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

enter image description here

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

enter image description here,

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:

enter image description here

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.

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

1 Answer 1

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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.

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  • $\begingroup$ Hi @DaftWullie , thanks a lot for your answer, it's very detailed and helpful. However, I still have a few doubts: $\endgroup$
    – MarcPN
    May 12, 2022 at 11:05
  • $\begingroup$ 1) Regarding the 1st excited state: If two operators change eigenvalue from -1 to +1, shouldn't the energy of the 1st excited state increase by 2*2J, that is 4J? 2 because there are two operators changing eigenvalue, and 2 for the difference of energy from -1 to +1 for each operator. 2) I don't understand the deg of the 1st excited state. For example, for N=2, does it mean that the 1st excited state is 28-fold degenerate, or, as you suggest, it should be multiplied by the deg of the ground state and be 112? Can this binomial method to calculate the degeneration be generalised for eachstate? $\endgroup$
    – MarcPN
    May 12, 2022 at 11:14
  • $\begingroup$ 3) Finally, for the energy of the other excited states, for every level 2 more operators need to change their eigenvalue. Correct? Thanks again, and sorry for so many questions. @DaftWullie $\endgroup$
    – MarcPN
    May 12, 2022 at 11:14
  • $\begingroup$ (1) Yes, you're right. My mistake. (3) Yes. $\endgroup$
    – DaftWullie
    May 12, 2022 at 12:15
  • $\begingroup$ (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. $\endgroup$
    – DaftWullie
    May 12, 2022 at 12:18

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