Boson condensation and instability in the tensor network representation of string-net states

This is the simplest TNR of the toric code state. We first copy each computational basis into two, as shown in the Fig. 2. That is, the labels $0$ and $1$ on every edge become $00$ and $11$ on the same edge. Now the local Hilbert space neighbouring each vertex is made out of three qubits. We associate a tensor with three physical indices/legs (throughout the paper we will use “indices” and “legs” interchangeably), and three virtual indices/legs to each vertex, represented algebraically as $(T^{0})^{ijk}_{\alpha\beta\gamma}$ where $i,j,k$ are the three physical indices and $\alpha,\beta,\gamma$ are the three virtual indices, as shown in Fig. 2. The components of the tensor are

where $\delta$ is the kronecker delta function. So, physical and virtual legs are identified and an even number of indices carry label $1$ out of every three edges neighbouring a vertex, i.e., we satisfy the vertex condition. The plaquette condition is also satisfied since every configuration is of equal weight. Therefore, the tensor network state constructed using the above local tensor leads to the toric code ground state given in Eq. (3).

It was shown by Chen et al. (2010) that single-line TNR of the toric code state is not stable in certain directions of variation. Before we explain what these unstable directions of variation are, we first note that the single-line TNR explained above has a virtual symmetry. If an operation on the virtual indices leaves the tensor invariant, we will call it a virtual symmetry of the tensor. Because the single-line tensor is non-zero only when virtual legs have even number of 1s, it has a natural $Z\otimes Z\otimes Z$ virtual symmetry (see Schuch et al. (2010) for TNR virtual symmetries of the quantum double models). That is, the tensor in (4) satisfies the relation,

$\displaystyle\centering\includegraphics[scale={0.3}]{SLTCsym}.\@add@centering$

(5)

where we have omitted the physical legs for visual clarity. (We would often omit physical legs from tensor diagrams throughout the paper when we are mostly concerned with the virtual space/indices.) It is a $Z_{2}$ symmetry with group elements $I\otimes I\otimes I$ and $Z\otimes Z\otimes Z$ acting on the virtual legs of the local tensor. Chen et al. (2010) showed that topological order is stable with any $Z_{2}$ respecting variations and unstable with any $Z_{2}$ violating variation. To illustrate this, we can consider two different directions of variation in single-line TNR. We can add an $X$ or $Z$ variation on one of the virtual indices of the tensor,

$\displaystyle\includegraphics[scale={0.3}]{SLTCpert}.$

(6)

More explicitly, these tensor components are given by,

$T^{(X)}$ variation violates the $Z_{2}$ symmetry while $T^{(Z)}$ does not. That is,

$\displaystyle\includegraphics[scale={0.3}]{SLTCpertZ2},$

(9)

and it was shown that $T^{(X)}$ type variations cause an instability and while $T^{(Z)}$ type variations do not. Note that, though we chose variations only on the virtual indices for simple illustration, the same conclusion applies for any random variation including those on the physical indices. However, if a variation acts only on the physical indices, it cannot break the $Z_{2}$ virtual symmetry, and hence would always be stable.

We reproduce this known result with a new algorithm for calculating $S_{\text{topo}}$. This algorithm allows us to calculate $S_{\text{topo}}$ in more complicated examples to be dealt with later. Before we move on to other TNRs, we would like to explain the algorithm used here. Readers can skip this section if they are not interested in the details of the algorithm.

Here we explain the algorithm we use to calculate the topological entanglement entropy of any translation invariant tensor network state. We use the idea presented by Cirac et al. (2011) to calculate reduced density matrix on a region and hence its entanglement entropy. We consider honeycomb lattice, though it can easily be extended to other lattices. By translation invariant we mean that all vertices on the sublattice A and sublattice B are attached with the same tensors, $T_{A}$ and $T_{B}$, respectively. First we define certain notations for convenience of later discussion. The starting objects are given tensors $T^{I}_{\alpha}$, where $I$ and $\alpha$ denote the set of physical and virtual indices, respectively: $I=(i_{1},i_{2},..),\alpha=(\alpha_{1},\alpha_{2},..)$. The state represented by these tensors can be written as

We denote the tensor resulting from contracting the virtual indices of tensors $T$ on a region $R$ as $T(R)$. $\mathbb{T}$ denotes the ‘double tensor’ resulting from contracting the physical indices of $T$ with those of $T^{\dagger}$, that is, $\mathbb{T}=TT^{\dagger}=\sum_{I}T^{I}_{\alpha}\left(T^{I}_{\alpha^{\prime}}\right)^{*}$. Similar to $T(R)$, we denote the double tensor contracted on a region $R$ as $\mathbb{T}(R)$.

Now let us consider putting this tensor network state on a cylinder. We denote the left half of the cylinder as $L$ and the right half as $R$. The honeycomb lattice is placed in a way so that $L$ and $R$ divide it into exact halves. So the line between the two halves goes through the middle of the plaqeuttes as shown in the Fig. 3. We denote the tensors on the left and right boundaries as $T_{l}$ and $T_{r}$.

When we contract bulk double tensors with the boundary double tensors, we get a density matrix operator on the virtual indices,

Cirac et al. (2011) showed that the physical reduced density matrix on one of these halves, let’s say the left one, is related to the density operator on the virtual indices as,

where $U$ is an isometry. Hence $\rho_{L}$ and $\sqrt[]{\sigma_{L}^{T}}\sigma_{R}\sqrt{\sigma_{L}^{T}}$ have the same spectrum. In addition, under right symmetry conditions, $\sigma_{L}^{T}=\sigma_{R}=\sigma_{b}$. When this is true, up to change of basis, we find that $\rho_{L}\propto\sigma_{b}^{2}$. The normalized reduced density matrix is

It is known that the Rényi entropy with any Rényi index gives the same topological entanglement entropyFlammia et al. (2009). So we calculate Rényi entropy with Rényi index $\frac{1}{2}$,

In the limit of large cylinder, it should behave like

where $|C|$ is the circumference of the cylinder. This is how we calculate $S_{\text{topo}}$ starting with a tensor network state.

Before we move on to the next step, we would like to mention an important subtlety regarding computation of $S_{\text{topo}}$ on a cylinder. In Ref.Dong et al., 2008; Zhang et al., 2012 it has been shown that $S_{\text{topo}}$ calculated this way on a cylinder, in general, might depend on the boundary conditions. We choose a particular boundary condition for all our calculations and examine the dependence of $S_{\text{topo}}$ on boundary condition in the appendix G. Our findings are consistent with the conclusion in Ref.Zhang et al., 2012.

We first have to calculate $\mathbb{T}(R)\mathbb{T}_{r}(\partial R)$ for the above setup. The problem is, the computational complexity of exact tensor contraction grows exponentially with the size of $R$, so we need to use some approximate renormalization algorithm. We use an algorithm which is a slight modification of known tensor renormalization algorithms (Gu et al., 2008; Vidal, 2003, 2009). Consider double tensors contracted along a thin strip on the cylinder giving us a transfer matrix operator, $\mathbb{S}$. If $R$ includes $n$ of such strips, we have $\mathbb{T}(R)=\mathbb{S}^{n}$. Since the tensor network state under consideration are short range correlated along the cylinder, the spectrum of $\mathbb{S}$ is gapped. Consequently, for large $n$, only the highest eigenvalue and the corresponding eigenvector of $\mathbb{S}$ dominates. That is, in thermodynamic limit, $\mathbb{T}(R)$ only depends on the highest eigenvalue/eigenvector of the transfer matrix operator, $\mathbb{S}$. Moreover, we expect to approximate the eigenvector of highest eigenvalue with a Matrix Product State (MPS) with finite bond dimensions, since the tensor network state is short range correlated along the circumference of the cylinder. So we can start with a boundary MPS, apply the transfer matrix operator, and approximate the resulting state as an MPS with a fixed, finite bond dimensions. With each step, approximation to the eigenvector with highest eigenvalue improves and we do this recursively until we reach the fixed point giving us the desired eigenvector. Note that we require transfer matrix operators to be reflection symmetric for the condition $\sigma_{L}^{T}=\sigma_{R}=\sigma_{b}\Rightarrow\rho_{L}\propto\sigma_{b}^{2}$ to hold true.

The recursive algorithm is as following:

1. 1.

   Initiate the boundary double tensor $\mathbb{T}_{A^{\prime}}=\mathbb{T}_{r,A^{\prime}}$ and $\mathbb{T}_{B^{\prime}}=\mathbb{T}_{r,B^{\prime}}$. The tensor network to be contracted looks as

   (16)

2. 2.

   Contract the bulk double tensors, $\mathbb{T}_{A}$ and $\mathbb{T}_{B}$ with the boundary tensors $\mathbb{T}_{A^{\prime}}$ and $\mathbb{T}_{B^{\prime}}$ in the following way to make the 4 leg tensor $\mathbb{T}_{AB^{\prime}BA^{\prime}}$,

   (17)

3. 3.

   Reshape the tensor $\mathbb{T}_{AB^{\prime}BA^{\prime}}$ into a matrix $M$ where $M_{\alpha\beta^{\prime},\beta\alpha^{\prime}}=(\mathbb{T}_{AB^{\prime}BA^{\prime}})_{\alpha\beta^{\prime}\beta\alpha^{\prime}}$ Vidal (2003). Now we perform an SVD decomposition of $M$, $M=U\Lambda V^{\dagger}$ and the approximation step: we keep only the highest $D_{\textrm{cut}}$ singular values, and define the new tensors $\mathbb{S}_{A^{\prime}}$ and $\mathbb{S}_{B^{\prime}}$ as $(S_{A^{\prime}})_{\alpha\beta^{\prime}\gamma}=U_{\alpha\beta^{\prime},\gamma}\sqrt[]{\Lambda_{\gamma,\gamma}}$ and $(S_{B^{\prime}})_{\gamma\beta\alpha^{\prime}}=\sqrt[]{\Lambda_{\gamma,\gamma}}V^{\dagger}_{\gamma,\beta\alpha^{\prime}}$ where $\gamma$ takes values $1,2,\ldots,D_{\textrm{cut}}$. $\mathbb{S}_{A^{\prime}}$ and $\mathbb{S}_{B^{\prime}}$ form an approximate decomposition of $\mathbb{T}_{AB^{\prime}BA^{\prime}}$,

   $\displaystyle\sum_{\gamma=1}^{D_{\text{cut}}}(S_{A^{\prime}})_{\alpha\beta^{\prime}\gamma}(S_{B^{\prime}}^{\dagger})_{\gamma\beta\alpha^{\prime}}\approx(\mathbb{T}_{AB^{\prime}BA^{\prime}})_{\alpha\beta^{\prime}\beta\alpha^{\prime}}$

   (18)

   (19)

4. 4.

   Check convergence of $\Lambda$. $\eta\ll 1$ is the precison tolerance. Let $n$ denote the $n$th recursion step. If $||\Lambda_{n}-\Lambda_{n-1}||_{1}<\eta$ exit algorithm.

5. 5.

   Put $\mathbb{T}_{A^{\prime}}=\mathbb{S}_{A^{\prime}}$ and $\mathbb{T}_{B^{\prime}}=\mathbb{S}_{B^{\prime}}$ and go to step 2.

We use the algorithm described in the previous section to calculate $S_{\text{topo}}$ of the tensor network state constructed by a local tensor with random variations added to the fixed point tensor given in Eq. (4). $I_{V}=I^{\otimes 3}$ is projector onto the full virtual space. $\mathbb{M}=\frac{1}{2}(I^{\otimes 3}+Z^{\otimes 3})$ is a projector on to the space of variations that respect the $Z^{\otimes 3}$ symmetries. So, $I_{V}-\mathbb{M}$ is a projector on to the space of variations that break $Z^{\otimes 3}$ symmetries. We first calculate $S_{\text{topo}}$ in the state constructed by the fixed point tensor, $T^{0}$. Then we generate a random tensor $T^{r}$ on the full space, project it on to the subspace $I_{V}-\mathbb{M}$, add it to the fixed point value, $T^{0}\rightarrow T^{0}+\epsilon(I_{V}-\mathbb{M})T^{r}$ and calculate $S_{\textrm{topo}}(\epsilon)$. Similarly, we generate a random tensor $T^{r}$ on the full space, project it on to $Z^{\otimes 3}$ respecting subspace $\mathbb{M}$, add it to the fixed point value, $T^{0}\rightarrow T^{0}+\epsilon\mathbb{M}T^{r}$ and calculate $S_{\textrm{topo}}(\epsilon)$. We keep the value of variation strength $\epsilon=0.01$ (low enough) to make sure it is not near any phase transition point. The results are shown in Fig. 4.

We see that $Z^{\otimes 3}$ respecting variations lead to the same topological entanglement entropy as the fixed point state, while $Z^{\otimes 3}$ violating variations lead to zero topological entanglement entropy. This reproduces the result by Chen et al. (2010).

In the double-line TNR of the toric code state, we associate with each vertex a tensor with 3 physical legs and 6 virtual legs, $T^{ijk}_{\alpha\alpha^{\prime};\beta\beta^{\prime};\gamma\gamma^{\prime}}$, (see Fig. 5). We will refer to these virtual indices as ‘plaquette indices’ or ‘plaquette legs’ sometimes, because they carry the plaquette degree of freedom that comes from the local Hamiltonian term. All indices take values 0 and 1. We denote the TNR corresponding to the RG fixed point state as $T^{0}$. (We use the same notation for different fixed point tensors, but it should be clear from the context which fixed point tensor we are discussing.) First property of $T^{0}$ is that $(T^{0})^{ijk}_{\alpha\alpha^{\prime};\beta\beta^{\prime};\gamma\gamma^{\prime}}\propto\delta_{\alpha\alpha^{\prime}}\delta_{\beta\beta^{\prime}}\delta_{\gamma\gamma^{\prime}}$, that is, indices on the same plaquette assume the same values. The second property is that the physical indices can be considered as labeling the domain wall between the virtual indices. If the two virtual indices in the same direction have the same values (both either $00$ or $11$) then the physical index in the middle has value $0$, otherwise it is $1$. That is, $i=\beta+\gamma,j=\gamma+\alpha,k=\alpha+\beta$ (all additions are modulo 2). So we can write $T^{0}$ as

We can write all non-zero components explicitly,

Is double-line TNR unstable too? We find that it is unstable as well in certain directions of variations. Similar to the single-line case, a variation is stable or unstable depending on whether or not it violates certain virtual symmetries. So let’s first look at the symmetries of the double-line TNR. It has 6 virtual indices, so the virtual space dimension is $2^{6}=64$, while the physical space dimension is again 4. So we need a symmetry group with $|G|=64/4=2^{4}$. Indeed the tensor has a $Z_{2}\times Z_{2}\times Z_{2}\times Z_{2}$ virtual symmetries. First it has a $X^{6}$ symmetry. That is, if we flip all the six virtual indices, the tensor remains the same. Second, it has three $Z\otimes Z$ symmetries, where $Z\otimes Z$ are applied to the two virtual indices on the same plaquette. So the double-line tensor in (LABEL:DLTNReq) satisfies these symmetry equations:

$\displaystyle\includegraphics[width=173.44865pt]{DLTCsym}.$

(21)

Single-line TNR had only one such $Z_{2}$ symmetry and it turned out that breaking it results in phase transition. For double-line we have four $Z_{2}$ symmetries. So the question is, are all of them important? That is, is it the case that breaking any of them with a variation leads to instability? Indeed many different possible kinds of variations are possible:

$\displaystyle\includegraphics[width=151.76633pt]{DLTCpert}.$

(22)

A variation can violate $Z\otimes Z$ but not $X^{\otimes 6}$ (for example, 22(a) ), or it can violate $X^{\otimes 6}$ but not $Z\otimes Z$ (for example, 22(b) ), or it can violate both (for example, 22(c) ), or it can violate neither(for example, 22(d) ), etc. So to find out, we need to look at the unstable directions of variations of the fixed point tensor.

Our numerical calculation reveals an interesting result. We find that (see Fig. 6 )

1. 1.

   If a variation violates any of the $Z\otimes Z$ symmetries then it is stable.

2. 2.

   If a variation respects all $Z\otimes Z$ symmetres then there are two subcases

   1. (a)

      If it respects $X^{\otimes 6}$ symmetry then it is stable.

   2. (b)

      If it breaks $X^{\otimes 6}$ symmetries then it is unstable.

So we see that the relation between unstable variations and virtual symmetries of the double-line TNR is more complicated than that for single-line TNR. The $X^{\otimes 6}$ symmetry looks like the $Z^{\otimes 3}$ symmetry of the single-line TNR, as they both operate as a loop operators, and unstable variations in both TNR violates these loop symmetries. However, the crucial difference in double-line is that then there are extra symmetries (the 3 $Z^{\otimes 2}$ symmetries) which have an exact opposite relation with the unstable variations: a variation is actually stable when it violates these symmetries (irrespective of whether or not it violated the loop symmetry). It indicates that the sources of these two kinds of symmetries must be different.

How can we understand this phenomena? We will now show that the sources of these symmetries are indeed different, and it is the interplay between these two symmetries that determines the tensor instability phenomena. The symmetries whose violation causes instability comes from the so-called MPO-injective subspaceŞahinoğlu et al. (2014) of the virtual space, while the symmetries whose violation causes stability comes from what we define to be stand-alone subspace. We now define these subspaces precisely and put forward the conjecture regarding their relationship to TNR instability.

A generic tensor can be thought of as a linear map from virtual vector space to the physical vector space. Consider a generic tensor, $T^{I}_{\alpha}$ where $I$ is the collection of all physical indices and $\alpha$ is the collection of all virtual indices. Double tensor $\mathbb{T}$ of a tensor $T$ is defined as $\mathbb{T}=\sum_{I}T^{I}_{\alpha}(T^{*})^{I}_{\alpha^{\prime}}$. For example, for single-line TNR the double tensor can be represented diagrammatically as follows

In general we would think of double tensor as having two layers of virtual indices, lower and upper. Double tensor can be interpreted as a density matrix on the virtual space. Now consider the double tensor of a RG fixed point TNR, $\mathbb{T}^{0}$, contracted over some large region $R$. Let’s say we remove $\mathbb{T}^{0}$ from one site and replace it with some other double tensor, $\mathbb{T}$ as follows:

$\displaystyle\includegraphics[scale={0.4}]{standalone0}.$

(24)

What do we get? In particular, are there tensors $\mathbb{T}$ such that this replacement collapses the whole tensor network? By ‘collapse’, we mean that we simply get zero upon contraction. The answer turns out to be yes for tensors that represent topological order. In fact, as we will see later, most tensors $\mathbb{T}$ will collapse the fixed point tensor network upon replacement. It turns out that only the tensors supported on a particular subspace of the full virtual space can replace the fix point tensor without collapsing the whole tensor network. We will call this space the stand-alone subspace of the TNR. Now we will give a systematic way of calculating this subspace for a given fixed-point TNR.

Consider contracting the fixed-point double tensors $\mathbb{T}^{0}$ on a large disc with an open boundary. Now we remove the tensor at the origin. We get a tensor network with dangling virtual indices at the origin and at the boundary of the disc. We want to find out the space of tensors that can be put on the origin without collapsing the tensor network. We do not care what tensor at the boundary we get. So we trace out the indices at the outer boundary (i.e. contract upper and lower virtual indices with each other). This leaves us with a tensor at the origin. The support space of this tensor will be precisely the stand-alone space. Any tensor supported on this subspace can stand alone with the surrounding tensor being the fixed-point tensors.

Note that definition of stand-alone space implies that it is a vector space. If $T_{1}$ and $T_{2}$ doesn’t collapse the tensor network then $aT_{1}+bT_{2}$ will also not collapse the network.

Stand-alone subspace of double-line TNR: Now let’s first calculate the stand-alone subspace of double-line TNR of toric code as it is more interesting than that of single-line TNR. The double tensor of $T^{0}$ in (LABEL:DLTNReq) can be written as (ignoring an overall normalization factor)

where the double tensor is written as an operator between the lower virtual indices and upper virtual indices. The $Z^{\otimes 2}$ and $X^{\otimes 6}$ act in the way it is shown in Fig. 21. We need to contract this tensor on a disc with a hole at the origin. To contract two tensors given in an operator form, we need to multiply them and take a trace on the shared indices. A cumbersome but straight-forward calculation shows that double tensor contracted on a region $R$ give (ignoring an overall normalization factor)

where $\partial R$ denote the boundary of $R$, and $m=|\partial R|$ is the length of the boundary. $O(\partial R)$ means the operator $O$ is applied on the virtual legs along the boundary $\partial R$. We will omit this when it is clear from the context which leg the operator is being applied on. The region we want is a disc with a vertex removed, $R=D_{2m}-D_{6}$. $D_{n}$ denotes the disc with $n$ virtual legs at the boundary. It has two disconnected boundaries, one the boundary of $D_{2m}$ and other the boundary of $D_{6}$ (with opposite orientation).

As explained in Fig. 8, $X$ operators act on the two boundaries simultaneously, but $Z$ operators act independently. Now to get the stand-alone space at the origin, we need to trace out the virtual legs at the boundary of $D_{2m}$. If we expand the expression for $\mathbb{T}(D_{2m}-D_{6})$ above and apply trace on the operators on the outer boundary, only the terms with identity on the outer boundary survive. $X$ operator does not have such a term, but $Z$ does. So finally, tracing out the outer boundary leaves only $Z^{\otimes 2}$ on the inner boundary. That is, we get the following tensor on the 6 virtual indices incident on a singe vertex

$B_{0}^{2}=8B_{0}$, so

is a projector on to the support space of $B_{0}$. $M_{0}$ defines the stand-alone space of double-line TNR of toric code. Any tensor $T$ that satisfies $M_{0}T\neq 0$ can ‘stand alone’. $M_{0}$ will be used to denote the projector on to stand-alone space throughout the paper. So we see that only the tensors that respect the $Z\otimes Z$ symmetry can stand alone. The $X^{\otimes 6}$ symmetry, however, is not required to define the stand-alone space.

Stand-alone subspace of single-line TNR: We can also calculate the stand-alone subspace of single-line TNR of toric code. One can calculate the double tensor of single-line TNR in Fig. 4 to be

This double tensor upon contraction on the disc with a hole at the origin ($D_{m}-D_{3}$) gives (up to an overall normalization)

Now contracting the outer circle gives us

So we see that, for single-line TNR, the stand-alone subspace is actually all of the virtual space. That is, there are no tensors that cannot stand alone.

Here we repeat the definition of MPO-injective subspace given in Ref. Şahinoğlu et al., 2014 for convenience.

As explained above, stand-alone subspace is the maximal virtual subspace such that any tensor supported on this subspace can be inserted into the tensor network without collapsing it. Therefore, the virtual space of the RG fixed point tensor, $T^{0}$ itself must be inside the stand-alone subspace. This virtual space of $T^{0}$, which is by definition a subspace of the stand-alone space, is what we would call the MPO-injective subspace. The reason for calling it an MPO-injective space is that, as it turns out, this space is protected by symmetry operators which are Matrix Product Operators (MPO). Let’s make the notion of virtual space of $(T^{0})^{I}_{\alpha}$ precise. We can think of it as a matrix with indices $\alpha$ and $I$ and perform an SVD decomposition,

where $V$ and $P$ are unitary matrices in virtual and physical space, and $\Lambda$ is the diagonal matrix containing the singular values.

Another way to think about this is to again consider the double tensor $\mathbb{T}^{0}_{\alpha,\alpha^{\prime}}=\sum_{I}(T^{0})^{I}_{\alpha}(T^{0;*}))^{I}_{\alpha}$ which is a matrix in the virtual space. An equivalent but more useful definition is,

Using Eq. (33) we can write

where $\lambda_{j}$ are the singular values and $v_{j}$ and $p_{j}$ are the corresponding vectors in virtual and physical space. MPO-injective space is the space spanned by $v_{j}$, so the projector on this space is

The mathematical understanding of the MPO-injective space is that it is the virtual subspace which is isomorphic to the ground state physical subspace and $T^{0}$ is the isomorphism. MPO-injective subspace by definition nested inside the stand-alone subspace, which by definition is nested inside the full virtual space. Similarly, the ground-state physical space is by definition a subspace of the full physical space. These spaces are represented visually in Fig. 9 for clarity.

MPO-injective subspace of single-line TNR: We can write the single-line tensor in Eq. (4) in the Eq. (33) form,

Of course, this happened to be already written in SVD decomposed form. So the MPO-injective space is spanned by vectors $\{|000\rangle,|011\rangle,|101\rangle,|110\rangle\}$. So the MPO projector is

We see that this projector can be written as a translation invariant superposition of tensor product of matrices. That’s why we call it the MPO-injective subspace. Remember that stand-alone space of single-line TNR was determined to be all of virtual space, $M_{0}=I_{V}$. So as expected, $\mathbb{M}\subset M_{0}$.

MPO-injective subspace of double-line TNR: For calculation of the MPO-injective subspace of double-line tensor in Eq. (LABEL:DLTNReq) we use the second definition in 2 above to avoid a cumbersome but straight forward calculation. We already calculated the double tensor of double-line TNR in Eq. (25). We ignored the normalization factor there. If we use a normalization factor of $\frac{1}{16}$ and write

Then $\mathbb{M}$ is a projector, that is, it satisfies $\mathbb{M}^{2}=\mathbb{M}$, and has the same support as $\mathbb{T}^{0}$ hence this is the desired MPO projector. Remember that stand-alone projector was calculated to be $M_{0}=\frac{1}{8}(I^{\otimes 2}+Z^{\otimes 2})^{\otimes 3}$, and hence $\mathbb{M}=M_{0}\frac{1}{2}(I^{\otimes}+X^{\otimes 6})\subset M_{0}$, as expected.

Now that we have defined the stand-alone and MPO-injective subspaces precisely, we are ready to state the central conjecture of this work.

It implies that the projector onto the stable space is $P_{S}=I_{V}-(M_{0}-\mathbb{M})$.

Or in simple words, a variation is unstable if and only if it has a component in the stand-alone space that is outside the MPO-injective subspace. A pictorial representation of the decomposition of the virtual space through these projectors is shown in Fig. 9. We will denote $(M_{0}-\mathbb{M})$ as $P_{U}$ for convenience. Note that $P_{U}$ shouldn’t be thought of as ‘the projector onto unstable subspace’ because unstable variations do not form a vector space, as opposed to stable variations that do form a vector space. It is because $P_{U}T^{1}\neq 0$ and $P_{U}T^{2}\neq 0$ does not imply $P_{U}(T^{1}+T^{2})\neq 0$.

Let’s first see how this conjecture is true for the single-line and double-line TNR of the toric code. For single-line we have already calculated the stand-alone and MPO-injective subspaces and found $M_{0}=I_{V}$ and $\mathbb{M}=\frac{1}{2}\left(I^{\otimes 3}+Z^{\otimes 3}\right)$. So,

So for a tensor $P_{U}T\neq 0$ if and only if it violates the $Z^{\otimes 3}$ symmetry. Indeed, this is exactly what we saw numerically in Fig 2. All variations can be summarized visually using Fig. 9 as follows:

Variations supported on the red region are unstable, while those on blue and white are stable. The dimension of each space is indicated. So we see that the space $(M_{0}-\mathbb{M})$ is 4 dimensional space spanned by basis

It is wrong to think that these basis set span the space of unstable variations, because unstable variations do not form a vector space. All we can say is if a variations has overlap with any of these basis, it would cause instabilty.

For double-line TNR we found,

So,

So $P_{U}T\neq 0$ if and only if $T$ satisfies the three $Z^{\otimes 2}$ symmetries but violates the $X^{\otimes 6}$ symmetry. Indeed this is precisely what we found numerically as shown in Fig. 5. All variations can be summarized visually as follows

Variations supported on the red region are unstable, while those on blue and white are stable. The dimension of each space is indicated. So we see that the space $(M_{0}-\mathbb{M})$ is 4 dimensional space spanned by basis

As claimed, the physical significance of stand-alone space is that it contains proliferatable bosonic excitations.

As claimed above, the physical significance of MPO-injective subspace is that the variations in this subspace are lifted to local physical variations, which have to be topologically trivial bosons since they can be removed by a local operation. Hence the physical significance of MPO-injective subspace is that it contains all the topologically trivial excitations.

To understand it better, let us look at concrete examples of variations that are lifted to local physical variation. Consider a $Z$ variation on the virtual leg of the fixed point single-line TNR. If we lift it to the physical level, what do we get? Since the virtual legs are just copies of the physical legs, we get

$\displaystyle\includegraphics[scale={0.3}]{SLZVZP}.$

(50)

So $Z$ virtual variation is lifted to a $Z$ physical variation, which is local. According to our claim, it should be in the MPO-injective subspace. And indeed it is, since it respects the MPO symmetry of the single-line TNR, $Z^{\otimes 3}$. Also note that a $Z$ physical variation corresponds to a pair of $m$ particles sitting next to each other, not a single $m$ particle. It is a trivial excitation and can be removed by applying one $Z$ operation on the state, so it matches our claim. Contrast this with the $X$ variation on the virtual level. Can we find any local physical operator $O$ such that

$\displaystyle\includegraphics[scale={0.3}]{XO}?$

(51)

One can try and see that there is no such local operator $O$ for which this equation holds. (We will later show that $X$ can be lifted to the physical level, but it results in a non-local operator. )

A similar phenomena occurs in double-line TNR. The $X\otimes X$ variation can be lifted to a local physical operator,

$\displaystyle\includegraphics[scale={0.3}]{DLXVXP},$

(52)

but a $Z$ variation cannot be. (We will later show that $Z$ can be lifted to the physical level, but it results in a non-local operator.) It is again consistent with the claim as $X\otimes X$ variation respects the double-line MPO symmetry $(X^{\otimes 6})$ but $Z$ variation breaks it. Note that $X\otimes X$ variation on the physical level corresponds to a pair of $e$ particles sitting across a plaquette. It is a topologically trivial excitation and can be removed by an $X\otimes X$ operation on the state. So again, this matches our claim.

Now we prove that these examples are no coincidence, and in fact any variation in the MPO-injective subspace is a local physical variation.

Let us repeat the definition of the MPO-injective subspace here for convenience. We SVD decomps the fixed point RG tensor $T^{0}$ as a matrix between virtual and physical legs

where $\lambda_{j}$ are the singular values, and $v_{j}$ and $p_{j}$ are orthonormal vectors in the virtual and ground-state physical spaces respectively. Then the MPO-injective subspace is the virtual subspace spanned by vectors $v_{j}$ such that corresponding singular values $\lambda\neq 0$. So the MPO projector is

A mathematically inclined reader would note that MPO-injective subspace is nothing but the virtual subspace which is isomorphic to the image of the tensor as a map from virtual to ground state physical space. That is, if we restrict the domain of the tensor to this subspace, then tensor is an injective map from the virtual to the physical space, and a bijective map from MPO-injective subspace to ground-state physical subspace. Since these spaces are isomorphic, any operator in MPO-injective subspace can be mapped to an operator in the ground-state physical space and vice-versa, and this mapping would be bijective (one-to-one) as well. Let us make it precise,

With this lemma, we see why in general variations in the MPO-injective subspace are trivial excitations. They are nothing but a local variation on the physical level, which is a local physical operator and can be removed by another local operator. In fact notice that if $A$ is unitary (within the space $\mathbb{M}$ ) then so is $B$ and vice-versa. Since all trivial excitations are obtained by local unitaries, or their linear combinations, we conclude that MPO-injective subspace should contain all trivial virtual excitations as well.

This completes the study of first kind of variations (those that are lifted to local physical variations) mentioned above. Now we study the second kind of variations.

The physical significance of subspace $M_{0}-\mathbb{M}$ is that it contains the second kind of variations: the virtual variations that are lifted to a non-local physical operator. So these variations cannot be removed by a local physical operation on the state, and hence represent a topologically non-trivial excitation. And this excitation has to be a boson, as all excitations in stand-alone space are. So it means that the physical significance of $M_{0}-\mathbb{M}$ space is that it contains condensable excitations that are topologically non-trivial bosons, and that’s why these variations cause a topological phase transition.

Let us first look at some concrete examples to understand this phenomena. We saw how a virtual $X$ variation on the single-line TNR couldn’t be lifted to a local physical variation. But, the question is, can it be lifted to a non-local physical variation? The answer is, yes. To see it, first note that although a single $X$ variation cannot be lifted locally, two $X$ variations can be. That is, the fixed point single-line tensor satisfies

$\displaystyle\includegraphics[scale={0.3}]{SLXXVP}.$

(60)

And also, we have the usual guage symmetry

$\displaystyle\includegraphics[scale={0.4}]{SLinsertXX}.$

(61)

Using these two relations, we see that a single $X$ virtual variation on can be moved to another tensor on the same sublattice, and this transfer produces an $X\otimes X$ operation on the physical level,

$\displaystyle\includegraphics[width=173.44865pt]{SLmovingX}.$

(62)

In the first equality, Eq. (61) is used while in the second equality Eq. (60) is used. We see that the $X$ variation moved from site 1 to site 3 while leaving operator $X\otimes X$ along the path (on site 2). We can repeat this process and move $X$ to the next tensor and so on. After $X$ is moved from site 1 to $n$ there will be an $X$-string operator applied on the physical level along the path. Finally, if there is already an $X$ variation present at site $n$, the two will cancel and we will be left with an $X$-string operator only,

Of course the particular path between site 1 and $n$ chosen is completely arbitrary. We can choose any path between them as we like. So we have successfully shown that though a single $X$ variation cannot be completely lifted to the physical level, two such variations sitting far apart can be, and they are lifted to a non-local physical operator between them. It implies that a $X$ variation on the single-line TNR cannot be removed locally on a physical level. Only two of them can be removed by applying a non-local operator between them. In fact, it is easy to recognize what this excitation is. Since $X$-string operators correspond to creation or annihilation of $e$-particles in the toric code, it is clear that the $X$ virtual variation actually is an $e$ particle excitation. It is topologically non-trivial, which is in line with our claim. Condensation of $X$ variations is actually the condensation of $e$ particles, and that is why it leads to topologically phase transition.

A similar analysis can be carried out for the $Z$ variation in the double-line tensor. We noted that it cannot be lifted to a local physical operator. But two $Z$ operators can be lifted to a non-local physical operator,

where in the first equality, the following relation (similar to Eq. (61)) has been used,

$\displaystyle\includegraphics[width=86.72267pt]{DLinsertZZ}.$

(65)

And in the second equality, the following property of the fixed point double-line tensor is used.

$\displaystyle\includegraphics[width=108.405pt]{DLZZVP}.$

(66)

So we see that two $Z$ variations on the virtual level, sitting far apart cannot be removed by local operations on the physical level. They can only be removed by a non-local operator, the $Z$-string operator. This suggests that the $Z$ variation is a topologically non-trivial excitation. Indeed, it is easy to see that it is nothing but the $m$ particle excitation, since $Z$-string operator creates and annihilates $m$-particles. This is in line with all our claims: $Z$ variation is in the $M_{0}-\mathbb{M}$ space; it cannot be removed locally, and that it is a topological boson.

It may look a little puzzling that a local virtual operator in $M_{0}-\mathbb{M}$ can create a non-trivial physical excitation. We analyzed what these variations correspond to by lifting them up to the physical level. To understand the phenomena better we can ask the opposite question: what happens when we ’bring down’ a non-trivial quasi-particle excitation on the physical level to the virtual level? Since such an excitation is created by a physical string-operator, the equivalent question is, what happens to string-operators of the model when we bring them down to the virtual level? We can look at the specific examples considered above. For example, if we look at Eq. (63) in the opposite way, we see that the physical $X$-string operator, which creates $e$-particles, becomes a gauge-string operator on the virtual level, which subsequently creates a variation in the $M_{0}-\mathbb{M}$ space. Contrast this with $Z$-string operator, that creates $m$-particles. This operator does not map to gauge-string operator,

$\displaystyle\includegraphics[scale={0.3}]{SLZnongauge}.$

(67)

Similarly, Eq. (64) shows that that the physical $Z$-string operator, which creates $m$-particles, becomes a gauge-string operator on the virtual level. So we see that if a physical anyonic string operator maps to a gauge-string operator on the virtual level, it creates an excitation in the $M_{0}-\mathbb{M}$ space. This property of the tensor network state in general is the reason why a local virtual variation can actually correspond to non-local variation on the physical level. In other words, certain gauge-string operators are non-trivial because they come from a non-trivial string operator on the physical level. We would call such physical string operators that map to gauge-string operator on the virtual level a zero-string operator. The reason behind this terminology will become clearer in the next chapter. So we conclude that the $m$-particle operator is the zero-string operator of the double-line TNR, while the $e$-particle string operator is the zero-string operator of the single-line TNR.

Since variations in the unstable space $M_{0}-\mathbb{M}$ are created by zero-string operators, it implies that, if, in a given TNR, none of the physical string operators map to a gauge-string operator then it will have no variation in $M_{0}-\mathbb{M}$. In that case, we would simply have $M_{0}=\mathbb{M}$. Such a TNR will have no instabilities.

Before going to the physical explanation of instabilities, we would like to mention that there is one more way of decomposing the stand-alone space in trivial and non-trivial excitations: using Wilson-loops of anyonic string-operators. We can use these operators to detect whether a nontrivial excitation is sitting at a site, hence can potentially differentiate between $M_{0}-\mathbf{M}$ and $\mathbf{M}$. This has been explored in appendix A for readers who are interested in this perspective.

Now we put together the physical understanding of all the relevant subspaces ($M_{0},\mathbb{M}$ and $M_{0}-\mathbb{M}$) to make the coherent picture of why variations in the subspace $M_{0}-\mathbb{M}$ are unstable, and, in particular, explain the numerical results shown in Fig. 4 and Fig. 6. The general explanation has already been stated in form of conjecture 2 but we repeat it again informally going through all possible variations one by one.

• •

  The variations on the physical indices are of course not stable because they are topologically trivial and can be removed with local operations.

• •

  Variations outside the stand-alone space, $I_{V}-M_{0}$ are not unstable because they cannot proliferate.

• •

  Every variation inside $M_{0}$ is ‘bosonic’ and does proliferate and the varied wave function is a condensate of that ‘boson’. But when the variations is inside $\mathbb{M}$, it was a topologically trivial boson and hence does not cause a topological phase transition. Or, equivalently, every variation inside MPO-injective subspace was nothing but a variation on the physical level hence stable.

• •

  Finally, when the variation was inside stand-alone, but outside MPO-injective subspace then it can condense and is a topologically non-trivial boson. Hence it causes a topological phase transition, resulting in a TNR instability.

Now we explain this boson condensation specifically for the single-line and double-line TNR considering specific variation.