Recently I was thinking how small local perturbations keep the topological order invariant in some Lattice Gauge Theories, like for the simplest one - Toric Code mode, which is defined as $$H_{TC} = H_0 = -J_s \sum_{site} A_s -J_p \sum_{plaquette} B_p.$$ Now consider Perturbation of the kind, such that: $$H = -J_s \sum_{site} A_s -J_p \sum_{plaquette} B_p -\lambda_x \sum_{edge} \sigma_i^{x} -\lambda_z \sum_{edge} \sigma_i^{z}. \tag 1$$ Where I have only added Uniform Magnetic Field at each edge of the lattice, where the spin degrees of freedom (that I started with) reside.

It has been extensively discussed in contemporary literature, "how big" the fields need to be in order to break the topological order of the system, and since there is no local order parameter here, on hinges on the fact that, after fixing the direction of external-fields, at one value of strength of the fields the mutual-semionic nature of the elementary excitations [pair of electric charge or magnetic fluxes] vanish.

For this post, the 2 questions are the following

Question 1: Whether the Hamiltonian above is exactly equivalent to the 2D Transverse Field Ising model, as reported in Breakdown of a Topological Phase: Quantum Phase Transition in a Loop Gas Model with Tension ?

The arguments and confusions are:

i) consider a plaquette flip operator whose eigenvalue $\mu^{z}_{p} = (-1)^{n_p}$ with $n_p$ as number of flips performed on the plaquette $p$, and since a $\sigma^x_i$ flips a $\sigma_z$ located at $i$ and that in turn flips $\mu^{z}_{p_{+}(i)}$ and $\mu^{z}_{p_{-}(i)}$, i.e. the 2 adjacent plaquettes to $i-th$ site.

ii) the above argument imples (adding the fact that $\sigma^z_i$) flips 2 adjacent $A_s$, where $\mu^{x}_{s}$ is a vertex-flip operation, or the Gauge Transformation of this theory.

$\implies \sigma^x_i = \mu^{z}_{p_{+}(i)}\mu^{z}_{p_{-}(i)}, \ \ \ \sigma^z_i = \mu^{x}_{s_{+}(i)}\mu^{x}_{s_{-}(i)}$

iii) Further define $\Pi_{i \ \in \ s}\sigma^x_i = \mu^z_{s}, \ \ \ \Pi_{i \ \in \ p}\sigma^z_i = \mu^x_{p}$

iv) Thus the Hamiltonian in $(1)$ becomes: {written termwise}

$$H = -J_s \sum_{s} \mu^z_s - J_p \sum_{p}\mu^x_{p} - \lambda_{x}\sum_{\langle ij \rangle} \mu^{z}_{p(i)}\mu^{z}_{p(j)} - \lambda_{z}\sum_{\langle ij \rangle} \mu^{x}_{s(i)}\mu^{x}_{s(j)} \\ = -J_s \sum_{s} \mu^z_s - J_p \sum_{p}\mu^x_{p} - \lambda_{x}\sum_{\langle p \ p' \rangle} \mu^{z}_{p}\mu^{z}_{p'} - \lambda_{z}\sum_{\langle s \ s' \rangle} \mu^{x}_{s}\mu^{x}_{s'} \tag 2$$

  • Is this correct? [the most prior part of the question]. If so, the standard Transverse Field Ising model [with effectibe spins residing on plaquttes and sites] has one nearest neighbour interaction term for either $x$ or $z$ spin variable, but this Hamiltonian $(2)$, has both $\lambda_{x}\sum_{\langle p \ p' \rangle} \mu^{z}_{p}\mu^{z}_{p'}$ and $\lambda_{z}\sum_{\langle s \ s' \rangle} \mu^{x}_{s}\mu^{x}_{s'}$.

  • Can we still call it Transverse Field Ising model?

  • What are the differences in the Spectrum (ground state + Elementary Excitations) between standard 2D TFIM and this Hamiltonian $(2)$, and how to obtain them. [pointing to the literature would be fine enough.]

Question 2: In the literature I have found expressions for series expansion of ground state energy per spin and the dispersion relation for elementary excitations (the Gap for lowest lying eigenstate after Ground State), vanishing of the latter determining the critical point where mutual-anyonic statistics fails to persist under a certain strength of External Field. They have either used Linked Cluster Method, or Perturbative Continuous Unitary Transform to map the Hamiltonian in an effective Hamiltonian [for (1)], but the result of series expansions are provided using computer programs.

  • Is there a way to compute this by hand?
  • Please direct towards suitable literature where the technique is explicitly demonstrated. [2nd most prior question]
  • $\begingroup$ Please focus on one question. $\endgroup$ May 3, 2023 at 12:42
  • $\begingroup$ That's why I have ordered them by priority. $\endgroup$ May 4, 2023 at 5:33
  • $\begingroup$ That's not the same. What if someone only knows the answer to one of the questions - does this count as an answer? $\endgroup$ May 4, 2023 at 9:54
  • 1
    $\begingroup$ If this could be computed by hand, why would people have used cluster expansions? (Though I believe that cluster expansions are "by hand", they are just not exact.) $\endgroup$ May 4, 2023 at 22:16
  • $\begingroup$ I had this exact confusion. Makes sense. $\endgroup$ May 6, 2023 at 5:09

2 Answers 2


For the Hamiltonian you cited, the toric code is only dual to the 2d transverse field Ising model (TFIM) in the limit $J_s \to \infty$. Otherwise, they are not dual. The reason is simple: under the Kramers-Wannier duality (more correctly here, the Wegner duality), the 2d transverse field Ising model is dual to the 2d $\mathbb{Z}_2$ lattice gauge theory (LGT) without dynamical matter. The intuition here is that you should think of the domain walls in the ferromagnetic phase of the TFIM as the electric field lines in the confined phase of the LGT, while the visons of the LGT in the deconfined phase correspond to the spin-flip excitations in the paramagnetic phase of the TFIM. The fact that there is no electric charges in the LGT system implies, by the $\mathbb{Z}_2$ version of Gauss's law, that the electric field lines do not terminate in space; this is to say, the domain walls in the TFIM must form closed loops. (I'm only providing the intuition here, but you can find more explicit details on the duality in, for example, McGreevy's notes.)

Once you add dynamical matter, electric field lines can end on charges, and the direct equivalence to the TFIM is broken. Nevertheless, in the paper you referenced, the perturbation added by the authors commutes with the $A_s$ terms. You can therefore fix $A_s = +1$ in the ground state and focus on the $B_p$ excitations, effectively freezing out the matter degrees of freedom in the equivalent LGT. This problem then becomes dual to a 2d TFIM once again.

(Finally, a small word of caution: I have been using the word 'dual' instead of 'equivalent', because strictly speaking, the spectra of the TFIM and the LGT do not match up one-to-one. This is due to global issues in the equivalence, and is quite common among these sorts of dualities. But all the features that depend on local properties, namely the excitation spectra and phase transitions, are equivalent between the two.)

  • $\begingroup$ In the context of the question, it seems more relevant when there is an equivalence of the ground states (topo order is really a property of the state, not of the Hamiltonian). In that case, $J_s\to\infty$ is not required, but one must restrict to either an X or a Z field. $\endgroup$ May 3, 2023 at 12:44
  • $\begingroup$ These two limits give the same ground state, right? $J_s \to \infty$ and $\lambda_Z \to 0$ both result in $A_s = 1$ in the ground state. $\endgroup$
    – Zack
    May 3, 2023 at 14:38
  • $\begingroup$ this clarifies a bit. But I am interested in computing excitation spectra and locate the transitions, and also the point where the anyonic excitations fail to exist but the charge and fluxes remain. Can you refer me to suitable texts where the the computation (possibly for different model) is demonstrated. $\endgroup$ May 4, 2023 at 5:38
  • $\begingroup$ @prikarsartam The model including its excitations can only be mapped to a 2D transverse field Ising model if you only look at one field (X or Z); then it will map to an Ising model, possibly with some twists for the excited sectors. $\endgroup$ May 4, 2023 at 9:57
  • 2
    $\begingroup$ If you're looking for a calculation giving the exact location of the transition, then I don't believe it can be done, for the same reason that the 3d classical Ising model's exact transition location can't be located analytically. The best you can do is perturbative arguments and numerics. $\endgroup$
    – Zack
    May 4, 2023 at 20:16

One more interesting thing about this, is that -

If we only allow Gauge Invariant[*] perturbations, then $-\lambda_z \sum_{edge} \sigma_i^{z} $ term drops out since, the Gauge Transformation:-$$\mathcal{G}(i): \{\sigma^z, \sigma^z, \sigma^z, \sigma^z \}_{around \ site(i)} \to \{-\sigma^z, -\sigma^z, -\sigma^z, -\sigma^z \}_{around \ site(i)} \ \forall \ site \\ \mathcal{G}(i):\{\sigma^x\} \to \{\sigma^x\} $$ Makes $\langle \sigma_z \rangle = 0$ due to Elitzur's Theorem, and only $ -\lambda_x \sum_{edge} \sigma_i^{x}$ is a valid perturbation. In that case $\lambda_z=0$, which in the Hamilton $(2)$ of op removes the $- \lambda_{z}\sum_{\langle s \ s' \rangle} \mu^{x}_{s}\mu^{x}_{s'}$ term, and we will have -

$$H = -J_s \sum_{s} \mu^z_s - J_p \sum_{p}\mu^x_{p} - \lambda_{x}\sum_{\langle p \ p' \rangle} \mu^{z}_{p}\mu^{z}_{p'} \tag 3$$

Which exactly is the 2d Transverse Field Ising model.

[*] The perturbations being Gauge invariant, means they are observables, and probably that's why the true robust nature of the topological order, against local OBSERVABLE(=physical) perturbations, can be demonstrated by analysing this model $(1)$. But this part is a plausible speculation for why it is the case - that uniform gauge invariant perturbations in the toric code model, makes it a 2d transverse field ising model. Nevertheless the statement is true, if $(2)$ is correct.

Yet there are issues, since $\langle \sigma^z \rangle = 0$ only in the exact ground state of the system, since whenever there are some excitation, say two electric charges, characterised by 2 $A_p$ operators being -1 at the end of an open string - the gauge symmetry anyhow is broken and Elitzur's theorem cannot vanish $-\lambda_z \sum_{edge} \sigma_i^{z} $ and it contributes - and we don't get exact (3) but (2).


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