The Amorphous Kitaev Model - Introduction part 2

Gauge Fields

The bond operators $u_{ij}$ are useful because they label a bond sector $\mathcal{\tilde{L}}_u$ in which we can easiy solve the Hamiltonian. However the gauge operators move us between bond sectors. Bond sectors are not gauge invariant!

Let’s consider instead the properties of the plaquette operators $\hat{\phi}_i$ that live on the faces of the lattice.

We already showed that they are conserved. And as one might hope and expect, the plaquette operators map cleanly on to the bond operators of the Majorana representation:

\[\begin{aligned} \tilde{W}_p &= \prod_{\mathrm{i,j}\; \in\; p} \tilde{K}_{ij}\\ &= \prod_{\mathrm{i,j}\; \in\; p} \tilde{\sigma}_i^\alpha \tilde{\sigma}_j^\alpha\\ &= \prod_{\mathrm{i,j}\; \in\; p} (ib^\alpha_i c_i)(ib^\alpha_j c_j)\\ &= \prod_{\mathrm{i,j}\; \in\; p} i u_{ij} c_i c_j\\ &= \prod_{\mathrm{i,j}\; \in\; p} i u_{ij} \end{aligned}\]

Where the last steps holds because each $c_i$ appears exactly twice and adjacent to its neighbour in each plaquette operator. Note that this is consistent with the observation from earlier that each $W_p$ takes values $\pm 1$ for even paths and $\pm i$ for odd paths.

Vortices and their movements

Let’s imagine we started from the ground state of the model and flipped the sign of a single bond. In doing so we will flip the sign of the two plaquettes adjacent to that bond. I’ll call these disturbed plaquettes vortices. I’ll refer to a particular choice values for the plaquette operators as a vortex sector.

If we chain multiple bond flips we can create a pair of vortices at arbitrary locations. The chain of bonds that we must flip corresponds to a path on the dual of the lattice.

Something else we can do is create a pair of vortices, move one around a loop and then anhilate it with its partner. This corresponds to a closed loop on the dual lattice and applying such a bond flip leaves the vortex sector unchanged.

Notice that the $D_j$ operators flip three bonds around a vertex. This is the smallest closed loop around which one can move a vortex pair and anhilate it with itself.

Such operations compose in the sense that we can build any larger loop by applying a series of $D_j$ operations. Indeed the symetrisation procedure $\prod_i \left( \frac{1 + D_i}{2}\right)$ that maps from the bond sector to a physical state is applying constructing a superposition over every such loop that leaves the vortex sector unchanged.

The only loops that we cannot build out of $D_j$s are non-contractible loops, such as those that span the major or minor circumference of the torus.

The plaquette operators are the gauge invariant quantity that determines the physics of the model

Composition of $u_{jk}$ loops

Figure 1: In the product of individual plaquette operators shared bonds cancel out. The product is equal to the enclosing path.

Second it is now easy to show that the loops and plaquettes satisfy nice composition rules, so long as we stick to loops that wind in a particular direction.

Consider the product of two non-overlapping loops $W_a$ and $W_b$ that share an edge $u_{12}$. Since the two loops both wind clockwise and do not overlap, one will contain a term $i u_{12}$ and the other $i u_{21}$. Since the $u_{ij}$ commute with one another, they square to $1$ and $u_{ij} = -u_{ji}$ we see have $i u_{12} i u_{21} = 1$ and we can repeat this for any number of shared edges. Hence, we get a version of Stokes’ theorem: the product of $i u_{jk}$ around any closed loop $\partial A$ is equal to the product of plaquette operators $\Phi$ that span the area $A$ enclosed by that loop: \[\prod_{u_{jk} \in \partial A} i \; u_{jk} = \prod_{\phi_i \in A} \phi_i\]

Figure 2: The loop composition rule extends to arbitrary numbers of vortices giving a discrete version of Stoke’s theorem.

Wilson loops can always be decomposed into products of plaquettes operators unless they are non-contractable

Gauge Degeneracy and the Euler Equation

We can check this analysis with a counting argument. For a lattice with $B$ bonds, $P$ plaquettes and $V$ vertices we can count how many bond sectors, vortices sectors and gauge symmetries there are and check them against Euler’s polyhedra equation.

Euler’s equation states for a closed surface of genus $g$, i.e that has $g$ holes so $0$ for the sphere, $1$ for the torus and $g$ for $g$ tori stuck together \[B = P + V + 2 - 2g\]

Figure 3: In periodic boundary conditions the Kitaev model is defined on the surface of a torus. Topologically the torus is distinct from the sphere in that it has a hole that cannot be smoothly deformed away. Associated with each such hole are two non-contractible loops on the surface, here labeled A and B, that cannot be smoothly deformed to a point. These two non-contracible loops can. be used to construct two symmetry operators \hat{A} and \hat{A} that flip u_{jk}s along their paths. — Figure 3: In periodic boundary conditions the Kitaev model is defined on the surface of a torus. Topologically the torus is distinct from the sphere in that it has a hole that cannot be smoothly deformed away. Associated with each such hole are two non-contractible loops on the surface, here labeled A and B, that cannot be smoothly deformed to a point. These two non-contracible loops can. be used to construct two symmetry operators $\hat{A}$ and $\hat{A}$ that flip $u_{jk}$s along their paths.

For the case of the torus where $g = 1$ we can rearrange this to read: \[B = (P-1) + (V-1) + 2\]

Each $u_{ij}$ takes two values and there is one associated with each bond so there are exactly $2^B$ distinct configurations of the bond sector. Let’s see if we can factor those configurations out into the cartesian product of vortex sectors, gauge symmetries and non-contractible loop operators.

Vortex sectors: each plaquette operator $\phi_i$ takes two values ($\pm 1$ or $\pm i$) and there are $P$ of them so naively one would think there are $2^P$. However vortices can only be created on pairs so there are really $\tfrac{2^P}{2} = 2^{P-1}$ vortex sectors.

Gauge symmetries: As discussed earlier these correspond to the all possible compositions of the $D_j$ operators. Again there are only $2^{V-1}$ of these because, as we will see in the next section, $\prod_{j} D_j = \mathbb{1}$ in the physical space, and we enforce this by chooising the correct product of single particle fermion states. You can get an intuitive picture for why $\prod_{j} D_j = \mathbb{1}$ by imagining larger and larger patches of $D_j$ operators on the torus. These patches correspond to transporting a vortex pair around the edge of the patch. At some point the patch wraps around and starts to cover the entire torus, as this happens the bounday of the patch disappears and hence it which corresponds to the identity operation. See Fig ?? (animated in the HTML version).

Finally the torus has two non-contractible loop operators asscociated with its major and minor diameters.

Putting this all together we see that there are $2^B$ bond sectors a space which can be decomposed into the cartesian product of $2^{P-1}$ vortex sectors, $2^{V-1}$ gauge symmetries and $2^2 = 4$ topological sectors associated with the non-contractible loop operators. This last factor forms the basis of proposals to construct topologically protected qubits since the 4 sectors cold only be mixed by a highly non-local perturbation, ref ?????.

$A honeycomb lattice with edges in light grey, along with its dual, the triangle lattice in light blue. The vertices of the dual lattice are the faces of the original lattice and hence are the locations of the vortices. (Left) The action of the gauge operator D_j at a vertex is to flip the value of the three u_{jk} variables (black lines) surounding site j. The corresponding edges of the dual lattice (blue lines) form a closed triangle. (middle) Composing multiple adjacent D_j operators produces a large closed loop or multiple disconnected loops. These loops are not directed as they are in the case of the Wilson loops. (right) A non-contractable loop which cannot be produced by composing D_j operators. All three operators can be thought of as the action of a vortex-vortex pair that is created, one of them is transported around the loop and then the two anhilate again. Note that every plaquette has an even number of u_{ij}s flipped on it’s edge and hence all retain the same value. This all works the same way for the amorphous lattice but is much harder to read visually.$ $In both figures a honeycomb lattice is shown in grey along with its dual in light blue. (Left) Taking a larger and larger set of D_j operators leads to an outward expanding boundary line shown in blue on the dual lattice. Eventually every lattice on the torus is included and the boundary dissapears. This is a visual proof that \prod_i D_i = \mathbb{1}. (Right) In black and blue the edges and dual edges that must be flipped to add vortices at the sites highlighted in orange. Flipping all the plaquettes in the system is not equivalent to the identity.$

Counting edges, plaquettes and vertices

It will be useful to know how the trivalent structre of the lattice constraints the number of bonds $B$, plaquettes $P$ and vertices $V$ it has.

We can immediately see that the lattice is built from vertices that each share 3 edges with their neighbours. This means each vertex comes with $\tfrac{3}{2}$ bonds i.e $3V = 2B$. This is consistent with the fact that in the Majorana representation on the torus each vertex brings three $b^\alpha$ operators which then pair along bonds to give $3/2$ bonds per vertex.

If we define an integer $N$ such that $V = 2N$ and $B = 3N$ and substitite this into the polyhedra equation for the torus we see that $P = N$. So if is a trivalent lattice on the torus has $N$ plaquettes, it has $2N$ vertices and $3N$ bonds.

We can also consider the sum of the number of bonds in each plaquette $S_p$, since each bond is a member of exactly two plaquettes \[S_p = 2B = 6N\]

The mean size of a plaquette in a trivalent lattice on the torus is exactly 6. Since the sum is even, this also tells us that all odd plaquettes must come in pairs.

Figure 4: The relationship between the different Hilbert spaces used in the solution is slightly complex.

The Projector

It will turn out that the projection from the extended space to the physical space is not actually that important for the results that I will present. However it it useful to go through the theory of it to explain why this is.

The physicil states are defined as those for which $D_i |\phi\rangle = |\phi\rangle$ for all $D_i$. Since $D_i$ has eigenvalues $\pm1$, the quantity $\tfrac{(1+D_i)}{2}$ has eigenvalue $1$ for physical states and $0$ for extended states so is the local projector onto the physical subspace.

The global projector is therefore \[ \mathcal{P} = \prod_{i=1}^{2N} \left( \frac{1 + D_i}{2}\right)\]

for a toroidal trivalent lattice with $N$ plaquettes $2N$ vertices and $3N$ edges. As I pointed out before the product over $(1 + D_j)$ can also be thought of as the sum of all possible subsets $\{i\}$ of the $D_j$ operators, which is the set of all possible gauge symmetry operations.

\[ \mathcal{P} = \frac{1}{2^{2N}} \sum_{\{i\}} \prod_{i\in\{i\}} D_i\]

Since the gauge operators $D_j$ commute and square to one, we can define the complement operator $C = \prod_{i=1}^{2N} D_i$ and see that it take each set of $\prod_{i \in \{i\}} D_j$ operators and gives us the complement of that set. I said earlier that $C$ is the identity in the physical subspace and we will shortly see why.

We use the complement operator to rewrite the projector as a sum over half the subsets of $\{i\}$ let’s call that $\Lambda$. The complement operator deals with the other half

\[ \mathcal{P} = \left( \frac{1}{2^{2N-1}} \sum_{\Lambda} \prod_{i\in\{i\}} D_i\right) \left(\frac{1 + \prod_i^{2N} D_i}{2}\right) = \mathcal{S} \cdot \mathcal{P}_0\]

To compute $\mathcal{P}_0$ the main quantity needed is the product of the local projectors $D_i$ \[\prod_i^{2N} D_i = \prod_i^{2N} b^x_i b^y_i b^z_i c_i \] for a toroidal trivalent lattice with $N$ plaquettes $2N$ vertices and $3N$ edges.

First we reorder the operators by bond type, this doesn’t require any information about the underlying lattice.

\[\prod_i^{2N} D_i = \prod_i^{2N} b^x_i \prod_i^{2N} b^y_i \prod_i^{2N} b^z_i \prod_i^{2N} c_i\]

The product over $c_i$ operators reduces to a determinant of the Q matrix and the fermion parity, see¹. The only difference from the honeycomb case is that we cannot explicitely compute the factors $p_x,p_y,p_z = \pm\;1$ that arise from reordering the b operators such that pairs of vertices linked by the corresponding bonds are adjacent.

\[\prod_i^{2N} b^\alpha_i = p_\alpha \prod_{(i,j)}b^\alpha_i b^\alpha_j\]

However they are simply the parity of the permutation from one ordering to the other and can be computed in linear time with a cycle decomposition^{app:cycle_decomp?}.

We find that \[\mathcal{P}_0 = 1 + p_x\;p_y\;p_z\; \hat{\pi} \; \mathrm{det}(Q^u) \; \prod_{\{i,j\}} -iu_{ij}\]

where $p_x\;p_y\;p_z = \pm 1$ are lattice structure factors. $det(Q^u)$ is the determinant of the matrix mentioned earlier that maps $c_i$ operators to normal mode operators $b'_i, b''_i$. These depend only on the lattice structure.

$\hat{\pi} = \prod{i}^{N} (1 - 2\hat{n}_i)$ is the parity of the particular many body state determined by fermionic occupation numbers $n_i$. As discussed in +¹ is $\hat{\pi}$ is gauge invariant in the sense that $[\hat{\pi}, D_i] = 0$.

This implies that $det(Q^u) \prod -i u_{ij}$ is also a guage invariant quantity. In translation invariant models this quantity which can be related to the parity of the number of vortex pairs in the system². However it is not so simple to evaluate in the amorphous case.

More general arguments^3,4 imply that $det(Q^u) \prod -i u_{ij}$ has an interesting relationship to the topological fluxes. In the non-Abelian phase we expect that it will change sign in exactly on of the four topological sectors. This forces that sector that contain a fermion and hence gives the model a three-fold degerenate ground state. In the Abelian phase this doesn’t happen and we get a fourfold degerate ground state. Whether this analysis generalises to the amorphous case in unclear.

An alternate way to view this is to consider the adiabatic insertion of the fluxes $\Phi_{x,y}$ as the operations that undo vortex transport around the lattice. In this picture the three fold degeneracy occurs because transporting a vortex around both the major and minor axes of the torus changes its fusion channel such that the two vortices fuse into a fermion excition rather than the vacuum.

All these factors take values $\pm 1$ so $\mathcal{P}_0$ is 0 or 1 for a particular state. Since $\mathcal{S}$ corresponds to symmetrising over all the gauge configurations and cannot be 0, this tells use that once we have determined the single particle eigenstates of a bond sector, the true many body ground state has the same energy as either the empty state with $n_i = 0$ or a state with a single fermion in the lowest level.

Let’s think about where are with the model now. We can map the spin Hamiltonian to a Majorana Hamiltonian in an extended Hilbert space. Along with that mapping comes a gauge field $u_{jk}$ defining bond sectors. The gauge symmetries of $u_{jk}$ are generated by the set of $D_j$ operators. The gauge invariant and therefore physically relevant variables are the plaquette operators $\phi_i$ which define as a vortex sector. In order to practically solve the Majorana Hamiltonian we must remove hats from the gauge field by restricting ourselves to a particular bond sector. From there the Majorana Hamiltonian becomes non-interacting and we can solve it like any quadratic theory. This lets us construct the single particle eigenstates from which we can also construct many body states. However the many body states constructed this way are not in the physical subspace!

However for the many body states within a particular bond sector, $\mathcal{P}_0 = 0,1$ tells us which of those have some overlap with the physical sector.

We see that finding a state that has overlap with a physical state only ever requires the addition or removal of one fermion. There are cases where this can make a difference but for most observables such as ground state energy this correction scales away as the number of fermions in the system grows.

If we wanted to construct a full many body wavefunction in the spin basis we would need to include the full symmetrisation over the gauge fields. However this was not necessary for any of the results that will be presented here.

The Ground State

As we have shown that the Hamiltonian is gauge invariant, only the flux sector and the two topological fluxes affect the spectrum of the Hamiltonian. Thus we can label many body ground state by a combination of flux sector and fermionic occupation numbers.

By studying the projector we saw that the fermionic occupation numbers of the ground state will always be either $n_m = 0$ or $n_0 = 1, n_{m>1} = 0$ because the projector really just enforces vortex and fermion parity.

I refer to the flux sector that contains the ground state as the ground state flux sector. Recall that we call the excitations of the fluxes away from the ground ground state configuration vortices, so that the ground state flux sector is the vortex free sector by definition.

On the Honeycomb, Lieb’s theorem implies that the the ground state corresponds to the state where all $u_{jk} = 1$ implying that the flux free sector is the ground state sector⁵.

Lieb’s theorem does not generalise easily to the amorphous case. However we can get some intuition by examining the problem that will lead to a guess for the ground state. We will then provide numerical evidence that this guess is in fact correct.

Let’s consider the partition function of the Majorana hamiltonian: \[ \mathcal{Z} = \mathrm{Tr}\left( e^{-\beta H}\right) = \sum_i \exp{-\beta \epsilon_i}\] At low temperatures $\mathcal{Z} \approx \beta \epsilon_0$ where $\epsilon_0$ is the lowest energy fermionic state.

How does the $\mathcal{Z}$ depend on the Majorana hamiltonian? Expanding the exponential out gives: \[ \mathcal{Z} = \sum_n \frac{(-\beta)^n}{n!} \mathrm{Tr(H^k)} \]

Now there’s an interesting observation to make here. The Hamiltonian is essentially a scaled adjacency matrix. An adjacency matrix being a matrix $g_{ij}$ such that $g_{ij} = 1$ if vertices $i$ and $j$ and joined by an edge and 0 otherwise.

Powers of adjacency matrices have the property that the entry $(g^n)_{ij}$ corresponds to the number of paths of length n on the graph that begin at site $i$ and end at site $j$. These include somewhat degenerate paths that go back on themselves etc.

The trace of an adjacency matrix \[\mathrm{Tr}(g^n) = \sum_i (g^n)_{ii}\] therefore counts the number number of loops of size $n$ that can be drawn on the graph.

Applying the same treatment to our Majorana Hamiltonian, we can interpret $u_ij$ to equal 0 if the two sites are not joined by a bond and we put ourselves in the isotropic phase where $J^\alpha = 1$ \[ \tilde{H}_{ij} = \tfrac{1}{2} i u_{ij}\]

We then see that the trace of the nth power of H is a sums over Wilson loops of size $n$ with an additional factor of $2^{-n}$. We showed earlier that the Wilson loop operators can always be written as products of the plaquette operators that they enclose.

Lumping all the prefactors together, we can write: \[ \mathcal{Z} = c_A \hat{A} + c_B \hat{B} + \sum_i c_i \hat{\phi}_i + \sum_{ij} c_{ij} \hat{\phi}_i \hat{\phi}_j + \sum_{ijk} c_{ijk} \hat{\phi}_i \hat{\phi}_j \hat{\phi}_k + ...\]

Where the $c$ factors would be something like \[c_{ijk...} = \sum_n \tfrac{(-\beta)^n}{n!} \tfrac{1}{2^n} K_{ijk...}\] which is a sum over all loop lengths $n$ and for each we have a combinatoral factor $K_{ijk...}$ that counts how many ways there are to draw a loop of length $n$ that only encloses plaquettes $ijk...$.

We also have the pesky non-contractible loop operators $\hat{A}$ and $\hat{B}$. Again the prefactors for these are very complicated but we can intuitively see that for larger and larger loops lengths there will be a combinatorial explosion of possible ways that they appear in these sums. These are suppressed exponentially with system size but at practical lattice sizes they cause significant finite size effects. The main evidence of this is that the 4 loop sectors spanned by the $\hat{A}$ and $\hat{B}$ operators are degenerate in the infinite system size limit, while that degeneracy is lifted in finite sized systems.

We don’t have much hope of actually evaluating this for an amorphous lattice. However it lead us to guess that the ground state vortex sector might be a simple function of the side length of each plaquette.

The ground state of the Amorphous Kitaev Model is found by setting the flux through each plaquette $\phi$ to be equal to $\phi^{\mathrm{g.s.}}(n_{\mathrm{sides}})$

\[\begin{aligned} \phi^{\mathrm{g.s.}}(n_{\mathrm{sides}}) = -(\pm i)^{n_{\mathrm{sides}}}, \end{aligned}\] where $n_{\mathrm{sides}}$ is the number of edges that form each plaquette and the choice of sign gives a twofold chiral ground state degeneracy.

This conjecture is consistent with Lieb’s theorem on regular lattices⁵ and is supported by numerical evidence. As noted before, any flux that differs from the ground state is an excitation which I call a vortex.

Finite size effects

This guess only works for larger lattices because of the finite size effects. In order to rigorously test it we would like to directly enumerate the $2^N$ vortex sectors for a smaller lattice and check that the lowest state found is the vortex sector predicted by ???.

To do this we tile use an amorphous lattice as the unit cell of a periodic $N\times N$ system. Bonds that originally crossed the periodic boundaries now connect adjacent unit cells. Using Bloch’s theorem the problem then essnetially reduces back to the single amorphous unit cell but now the edges that cross the periodic boundaries pick up a phase dependent on the crystal momentum $\vex{q} = (q_x, q_y)$ and the lattice vector of the bond $\vec{x} = (+1, 0, -1, +1, 0, -1)$. Assigning these lattice vectors to each bond is also a very conveninent way to store and plot toroidal graphs.

This can then be solved using Bloch’s theorem. For a given crystal momentum $\textbf{q} \in [0,2\pi)^2$, we are left with a Bloch Hamiltonian, which is identical to the original Hamiltonian aside from an extra phase on edges that cross the periodic boundaries in the $x$ and $y$ directions, \[\begin{aligned} M_{jk}(\textbf{q}) = \frac{i}{2} J^{\alpha} u_{jk} e^{i q_{jk}},\end{aligned}\] where $q_{jk} = q_x$ for a bond that crosses the $x$-periodic boundary in the positive direction, with the analogous definition for $y$-crossing bonds. We also have $q_{jk} = -q_{kj}$. Finally $q_{jk} = 0$ if the edge does not cross any boundaries at all – in essence we are imposing twisted boundary conditions on our system. The total energy of the tiled system can be calculated by summing the energy of $M( \textbf{q})$ for every value of $\textbf{q}$. The use of a fourier series then allows us to compute the diagonalisation with a penalty only linear in the number of tiles used compared to diagonalising a single lattice. With this technique the finite size effects related to the non-contractible loop operators are removed with only a linear penalty in computation time compared to the exponential penalty paid by simply simply diagonalising larger lattices.

Using this technique we verified that $\phi_0$ correctly predicts the ground state for hundreds of thousands of lattices with upto 20 plaquettes. For larger lattices we verified that random perturbations around the predicted ground state never yield a lower energy state.

Chiral Symmetry

In the discussion above we see that the ground state has a twofold chiral degeneracy that comes about because the global sign of the odd plaquettes does not matter.

This happens because by adding odd plaquettes we have broken the time reversal symmetry of the original model^6–13.

Similar to the behaviour of the original Kitaev model in response to a magnetic field, we get two degenerate ground states of different handedness. Practicaly speaking, one ground state is related to the other by inverting the imaginary $\phi$ fluxes⁷.

Phases of the Kitaev Model

discuss the abelian A phase / toric code phase / anisotropic phase

the isotropic gapless phase of the standard model

The isotropic gapped phase with the addition of a magnetic field

What’s so great about two dimensions?

Topology, chirality and edge modes

Most thermodynamic and quantum phases studied can be characterised by a local order parameter. That is, a function or operator that only requires knowledge about some fixed sized patch of the system that does not scale with system size.

However there are quantum phases that cannot be characterised by such a local order parameter. These phases are intead said to posess ‘topological order’.

One property of topological order that is particularly easy to observe that the ground state degeneracy depends on the topology of the manifold that we put the system on to. This is referred to as topological degeneracy to distinguish it from standard symmetry breaking.

The Kitaev model will be a good example of this, we have already looked at it defined on a graph that is embedded either into the plane or onto the torus. The extension to surfaces like the torus but with more than one handle is relatively easy.

Anyonic Statistics

In dimensions greater than two, the quantum state of a system must pick up a factor of $-1$ or $+1$ if two identical particles are swapped. We call these Fermions and Bosons.

This argument is predicated on the idea that performing two swaps is equivalent to doing nothing. Doing nothing should not change the quantum state at all, so doing one swap can at most multiply it by $\pm 1$.

However there are many hidden parts to this argument. Firstly, this argument just isn’t the whole story, if you want to know why Fermions have half integer spin, for instance, you have to go to field theory.

There is also a second niggle, why does this argument only work in dimensions greater than two? What we’re really saying when we say that two swaps do nothing is that the world lines of two particles that have been swapped twice can be untangled without crossing. Why can’t they cross? Well because if they cross then the particles can interact and the quantum state could change in an arbitrary way. We’re implcitly using the locality of physics here to argue that if the worldlines stay well separated then the overall quantum state cannot too much.

In two dimensions we cannot untangle the worldlines of two particles that have swapped places, they are braided together. See fig. 5 for a diagram.

From this fact flows a whole new world of behaviours, now the quantum state can aquire a phase factor $e^{i\phi}$ upon exchange of two identical particles, which we now call Anyons.

The Kitaev Model is a good demonstration of the connection beween Anyons and topological degeneracy. In the Kitaev model we can create a pair of vortices, move one around a non-contractable loop $\mathcal{T}_{x/y}$ and then anhilate them together. Without topology this should leave the quantum state unchanged. Instead it moves us to another ground state in a topologically degenerate ground state subspace. Practically speaking it flips a dual line of bonds $u_{jk}$ going around the loop which we cannot undo with any gauge transformation made from $D_j$ operators.

If the ground state subspace is multidimensional, quasiparticle exchange can move us around in the space with an action corresponding to a matrix. These matrices do not in general commmute and so these are known as non-Abelian anyons.

From here things get even more complex, the Kitaev model has a non-Abelian phase when exposed to a magnetic field, and the amorphous Kitaev Model has a non-Abelian phase because of its broken chiral symmetry.

The way that we have subdivided the Kitaev model into vortex sectors, we have a neat separation beween vortices and fermionic excitations. However if we looked at the full many body picture we would see that a vortex caries with it a cloud of bound majorana states.

Figure 6: (Left) A large amorphous lattice in the ground state save for a single pair of vortices shown in red, separated by the string of bonds that we flipped to create them. (Right) The density of the lowest energy Majorana state in this vortex sector. The state is clearly bound to the vortices.

Consider two processes

We transport one half of a vortex pair around either the x or y loops of the torus before anhilating back to the ground state vortex sector $\mathcal{T}_{x,y}$.
We flip a line of bond operators coresponding to measuring the flux through either the major or minor axes of the torus $\mathcal{\Phi}_{x,y}$

Figure 7: (Left) The two topological flux operators of the toroidal lattice, these don’t correspond to any face of the lattice, but rather measure flux that threads through the major and minor axes of the torus. This shows a particular choice but any loop that crosses the boundary is gauge equivalent to one of or the sum of these two loop. (Right) The two ways to transport vortices around the diameters. These correspond to creating a vortex pair, transporting one of them around the major or minor diameters of the torus and then anhilating them again.

The plaquette operators $\phi_i$ are associated with fluxes. Wilson loops that wind the torus are associated with the fluxes through its two diameters $\mathcal{\Phi}_{x,y}$.

In the Abelian phase we can move a vortex along any path we like and then when we bring them back together they will anhilate back to the vacuum, where we understand ‘the vacuum’ to refer to one of the ground states, though not necesarily the same one we started in. We can use this to get from the $(\Phi_x, \Phi_y) = (+1, +1)$ ground state and construct the set $(+1, +1), (+1, -1), (-1, +1), (-1, -1)$.

Figure 8: Wilson loops that wind the major or minor diameters of the torus measure flux winding through the hole of the donut/torus or through the filling. If they made donuts that had both a jam filling and a hole this analogy would be a lot easier to make14. — Figure 8: Wilson loops that wind the major or minor diameters of the torus measure flux winding through the hole of the donut/torus or through the filling. If they made donuts that had both a jam filling and a hole this analogy would be a lot easier to make¹⁴.

However in the non-Abelian phase we have to wrangle with monodromy^3,4. Monodromy is behaviour of objects as they move around a singularity. This manifests here in that the identity of a vortex and cloud of Majoranas can change as we wind them around the torus in such a way that rather than anhilating to the vacuum the anhilate to create an excited state instead of a ground state. This means we end up with only three degenerate ground states in the non-Abelian phase $(+1, +1), (+1, -1), (-1, +1)$¹⁵. The way that this shows up concretly is that the projector enforces both flux and fermion parity. When we wind a vortex around both non-contractible loops of the torus, it flips the flux parity which forces means we have to introduce a fermionic excitation to make the state physical. Hence the process does not give a fourth ground state.

$(Bond Sector) A state in the bond sector is specified by assigning \pm 1 to each edge of the lattice. However this description has a substantial gauge degeneracy. We can simplfy things by decomposing each state into the product of three kinds of objects: (Vortex Sector) Only a small number of bonds need to be flipped (compared to some arbitrary reference) to reconstruct the vortex sector. The edges here are chosen from a spanning tree of the dual lattice, so there are no loops. (Gauge Field) The ‘loopiness’ of the bond sector can be factored out giving a network of loops that can always be written as a product the of the gauge operators D_j. (Topolical Sector) Finally there are two loops that have no effect on the vortex sector, nor can they be constructed from gauge symmetries. These can be thought of as two fluxes \Phi_{x/y} that thread through the major and minor axes of the torus. Measuring \Phi_{x/y} corresponds to constructing Wilson loops around the axes of the torus. We can flip the value of \Phi_{x} by transporting a vortex pair around the torus in the y direction and that is what is shown here. In each of the three figures on the right, black bonds correspond to those that must be flipped, composing the three together gives back the original bond sector on the left.$

One reason the topology has gained interest recently is there have proposals to use this ground state degeneracy to implement both passively fault tolerant and actively stabilised quantum computations [¹⁶;¹⁷; hastingsDynamicallyGeneratedLogical2021].

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