# Bifurcation in entanglement renormalization group flow of a gapped spin
model

Jeongwan Haah  Department of Physics, Massachusetts Institute of Technology,
Cambridge, MA 02139  Institute for Quantum Information and Matter, California
Institute of Technology, Pasadena, CA 91125

(5 February 2014)

###### Abstract

We study entanglement renormalization group transformations for the ground
states of a spin model, called cubic code model $H_{A}$ in three dimensions,
in order to understand long-range entanglement structure. The cubic code model
has degenerate and locally indistinguishable ground states under periodic
boundary conditions. In the entanglement renormalization, one applies local
unitary transformations on a state, called disentangling transformations,
after which some of the spins are completely disentangled from the rest and
then discarded. We find a disentangling unitary to establish equivalence of
the ground state of $H_{A}$ on a lattice of lattice spacing $a$ to the tensor
product of ground spaces of two independent Hamiltonians $H_{A}$ and $H_{B}$
on lattices of lattice spacing $2a$. We further find a disentangling unitary
for the ground space of $H_{B}$ with the lattice spacing $a$ to show that it
decomposes into two copies of itself on the lattice of the lattice spacing
$2a$. The disentangling transformations yield a tensor network description for
the ground state of the cubic code model. Using exact formulas for the
degeneracy as a function of system size, we show that the two Hamiltonians
$H_{A}$ and $H_{B}$ represent distinct phases of matter.

##  I Introduction

The renormalization group (RG) is a collection of transformations that select
out quantities relevant to long-distance physics. Wilson1975RG  It generally
consists of averaging out short-distance fluctuations and rescaling of the
system in order to recover the original picture. In practice, however, details
of RG transformations are context-dependent. When an action is given and the
corresponding partition function is of interest, the RG transformation
concerns the effective parameters (e.g., coupling constants, temperature) of
the theory as a function of probing length/energy scale. When a wave-function
is of interest, the RG transformation takes place in a parametrization space
of the wave functions such that the transformed wave-function recovers
correlations at long distance.

This paper is on the wave function renormalization, focusing on long-range
entanglement structure. As the entanglement of many body system is not
characterized by a single number, our general goal is to compare states with
well-known states or to classify them under a suitable RG scheme.
VerstraeteCiracLatorreEtAl2005Renormalization ; Vidal2007ER ;
ChenGuWen2010transformation  The entanglement between any adjacent pair of
spins can be arbitrary since it can be changed simply by applying a local
unitary operator, which will certainly not affect the long-range behavior in
any possible way. This means that we should allow local unitary
transformations in our definition of equivalence of long-range entanglement,
and the block of spins on which the local unitary is acting should generally
be regarded as a single degree of freedom; the long-range entanglement will
only depend on the entanglement among the coarse-grained blocks. In the case
where the state is represented by some fixed network of tensors,
VerstraeteCirac2004PEPS  this observation has been used to choose the most
relevant part of the tensors VerstraeteCiracLatorreEtAl2005Renormalization ;
ChenGuWen2010transformation  and to speed up certain numerical calculations.
LevinNave2007Tensor

Here, we study long-range entanglement of the ground states of a particular
three-dimensional gapped spin model, via local unitary transformations that
simplify the entanglement pattern. This model, called the cubic code model,
Haah2011Local  shares an important property with intrinsically topologically
ordered systems, Wen1991SpinLiquid  namely the _local indistinguishability_
BravyiHastings2011short  of ground states. However, there are two crucial
differences: One is that the degeneracy under periodic boundary conditions is
a very sensitive function of the system size. The other is that it only admits
point-like excitations whose hopping amplitude is exactly zero in presence of
any small perturbation. Although the cubic code model as presented is exactly
solvable, it is important to ask for a corresponding continuum theory. This is
one of the main motivations of this work.

Our result is as follows. Let $H_{A}(a)$ be the Hamiltonian of the cubic code
model. (See Eq. (4).) $H_{A}(a)$ lives on a simple cubic lattice with two
qubits per site where the lattice spacing is $a$. (We will mostly use the term
‘qubit’ in place of ‘spin-$1/2$’ from now on, since only the fact that each
local degree of freedom is two-dimensional is important.) Let $H_{B}(a)$
denote another gapped spin Hamiltonian on a three-dimensional simple cubic
lattice with four qubits per site where the lattice spacing is $a$. $H_{B}(a)$
will be given explicitly later in Eq. (14). We find a constant number of
layers of local unitary transformations (finite-depth quantum circuit) $U$
such that for any ground state $\left|\psi_{A}(a)\right\rangle$ of $H_{A}(a)$,
we have

$U\left|\psi_{A}(a)\right\rangle=\sum_{i}c_{i}\left|\psi_{A}^{i}(2a)\right\rangle\otimes\left|\psi_{B}^{i}(2a)\right\rangle\otimes\left|\uparrow\cdots\uparrow\right\rangle$
(1)

where $c_{i}$ are complex numbers that depend on
$\left|\psi_{A}(a)\right\rangle$, and
$\left|\psi_{A}(2a)^{i}\right\rangle,~{}\left|\psi_{B}(2a)^{i}\right\rangle$
are ground states of $H_{A}(2a),~{}H_{B}(2a)$, respectively. Note that on the
right-hand side the wave function lives on the coarser lattice with lattice
spacing $2a$. The coarser lattice is depicted in Fig. 1. The unit cell of the
coarser lattice has 16 qubits per Bravais lattice point. 10 qubits in each
unit cell are in the trivial state, disentangled from the rest. The
Hamiltonian $H_{A}$ and $H_{B}$ live on the disjoint systems of qubits
designated by $A$ and $B$ in Fig. 1, respectively.

Figure 1: Simple cubic lattice of lattice spacing of $2a$ and the unit cell.
There are 16 qubits labeled by $0$, $A$, or $B$ in the unit cell. Those that
are labeled by $0$ are in the trivial product state.
$\left|\psi_{A}\right\rangle$ and $\left|\psi_{B}\right\rangle$ in Eq. (1) are
states of the system of the qubits labeled by $A$ and $B$, respectively.

Furthermore, we find another finite-depth quantum circuit $V$ such that for
any ground state $\psi_{B}(a)$ of $H_{B}(a)$,

$V\left|\psi_{B}(a)\right\rangle=\sum_{i}c^{\prime}_{i}\left|\psi_{B}^{i}(2a)\right\rangle\otimes\left|\psi_{B}^{i}(2a)\right\rangle\otimes\left|\uparrow\cdots\uparrow\right\rangle$
(2)

for some numbers $c^{\prime}_{i}$. Again, on the right-hand side the wave
functions live on the coarser lattice. The qubits in the trivial state in Eq.
(2) are uniformly distributed throughout the lattice, similar to Fig. 1. The
first and second $\left|\psi_{B}\right\rangle$ in Eq. (2) are states of
disjoint systems of qubits, similar to $A$ and $B$ of Fig. 1.

The result can be written suggestively as

$\mathcal{R}(H_{A})=H_{A}\oplus H_{B},\quad\mathcal{R}(H_{B})=H_{B}\oplus
H_{B}$ (3)

where $\mathcal{R}$ denotes the disentangling transformation followed by the
scaling transformation by a factor of 2. This is rather unexpected and should
be contrasted with the previous results. AguadoVidal2007Entanglement ;
ChenGuWen2010transformation ; Aguado2011 ;
BuerschaperMombelliChristandlEtAl2013 ; SchuchCiracPerez-Garcia2010G-injective
It has been known that Levin-Wen string-net model LevinWen2005String-net  and
Kitaev quantum double model Kitaev2003Fault-tolerant  are entanglement RG
fixed points. Those results would have been summarized as $\mathcal{R}(H)=H$.
The ground-state subspace is retained at the coarse-grained lattice. There was
no splitting. We will comment further on it later.

The present paper is organized as follows. We begin by defining the model and
reviewing its properties in Sec. II. We give details on the entanglement RG in
Sec. III. The actual unitary operators appearing in Eqs. (1),(2) will not be
displayed in the text, but in a Mathematica script in Supplementary Material.
SM  Next, we argue in Sec. IV that the newly found Hamiltonian $H_{B}$
represents a different phase of matter, based on the degeneracy formulas of
the models on periodic lattices. In Sec. V, we point out the relevance of so-
called branching MERA EvenblyVidal2012RG  description for the ground states of
the cubic code model. In Sec. VI, we describe a special representation of the
models, exploiting the translation symmetry and properties of Pauli matrices.
The special representation simplifies the calculation of the unitaries of Eqs.
(1),(2) significantly. Sec. VII builds on the preceeding section, giving an
algebro-geometric criterion and some intuition behind the entanglement RG
calculations. We conclude with a short discussion in Sec. VIII. Appendix A
contains a direct bound EvenblyVidal2013bMERAentropy  on the entanglement
entropy of a branching MERA state for a box region.

##  II Model

The spin model primarily considered in this paper is described by an
unfrustrated translation-invariant Hamiltonian on the simple cubic lattice
$\Lambda=\mathbb{Z}^{3}$ with two qubits per lattice site. Haah2011Local

$\displaystyle H_{A}=-J\sum_{i\in\Lambda}\left(G^{x}_{i}+G^{z}_{i}\right)$ (4)

where $J>0$ and

$\displaystyle G^{x}_{i}$
$\displaystyle=\sigma^{x}_{i,1}\sigma^{x}_{i,2}\sigma^{x}_{i+\hat{x},1}\sigma^{x}_{i+\hat{y},1}\sigma^{x}_{i+\hat{z},1}\sigma^{x}_{i+\hat{y}+\hat{z},2}\sigma^{x}_{i+\hat{z}+\hat{x},2}\sigma^{x}_{i+\hat{x}+\hat{y},2}$
(5) $\displaystyle G^{z}_{i}$
$\displaystyle=\sigma^{z}_{i,1}\sigma^{z}_{i,2}\sigma^{z}_{i-\hat{x},2}\sigma^{z}_{i-\hat{y},2}\sigma^{z}_{i-\hat{z},2}\sigma^{z}_{i-\hat{y}-\hat{z},1}\sigma^{z}_{i-\hat{z}-\hat{x},1}\sigma^{z}_{i-\hat{x}-\hat{y},1}$
(6)

are eight-qubit interaction terms consisted of Pauli matrices. The index $i$
runs over all elementary cubes. The terms $G^{x}_{i}$ and $G^{z}_{i}$ are
visually depicted as

---
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---
$\textstyle{ZI\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{IZ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{IZ\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{ZZ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{II\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{ZI\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{ZI\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{IZ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\circ\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{z}}$$\scriptstyle{\hat{x}}$$\scriptstyle{\hat{y}}$

For the arrangement of the Pauli matrices on the vertices of the unit cube,
this is called cubic code model. (It is a quantum error correcting code, but
we will not discuss the theory of quantum error correction.) One can easily
verify that each term $G^{x}_{i}$ or $G^{z}_{i}$ commutes with any other term
$G^{x}_{j}$ or $G^{z}_{j}$ in the Hamiltonian $H_{A}$. A ground state
$\left|\psi\right\rangle$ of $H_{A}$ can be written as

$\displaystyle\left|\psi\right\rangle=\sum_{G\in\mathcal{G}}G\left|\uparrow\uparrow\cdots\uparrow\right\rangle$
(7)

where $\mathcal{G}$ is the abelian group generated multiplicatively by terms
$G^{x}_{i}$’s and $G^{z}_{i}$’s. Since
$\left|\uparrow\uparrow\cdots\uparrow\right\rangle$ is an eigenstate of
$G^{z}_{i}$ for any $i$ with eigenvalue $+1$, the group $\mathcal{G}$ can be
replaced by a smaller group consisting all products of $G^{x}_{i}$’s. The
ground state is degenerate (ground space). This will not concern us.

The energy spectrum can be understood by commutation relations among Pauli
matrices, since the Hamiltonian Eq. (4) is a sum of commuting tensor products
of Pauli matrices. Let us call a tensor product of Pauli matrices
$\sigma^{x},\sigma^{y},\sigma^{z}$ a Pauli operator. If
$\left|\psi\right\rangle$ is a ground state and $P$ is any Pauli operator,
then $P\left|\psi\right\rangle$ is also an energy eigenstate. In fact, it is a
common eigenstate for $G^{x}_{i}$ and $G^{z}_{i}$. This is because any term
$G^{x}_{i}$ or $G^{z}_{i}$ in the Hamiltonian, being a Pauli operator, either
commutes or anticommutes with $P$ ($PG^{x,z}=\pm G^{x,z}P$) and the ground
state $\left|\psi\right\rangle$ is stabilized by any $G^{x}$ and $G^{z}$
($G^{x,z}\left|\psi\right\rangle=\left|\psi\right\rangle$).

To understand the (excited) state $P\left|\psi\right\rangle$ better, imagine
that we measure all $G^{x}$ and $G^{z}$ simultaneously. This is possible since
they are pairwise commuting. The measurement outcomes on
$P\left|\psi\right\rangle$ are definite and take values $\pm 1$. Let us say
that there is a $X$-type _defect_ at $i$ if the expectation value of
$G^{x}_{i}$ is $-1$. Likewise, we define $Z$-type defects. Each defect has
energy $2J$, and a state with no defect is a ground state. A configuration of
the defects characterizes an excited state effectively, but not uniquely due
to the ground state degeneracy; for orthogonal ground states
$\left|\psi\right\rangle$ and $\left|\phi\right\rangle$, orthogonal states
$P\left|\psi\right\rangle$ and $P\left|\phi\right\rangle$ give the same
configuration of defects. Note that the whole Hilbert space is spanned by
states of form $P\left|\psi\right\rangle$ for some Pauli operator $P$ and some
ground state $\left|\psi\right\rangle$.

An exotic property of the cubic code model is that the excitations are
_pointlike_ but _immobile_. They are pointlike because a single isolated
defect is a valid configuration, but are immobile because they are not allowed
to hop to other position by a local operator. Here, the locality is important.
There indeed exists a non-local operator that annihilates a defect and create
another at a different place. The statement remains true even if we loosen our
restriction that there be exactly one defect at $p$. In a general case, one
should distinguish a cluster of defects that is locally created, in which case
we call the cluster _neutral_ , from a cluster that is not locally created, in
which case we call the cluster _charged_. (Since the charged cluster has
nothing to do with any symmetry, it is termed “topologically charged.”) The
immobility asserts that any charged cluster cannot be transported by any
operator of finite support.

Rigorously, the immobility is stated as follows. Suppose
$\left|\psi\right\rangle$ is a state with a single defect, or more generally
any charged cluster of defects, contained in a box of linear dimension $w$.
Let $\mathbb{T}$ denote a translation operator by one unit length in the
lattice along arbitrary direction. Then, for any operator $O$ of finite
support, (i.e., $O$ is local,) one has
$\langle\psi|O\mathbb{T}^{n}|\psi\rangle=0$ whenever $n>15w$. The number $15$
is merely a convenient number to make an argument smooth. Important is that
there is some finite $n=n(w)$ such that the transition amplitude becomes
_exactly_ zero. See Ref. Haah2011Local,  for proofs.

The cubic code model is _topologically ordered_ Wen1991SpinLiquid  in the
sense that the ground state subspace is degenerate and no local operator is
capable of distinguishing any two ground states; Haah2011Local  if $O$ is an
arbitrary local operator and $\left|\psi_{1}\right\rangle$ and
$\left|\psi_{2}\right\rangle$ are two arbitrary ground states, then one has

$\displaystyle\langle\psi_{1}|O|\psi_{1}\rangle=c(O)\langle\psi_{1}|\psi_{2}\rangle$
(8)

for some number $c(O)$ that only depends on the operator $O$ but not on the
states $|\psi_{1,2}\rangle$. In addition, the model satisfies the so-called
“local topological order” condition, MichalakisZwolak2013Stability  which
implies that the degeneracy is exact up to an error that is exponentially
small in the system size. BravyiHastings2011short  In other words, all ground
states have exactly the same local reduced density matrices, and this property
does not require a fine-tuning. For an application of the model in robust
quantum memory, see Ref. BravyiHaah2011Memory, .

The actual degeneracy and questions on non-local operators that distinguish
different ground states are fairly technical. One can show Haah2012PauliModule
that the degeneracy of the cubic code model on a $L\times L\times L$ lattice
with periodic boundary conditions is equal to $2^{k}$ where

$\displaystyle\frac{k+2}{4}$
$\displaystyle=\mathrm{deg}_{x}~{}\gcd\begin{bmatrix}1+(1+x)^{L},\\\
1+(1+\omega x)^{L},\\\ 1+(1+\omega^{2}x)^{L}\end{bmatrix}_{\mathbb{F}_{4}}$
(9) $\displaystyle=\begin{cases}1&\text{if $L=2^{p}+1~{}(p\geq 1)$},\\\
L&\text{if $L=2^{p}~{}(p\geq 1)$}\end{cases}$ (10)

That is, one computes three polynomials over the field of four elements
$\mathbb{F}_{4}=\\{0,1,\omega,\omega^{2}\\}$ and takes the greatest common
divisor polynomial and reads off the degree in $x$. The proof of this formula
contained in Ref. Haah2012PauliModule,  is based on an algebraic
representation of the Hamiltonian Eq. (4), which will be reviewed in Sec. VI
below.

The cubic code Hamiltonian Eq. (4) belongs to a class of so-called stabilizer
(code) Hamiltonians, as it is defined as a sum of commuting Pauli operators.
The Kitaev toric code model Kitaev2003Fault-tolerant  and the Wen plaquette
model Wen2003Plaquette  are well-known examples of stabilizer Hamiltonians.
The ground states in these models have a nice geometric interpretation in
terms of string-nets, LevinWen2005String-net  whereas, unfortunately, there is
no known geometric interpretation for the ground state of the cubic code
model, other than the trivial expression Eq. (7).

##  III Entanglement renormalization and bifurcation

It will be useful to recall the notion of _finite depth quantum circuit_. A
depth-1 quantum circuit is a product of local unitary operators of disjoint
support. We do not restrict the number of the unitary operators participating
in the product, but each unitary operator must be local, that is, its support
can be covered by some ball of fixed radius. This radius is referred to as the
range of the circuit. A finite depth quantum circuit is a finite product of
depth-1 quantum circuits. The number of layers must be independent of system
size. The finite depth quantum circuit is a discrete version of the unitary
evolution $e^{-it\mathcal{H}}$ by a sum $\mathcal{H}$ of local Hermitian
operators for $t=O(1)$.

The _entanglement renormalization group_ transformation is a procedure where
one disentangles some of degrees of freedom by local unitary transformations,
and compares the transformed state to the original state. The purpose is to
understand “long range” entanglement. Given a many-qubit quantum state
$\left|\psi\right\rangle$ and a finite depth quantum circuit $U$ such that
$U\left|\psi\right\rangle=\left|\phi\right\rangle\otimes\left|\uparrow\right\rangle\otimes\cdots\otimes\left|\uparrow\right\rangle$,
we discard the qubits in the trivial state $\left|\uparrow\right\rangle$ from
$U\left|\psi\right\rangle$. Then we proceed with $\left|\phi\right\rangle$ in
the next stage of entanglement RG transformations.

The entanglement RG analysis can be done in the Heisenberg picture when we are
interested in a state that is a common eigenstate of a set of operators.
Suppose $\left|\psi\right\rangle$ is defined by equations

$\displaystyle
G_{i}\left|\psi\right\rangle=\left|\psi\right\rangle\quad\text{for any }i$
(11)

where $i$ is some index. Then, the transformed state
$U\left|\psi\right\rangle$ is described by equations

$(UG_{i}U^{\dagger})U\left|\psi\right\rangle=U\left|\psi\right\rangle.$

If $UG_{i}U^{\dagger}$ happens to be an operator, say $\sigma^{z}$ on a single
qubit, then that qubit must be in the state $\left|\uparrow\right\rangle$,
disentangled from the others. This is the criterion by which we identify
disentangled qubits in the calculation below. In addition, we can use this
information to restrict other $G_{j}$ in the next stage of entanglement
renormalization.

The ground state subspace of our model Eq. (4) is described by the stabilizer
equation (11) where the stabilizers $G_{i}$ are just $G^{x}_{i}$ and
$G^{z}_{i}$. Here, observe that the stabilizers $G_{i}$ in Eq. (11) are
invertible operators; $G_{i}$’s form an abelian group $\mathcal{G}=\langle
G_{i}\rangle$, called the stabilizer group. Then, the disentangling criterion
is that for some element $G$ of the stablizer group $\mathcal{G}$,
$UGU^{\dagger}$ acts on a single qubit, where $G$ can be a product of several
$G_{i}$’s.

In fact, only the group $\mathcal{G}$ is important. Consider two gapped
Hamitonians

$H=-J\sum_{i}G_{i},\quad\quad H^{\prime}=-J\sum_{j}G^{\prime}_{j}$

where the terms $G_{i}$ and $G^{\prime}_{j}$ generate the same multiplicative
group $\mathcal{G}=\langle G_{i}\rangle=\langle G^{\prime}_{i}\rangle$. The
ground-state subspace of the two gapped Hamiltonians are identical and they
represent the same quantum phase of matter, in which case we will write

$H\cong H^{\prime}.$ (12)

One can say that $H^{\prime}$ is another parent gapped Hamiltonian of the
ground-state subspace of $H$.

Since the ground state is degenerate, the stabilizer equation (11) does not
pick out a particular state. Nevertheless, the disentanglement criterion in
the Heisenberg picture determines a qubit in the trivial state unambiguously
for any ground state. Thus, even after discarding disentangled qubits, the
transformed Hamiltonian $UHU^{\dagger}$ has a ground-state subspace that is
isomorphic to that of $H$. Our entanglement RG transformation preserves the
ground-state subspace.

We can now state our main result. Let $H_{A}(a)$ be the cubic code Hamiltonian
defined in Eq. (4). Here, the lattice spacing constant $a$ is specified for
notational clarity. We find a finite depth quantum circuit $U$ such that

$\displaystyle UH_{A}(a)U^{\dagger}\cong H_{A}(2a)+H_{B}(2a)$ (13)

where no qubit is involved in both $H_{A}(2a)$ and $H_{B}(2a)$. In Eq. (13),
we have suppressed disentangled qubits; single $\sigma^{z}$ terms are dropped.
The new model $H_{B}$ is defined on a simple cubic lattice with four qubits
per site, with the Hamiltonian

$\displaystyle
H_{B}=-J\sum_{i\in\Lambda}\left(S^{x,1}_{i}+S^{x,2}_{i}+S^{z,1}_{i}+S^{z,2}_{i}\right)$
(14)

where

$\displaystyle S^{x,1}_{i}$
$\displaystyle=\sigma^{x}_{i+\hat{x},1}\sigma^{x}_{i+\hat{z},1}\sigma^{x}_{i,2}\sigma^{x}_{i+\hat{x},2}\sigma^{x}_{i+\hat{x},3}\sigma^{x}_{i+\hat{y},3}\sigma^{x}_{i,4}\sigma^{x}_{i+\hat{y},4}$
$\displaystyle S^{x,2}_{i}$
$\displaystyle=\sigma^{x}_{i,1}\sigma^{x}_{i+\hat{x},1}\sigma^{x}_{i,2}\sigma^{x}_{i+\hat{z},2}\sigma^{x}_{i,3}\sigma^{x}_{i+\hat{y},3}\sigma^{x}_{i,4}\sigma^{x}_{i+\hat{x},4}$
$\displaystyle S^{z,1}_{i}$
$\displaystyle=\sigma^{z}_{i,1}\sigma^{z}_{i-\hat{y},1}\sigma^{z}_{i,2}\sigma^{z}_{i-\hat{x},2}\sigma^{z}_{i,3}\sigma^{z}_{i-\hat{x},3}\sigma^{z}_{i,4}\sigma^{z}_{i-\hat{z},4}$
$\displaystyle S^{z,2}_{i}$
$\displaystyle=\sigma^{z}_{i-\hat{x},1}\sigma^{z}_{i-\hat{y},1}\sigma^{z}_{i,2}\sigma^{z}_{i-\hat{y},2}\sigma^{z}_{i-\hat{x},3}\sigma^{z}_{i-\hat{z},3}\sigma^{z}_{i,4}\sigma^{z}_{i-\hat{x},4}$

are eight-qubit interactions. The interaction terms are visually depicted as

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We perform a similar entanglement RG transformation for $H_{B}$, and find a
finite depth quantum circuit $V$ such that

$\displaystyle VH_{B}(a)V^{\dagger}\cong H_{B}(2a)+H_{B}^{\prime}(2a)$ (15)

with no qubit is involved in both $H_{B}(2a)$ and $H_{B}^{\prime}(2a)$.
$H_{B}$ and $H_{B}^{\prime}$ are the _same_ but act on disjoint sets of
qubits. We have dropped single qubits in the disentangled states on the right-
hand side of Eq. (15). The proof of these formulas and a compact
representation of the models are given in Sec. VI.

The new model $H_{B}$ is different from the original cubic code model $H_{A}$.
We will argue in the next section that they represent different quantum phases
of matter. However, they resemble each other in many ways because they are
related by the finite depth quantum circuit of Eq. (13). Recall that under a
finite depth quantum circuit, any local operator is mapped to a local
operator, and the corresponding operator algebras are isomorphic. In
particular, the two models admit pointlike excitations, which are immobile in
both cases. They have degenerate ground states that are locally
indistinguishable.

##  IV Model A and B are different

By the quantum phase of matter, we mean an equivalence class of gapped
Hamiltonians where the equivalence is defined by adiabatic paths in the space
of gapped Hamiltonians and finite depth quantum circuits assisted with some
ancillary qubits. ChenGuWen2010transformation  The equivalence may be observed
at some different length scale, so one might have to coarse-grain the lattice
in order to see the equivalence. The nonequivalence, on the other hand, must
be proved by contrasting some invariants. We focus on the degeneracy of the
ground states for this purpose.

Suppose the two models $A$ and $B$ represent the same quantum phase of matter.
They must have the same ground-state subspace structure, and in particular the
dimension of the ground-state subspace must be the same. In view of the
fluctuating degeneracy as in Eqs. (9),(10), it means that the degeneracy is
given by the same function of the system size under the same boundary
conditions. Let $k_{A}(L)$ be $\log_{2}$ of the ground-state subspace
dimension of the model $A$, the original cubic code model, on $L\times L\times
L$ periodic lattice, and let $k_{B}(L)$ be that of the model $B$, the new
model discovered by the entanglement RG transformation. From Eq. (13), we have

$k_{A}(2L)=k_{A}(L)+k_{B}(L).$

Eq. (10) implies that

$k_{A}(2L)=2k_{A}(L)+2.$ (16)

Then, it follows that

$k_{B}(2L)=2k_{B}(L),$ (17)

which can also be shown by Eq. (15). It is clear that the function $L\mapsto
k_{A}(L)$ is _different_ from the function $L\mapsto k_{B}(L)$. This is the
basis of the argument for distinctness of the two phases.

We need to take into account the possibility of the equivalence at different
length scales or on distorted lattices. For example, we know that the Wen
plaquette model Wen2003Plaquette  exhibits the same phases of matter as the
toric code model. Kitaev2003Fault-tolerant  However, the Wen plaquette model

$H_{\text{Wen}}=-\sum_{i}\sigma^{z}_{i}\sigma^{x}_{i+\hat{x}}\sigma^{x}_{i+\hat{y}}\sigma^{z}_{i+\hat{x}+\hat{y}}$

has one qubit per lattice site, whereas the toric code model has two. The
degeneracies as functions of system size are different, too.

$k_{\text{toric}}(L)=2,\quad k_{\text{Wen}}(L)=\begin{cases}1&\text{if $L$ is
odd,}\\\ 2&\text{if $L$ is even.}\end{cases}$

To see the equivalence, one has to take a new Bravais lattice for the Wen
plaquette model such that the new unit cell now contains two qubits, and the
unit vectors for the coarser lattice are in the diagonal directions of the
original lattice. The toric code model is recovered once we make local unitary
transformations $\sigma^{z}\leftrightarrow\sigma^{x}$ on every, say, first
qubit in each new unit cell. See Fig. 2.

Figure 2: Equivalence between the Wen plaquette model and the toric code model
can be observed, only when a unit cell is properly chosen.

For the most general choice of new Bravais lattice (smaller translation group)
in the cubic lattice, the new unit translation vectors have integer
coordinates such that the $3\times 3$ matrix $M$ of the cooridnates in the
columns is nonsingular. The unit vectors define a rhombohedron unit cell.
Conversely, given a $3\times 3$ nonsingular integer matrix $M$, one can
introduce a new Bravais lattice to the original cubic lattice by declaring the
columns of $M$ to be new unit translation vectors. Imposing periodic boundary
conditions amounts to specifying the number of translations in each new
direction $\vec{L}^{\prime}=(L_{x}^{\prime},L_{y}^{\prime},L_{z}^{\prime})$
before the translations become the identity translation. Hence, the degeneracy
under periodic boundary conditions is a function of $M$ and the lattice
dimension vector $\vec{L}^{\prime}$; $k=k(M,\vec{L}^{\prime})$.

Suppose now that two models $H_{A}$ and $H_{B}$ are equivalent, and the
equivalence is made explicit at coarser lattices $\Lambda^{\prime}_{A}$ and
$\Lambda^{\prime}_{B}$ defined by nonsingular integer matrices $M_{A}$ and
$M_{B}$, respectively, with respect to the original cubic lattice $\Lambda$.
In particular, we must have

$k_{A}\left(M_{A},\vec{L^{\prime}}\right)=k_{B}\left(M_{B},\vec{L^{\prime}}\right)$

for any lattice dimension vector $\vec{L^{\prime}}$. Consider an even coarser
lattice $\Lambda^{\prime\prime}_{A}$ defined by a nonsingular integer matrix
$N$ with respect to $\Lambda^{\prime}_{A}$, and $\Lambda^{\prime\prime}_{B}$
defined by the same $N$ with respect to $\Lambda^{\prime}_{B}$. We must have

$k_{A}\left(M_{A}N,\vec{L^{\prime\prime}}\right)=k_{B}\left(M_{B}N,\vec{L^{\prime\prime}}\right)$

for any lattice dimension vector $\vec{L^{\prime\prime}}$. Note that $N$ was
arbitrary.

Set the matrix $N$ to be the adjugate matrix of $M_{A}$ so that
$M_{A}N=\det(M_{A})I_{3\times 3}$. $N$ is nonsingular and integral. For
$\vec{L^{\prime\prime}}=(\ell,\ell,\ell)$, we have

$\displaystyle k_{A}\left(\det(M_{A})I_{3\times 3},(\ell,\ell,\ell)\right)$
$\displaystyle=k_{B}\left(M_{B}N,(\ell,\ell,\ell)\right)$
$\displaystyle=k_{A}\left(\det(M_{A})\ell\right),$

where the last function is one that has appeared in Eq. (16). The function
$\phi_{B}:\ell\mapsto k_{B}\left(M_{B}N,(\ell,\ell,\ell)\right)$ has a
property that $\phi_{B}(2\ell)=2\phi_{B}(\ell)$ because of Eq. (15),
regardless of how $M_{B}$ or $N$ is chosen. However, we know from Eq. (16)
that the function $\phi_{A}:\ell\mapsto k_{A}\left(\det(M_{A})\ell\right)$ has
a property that $\phi_{A}(2\ell)=2\phi_{A}(\ell)+2$. This is a contradiction,
and therefore the model $H_{A}$ and $H_{B}$ represents different phases of
matter.

##  V A tensor network description: Branching MERA

The entanglement RG transformation yields a tensor network description for the
state. If one reverses the transformation starting from, say, a state on
$L^{3}$ lattice, one gets a state on $(2L)^{3}$ lattice. After many iterations
one obtains a state on an infinite lattice. It will be an exact description
since our finite depth quantum circuits $U,V$ are exact. In this section we
will refer to local degrees of freedom as qudits.

Let us review Multi-scale Entanglement Renormalization Ansatz (MERA) states.
Vidal2007ER ; Vidal2008MERA  The MERA state is a many-qudit state that is
obtained by reversing the entanglement RG transformations as follows. One
starts with a qudit system on some lattice. (Step 1) Apply a finite depth
quantum circuit with some ancillary qudits in a fixed state
$\left|\uparrow\right\rangle$. Due to the insertion of the ancillary qudits,
the number density of qudits is increased. In order to retain the number
density, (Step 2) one expands the lattice. Then, (Step 3) Iterate Step 1 and
2. In a scale invariant system, one expects that the quantum circuit in Step 1
is the same for every level of the iterations. The class of states that can be
written as a MERA is proposed to describe ground states of some critical
systems, and is shown to admit efficient classical algorithms.

Since the ground state of the toric code model for example is an entanglement
RG fixed point, it naturally has a scale-invariant MERA description. On the
other hand, the cubic code model is not a usual fixed point. At a coarse-
grained level, the ground-state subspace is a tensor product of two
independent ground-state subspaces (Eq. (13)), each of which is again a tensor
product of two independent ground-state subspaces (Eqs. (13),(15)). Reversing
the entanglement RG flow, we see that the final state is obtained by
entangling two states, each of which is again obtained by entangling two
states, and so on.

The “branching MERA,” recently introduced by Evenbly and Vidal,
EvenblyVidal2012RG  is a variant of MERA that captures this scenario. In a
branching MERA, the ancillary trivial qudits in the Step 1 of the usual MERA
are allowed to be in branching MERA states. The self-referential nature is
essential. The total number of branches would grow exponentially with the
coarse-graining level.

The branching structure usually yields very highly entangled states. For
example, in a 1D spin chain, a typical branching MERA state with the total
number of branches being $b_{n}=2^{n}$ at coarse-graining level $n$, obeys a
“volume” law of entanglement entropy. In general, the entanglement entropy of
a ball-like region of linear dimension $L$, for a branching MERA state in a
$D$-dimensional lattice scales like

$S\leq O(1)\sum^{\log_{2}L}_{n=0}b_{n}\left(\frac{L}{2^{n}}\right)^{D-1}$ (18)

where $b_{n}$ is the total number of branches at RG level $n$.
EvenblyVidal2013bMERAentropy  A proof of the formula is given in Appendix A.
In case of our cubic code model where $b_{n}=2^{n}$, it gives an area law. It
is consistent with the fact that it is a stabilizer code Hamiltonian.
HammaIonicioiuZanardi2005

It should be noted that the entanglement entropy scaling alone does not
necessitate the branching structure; it does not nullify the possibility of a
description by the usual unbranched MERA. Our bound in Eq. (18) merely
illustrates that the branching MERA description of the cubic code model is
consistent in view of the entanglement entropy scaling, despite the intuition
that the branching MERA yields much more entanglement.

Rather, the necessity of the branching structure relies on the ground state
degeneracy. If a usual MERA description were possible, the ground space of the
cubic code model on $L^{3}$ (with $L=2^{n}$) lattice would have a one-to-one
correspondence with the Hilbert space of $O(1)=O(L^{0})$ qubits in the top
level of the MERA, and therefore would be of a constant dimension. This would
contradict Eq. (10).

##  VI Calculation method

The finite depth quantum circuits $U$ and $V$ are complicated and not very
enlightening. Explicit circuits and calculation can be found in a Mathematica
script in Supplementary Material. SM  In this section, we explain a machinary
to compute $U$ and $V$. It heavily depends on a special structure of the
Hamiltonians $H_{A}$ and $H_{B}$. The content here is essentially presented in
Ref. Haah2012PauliModule, , so we will be brief.

###  VI.1 Laurent polynomial matrix description

The Pauli $2\times 2$ matrices $\sigma^{x},\sigma^{y},\sigma^{z}$ have a
special property that (i) they square to identity, (ii) the product of any
pair of the matrices results in the third up to a phase factor $\pm 1,\pm i$,
and (iii) they anticommute with one another. In other words, they form an
abelian group under multiplication up to the phase factors. This group,
ignoring the phase factors, is just $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. A
conventional correspondence is given by

$\displaystyle(\sigma^{x})^{n}(\sigma^{z})^{m}$
$\displaystyle\in\langle\sigma^{x},\sigma^{y},\sigma^{z}\rangle/\\{\pm 1,\pm
i\\}$ $\displaystyle\Updownarrow$ (19) $\displaystyle(n,m)$
$\displaystyle\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$

The correspondence easily generalizes to Pauli operators (tensor products of
Pauli matrices). An $n$-qubit Pauli operator corresponds to a bit $\\{0,1\\}$
string of length $2n$: The first half of the bit string expresses
$\sigma^{x}$, while the second half expresses $\sigma^{z}$.

If a qubit system admits translations, e.g. one-dimensional spin chain, the
corresponding bit string can be written in a compact way: Write the bits in
the coefficients of the translation group elements in a formal linear
combination. For example,

$\displaystyle\cdots\otimes\sigma^{x}\otimes\sigma^{z}\otimes
I\otimes\sigma^{y}\otimes\cdots$
$\displaystyle\Leftrightarrow\begin{pmatrix}\cdots&1&0&0&1&\cdots\\\
\cdots&0&1&0&1&\cdots\end{pmatrix}$ (20)
$\displaystyle\Leftrightarrow\begin{pmatrix}\cdots+1t^{-1}+0t^{0}+0t^{1}+1t^{2}+\cdots\\\
\cdots+0t^{-1}+1t^{0}+0t^{1}+1t^{2}+\cdots\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\cdots+t^{-1}+t^{2}+\cdots\\\
\cdots+1+t^{2}+\cdots\end{pmatrix}$

where $t$ denotes the translation by one unit length to the right. This is
merely a change of notation. It yields a particularly simple expression for
translation-invariant Hamiltonians whose terms are Pauli operators, because
one only has to keep a few polynomials that express different types of local
terms. Local terms are expressed not by an infinite Laurent series, but by a
finite linear combination of the translation group elements. Summarizing, we
have introduced a notation for Hamiltonians of Pauli operators using the
translation group algebra with coefficients in $\mathbb{Z}_{2}$.

The cubic code model $H_{A}$ in Eq. (4) can now be written as

$\displaystyle G^{x}=\begin{pmatrix}1+x+y+z\\\ 1+xy+yz+zx\\\ 0\\\
0\end{pmatrix},~{}~{}G^{z}=\begin{pmatrix}0\\\ 0\\\
1+\bar{x}\bar{y}+\bar{y}\bar{z}+\bar{z}\bar{x}\\\
1+\bar{x}+\bar{y}+\bar{z}\end{pmatrix}.$ (21)

where $x,y,z$ are translations along $+\hat{x},+\hat{y},+\hat{z}$-direction,
respectively, and $\bar{x}=x^{-1}$, etc. Since the unit cell of the cubic code
model contains two qubits, we need $2\times 2=4$ rows in the matrix. The first
row expresses $\sigma^{x}$ in the first qubit at each site, the second row
$\sigma^{x}$ the second qubit, the third row $\sigma^{z}$ in the first qubit,
and the fourth row $\sigma^{z}$ in the second qubit. It is the most convenient
to write two matrices in a single matrix where each type of term is written in
each column.

$\displaystyle\sigma=\begin{pmatrix}1+x+y+z&0\\\ 1+xy+yz+zx&0\\\
0&1+\bar{x}\bar{y}+\bar{y}\bar{z}+\bar{z}\bar{x}\\\
0&1+\bar{x}+\bar{y}+\bar{z}\end{pmatrix}$ (22)

We refer to this matrix $\sigma$ as a _generating matrix_ of $H_{A}$.

###  VI.2 Applying periodic local unitary operators

A subclass of finite depth quantum circuits is effectively implemented using
this Laurent polynomial description. It consists of unitaries that respect the
translation symmetry and map Pauli operators to Pauli operators. More
specifically, they are compositions of so-called CNOT, Hadamard, and Phase
gates. For example, Hadamard gate

$U_{\mathrm{Hadamard}}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\\
1&-1\end{pmatrix}\begin{matrix}\left|\uparrow\right\rangle\\\
\left|\downarrow\right\rangle\end{matrix}$

swaps $\sigma^{x}$ and $\sigma^{z}$:

$U_{H}\sigma^{x}U_{H}^{\dagger}=\sigma^{z},~{}~{}U_{H}\sigma^{z}U_{H}^{\dagger}=\sigma^{x}$

If the Hadamard is applied for every qubit on the lattice, then the upper half
and the lower half of the Laurent polynomial matrix will be interchanged.
Similarly, one can work out the action of the CNOT gate

$U_{\mathrm{CNOT}}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&0&1\\\
0&0&1&0\end{pmatrix}\begin{matrix}\left|\uparrow\uparrow\right\rangle\\\
\left|\uparrow\downarrow\right\rangle\\\
\left|\downarrow\uparrow\right\rangle\\\
\left|\downarrow\downarrow\right\rangle\end{matrix}$

and Phase gate

$U_{\mathrm{Phase}}=\begin{pmatrix}1&0\\\
0&i\end{pmatrix}\begin{matrix}\left|\uparrow\right\rangle\\\
\left|\downarrow\right\rangle\end{matrix}$

on the Laurent polynomial matrix. The result is that they correspond to _row
operations_ on the Laurent polynomial matrix. That is, any elementary row
operation $E$, viewed as a left matrix multiplication $\sigma\mapsto E\sigma$,
is admissible as long as $E$ satisfies the symplectic condition

$\bar{E}^{T}\begin{pmatrix}0&I_{q}\\\
I_{q}&0\end{pmatrix}E=\begin{pmatrix}0&I_{q}\\\ I_{q}&0\end{pmatrix}\mod 2.$
(23)

where the bar means the antipode map under which $x\mapsto x^{-1}$, $y\mapsto
y^{-1}$, and $z\mapsto z^{-1}$. Here, $q$ is the number of qubits per unit
cell. $I_{q}$ is the $q\times q$ identity matrix. For a proof, see Ref.
Haah2012PauliModule, .

Note that when the two-qubit unitary operator CNOT above acts within a unit
cell, the antipode map is trivial since $E$ in Eq. (23) will not involve any
variable $x,y,z$, etc; the antipode map does not do anything to coefficients.
When the CNOT acts on a pair of qubits across the unit cells, which is allowed
only if the unit cell contains two or more qubits, the antipode map is
nontrivial. Of course, in any case, the overall unitary must have the same
periodicity with the lattice.

Using the above row operations, one can only generate a finite depth quantum
circuit whose periodicity is $1$. If one wishes to apply, say, Hadamard gates
on every other qubits (periodicity 2), one has to choose a subgroup
$\mathcal{T}^{\prime}$ of the original translation group $\mathcal{T}$, so
that one unit of translation under $\mathcal{T}^{\prime}$ is the translation
by two units under $\mathcal{T}$. Then, one can implement the periodicity $2$
quantum circuit, using the prescription above. Under such a coarse translation
group, our matrix representation of the Hamiltonian must be different.
Computing a new representation is easy, and a prescription is as follows. If
one wishes to take the coarse translation group to be

$\mathcal{T}^{\prime}=\langle x^{\prime},y,z\rangle\leq\langle
x,y,z\rangle=\mathcal{T}$

where $x^{\prime}=x^{2}$, one simply replaces each Laurent polynomial
$f(x,y,z)$ of $\sigma$ with the matrix

$\displaystyle f\left(\begin{pmatrix}0&x^{\prime}\\\
1&0\end{pmatrix},\begin{pmatrix}y&0\\\ 0&y\end{pmatrix},\begin{pmatrix}z&0\\\
0&z\end{pmatrix}\right)$ (24)

If the old generating matrix $\sigma$ was $2q\times m$, then the new
generating matrix is $4q\times 2m$. Again, a proof of this claim can be found
in Ref. Haah2012PauliModule, .

###  VI.3 Example: Toric code model

Let us perform an entanglement RG for the toric code model (Ising gauge
theory). Kitaev2003Fault-tolerant  As we call for strict translation-
invariance, we take the square lattice with the unit cell at a vertex
consisting of one horizontal edge on the east (1) and one vertical edge on the
north (2). The Hamiltonian is

$\displaystyle H_{\mathrm{toric}}=$
$\displaystyle-\sum_{i}\sigma^{x}_{i,1}\sigma^{x}_{i-\hat{x},1}\sigma^{x}_{i,2}\sigma^{x}_{i-\hat{y},2}$
$\displaystyle-\sum_{i}\sigma^{z}_{i,1}\sigma^{z}_{i+\hat{y},1}\sigma^{z}_{i,2}\sigma^{z}_{i+\hat{x},2}$

Following the correspondence Eq. (20), the generating matrix is

$\displaystyle\sigma_{\mathrm{toric}}=\begin{pmatrix}1+\bar{x}&0\\\
1+\bar{y}&0\\\ 0&1+y\\\ 0&1+x\end{pmatrix}.$ (25)

Let us take a smaller translation group $\mathcal{T}^{\prime}=\langle
x^{\prime},y\rangle\leq\langle x,y\rangle$ where $x^{\prime}=x^{2}$.
Accordingto the prescription Eq. (24), the new generating matrix with respect
to $\mathcal{T}^{\prime}$ becomes

$\displaystyle\sigma^{\prime}_{\mathrm{toric}}=\begin{pmatrix}1&1&&\\\
\bar{x}^{\prime}&1&&\\\ 1+\bar{y}&0&&\\\ 0&1+\bar{y}&&\\\ &&1+y&0\\\
&&0&1+y\\\ &&1&x^{\prime}\\\ &&1&1\end{pmatrix}$ (26)

Some zeros are not shown. Now we apply row operations that satisfy Eq. (23).

$\displaystyle\left(\begin{array}[]{cccccccc}1&0&0&0&&&&\\\
\bar{x}^{\prime}&1&0&0&&&&\\\ \bar{y}+1&0&1&0&&&&\\\ \bar{y}+1&0&1&1&&&&\\\
&&&&1&x^{\prime}&1+y&0\\\ &&&&0&1&0&0\\\ &&&&0&0&1&1\\\ &&&&0&0&0&1\\\
\end{array}\right)\sigma^{\prime}_{\mathrm{toric}}$ (35)
$\displaystyle=\left(\begin{array}[]{cccc}1&1&&\\\ 0&\bar{x}^{\prime}+1&&\\\
0&\bar{y}+1&&\\\ 0&0&&\\\ &&0&0\\\ &&0&1+y\\\ &&0&1+x^{\prime}\\\ &&1&1\\\
\end{array}\right)$ (44)

Let us recover the Hamiltonian. We have found a finite depth quantum circuit
$U$ from Eq. (44) such that

$\displaystyle UH_{\mathrm{toric}}U^{\dagger}$
$\displaystyle=-\sum_{i^{\prime}}\sigma^{x}_{i^{\prime},1}-\sum_{i^{\prime}}\sigma^{x}_{i^{\prime},1}\sigma^{x}_{i^{\prime},2}\sigma^{x}_{i^{\prime}-\hat{x}^{\prime},2}\sigma^{x}_{i^{\prime},3}\sigma^{x}_{i^{\prime}-\hat{y},3}$
$\displaystyle-\sum_{i^{\prime}}\sigma^{z}_{i^{\prime},4}-\sum_{i^{\prime}}\sigma^{z}_{i^{\prime},2}\sigma^{z}_{i^{\prime}+\hat{y},2}\sigma^{z}_{i^{\prime},3}\sigma^{z}_{i^{\prime}+\hat{x}^{\prime},3}\sigma^{z}_{i^{\prime},4}.$

Since the Hamiltonian is frustration-free, it is clear that the first and
fourth qubits in each unit cell are in a trivial state and are disentangled
from the rest. As noted above in Sec. III, only the multiplicative group
generated by the terms in the Hamiltonian is important, and we recover
$H_{\mathrm{toric}}$ we started with at a coarse-grained lattice
$\mathcal{T}^{\prime}$. The example demonstrates that _any column operation on
the generating matrix $\sigma$ is allowed_ in view of equivalence Eq. (12).
This shows that the ground state of the toric code model is a fixed point in
an entanglement RG flow. AguadoVidal2007Entanglement

In Supplementary Material, we perform similar calculations for 3D and 4D toric
code models. (3D toric code model is also known as 3D Ising gauge theory.
Wegner1971IsingGauge  4D toric code is similar; qubits live on plaquettes, and
the gauge transformation flips qubits around an edge.
DennisKitaevLandahlEtAl2002Topological ) We verify that they are all
entanglement RG fixed points.

##  VII An algebro-geometric test on entanglement RG

Our example of the bifurcation is very specific to the cubic code model, and
general criteria for the bifurcation to happen are not well understood.
However, we can rule out certain possibilities as follows. We have found an
equivalence by a finite depth quantum circuit between the ground space of
$H_{A}(a)$, where $a$ in the parentheses is the lattice spacing, and that of
$H_{A}(2a)\oplus H_{B}(2a)$. Can we find a similar relation between the ground
space of $H_{A}(a)$ and that of, say, $H_{A}(3a)\oplus H^{\prime}$ for some
Hamiltonian $H^{\prime}$? Put differently, how coarse should a new Bravais
lattice be, if one wishes to find a copy of $H_{A}$ on the new Bravais lattice
by a finite depth quantum circuit?

In this section, we give a _necessary_ condition for this question to be
answered positively by exploiting our Laurent polynomial matrix descriptions.
The condition will detect cases when one will not find a copy of the original
model one started with on a coarser lattice. Our choice of new Bravais lattice
of lattice spacing $2a$ for the cubic code model and the toric code model
satisfies the condition, as it must do.

Let us restrict ourselves to the simplest situation where the generating
matrix $\sigma$ is $2q\times q$, where $q$ is even, and block-diagonal, as in
Eq. (22) and Eq. (25). This is the case when the number of qubits in the unit
cell is the same as the number of interaction types in the Hamiltonian. Note
that in either Eq. (22) or Eq. (25), the upper-left block is described by two
polynomials $f,g$: For the cubic code model, they are $1+x+y+z$ and
$1+xy+yz+zx$. For the toric code model, they are $1+x^{-1}$ and $1+y^{-1}$.
The lower-right blocks in both cases are related to the upper-left blocks by
the antipode map, so we can focus only on the upper-left blocks.

Consider all $q/2\times q/2$ submatrices of the upper-left block of the
generating matrix $\sigma$, and take the determinants of them. Let
$I(\sigma)=\\{f_{i}\\}$ be the set of all such determinants. For example,
$I(\sigma_{\text{toric}})=\\{1+x^{-1},1+y^{-1}\\}$, and
$I(\sigma_{\text{cubic}})=\\{1+x+y+z,1+xy+yz+zx\\}$. Let $V(\sigma)$ be the
set of solutions of the polynomial equations $f_{i}=0$. For example,
$V(\sigma_{\text{toric}})=\\{(x,y)|1+x^{-1}=0,~{}1+y^{-1}=0\\}=\\{(1,1)\\}$.
It is shown in Ref. Haah2012PauliModule,  that $V(\sigma)$ is invariant under
a class of local unitary transformations such that the transformed Hamiltonian
still admits a description by a Laurent polynomial matrix. $V(\sigma)$ is the
object for our algebro-geometric test.

$V(\sigma)$ is a variety, a rather abstract geometric set. In our Laurent
polynomial matrix description, the variables $x,y$, etc. were directly related
to translations. But, now we are treating them as unknown variables and
furthermore equating the polynomials in those variables with zero! Indeed, it
requires good deal of preparation before defining the variety properly, which
is out of the scope of the present paper. We will state facts that are useful
for our purpose. Interested readers are referred to Ref. Haah2012PauliModule,
.

We have seen in Sec. VI.3 that the generating matrix $\sigma$ takes a
different form $\sigma\to\sigma^{\prime}$ depending on our choice of
translation group. Upon taking a coarse translation group, the variety is
changed to $V(\sigma)\to V(\sigma^{\prime})$. Interestingly, one can show that
the change is again given by a nice algebraic map. For example, if we take

$\mathcal{T}^{\prime}=\langle
x^{\prime},y^{\prime},z^{\prime}\rangle\leq\langle x,y,z\rangle=\mathcal{T}$

where $x^{\prime}=x^{n}$, $y^{\prime}=y^{n}$, and $z^{\prime}=z^{n}$ in three-
dimensional lattice, which means $n^{3}$ sites are blocked to form a single
new site, then the change is given by an almost surjective map111Rigorously
speaking, the image of the map is dense in the target variety under Zariski
topology. See e.g. Hartshorne, Algebraic Geometry, Springer

$V(\sigma)\ni(a,b,c)\mapsto(a^{n},b^{n},c^{n})\in V(\sigma^{\prime}).$ (45)

The variety $V(\sigma_{1}\oplus\sigma_{2})$ for the juxtaposition of two
independent systems $\sigma_{1}$ and $\sigma_{2}$ as in Eq. (13), is given by
the union $V(\sigma_{1})\cup V(\sigma_{2})$ of respective varieties.

We have noted that $V(\sigma)$ is invariant under local unitary
transformations. The entanglement RG is a combination of local unitary
transformations after a choice of a smaller translation group. Hence, if a
copy of the original model is to be found in the coarse lattice, _the new
variety $V(\sigma^{\prime})$ must contain the original $V(\sigma)$._ This is a
criterion by which the bifurcation, or an occurrence of the original model at
a coarse lattice _may_ happen. It is unknown if the criterion is a sufficient
condition.

###  VII.1 Examples

Let us apply the criterion to the toric code model and the cubic code model.
As we have seen above, $V(\sigma_{\text{toric}})=\\{(1,1)\\}$. Upon a choice
of a coarser lattice, blocking $2\times 2$ sites as a new one site, the
variety is transformed by the map $x\mapsto x^{2}$ and $y\mapsto y^{2}$.
Obviously, the point $(1,1)$ is invariant under this map, which is consistent
with the fact that the toric code is a RG fixed point.
AguadoVidal2007Entanglement  (See Sec. VI.3.) The readers are encouraged to
compute $V(\sigma^{\prime}_{\text{toric}})$ from Eq. (26) and Eq. (44):
Compute the determinants of all possible $2\times 2$ submatrices of the upper-
left block of $\sigma^{\prime}_{\text{toric}}$, equate them with zero, and
decide the set of solutions.

For the cubic code, the variety is also simple. It consists of two lines each
of which is parametrized by an auxiliary variable $s$:

$\displaystyle\begin{cases}x&=1+s\\\ y&=1+\omega s\\\
z&=1+\omega^{2}s\end{cases},\quad\begin{cases}x&=1+s\\\ y&=1+\omega^{2}s\\\
z&=1+\omega s\end{cases}.$

where $\omega$ is a third root of unity satisfying $\omega^{2}+\omega+1=0$.
(It should be noted that the numbers are not complex numbers; they belong to
extension fields of the binary field $\mathbb{F}_{2}$.) On a coarser lattice
blocking $2^{3}$ sites together, the variety is transformed by the squaring
map. See Eq. (45). Over the binary field,
$(a+b)^{2}=a^{2}+2ab+b^{2}=a^{2}+b^{2}$ for any $a,b$. Hence, the image of the
squaring map is the union of two lines

$\displaystyle\begin{cases}x&=1+s^{2}\\\ y&=1+\omega^{2}s^{2}\\\ z&=1+\omega
s^{2}\end{cases}\quad\begin{cases}x&=1+s^{2}\\\ y&=1+\omega s^{2}\\\
z&=1+\omega^{2}s^{2}\end{cases}$

This is indeed the original variety, although the two lines are interchanged
by the squaring map. This is consistent with the fact that we have found the
original copy $H_{A}$ in the coarse lattice.

Note that the varieties for $H_{A}$ and $H_{B}$ are the same. They do not
distinguish two different phases of matter; the variety is a crude algebro-
geometric object associated to the Hamiltonian.

Before concluding the section, we illustrate an example where the test helps
to choose a correct new unit cell. The color code model,
BombinMartinDelgado2006ColorCode  which is known to be equivalent to two
copies of the toric code model, BombinDuclosCianciPoulin2012  lives on a
honeycomb lattice with one qubit at each vertex. Being a hexagon, any
plaquette $p$ has six vertices $v$. The color code model is defined by the
Hamiltonian

$H=-J\sum_{p}\left(\prod_{v\in p}\sigma^{z}_{v}+\prod_{v\in
p}\sigma^{x}_{v}\right),$

where the sum is over all hexagons. This is expressed with Pauli matrices and
each term commutes with any other, and thus our Laurent polynomial matrix
description is applicable. Since the honeycomb lattice has two vertices in the
conventional unit cell (Fig. 3), our generating matrix $\sigma_{\text{color}}$
is $4\times 2$, as in the toric code model. Explicitly,

$\sigma_{\text{color}}=\begin{pmatrix}1+x+y&0\\\ x+y+xy&0\\\ 0&1+x+y\\\
0&x+y+xy\end{pmatrix}.$

The associated variety is

$\displaystyle V(\sigma_{\text{color}})$
$\displaystyle=\\{(x,y)~{}|~{}1+x+y=0,~{}x+y+xy=0\\}$
$\displaystyle=\\{(\omega,\omega^{2}),~{}(\omega^{2},\omega)\\},$

where $\omega$ is a third root of unity over the binary field.

Figure 3: Honeycomb lattice with qubits numbered within a unit cell.

Suppose one tries to find a copy of itself at a coarser lattice, to see if the
model is an entanglement RG fixed point. One could choose a new Bravais
lattice $\Lambda^{\prime}$ by saying that $x^{\prime}=x^{3}$ and
$y^{\prime}=y^{3}$ are new unit translations. According to Eq. (45), the new
variety $V(\sigma^{\prime}_{\text{color}})$ would be a single point $(1,1)$
since $\omega^{3}=(\omega^{2})^{3}=1$. The original variety is not contained
in the new variety, and therefore one will not find a copy of the original
model on the coarse Bravais lattice $\Lambda^{\prime}$.

On the other hand, if one tried to show the equivalence of the color code
model and the toric code model, then one should take the mentioned Bravais
lattice $\Lambda^{\prime}$; otherwise, the variety of the transformed color
code model would not match that of the toric code model, and the equivalence
would never be explicit.

##  VIII Discussion

We have shown that under the entanglement renormalization group flow the cubic
code model bifurcates. The cubic code model $A$ does not simply produce
exactly the same two copies of itself, but yields a different model $B$. In
order to complete the entanglement RG, we have further shown that the model
$B$ bifurcates into two copies of itself.

The bifurcation alone, as seen in phase B, can be observed in a trivial and
rather ad hoc example: An infinite stack of toric codes. We need to be a
little formal because the example is too trivial. Let $H_{\text{toric}}(a)$ be
the Hamiltonian of the toric code model on a 2D square lattice with qubits on
edges, where lattice spacing is $a$. The entanglement RG transformation
reveals that there is a finite depth quantum circuit $U$ such that

$UH_{\text{toric}}(a)U^{\dagger}\cong H_{\text{toric}}(2a)$

Consider an infinite stack of 2D square lattices with qubits on the edges.
Suppose each layer is parallel to $xy$-plane, and the total system is stacked
in $z$-direction. Our ad hoc Hamiltonian is

$H_{\text{stack}}(a)=\sum_{z=-\infty}^{\infty}H_{\text{toric}}(a)_{z},$

where the subscript $z$ designates the layer that $H_{\text{toric}}(a)$ lives
on. Choosing a new Bravais lattice such that $(0,0,2)$ is a new unit
translation vector, we have

$H_{\text{stack}}(a)=\sum_{z^{\prime}=-\infty}^{\infty}H_{\text{toric}}(a)_{2z^{\prime}}+H_{\text{toric}}(a)_{2z^{\prime}+1}.$

Let $V=\bigotimes_{z=-\infty}^{\infty}U_{z}$ be a finite depth quantum circuit
where $U_{z}$ is just $U$ acting on the layer $z$. Then,

$\displaystyle VH_{\text{stack}}(a)V^{\dagger}$
$\displaystyle=\sum_{z^{\prime}=-\infty}^{\infty}U_{2z^{\prime}}H_{\text{toric}}(a)_{2z^{\prime}}U_{2z^{\prime}}^{\dagger}$
$\displaystyle~{}~{}+\sum_{z^{\prime}=-\infty}^{\infty}U_{2z^{\prime}+1}H_{\text{toric}}(a)_{2z^{\prime}+1}U_{2z^{\prime}+1}^{\dagger}$
$\displaystyle\cong\sum_{z^{\prime}=-\infty}^{\infty}H_{\text{toric}}(2a)_{2z^{\prime}}$
$\displaystyle~{}~{}+\sum_{z^{\prime}=-\infty}^{\infty}H_{\text{toric}}(2a)_{2z^{\prime}+1}$
$\displaystyle=H_{\text{stack}}(2a)_{\text{even}}+H_{\text{stack}}(2a)_{\text{odd}}.$

In contrast, our model cannot be written as a stack of lower dimensional
systems. If it were possible, the ground state degeneracy could not have such
complicated dependence on the system size; at least one parameter, say $L_{z}$
must be factored out from Eq. (9). The fact that the model A and the model B
are different gives a more direct proof that the model $A$ cannot be described
in terms of 2D systems. If the model $A$ was a stack of lower dimensional
ones, the entanglement RG would have yielded the same two copies of itself.

In our tensor network description, the branching MERA, one parametrizes states
by a network of tensors. The topology of the network is fixed and the
entanglement RG changes the values of components of the tensors — It is the
space of tensors where the entanglement RG flows. It should be pointed out,
however, that in our calculation of entanglement RG the disentangling
transformations are obtained accidentally. The calculation was not guided by
any equation, but we just tried to disentangle as many qubits as possible and
discovered that the state belongs to the ground space of two independent
systems. (In fact, the only guide was the consistent behavior of the algebraic
variety under a choice of a new Bravais lattice.) This motivates us to
establish RG equations that incorporates the branching structure. In previous
studies in this direction, VerstraeteCiracLatorreEtAl2005Renormalization ;
GuLevinWen2008Tensor-entanglement  it was implicitly assumed that there is no
branching at the coarse-grained level.

Recently, Swingle Swingle2013  has shown several examples where entanglement
entropy does not decrease under renormalization group transformations, and
argued that the so-called $c$-theorem Zamolodchikov1986cthm  and its higher
dimensional analogs JafferisKlebanovPufuEtAl2011 ; KomargodskiSchwimmer2011
can be violated if Lorentz symmetry is broken. In other words, he argues that
the entanglement entropy is not a RG-monotone in non-Lorentz-invariant
theories. Our example is a yet different (counter)example to those RG-monotone
theorems. The picture that the number density of effective degrees of freedom
should decrease under RG, is manifestly broken. Although it is not
straightforward to directly relate our entanglement RG and the field-theoretic
RG, it will not be the case that in any renormalizable field theory the number
of distinct fields increases as the probing energy scale decreases. This
suggests that the model admits no conventional field theory description that
gives the correct ground space.

###### Acknowledgements.

The author would like to thank Guifre Vidal for raising a question that has
resulted in this work. The author also thanks John Preskill and Glen Evenbly
for numerous helpful discussions. A part of this work was done at IBM Watson
Research Center, Yorktown Heights, New York, where the author was a summer
research intern. The author is supported in part by Caltech Institute for
Quantum Information and Matter, an NSF Physics Frontier Center with support
from Gordon and Betty Moore Foundation, and by MIT Pappalardo Fellowship in
Physics.

##  Appendix A Entanglement entropy of branching MERA states

In this section, we bound the entanglement entropy of a branching MERA
EvenblyVidal2012RG  state between some ball-like region and its complement by
a function of the region’s size. The proof here will be a simplified version
of Ref. EvenblyVidal2013bMERAentropy, . We will relate the entropy scaling
with spatial dimension and the number of branches. A simple lemma will be
useful. Each qudit has Hilbert space dimension $\chi$.

Lemma. Let $A,B,C,D$ be disjoint sets of qudits of dimension $\chi$, and $U$
be a unitary operator acting on $B$ and $C$. Let
$S_{AB}(\rho)=S(\mathop{\mathrm{Tr}}\nolimits_{(AB)^{c}}\rho)$ be the von
Neumann entropy. Then, we have

$\displaystyle|S_{AB}(U\rho U^{\dagger})-S_{AB}(\rho)|\leq(2\log\chi)|C|$ (46)

where $|C|$ is the number of qudits in $C$.

###### Proof.

Let $\rho^{\prime}=U\rho U^{\dagger}$.

$\displaystyle|S_{AB}(\rho^{\prime})-S_{AB}(\rho)|$
$\displaystyle=|S_{AB}(\rho^{\prime})-S_{ABC}(\rho^{\prime})+S_{ABC}(\rho^{\prime})-S_{AB}(\rho)|$
$\displaystyle=|S_{AB}(\rho^{\prime})-S_{ABC}(\rho^{\prime})+S_{ABC}(\rho)-S_{AB}(\rho)|$
$\displaystyle\leq|S_{AB}(\rho^{\prime})-S_{ABC}(\rho^{\prime})|+|S_{ABC}(\rho)-S_{AB}(\rho)|$
$\displaystyle\leq S_{C}(\rho^{\prime})+S_{C}(\rho)\leq(2\log\chi)|C|$

In the second inequality, we used the subadditivity of entropy. ∎

The inequality is saturated by the swap operator. If $A,B,C,D$ are single
qubits, respectively, and $\psi$ consists of two pairs of singlets in $AB$ and
$CD$, then $S_{AB}(\psi)=0$. Swapping $B$ and $C$, we have
$S_{AB}(\psi^{\prime})=2\log 2$. The lemma implies that a finite depth quantum
circuit can only generate entanglement between two regions along the boundary.

We wish to consider the entanglement entropy
$S_{0}(\left|\psi\right\rangle)=S(\rho)$, where
$\rho=\mathop{\mathrm{Tr}}\nolimits_{B^{c}}(\left|\psi\right\rangle\left\langle\psi\right|)$,
between a (hyper)cubic region $B$ of linear size $L$ and its complement of a
branching MERA state $\left|\psi\right\rangle$.

By definition, $\left|\psi\right\rangle$ accompanies entanglement RG
transformations $U_{\tau}$ ($\tau=1,2,\ldots$). $U_{1}\left|\psi\right\rangle$
is either a tensor product of one or more states
$\left|\psi_{1}^{1}\right\rangle,\left|\psi_{1}^{2}\right\rangle,\ldots,\left|\psi_{1}^{b}\right\rangle$
($b\geq 1$) each of which is living on a coarser lattice (branch), or some
entangled state of those. To be concrete, suppose the density of degrees of
freedom decreases by a factor of $2^{D}$ on the coarser lattice. The number
$b$ of branches should be $\leq 2^{D}$.

Let $\rho_{1}^{(1)},\ldots,\rho_{1}^{(b)}$ be reduced density matrices of
$U_{1}\left|\psi\right\rangle$ for the corresponding region $B_{1}^{i}$ on
each branch. Each $B_{1}^{i}$ contains $(L/2)^{D}$ qudits. By the lemma and
the subadditivity of entropy, we have

$\displaystyle S(\rho)$ $\displaystyle\leq
S(\mathop{\mathrm{Tr}}\nolimits_{B^{c}}U_{1}\left|\psi\right\rangle\left\langle\psi\right|U_{1}^{\dagger})+c|\partial
B|$ $\displaystyle\leq S(\rho_{1}^{(1)})+\cdots+S(\rho_{1}^{(b)})+c|\partial
B|$ (47)

where $c$ is a constant depending only on the detail of the circuit $U_{1}$’s
locality property. Here, $|\partial B|$ is the number of qudits outside $B$
but within the range of $U_{1}$ from $B$. So, $c|\partial
B|\leq(2\log\chi)2D(L+2)^{D-1}$ if $U_{1}$ is of depth 1 and range 2. One can
iterate the inequality Eq. (47) with $B_{1}^{i}$ in place of $B$.

$S(\rho)\leq\sum_{i=1}^{b_{N}}S(\rho_{N}^{(i)})+c^{\prime}\sum_{n=0}^{N-1}b_{n}\left(\frac{L}{2^{n}}\right)^{D-1}$
(48)

for any $N\geq 0$ where $b_{n}$ is the total number of all branches, and
$\rho_{N}^{(i)}$ is the reduced density matrix of $U_{N}U_{N-1}\cdots
U_{1}\left|\psi\right\rangle$ for the region $B_{N}^{(i)}$ of linear size
$L/2^{N}$ on branch $i$. In particular, $b_{0}=1$ and $b_{1}=b$ above. In a
usual MERA, we have $b_{n}=1$ for all $n$. The constant $c^{\prime}$ only
depends on $\chi$ and the details of the depth and range of circuits
$U_{1},\ldots,U_{N}$.

An appropriate $N$ must be chosen in order for Eq. (48) to be useful. A
straightforward choice is such that $B_{N}^{(i)}$ contains a constant number
of qudits, i.e., $N=\lfloor\log_{2}L\rfloor$. Then, $\rho_{N}^{(i)}$ is a
density matrix of a constant number of qudits, so
$S(\rho_{N}^{(i)})=O(\log\chi)$. Eq. (48) finally implies

$\displaystyle S(\rho)$ $\displaystyle\leq
O(\log\chi)\sum_{n=0}^{\lfloor\log_{2}L\rfloor}b_{n}\left(\frac{L}{2^{n}}\right)^{D-1}.$
(49)

Specializing, we get

$\displaystyle S(\rho)$ $\displaystyle=\begin{cases}O(L^{D-1})&\text{if
$b_{n}=b^{n}<(2^{D-1})^{n}$},\\\ O(L^{D-1}\log L)&\text{if
$b_{n}=(2^{D-1})^{n}$},\\\ O\left(L^{\log_{2}(b/2^{D-1})}\right)&\text{if
$b_{n}=b^{n}>(2^{D-1})^{n}$}.\end{cases}$ (50)

The number $2$ is of course the linear size of a superblock, and can be
replaced by any positive integer.

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