9
$\begingroup$

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.

Improve this question
$\endgroup$
15
  • 1
    $\begingroup$ 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. $\endgroup$
    – Tengen
    Commented Mar 23, 2013 at 6:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – Tengen
    Commented Mar 23, 2013 at 7:10
  • $\begingroup$ @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. $\endgroup$
    – Kai Li
    Commented Mar 23, 2013 at 9:36
  • $\begingroup$ 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. $\endgroup$
    – Tengen
    Commented Mar 23, 2013 at 11:01
  • $\begingroup$ @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? $\endgroup$
    – Kai Li
    Commented Mar 23, 2013 at 12:06

2 Answers 2

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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 ? $\endgroup$
    – Kai Li
    Commented Apr 1, 2013 at 18:56
  • 2
    $\begingroup$ I've modified the answer, please have a look. $\endgroup$
    – Tengen
    Commented Apr 2, 2013 at 12:26
  • $\begingroup$ 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. $\endgroup$
    – Kai Li
    Commented Apr 2, 2013 at 13:35
  • $\begingroup$ 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. $\endgroup$
    – Kai Li
    Commented Apr 2, 2013 at 15:03
  • $\begingroup$ 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. $\endgroup$
    – Tengen
    Commented Apr 3, 2013 at 4:48
4
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Great, thank you. $\endgroup$
    – Kai Li
    Commented Mar 27, 2020 at 17:31
  • $\begingroup$ Here maybe other simple examples. Consider a lattice fermion Hamiltonian $H=\sum_{i}\mu _i \hat{n}_i$, where the site-dependent $\mu _i>0$ for each site. Then its unique ground state (the vacuum state) $\left | 0\right \rangle$ has an emergent translation symmetry. Similarly, we can add time-reversal breaking hopping terms to the Hamiltonian and choose a strong enough (uniform) chemical potential such that $\left | 0\right \rangle$ is the unique ground state with emergent time-reversal symmetry. But I think these examples are somewhat trivial. $\endgroup$
    – Kai Li
    Commented Apr 25, 2020 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.