# A simple model that exhibits emergent symmetry?

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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?

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

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

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

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

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

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Tengen
@Tengen:Oh,I got what you mean now. You're right,the two ground states are definitely eigenstates of the symmetry operator you defined,but with eigenvalues zeros, which means that the ground states would disappear once the symmetry operator acts on them.So in this case can we still say that the ground states are symmetric under the operation you defined?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
@Tengen：What more important, I think the "symmetry" operator you defined in fact doesn't represent any symmetry, because it can not even be inversed, which means that it can not be an unitary or antiunitary operator. And its physical meaning is unclear.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
@Tengen:Maybe I misunderstood your explanation.Let A be the operator you defined, then exp(i A ) can be a symmetry operator, did you mean this? If so ,then what's the physical meaning of it?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
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
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
@Tengen:Yeah, I agree with you now, I made a misunderstanding. So do you know the physical meaning of it?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
@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
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
@Tengen: Yeah, that's right

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
@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
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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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
answered Apr 1, 2013 by (105 points)
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
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.
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.