned, I didn't know much about it. But I want to know if you take periodic boundary conditions, are the ground states exactly degenerate for finite lattice sites? What causes the ground state degeneracy in Haldane chain, topology or symmetry of the system? If it's the topology, whether these topological ground states are protected by some symmetries?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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The ground state is nondegenerate if you take periodic boundary condition. It is an example of symmetry protected topological state. The symmetry is the spin rotational symmetry SO(3) for integral spin.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

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We say a subspace has a symmetry if it support a representation of the symmetry group. All we are talking about here is the generator of the symmetry transformation, not the group itself. Zero eigenvalue is not a problem (you can plus any nonzero number to the symmetry operator if you like). Think about the spin rotational symmetry: the total spin can also be zero.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

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Yes, you are right. No physical meanings maybe. I just constructed it to fulfil your requirement.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

Answer 1

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.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

Comments

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 ?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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I've modified the answer, please have a look.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen

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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.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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And I like your last sentence "It is exactly those emergent symmetries that make this model soluble" very much, since this maybe one of the facts that would make emergent symmetries important in physics.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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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.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen