# Energy and degeneracy of the ground state and excitations of the toric code

Recall the hamiltonian of the toric code: (information mainly extracted from https://arxiv.org/pdf/1610.09260.pdf)

Consider Je=Jm=1. I've been trying to get the exact energies and degeneracies of the ground state but also for all the excited states of the toric code, depending on the size NxN of the lattice (we can do so as the hamiltonian is finite). For the degeneracy of the ground state it is well-known that it is 4 in a torus, yet for the excitations I haven't been able to find a satisfying answer. Also, from what I've understood, the ground state energy should be

,

and for each excitation, the energy should be increased by 4J. However, how do I find the degeneracies of these states, as well as the exact number of excited states?

I haven't found any analytical solution myself, so I've been trying to find it numerically. The main issue is that due to the amount of Kronecker/tensor products that need to be done (the dimension of the hamiltonian increases like 2^(2*N^2)), I've just been able to compute, using Python, the answer for N=2 and N=3 (for N=4 the computer still works, but the results I obtain cannot be correct). In particular, for N=2 and N=3 I've found:

Summing up, I have a few questions:

1. Could anyone provide some directions on how to get the exact energies and degeneracies for the ground state and the excitations, independently of N? (I believe that the fact that A and B commute with H might help me, but I don't really know how). Ideally I would like an analytical answer, but if I knew how to compute the results correctly for larger N's, I'd also be satisfied.

2. Regarding my table, are the numbers on the table correct? I don't understand how do I only get 512 states for N=3, and also, how is it possible that the degeneracy of the ground state for N=3 is 8 instead of 4.

3. A more general question. I know that excitations behave like anyons. Is there any relation between the behaviour of anyons andthe energies and degeneracies of the toric code?

Thanks a lot in advance and sorry for the long question.

• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Commented May 12, 2022 at 6:00

One of the major points of the Toric code Hamiltonian is that all the terms commute, each of which as $$\pm 1$$ eigenvalues. So, to find the ground state, you need something that is the $$+1$$ eigenstate of every term. So, if there are $$2N^2$$ $$A_v$$ terms and $$2N^2$$ $$B_p$$ terms, you get a total energy $$-2N^2J_e-2N^2J_m$$
Now, to find the first excited state, you might think that you just have to change one of those $$+1$$ eigenvalues into a $$-1$$ eigenvalue. However, you cannot do this: notice that $$\prod_vA_v=I=\prod_pB_p$$. Hence, the product of all eigenvalues on each of the two types of stabilizer must be +1. Hence, the first excited state has two of the eigenvalues at -1. Hence, there are terms like $$(4-2N^2)J_e-2N^2J_m,\qquad -2N^2J_2+(4-2N^2)J_m.$$ These are repeated as many times as there are pairs of eigenvalues to change, i.e. $$\binom{2N^2}{2}$$ each (multiplied by the degeneracy of the ground state).
What is the degeneracy of the ground state? Each stabilizer taking on either a $$\pm 1$$ value splits the space in half. There are $$4N^2$$ qubits, so the initial space is of dimension $$2^{4N^2}$$. However, there are only $$4N^2-2$$ distinct stabilizer values (because of that product of the identity again), so the remaining dimension is 4.
• (2) For N=2, and assuming $J_e=J_m$, then the degeneracy is $4\times 2\times\binom{8}{2}=224$. Basically, you can have any pair of face-type errors or any pair of vertex-type errors on top of any of the 4 ground states. Commented May 12, 2022 at 12:18