# On The Stabilizer Formalism And Its Generalization

Éloi DESCAMPS1,2, Borivoje DAKIĆ3,4 1Département de Physique de l’École
Normale Supérieure - PSL, 45 rue d’Ulm, 75230 Paris, France  2Universitée
Paris Cité, Laboratoire Matériaux et Phénomènes Quantiques, 75013 Paris,
France  3University of Vienna, Faculty of Physics, Vienna Center for Quantum
Science and Technology, Boltzmanngasse 5, 1090 Vienna, Austria  4 Institute
for Quantum Optics and Quantum Information (IQOQI) Vienna, Austrian Academy of
Sciences, Boltzmanngasse 3, A-1090 Vienna, Austria

###### Abstract

The standard stabilizer formalism provides a setting to show that quantum
computation restricted to operations within the Clifford group are classically
efficiently simulable: this is the content of the well-known Gottesman-Knill
theorem [1]. This work analyzes the mathematical structure behind this theorem
to find possible generalizations and derivation of constraints required for
constructing a non-trivial generalized Clifford group. We prove that if the
closure of the stabilizing set is dense in the set of $SU(d)$ transformations,
then the associated Clifford group is trivial, consisting only of local gates
and permutations of subsystems. This result demonstrates the close
relationship between the density of the stabilizing set and the simplicity of
the corresponding Clifford group. We apply the analysis to investigate
stabilization with binary observables for qubits and find that the formalism
is equivalent to the standard stabilization for a low number of qubits. Based
on the observed patterns, we conjecture that a large class of generalized
stabilizer states are equivalent to the standard ones. Our results can be used
to construct novel Gottesman-Knill-type results and consequently draw a
sharper line between quantum and classical computation.

††preprint: APS/123-QED

##  I Introduction

Quantum computation uses quantum systems to perform calculations beyond the
capabilities of classical (standard) computers [2]. Many quantum algorithms
solve problems seemingly intractable for classical computers. Prominent
examples include Shor’s factoring [3], Grover search [4], phase estimation
[5], quantum simulation algorithms [6] etc. A common belief is that only some
parts of quantum computation possess a significant speedup. Thus, it is
crucial to identify sets of gates that make quantum computation classically
(in)tractable to understand how this advantage arises. Some models do not
bring (exponential) speedups, and this question has been extensively studied
in the literature [7, 8] prominently with the aid of the so-called stabilizer
formalism. Based on this formalism, the Gottesman-Knill theorem identifies
a“non-trivial” portion of the quantum-mechanical computation that can be
efficiently simulated classically. The theorem has a pedagogical value in
showing substantial differences and similarities between classical and quantum
computation. This theorem also saw few generalizations obtained by varying
different parameters of the setting [9, 10] leading to a better insights into
the boundary between classical and quantum models.

In this work, we continue this search for possible generalizations of the
Gottesman-Knill theorem and the underlying stabilizer formalism. Our main
results are: a) we prove the constraint theorem for the generalized Clifford
group, b) we provide an exhaustive analysis of binary stabilization for two
and three qubits, and based on observed patterns, c) we conjecture that binary
stabilization for qubits is equivalent to the standard stabilizer formalism.
The paper is organized as follows. In Section II we discuss in more details
the stabilizer formalism and the Gottesman-Knill theorem and then proceed by
defining the generalized stabilizer states, stabilizer sets and Clifford
groups. In Section III, we focus on infinite stabilization sets, specifically
those that generate a dense set within the group of $SU(d)$ transformations.
In such case, we derive a theorem that demonstrates that the corresponding
Clifford group is trivial, composed solely of local gates and permutations of
subsystems. This finding highlights the direct relationship between the
density of the stabilization set and the triviality of the associated Clifford
group. Lastly, in Section IV we discuss the stabilization for qubits and
binary operators. We provide a complete analysis for the two and three qubits
cases and we show that in these cases the stabilizer formalism is equivalent
to the standard one. Based on this, we conjecture that this equivalence
extends to an arbitrary number of qubits.

##  II Towards a generalization of the stabilizer formalism

We start our analysis with the set of $N$ qudits residing in the Hilbert space
$(\mathbb{C}^{d})^{\otimes N}$. The basic idea of the stabilizer formalism is
to fix a state $\ket{\psi}$ with the set of operators $O_{1},\cdots,O_{k}$
such that:

  * •

$\ket{\psi}$ is a $+1$-eigenvector (with eigenvalue $+1$) for all $O_{i}$, and

  * •

$\ket{\psi}$ is unique (up to a complex factor).

These hypotheses allow for a full characterization of the state $\ket{\psi}$
by only specifying this list $O_{1},\cdots,O_{k}$ of operators. We will
consider (local) tensor operators $O_{i}\in\mathcal{A}^{\otimes N}$ with
$\mathcal{A}\subset{U(d)}$, where $U(d)$ is the group of $d\times d$ unitary
matrices. The set $\mathcal{A}$ is called the stabilizing set, and any state
$\ket{\psi}$ uniquely stabilized (i.e., satisfying the two conditions above)
is a stabilizer state.

We now introduce the generalized Clifford group $\mathcal{C}_{N}(\mathcal{A})$
as follows

$\mathcal{C}_{N}(\mathcal{A})=\\{U\in U(d^{N})|\forall
O\in\mathcal{A}^{\otimes N},~{}UOU^{\dagger}\in\mathcal{A}^{\otimes N}\\}.$
(1)

The definition ensures that $O\ket{\psi}$ is also a stabilizer state for any
stabilizer states $\ket{\psi}$ and $O\in\mathcal{C}_{N}(\mathcal{A})$.

A classic example of such stabilization setting is given by the set of Pauli
matrices and this is the (standard) stabilizer formalism [11]. In this case
$\mathcal{A}=\mathcal{P}$, and

$\mathcal{P}=\\{\pm 1,\pm
i\\}\cdot\\{\mathds{1},\sigma_{X},\sigma_{Y},\sigma_{Z}\\},$ (2)

where $\mathds{1}$ is the identity and $\sigma_{X}$, $\sigma_{Y}$ and
$\sigma_{Z}$ are the usual Pauli matrices. The corresponding Clifford group
is:

$\mathcal{C}_{N}=\\{U\in U(d^{N})|\forall
O\in\mathcal{P},~{}UOU^{\dagger}\in\mathcal{P}\\}.$ (3)

This Clifford group is generated by two single-qubit gates and one two-qubit
gate [12]:

  * •

Hadamard gate: $H=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\ 1&-1\\\
\end{array}\right)$

  * •

$\frac{\pi}{2}$-phase gate: $S=\left(\begin{array}[]{cc}1&0\\\ 0&i\\\
\end{array}\right)$

  * •

Controlled phase gate $\Lambda Z=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&-1\end{array}\right)$ or

  * •

Controlled not gate $\Lambda X=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\ 0&0&0&1\\\ 0&0&1&0\end{array}\right)$

Computation involving gates from the Clifford group can be realised as a
successive application of these three generators. With these definitions we
can give a precise statement of the Gottesman-Knill theorem [1]:

######  Theorem 1.

(Gottesman-Knill) Computation utilizing only:

  * •

Preparation of qubits in states of the computational basis,

  * •

Quantum gates from the Clifford group $\mathcal{C}_{N}$, and

  * •

Measurements in the computational basis,

can be simulated efficiently on a classical probabilistic computer.

The proof is mainly based on the stabilizer formalism [2], suggesting that the
general definition we gave for the stabilization may lead to a generalized
version of the Gottesman-Knill theorem. However, it may happen that the set
$\mathcal{C}_{N}(\mathcal{A})$ defined in eq. (1), is composed of local gates
only, reducing the model to computation by local gates only. In such
situation, the corresponding generalized theorem would be trivially true.
Thus, two interesting questions arise in this respect:

  * •

For which set $\mathcal{A}$ is the corresponding Clifford group not trivial?

  * •

What kind of states can be stabilized by different sets $\mathcal{A}$?

What is meant by a trivial Clifford group here is a group contained in
$U(d)^{\otimes N}P_{N}(d)$, where $P_{N}(d)$ is the set of permutation gates
on $N$ qudits, with elements $P_{\sigma}$ acting on the basis states in the
following way

$P_{\sigma}\ket{i_{1},\cdots,i_{N}}=\ket{i_{\sigma(1)},\cdots,i_{\sigma(N)}},$
(4)

for any permutation $\sigma$ of $N$ elements. In other words, a trivial
Clifford group is composed only of local and permutation gates. Such a group
does not generate entanglement and thus the computation involving only gates
from it is considered trivial.

##  III Generalized Clifford Group

We shall try to find examples of $\mathcal{C}_{N}(\mathcal{A})$ being non-
trivial, this means containing entangling gates. This problem was also studied
by [9] where the authors studied the Clifford group with $\mathcal{A}$ being a
finite group acting irreducibly on $\mathbb{C}^{d}$. Here we consider a more
general case of infinite groups being involved. We cover a first step in this
direction by considering the stabilizing set $\mathcal{A}$ to generate a group
$\langle\mathcal{A}\rangle$ dense inside $SU(d)$, i.e.,
$SU(d)\subset\overline{\langle\mathcal{A}\rangle}$, where $\overline{O}$
denote the topological closure of the set $O$.
Under these assumptions we show a series of results leading to a new necessary
condition for a generalized Clifford group to be non-trivial. We show the
following result.

######  Proposition 1.

If the set $\langle\mathcal{A}\rangle$ is dense inside $SU(d)$, then for all
integers $N$, the Clifford group $\mathcal{C}_{N}(\mathcal{A})$ verifies
$\mathcal{C}_{N}(\mathcal{A})\subset\mathcal{C}_{N}(U(d))$.

######  Proof.

See Appendix (A) ∎

Next, we show that a gate from $\mathcal{C}_{N}(U(d))$ is non-entangling,
which means that it maps product-states to product-states.

######  Proposition 2.

If $U\in\mathcal{C}_{N}(U(d))$, then for all product-state
$\ket{\psi}=\ket{\varphi_{1}}\cdots\ket{\varphi_{N}}$, we have
$U\ket{\psi}=\ket{\tilde{\varphi_{1}}}\cdots\ket{\tilde{\varphi_{N}}}$ for
some states $\ket{\tilde{\varphi_{1}}},\cdots,\ket{\tilde{\varphi_{N}}}$.

######  Proof.

See Appendix (B) ∎

An equivalent result can be obtained for product “bra” (dual) states. This is
simply obtained by varying the proof and looking at the left eigenvectors. We
thus have that all $U\in\mathcal{C}_{N}(\mathcal{A})$ stabilize the set of
product linear forms, i.e.,
$\bra{\psi}=\bra{\varphi_{1}}\cdots\bra{\varphi_{N}}$, and
$\bra{\psi}U=\bra{\tilde{\varphi_{1}}}\cdots\bra{\tilde{\varphi_{N}}}$.
With this, we can show an already known result [13], of factorizing non-
entangling gates into a tensor product of local gates up to a permutation
gate. However, in comparison to the proof provided in [13] our proof employs
elementary mathematics and provides the result in full generality (for
arbitrary $d$ and $N$).

######  Proposition 3.

The non-entangling set $\mathcal{C}_{N}(U(d))$ acting on N qudits is composed
of trivial gates , i.e.,

$\mathcal{C}_{N}(U(d))=U(d)^{\otimes N}P_{N}(d).$ (5)

######  Proof.

See Appendix (D). ∎

By combining these intermediary results, we arrive to the following theorem:

######  Theorem 2.

Let $\mathcal{A}\subset U(d)$ such that, the group $\langle\mathcal{A}\rangle$
generated by $\mathcal{A}$ is dense inside $SU(d)$. Then for all integers $N$,
the Clifford group of order $N$, $\mathcal{C}_{N}(\mathcal{A})$ is trivial,
i.e., verifies $\mathcal{C}_{N}(\mathcal{A})\subset U(d)^{\otimes N}P_{N}(d)$.

Theorem 2 simply means that if the set $\mathcal{A}$ generates a group which
is too big, then its corresponding generalized Clifford group
$\mathcal{C}_{N}(\mathcal{A})$ is only made of trivial gates. In this
situation, a computation starting with a product-state will keep a separable
form throughout the application of gates from the Clifford group. Given this,
we can keep track of each qudit individually and perform an efficient
classical simulation of the computation. This result gives a non-trivial
constraint on the stabilizing set $\mathcal{A}$, which is needed to generalize
the stabilizer formalism.

##  IV Exploring generalized stabilization for qubits

In the previous section we found some constraints on the stabilizing sets
$\mathcal{A}$. However, the Gottesman-Knill theorem does not only rely on the
fact that the standard Clifford group $\mathcal{C}_{N}$ is not trivial but
also that the set of all Pauli-stabilizer states (i.e., state of the standard
stabilizer formalism) posses a specific and rich structure. Thus, when
analyzing a potential generalized stabilizer setting, an important question to
address is determining the specific nature (e.g., entanglement structure) of
the stabilizer states. We focus our attention to the case of qubits ($d=2$).
To arrive at novel structures, it is thus important to understand when a set
$\mathcal{A}$ stabilizes states that are not locally equivalent to Pauli-
stabilized ones. We say that two $N$-qubits states $\ket{\psi}$ and
$\ket{\phi}$ are locally equivalent if there exist a local gate
$U=U_{1}\otimes\cdots\otimes U_{N}$ such that $\ket{\psi}=U\ket{\phi}$ with
$U_{i}\in U(2)$.

###  IV.1 Binary operator case

The problem in its full generality ($\mathcal{A}$ composed of arbitrary
unitary matrices) is hard, we will thus consider a simpler case of binary
observables. This choice comes naturally, as the Pauli matrices generating the
Pauli group (standard stabilizer formalism), are specific instance of binary
operators, i.e., operators of the form
$\overrightarrow{\mathbf{\sigma}}\cdot\overrightarrow{\mathbf{n}}$ where
$\overrightarrow{\mathbf{\sigma}}=(\sigma_{X},\sigma_{Y},\sigma_{Z})$ and
$\overrightarrow{\mathbf{n}}$ is a unit-norm vector. We first take a look at
the stabilization problem when the elements of $\mathcal{A}$ are general
binary operators.

Recall that a necessary condition to have stabilization by a list of
operators, is to have for each pair of stabilizers common $+1$-eigenvectors.
We thus investigate this pairwise condition in the case of binary observables
$\overrightarrow{\mathbf{\sigma}}\cdot\overrightarrow{\mathbf{n}}$. Since
stabilized sets are equivalent up to local unitary, without loss of
generality, we set the first stabilizing operator to be
$\sigma_{Z}\otimes\cdots\otimes\sigma_{Z}$ and the second one can to be
composed of operators in the $(XZ)$-plane (i.e., operators of the form
$\cos(\theta)\sigma_{Z}+\sin(\theta)\sigma_{X}$). To begin with, we introduce
some useful notation.

Notation:

  * •

For real angles $\theta$ and $\phi$ a general binary operator on the Bloch
sphere is defined as

$A_{\theta,\phi}=\cos(\theta)\sigma_{Z}+\sin(\theta)\Big{(}\cos(\phi)\sigma_{X}+\sin(\phi)\sigma_{Y}\Big{)}$
(6)

  * •

For operators in the $(XZ)$-plane we set $A_{\theta}=A_{\theta,0}$.

  * •

For $n$ binary operators $A_{1}$,…,$A_{n}$ the projector on the
$+1$-eigenspace is denoted as: $P_{A_{1}\otimes\cdots\otimes
A_{n}}=\frac{1}{2}(\mathds{1}+A_{1}\otimes\cdots\otimes A_{n})$

  * •

For operators in the $(XZ)$-plane, $A_{\theta_{1}}$,…,$A_{\theta_{n}}$ we set
notation $P_{A_{\theta_{1}}\otimes\cdots\otimes
A_{\theta_{n}}}=P_{\theta_{1},\dots,\theta_{n}}$

  * •

For indices $j_{1},\dots,j_{n}\in\\{-1,+1\\}$, we define the unormalized
states:

$\ket{\psi_{j_{1},\dots,j_{n}}}=\operatorname{Re}\left[\sum\limits_{k_{1},\dots,k_{n}=0,1}(ij_{1})^{k_{1}}\cdots(ij_{n})^{k_{n}}\ket{k_{1}\cdots
k_{n}}\right].$ (7)

From a technical standpoint, the last definition proves to be highly useful as
in our calculations, it frequently becomes necessary to retain terms within
sums that involve an even number of indices $k_{i}$, all of which are set to
the value $1$. We also see that
$\ket{\psi_{j_{1},\cdots,j_{n}}}=\ket{\psi_{-j_{1},\cdots,-j_{n}}}$ and this
is why we fix the first index to $+1$ to avoid counting twice the same state.
With these definitions in place, we can state a technical theorem which has
profound consequences on the stabilization with binary operators in the
$(XZ)$-plane.

######  Theorem 3.

For $n$ angles $\theta_{1},\dots,\theta_{n}$, the eigenvalues (with
multiplicity) of $P_{0,\cdots,0}P_{\theta_{1},\cdots,\theta_{n}}$ are the
following:

  * •

$2^{n-1}$ zeros,

  * •

$\frac{1}{2}(1+\cos(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n}))$ for
$j_{1},...,j_{n}\in\\{-1,+1\\}$ and $j_{1}=1$, with the corresponding
eigenvector $\ket{\psi_{j_{1},\cdots,j_{n}}}$.

######  Proof.

See Appendix (E) ∎

Recall that we are interested in the stabilization of a state by a family of
tensor product of binary operators. Two of them are chosen as a reference,
and, as already said, due to local equivalence, we can always set them to
$\sigma_{Z}\otimes\cdots\otimes\sigma_{Z}$ and
$A_{\theta_{1}}\otimes\cdots\otimes A_{\theta_{N}}$. Their common
$+1$-eigenvector is simply given by a state stabilized by the product of the
projectors $P_{0,\cdots,0}P_{\theta_{1},\cdots,\theta_{N}}$. Given Theorem 3,
we can conclude the following. As the eigenvalues are
$\frac{1}{2}(1+\cos(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n}))$, we can vary
parameters $\theta_{1},\dots,\theta_{n}$ to control the number of stabilized
states. Note that these eigenstates are independent of the value of
$\theta_{k}$. Thus stabilization with operators in the $(XZ)$-plane, we will
only lead to states given by equation (7). On the other hand, such states can
be stabilized by taking tensor product operator from
$\\{\sigma_{Z},\pm\sigma_{X}\\}$ only. Indeed,
$\ket{\psi_{j_{1},\cdots,j_{N}}}$ is uniquely stabilized by the operators

$O_{k}=\sigma_{X}\otimes\sigma_{Z}\otimes\cdots\otimes\sigma_{Z}\otimes\underbrace{(-j_{k}\sigma_{X})}_{\text{at
}k}\otimes\sigma_{Z}\otimes\cdots\otimes\sigma_{Z},$ (8)

for $k=2,\dots,n$. To see this, we can compute the eigenvalue of $O_{k}$ for
any state $\ket{\psi_{i_{1},\cdots,i_{N}}}$ and verify that it is $1$ if and
only if $i_{k}=j_{k}$.

This means all states stabilized by operators in the $(XZ)$-plane are locally
equivalent to Pauli stabilizer states. However, note that not all of Pauli
stabilizer states can be obtained in this way. For example the state
$\ket{00}$ cannot be stabilized by only binary operators in the $(XZ)$-plane,
because states in eq. (7) for two qubits are maximally entangled, i.e,
$\ket{\psi_{j_{1},j_{2}}}=\ket{00}\pm\ket{11}$. Since the identity operator
allows for the stabilization of product-states, we shall add the identity as
an element of our stabilizing set $\mathcal{A}$.

###  IV.2 Adding the identity

We have seen in the previous section that stabilization by operators
constructed from the $(XZ)$-plane only yields standard stabilizers states, but
states in eq. (7) do not span all possibilities, thus adding identity to the
set $\mathcal{A}$ is necessary. Therefore, we will explore stabilizing sets of
the form $\mathcal{A}=\\{A_{\theta,\phi},\mathds{1}_{2}\\}$. For a small
number of qubits, we can exhaustively search all possibilities, and from
there, we can get better insight into general constraints on the stabilizing
set. We can also compute the associated stabilizer states up to local
equivalence.

####  IV.2.1 Methods

To do an exhaustive search, firstly, we list all non-equivalent stabilization
patterns. A stabilization pattern simply corresponds to the list of
stabilizers:

$\displaystyle O_{1}$ $\displaystyle=A_{1,1}\otimes\cdots\otimes A_{1,n}$
$\displaystyle\cdots$ (9) $\displaystyle O_{k}$
$\displaystyle=A_{k,1}\otimes\cdots\otimes A_{k,n}$

We are interested in stabilization up to local rotation, thus, two
stabilization patterns are equivalent if they can be transformed to each other
via a) permutation of the qubits, b) permutation of the operators, and c)
local rotations. With this, we seek for

  * •

Unique stabilization. There is only one common $+1$\- eigenstate of all the
operators $O_{k}$.

  * •

Minimal stabilization. We cannot achieve a unique stabilization with a proper
subset of the operators $\\{O_{k}\\}$.

After listing all inequivalent stabilization patterns, we shall analyze each
one separately to identify parameters $\theta_{k}$ for which we get unique
stabilization. Finally, we shall identify stabilized states. To do so we will
use two different methods.

With the first method, which we call the determinant method, we consider
$\ket{\psi_{i}}$ ($1\leq i\leq m$) to be the eigenbasis basis of $O_{1}$. A
common eigenstate $\ket{\psi}$ is then expanded over this basis:

$\ket{\psi}=\sum_{i=1}^{m}\alpha_{i}\ket{\psi_{i}}.$ (10)

We write $O_{j}\ket{\psi_{i}}=\sum_{l=1}^{m}\beta_{ijl}\ket{\psi_{l}}$, and
thus the condition $O_{j}\ket{\psi}=\ket{\psi}$ implies for all $1\leq j\leq
k$ and $1\leq l\leq m$,

$\sum_{i=1}^{m}\alpha_{i}\beta_{ijl}=\alpha_{l}.$ (11)

This is a linear system in $m$ unknown $\alpha_{i}$’s with $km$ equations,
which can be summarized in matrix form as:

$M\cdot\begin{pmatrix}\alpha_{1}\\\ \vdots\\\ \alpha_{m}\end{pmatrix}=0,$ (12)

with $M$ being a $km\times m$ matrix. The solution of the system is then given
by the kernel $M$. To find the kernel, we impose the constraint
$\det(M^{\dagger}M)=0$ which gives us a necessary condition for the set of
coefficients $\beta_{ijl}$ and consequently on the set of angles
$\theta_{1},\dots,\theta_{N}$ that parameterize our stabilizers.

The second approach, called the projector method, involves examining the
projector $P_{i}=(\mathds{1}+O_{i})/2$ on the $+1$-eigenspace for each
stabilizing operator $O_{i}$. We consider their product $M=P_{1}P_{2}\dots
P_{k}$. A state $\ket{\psi}$ is stabilized by all $O_{i}$ if and only if
$M\ket{\psi}=\ket{\psi}$, therefore, we search for the eigenvalues and
eigenvectors of $M$. The expressions for the eigenvalues of $M$, combined with
the requirement that one of them equals $1$, impose constraints on the angles
$\theta_{1},\dots,\theta_{N}$. Furthermore computing the eigenbasis of $M$
directly provides the expression of the stabilizer states. This method
requires more computational resources, making it suitable only for simpler
scenarios characterized by a limited number of free parameters.

These two methods are applied in the analysis for two and three qubits.

####  IV.2.2 Two-qubits case

We have identified four inequivalent stabilization patterns with two
operators. Table 1 outlines the stabilization patterns along with the
necessary conditions on the free parameters to achieve unique stabilization,
as well as the corresponding stabilizer states.

Operators | Conditions | Stabilized states
---|---|---
$\begin{matrix}\sigma_{Z}\otimes\sigma_{Z}\\\ A_{\theta}\otimes A_{\phi}\end{matrix}$ | $\theta\pm\phi=0~{}[2\pi]$ | $\ket{\psi}=\ket{00}\pm\ket{11}$
$\begin{matrix}\sigma_{Z}\otimes\sigma_{Z}\\\ A_{\theta}\otimes\mathds{1}\end{matrix}$ | $\theta=0~{}[\pi]$ |  $\ket{\psi}=\ket{00}$ or $\ket{11}$
$\begin{matrix}\sigma_{Z}\otimes\mathds{1}\\\ A_{\theta}\otimes\mathds{1}\end{matrix}$ | $\theta=0~{}[2\pi]$ | Non unique
$\begin{matrix}\sigma_{Z}\otimes\mathds{1}\\\ \mathds{1}\otimes\sigma_{Z}\end{matrix}$ | None | $\ket{\psi}=\ket{00}$
Table 1: List of non-equivalent stabilization pattern for two qubits together
with the associated stabilizer states.

We see that the stabilizer states are either product-states or maximally
entangled states. These are all locally equivalent to Pauli stabilizer states.

####  IV.2.3 Three qubits-case

The three qubit stabilization analysis is more involved, since there are dozen
of inequivalent stabilization patterns. We start the analysis of
stabilizations with only two operators. This gives us some minimal unique
stabilization, and also a two-operators compatibility rule. We enumerate $9$
distinct stabilization patterns, but only one of them yields a unique
stabilization, where the stabilizer states correspond to those described in
eq. (7). Additionally, we identify $8$ patterns that stabilize a two-
dimensional space, which establishes a compatibility condition used in the
subsequent analysis. From there, we also provide an exhaustive search for
three operators. Details are provided in Appendix F.
We found that stabilization is locally equivalent to Pauli stabilizer states.
Table 2 and 3 shown in appendix F summarizes the analysis and findings.

####  IV.2.4 Conjecture

Our analysis for a low number of qubits shows that all stabilization with
binary operators (including identity operator) is locally equivalent to the
standard Pauli stabilization. This motivates us to put forward the following
conjecture:

######  Conjecture 1.

All states stabilized by the set $\mathcal{A}$ composed of binary operators
and the identity i.e., $\mathcal{A}=\\{A_{\theta,\phi},\mathds{1}_{2}\\}$, are
locally equivalent to standard stabilizer states (i.e., stabilized by the
Pauli group $\mathcal{P}=\\{\pm 1,\pm
i\\}\cdot\\{\mathds{1},\sigma_{X},\sigma_{Y},\sigma_{Z}\\}$).

One way to tackle this conjecture is to go back to stabilization on the
$(XZ)$-plane, where a weaker conjecture with
$\mathcal{A}=\\{A_{\theta},\mathds{1}\\}$ is also probably true. Furthermore,
on a small number of qubits, we observe that all Pauli stabilizations also
seem possible (up to local equivalence) without the operator $\sigma_{Y}$. So,
potentially, it is possible that by including the identity matrix as a
stabilizer, a way can be found to eliminate the $Y$ components.

It is interesting to point out that this conjecture is false for the case of
stabilization with general $2\times 2$ unitary matrices. Indeed, as shown in
[14], there is a setting - the so called XS-stabilizer formalism \- where one
can explicitly construct a stabilizer state on $6$ qubits which is not locally
equivalent to a Pauli stabilizer state. The set $\mathcal{A}$ of this setting
is the group generated by three operators
$\\{\sigma_{X},\operatorname{diag}(1,i),e^{i\pi/4}\mathds{1}\\}$. The
stabilizer state is given by

$\ket{\psi}=\sum\limits_{x_{i}=0,1}(-1)^{x_{1}x_{2}x_{3}}\ket{x_{1},x_{2},x_{3},x_{1}\oplus
x_{2},x_{2}\oplus x_{3},x_{3}\oplus x_{1}},$ (13)

(where $\oplus$ correspond to the addition modulo $2$) and it is stabilized by
the following set

$\displaystyle\begin{split}O_{1}&=\sigma_{X}\otimes S^{3}\otimes S^{3}\otimes
S\otimes\sigma_{X}\otimes\sigma_{X}\\\ O_{2}&=S^{3}\otimes\sigma_{X}\otimes
S^{3}\otimes\sigma_{X}\otimes S\otimes\sigma_{X}\\\ O_{3}&=S^{3}\otimes
S^{3}\otimes\sigma_{X}\otimes\sigma_{X}\otimes\sigma_{X}\otimes S.\end{split}$
(14)

One can see that this state is not a Pauli stabilizer state because the
exponents of the coefficients $(-1)^{x_{1}x_{2}x_{3}}$ are a cubic polynomial
in variables $x_{1},x_{2},x_{3}$. However, showing that this state is not
locally equivalent to a Pauli stabilizer state is more involved and requires a
full classification of standard stabilizer formalism states [14].

##  V Conclusion

We have investigated two important questions about stabilizer formalism in
this work. We have found a set of new constraints on the set of stabilization
operators used to generate non-trivial generalized Clifford groups. We have
also shown that such a set cannot generate a group dense inside $SU(d)$.
Finally, we explored the case of stabilization with binary operators, and we
observed, for a low number of qubits possibility of stabilization only by the
standard Pauli group (up to local equivalence). This led us to conjecture that
the same holds for any number of qubits. It is worth pointing out that there
exist many generalizations for the case of qudits ($d>2$) [15, 16]. It is thus
interesting to investigate equivalent versions of this conjecture in such
settings. Our findings offer promising perspectives regarding new Gottesman-
Knill-like theorems and potential applications in quantum error correction
[1].

##  VI Acknowledgments

Authors would like to thank Sebastian Horvat for his valuable feedback. E.D.
extends thanks to the University of Vienna and the Erasmus program for
enabling his fellowship and contributing to his stay in Vienna. This research
was funded in whole, or in part, by the Austrian Science Fund (FWF) [F7115]
(BeyondC). For the purpose of open access, the author(s) has applied a CC BY
public copyright licence to any Author Accepted Manuscript version arising
from this submission.

## References

  * [1] Daniel Gottesman.  The Heisenberg Representation of Quantum Computers, July 1998\.  Number: arXiv:quant-ph/9807006 arXiv:quant-ph/9807006.  URL: http://arxiv.org/abs/quant-ph/9807006.
  * [2] Michael A. Nielsen and Isaac L. Chuang.  Quantum computation and quantum information.  Cambridge University Press, Cambridge ; New York, 10th anniversary ed edition, 2010.
  * [3] Peter W. Shor.  Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer.  SIAM Journal on Computing, 26(5):1484–1509, October 1997.  arXiv:quant-ph/9508027.  URL: http://arxiv.org/abs/quant-ph/9508027, doi:10.1137/S0097539795293172.
  * [4] Lov K. Grover.  A fast quantum mechanical algorithm for database search, November 1996\.  Number: arXiv:quant-ph/9605043 arXiv:quant-ph/9605043.  URL: http://arxiv.org/abs/quant-ph/9605043.
  * [5] A. Yu Kitaev.  Quantum measurements and the Abelian Stabilizer Problem, November 1995.  Number: arXiv:quant-ph/9511026 arXiv:quant-ph/9511026.  URL: http://arxiv.org/abs/quant-ph/9511026.
  * [6] Richard P. Feynman.  Simulating physics with computers.  International Journal of Theoretical Physics, 21(6):467–488, June 1982.  doi:10.1007/BF02650179.
  * [7] Simon Anders and Hans J. Briegel.  Fast simulation of stabilizer circuits using a graph state representation.  Physical Review A, 73(2):022334, February 2006.  arXiv:quant-ph/0504117.  URL: http://arxiv.org/abs/quant-ph/0504117, doi:10.1103/PhysRevA.73.022334.
  * [8] Guifre Vidal.  Efficient classical simulation of slightly entangled quantum computations.  Physical Review Letters, 91(14):147902, October 2003.  arXiv:quant-ph/0301063.  URL: http://arxiv.org/abs/quant-ph/0301063, doi:10.1103/PhysRevLett.91.147902.
  * [9] Sean Clark, Richard Jozsa, and Noah Linden.  Generalised Clifford groups and simulation of associated quantum circuits, January 2007.  Number: arXiv:quant-ph/0701103 arXiv:quant-ph/0701103.  URL: http://arxiv.org/abs/quant-ph/0701103.
  * [10] Juan Bermejo-Vega, Cedric Yen-Yu Lin, and Maarten Van den Nest.  Normalizer circuits and a Gottesman-Knill theorem for infinite-dimensional systems, January 2015.  arXiv:1409.3208 [math-ph, physics:quant-ph].  URL: http://arxiv.org/abs/1409.3208.
  * [11] Daniel Gottesman.  Stabilizer Codes and Quantum Error Correction, May 1997.  arXiv:quant-ph/9705052.  URL: http://arxiv.org/abs/quant-ph/9705052.
  * [12] Peter Selinger.  Generators and relations for n-qubit Clifford operators.  Logical Methods in Computer Science, 11(2):10, June 2015.  arXiv:1310.6813 [quant-ph].  URL: http://arxiv.org/abs/1310.6813, doi:10.2168/LMCS-11(2:10)2015.
  * [13] Jean-Luc Brylinski and Ranee Brylinski.  Universal quantum gates, August 2001.  Number: arXiv:quant-ph/0108062 arXiv:quant-ph/0108062.  URL: http://arxiv.org/abs/quant-ph/0108062.
  * [14] Xiaotong Ni, Oliver Buerschaper, and Maarten Van den Nest.  A Non-Commuting Stabilizer Formalism.  Journal of Mathematical Physics, 56(5):052201, May 2015.  arXiv:1404.5327 [cond-mat, physics:quant-ph].  URL: http://arxiv.org/abs/1404.5327, doi:10.1063/1.4920923.
  * [15] Vlad Gheorghiu.  Standard Form of Qudit Stabilizer Groups.  Physics Letters A, 378(5-6):505–509, January 2014.  arXiv:1101.1519 [quant-ph].  URL: http://arxiv.org/abs/1101.1519, doi:10.1016/j.physleta.2013.12.009.
  * [16] Markus Grassl, Martin Roetteler, and Thomas Beth.  Efficient Quantum Circuits for Non-Qubit Quantum Error-Correcting Codes.  International Journal of Foundations of Computer Science, 14(05):757–775, October 2003.  arXiv:quant-ph/0211014.  URL: http://arxiv.org/abs/quant-ph/0211014, doi:10.1142/S0129054103002011.

##  Appendix A Proof of proposition 1

######  Proof.

$\blacktriangleright$ We start with $N=2$ qudits and a gate
$U\in\mathcal{C}_{2}(\mathcal{A})$. We define the set

$G=\\{u\otimes v\in U(d)\otimes U(d)|U(u\otimes v)U^{\dagger}\in U(d)\otimes
U(d)\\}.$ (15)

We show that $G=U(d)\otimes U(d)$. This is because of the following:

  * •

G is a subgroup of $U(d)\otimes U(d)$,

  * •

By definition of $U$, tensor product of elements of $\mathcal{A}$ are in $G$ ,
i.e., $\mathcal{A}\otimes\mathcal{A}\subset G$,

  * •

$G$ is closed,

  * •

$G$ remains stable when multiplied by a complex phase, i.e.,
$\forall\varphi,~{}e^{i\varphi}G\subset G$,

  * •

Since $SU(d)\subset\overline{\langle\mathcal{A}\rangle}$, for all $u\in U(d)$,
we have $u\otimes u\in G$.

Thus, for $A_{1},A_{2}\in\mathcal{A}$, $u,w\in U(d)$ and $e^{i\varphi}\in
U(1)$, we have:

$\displaystyle\begin{split}&e^{i\varphi}(uA_{1}w)\otimes(uA_{2}w)\in G\\\
&\Rightarrow
e^{i\varphi}(uA_{1}A_{2}^{\dagger}u^{\dagger})\otimes\mathds{1}\in
G~{}~{}(\text{ for }w=(uA_{2})^{\dagger}).\end{split}$ (16)

Now we introduce

$H=\overline{\langle\\{e^{i\varphi}uA_{1}A_{2}^{\dagger}u^{\dagger}|\varphi\in\mathbb{R},~{}u\in
U(d),~{}A_{1},A_{2}\in\mathcal{A}\\}\rangle}.$ (17)

$H$ is a closed normal subgroup of $U(d)$ containing $U(1)$, by definition.
For the subgroup $\tilde{H}$ of $H$ of determinant-one matrices, we see that
$\tilde{H}\triangleleft SU(d)$. Because $SU(d)$ is a simple Lie group, we only
have two possibilities (since $\tilde{H}$ is also closed):

  * •

$\tilde{H}\subset\operatorname{Z}(SU(d))=\\{e^{2i\pi
k/d}\mathds{1}|k=0,\cdots,d-1\\}$ (the center of $SU(d)$). This would imply
$H=U(1)$, and thus for all $A_{1},A_{2}\in\mathcal{A}$, $A_{1}\propto A_{2}$.
This is not possible since $\mathcal{A}$ generate the entire $SU(d)$ group.

  * •

Because of the previous point, we necessarily have $\tilde{H}=SU(d)$ and
$H=U(d)$.

This means that $U(d)\otimes\mathds{1}\subset G$. Similarly, we show that
$\mathds{1}\otimes U(d)\subset G$. And we finally get $G=U(d)\otimes U(d)$.
Thus $U\in\mathcal{C}_{2}(U(d))$ and
$\mathcal{C}_{2}(\mathcal{A})\subset\mathcal{C}_{2}(U(d))$.

$\blacktriangleright$ For the general case of $N$ qudits, we take
$U\in\mathcal{C}_{N}(\mathcal{A})$ and we define in a similar manner the group
$G$:

$G=\\{u_{1}\otimes\cdots\otimes u_{N}\in U(d)^{\otimes
N}|U(u1\otimes\cdots\otimes u_{N})U^{\dagger}\in U(d)^{\otimes^{N}}\\}.$ (18)

All the properties listed for $G$ (generalized on $N$ qudits) for the two-
qudits case are still valid. This means that for all
$A_{1},A_{2}\in\mathcal{A}$, $u,w\in U(d)$, $\varphi\in\mathbb{R}$ we have

$\displaystyle\begin{split}&(e^{i\varphi}uA_{1}w)\otimes\cdots\otimes(e^{i\varphi}uA_{1}w)\otimes(uA_{2}w)\in
G\\\
\Rightarrow&(e^{i\varphi}uA_{1}A_{2}^{\dagger}u^{\dagger})\otimes\cdots\otimes(e^{i\varphi}uA_{1}A_{2}^{\dagger}u^{\dagger})\otimes\mathds{1}\in
G.\end{split}$ (19)

As before, we introduce the set:

$H=\overline{\langle\\{e^{i\varphi}uA_{1}A_{2}^{\dagger}u^{\dagger}|\varphi\in\mathbb{R},~{}u\in
U(d),~{}A_{1},A_{2}\in\mathcal{A}\\}\rangle}$ (20)

The same argument applies here and we have $H=U(d)$. This means
$\\{u\otimes\cdots\otimes u\otimes\mathds{1}|u\in U(d)\\}\subset G$. By
recursive application of the argument we get $\\{u\otimes\cdots\otimes
u\otimes\mathds{1}\otimes\mathds{1}|u\in U(d)\\}\subset G$ all the way until
we get $U(d)\otimes\mathds{1}\cdots\otimes\mathds{1}\subset G$. Through
permutations of the qudits, we also obtain that
$\mathds{1}\otimes\cdots\otimes\mathds{1}\otimes
U(d)\otimes\mathds{1}\otimes\cdots\otimes\mathds{1}\subset G$. If we take the
product of all those sets, we finaly have that $U(d)^{\otimes N}\subset G$.
Thus $U\in\mathcal{C}_{N}(U(d))$ and
$\mathcal{C}_{N}(\mathcal{A})\subset\mathcal{C}_{N}(U(d))$. ∎

##  Appendix B Proof of proposition 2

######  Proof.

For the $i$-th qudit, we introduce an orthonormal basis
$(\ket{\varphi_{i,j}})_{j}$ with $\ket{\varphi_{i,1}}=\ket{\varphi_{i}}$.
We then define the following operator

$O=\bigotimes\limits_{k=1}^{N}\left(\sum\limits_{l=1}^{d}e^{i(l-1)\pi/d^{k}}\outerproduct{\varphi_{k,l}}{\varphi_{k,l}}\right).$
(21)

We have $O\in U(d)^{\otimes N}$. By expanding the tensor product, one can
verify that all the eigenvalues of $O$ are different. Furthermore if we
consider an operator $A$ with eigenvalue $\lambda$ and corresponding
eigenvector $\ket{\rho}$ we see that:

$A\ket{\rho}=\lambda\ket{\rho}\Rightarrow UAU^{\dagger}(U\ket{\rho})=\lambda
U\ket{\rho}.$ (22)

Thus eigenvectors of $A$ are sent to eigenvectors of $UAU^{\dagger}$ with same
eigenvalues. By hypothesis on $U$, we have $UOU^{\dagger}\in U(d)^{\otimes
N}$, which means that $UOU^{\dagger}$ has an eigenbasis composed of product-
states. Considering the uniqueness of the eigenvalues, it follows that
$U\ket{\psi}$ is a product-state. ∎

##  Appendix C Two lemmas

In this section, we present and prove two technical lemmas that will be
utilized in the proof of Proposition 3.

######  Lemma 1.

If the following conditions hold for
$u,v,w,x\in\mathbb{C}^{k}\setminus\\{0\\}$:

  * •

$u$ and $v$ are not collinear,

  * •

$w=\alpha u$ and $x=\beta v$,

  * •

$u+v\propto w+x$.

then $\alpha=\beta$.

######  Proof.

We represent vectors $w$ and $x$ in basis $\\{u,v\\}$, the condition
$u+v\propto w+x$ becomes:

$\left(\begin{matrix}1\\\
1\end{matrix}\right)\propto\left(\begin{matrix}\alpha\\\
\beta\end{matrix}\right),$ (23)

thus $\alpha=\beta$. ∎

######  Lemma 2.

Let $U$ be an invertible matrix of size $d^{n}$. There exist $d\times d$
unitary matrices $u_{1},\dots,u_{n}$ and $v_{1},\dots,v_{n}$ such that all the
coefficients of the matrix

$V=(u_{1}\otimes\cdots\otimes u_{n})U(v_{1}\otimes\cdots\otimes v_{n})$ (24)

are all non-zero.

######  Proof.

The matrices $u_{i}$ and $v_{j}$ will be constructed with the help of a
$d\times d$ matrix $M$ which depends on a real parameter $t$,

$M=(i+1+t^{2}+\cdots+t^{2(d-1)})\mathds{1}-2\left(\begin{matrix}1&t&t^{2}&\cdots&t^{d-1}\\\
t&t^{2}&t^{3}&\cdots&t^{d}\\\ \vdots&\vdots&\vdots&&\vdots\\\
t^{d-1}&t^{d}&t^{d+1}&\cdots&t^{2(d-1)}\end{matrix}\right).$ (25)

The general matrix element of $M$ is

$M_{k,l}=-2t^{k+l-2}+\delta_{kl}(i+\sum\limits_{s=0}^{d-1}t^{2s}).$ (26)

Firstly, let us check that the columns of $M$ are orthogonal to each other,
i.e., for $l_{0}\neq l_{1}$ we have

$\displaystyle\sum\limits_{k=1}^{d}M_{k,l_{0}}M_{k,l_{1}}^{\ast}$
$\displaystyle=\sum\limits_{k=1}^{d}\left(-2t^{k+l_{0}-2}+\delta_{kl_{0}}i+\delta_{k,l_{0}}\sum\limits_{s=0}^{d-1}t^{s}\right)\left(-2t^{k+l_{1}-2}-\delta_{kl_{1}}i+\delta_{k,l_{1}}\sum\limits_{s=0}^{d-1}t^{s}\right)$
(27)
$\displaystyle=\sum\limits_{k=1}^{d}\Bigg{[}4t^{2k+l_{0}+l_{1}-4}+2t^{k+l_{0}-2}\delta_{kl_{1}}i-2t^{k+l_{0}-2}\delta_{kl_{1}}\sum\limits_{s=0}^{d-1}t^{2s}-2t^{k+l_{1}-2}\delta_{kl_{0}}i-2t^{k+l_{1}-2}\delta_{kl_{0}}\sum\limits_{s=0}^{d-1}t^{2s}\Bigg{]}$
(28)
$\displaystyle=4t^{l_{0}+l_{1}-2}\sum\limits_{k=0}^{d-1}t^{2k}+\cancel{2t^{l_{0}+l_{1}-2}i}-\cancel{2t^{l_{0}+l_{1}-2}i}-2t^{l_{0}+l_{1}-2}\sum\limits_{s=0}^{d-1}t^{2s}-2t^{l_{0}+l_{1}-2}\sum\limits_{s=0}^{d-1}t^{2s}$
(29) $\displaystyle=0.$ (30)

Then we verify that all the columns of matrix $M$ have the same norm. So for a
fixed $l$ we compute

$\displaystyle\sum\limits_{k=1}^{d}M_{k,l}M_{k,l}^{\ast}$
$\displaystyle=\sum\limits_{k=1}^{d}\left(-2t^{k+l-2}+\delta_{kl}i+\delta_{k,l}\sum\limits_{s=0}^{d-1}t^{s}\right)\left(-2t^{k+l-2}-\delta_{kl}i+\delta_{k,l}\sum\limits_{s=0}^{d-1}t^{s}\right)$
(31)
$\displaystyle=\sum\limits_{k=1}^{d}\Bigg{[}4t^{2k+2l-4}-2t^{k+l-2}\delta_{kl}\sum\limits_{s=0}^{d-1}t^{2s}+\delta_{kl}-2t^{k+l-2}\delta_{kl}\sum\limits_{s=0}^{d-1}t^{2s}+\delta_{kl}\left(\sum\limits_{s=0}^{d-1}t^{2s}\right)^{2}\Bigg{]}$
(32)
$\displaystyle=4t^{2l-2}\sum\limits_{k=0}^{d-1}t^{2k}+1-4t^{2l-2}\sum\limits_{s=0}^{d-1}t^{2s}+\left(\sum\limits_{s=0}^{d-1}t^{2s}\right)^{2}$
(33) $\displaystyle=1+\left(\sum\limits_{s=0}^{d-1}t^{2s}\right)^{2}\neq 0.$
(34)

We see that the last expression is independent of the column index $l$. The
last two computations show that for all $t$ the matrix $M$ is proportional to
a unitary matrix. This means that we can take $\tilde{u_{i}}=M(t_{i})$ and
$\tilde{v_{j}}=M(t_{j}^{\prime})$ and show that there exist some values
$t_{i}$ and $t_{j}^{\prime}$ such that the requirements of the lemma are met.
At the final step, we simply normalize the matrices $\tilde{u}_{i}$ and
$\tilde{v}_{j}$. To prove the existence of $t_{i}$ and $t^{\prime}_{j}$, we
show the following statement by induction:

For all finite family $\\{A^{(i)}\\}_{i\in I}$ of non-zero $d^{n}\times d^{n}$
matrices (i.e., $A^{(i)}\in\mathcal{M}_{d^{n}}(\mathbb{C})\setminus\\{0\\}$),
there exist some real numbers $t_{1},\dots,t_{n}$ and
$t_{1}^{\prime},\dots,t_{n}^{\prime}$ such that

$\forall i\in I,~{}\Big{(}M(t_{1})\otimes\cdots\otimes
M(t_{n})\Big{)}A^{(i)}\Big{(}M(t_{1}^{\prime})\otimes\cdots\otimes
M(t_{n}^{\prime})\Big{)}\in\mathcal{M}_{d^{n}}(\mathbb{C}\setminus\\{0\\}),$
(35)

i.e., are matrices with only non zero coefficients.

If $n=0$, then the $A^{(i)}$’s are non zero $1\times 1$ matrix, i.e., single
non zero numbers. In this case there is nothing to show.

If $n\geq 1$, lets consider a finite set $\\{A^{(i)}\\}_{i\in I}$ of non zero
matrices. We write each of these matrices in block form as

$A^{(i)}=\left(\begin{matrix}A^{(i)}_{1,1}&\cdots&A^{(i)}_{1,d}\\\
\vdots&&\vdots\\\ A^{(i)}_{d,1}&\cdots&A^{(i)}_{d,d}\end{matrix}\right),$ (36)

where the $A^{(i)}_{k,l}$’s are $d^{n-1}\times d^{n-1}$ square matrices. We
shall firstly look at the product
$(M(t)\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})A^{(i)}$:

$\displaystyle\Big{[}(M(t)\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})A^{(i)}\Big{]}_{k,l}$
$\displaystyle=t^{k-1}(A^{(i)}_{1,l}+tA^{(i)}_{2,l}+t^{2}A^{(i)}_{3,l}+\cdots+t^{d-1}A^{(i)}_{d,l})-2A^{(i)}_{k,l}(i+1+t^{2}+\cdots+t^{2(d-1)})$
(37) $\displaystyle:=P^{(i)}_{k,l}(t),$ (38)

calculated at the square block at position $(k,l)$. Next, we multiply
expression (37) by
$M(t^{\prime})\otimes\mathds{1}\otimes\cdots\otimes\mathds{1}$ from the right-
hand side with the choice $t^{\prime}=t^{2d}$ and we get while considering the
block at position $(k,l)$

$\displaystyle\Big{[}(M(t)$
$\displaystyle\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})A^{(i)}(M(t^{2d})\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})\Big{]}_{k,l}$
$\displaystyle=(t^{2d})^{l-1}\left[P^{(i)}_{k,1}(t)+t^{2d}P^{(i)}_{k,2}(t)+(t^{2d})^{2}P^{(i)}_{k,3}(t)+\cdots+(t^{2d})^{d-1}P^{(i)}_{k,d}(t)\right]$
(39)
$\displaystyle~{}~{}~{}~{}~{}~{}-2P^{(i)}_{k,l}(t)(i+1+(t^{2d})^{2}+\cdots+(t^{2d})^{2(d-1)}).$
$\displaystyle:=Q^{(i)}_{k,l}(t)$ (40)

This complicated expression is a polynomial function in $t$ with matrix
coefficients. By this, we mean a function of the form $\sum_{j}B_{j}t^{j}$
where the coefficients $B_{j}$ are matrices. We can verify that the polynomial
of equation (C) is non zero, i.e., at least one of the matrix coefficient is
non-zero. To see why this is the case, lets assume by contradiction that all
coefficients of eq. (C) are zero. Since the degree of all
$P^{(i)}_{\alpha,\beta}(t)$ ($1\leq\alpha\leq d$ and $1\leq\beta\leq d$) is
less than $2d$, by looking at the term of degree less that $2d$, it would
imply that all the coefficients of $P^{(i)}_{k,l}(t)$ are zero. And then that
the coefficients of all the $P^{(i)}_{k,\beta}(t)$ are also zero for
$1\leq\beta\leq d$. Finally, by considering eq. (37), it would imply that all
the $A^{(i)}_{\alpha,\beta}$ are also zero. This is clearly a contradiction,
since we assumed that the matrix $A^{(i)}$ is non-zero. So for every $k$ and
$l$, the polynomial with matrix coefficient $Q^{(i)}_{k,l}(t)$, is non zero.
Such polynomials have at most a finite number of roots, and since the set $I$
is also finite, there must exist a value $t_{1}$ such that

$\forall i\in I,~{}\forall 1\leq k\leq d,~{}\forall 1\leq l\leq
d,~{}Q_{k,l}^{(i)}(t_{1})\neq 0,$ (41)

i.e., are not the zero matrix. We can thus write, with
$t_{1}^{\prime}=t_{1}^{2d}$:

$(M(t_{1})\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})A^{(i)}(M(t_{1}^{\prime})\otimes\mathds{1}\otimes\cdots\otimes\mathds{1})=\left(\begin{matrix}Q^{(i)}_{1,1}(t_{1})&\cdots&Q^{(i)}_{1,d}(t_{1})\\\
\vdots&&\vdots\\\
Q^{(i)}_{d,1}(t_{1})&\cdots&Q^{(i)}_{d,d}(t_{1})\end{matrix}\right).$ (42)

Applying the induction hypothesis to the set
$\Big{\\{}Q^{(i)}_{j,k}(t_{1})\Big{|}i\in I,~{}1\leq j\leq d,~{}1\leq k\leq
d\Big{\\}}$, which is a finite set of $d^{n-1}\times d^{n-1}$ non-zero
matrices, we know that there exist $2(n-1)$ real parameters
$t_{2},\dots,t_{n}$ and $t_{2}^{\prime},\dots,t_{n}^{\prime}$ such that

$\Big{(}M(t_{2})\otimes\cdots\otimes
M(t_{n})\Big{)}Q^{(i)}_{j,k}(t_{1})\Big{(}M(t_{2}^{\prime})\otimes\cdots\otimes
M(t_{n}^{\prime})\Big{)}:=\tilde{A}^{(i)}_{k,l}$ (43)

are matrices with only non-zero coefficients, for all values of $i$, $k$ and
$l$. Since we have

$\Big{(}\mathds{1}\otimes M(t_{2})\otimes\cdots\otimes
M(t_{n})\Big{)}\left(\begin{matrix}Q^{(i)}_{1,1}(t_{1})&\cdots&Q^{(i)}_{1,d}(t_{1})\\\
\vdots&&\vdots\\\
Q^{(i)}_{d,1}(t_{1})&\cdots&Q^{(i)}_{d,d}(t_{1})\end{matrix}\right)\Big{(}\mathds{1}\otimes
M(t_{2}^{\prime})\otimes\cdots\otimes
M(t_{n}^{\prime})\Big{)}=\left(\begin{matrix}\tilde{A}^{(i)}_{1,1}&\cdots&\tilde{A}^{(i)}_{1,d}\\\
\vdots&&\vdots\\\
\tilde{A}^{(i)}_{d,1}&\cdots&\tilde{A}^{(i)}_{d,d}\end{matrix}\right),$ (44)

we finally have for each $i$,

$\Big{(}M(t_{1})\otimes\cdots\otimes
M(t_{n})\Big{)}A^{(i)}\Big{(}M(t_{1}^{\prime})\otimes\cdots\otimes
M(t_{n}^{\prime})\Big{)}=\left(\begin{matrix}\tilde{A}^{(i)}_{1,1}&\cdots&\tilde{A}^{(i)}_{1,d}\\\
\vdots&&\vdots\\\
\tilde{A}^{(i)}_{d,1}&\cdots&\tilde{A}^{(i)}_{d,d}\end{matrix}\right),$ (45)

which are indeed a matrices with only non-zero coefficients. This conclude the
proof by induction.

The proof of the Lemma 2 follow from the result shown by induction, by
applying it to the set $\\{U\\}$ containing the single element $U$ (which is
indeed a non-zero matrix, because it is invertible). This tells us that there
exist some real number $t_{1},\dots,t_{n}$ and
$t^{\prime}_{1},\dots,t^{\prime}_{n}$ such that

$\Big{(}M(t_{1})\otimes\cdots\otimes
M(t_{n})\Big{)}U\Big{(}M(t_{1}^{\prime})\otimes\cdots\otimes
M(t_{n}^{\prime})\Big{)}$ (46)

is a matrix with only non-zero coefficients. Defining the unitary matrices

$\displaystyle
u_{i}=M(t_{i})\Bigg{/}\sqrt{1+\left(\sum\limits_{s=0}^{d-1}t_{i}^{2s}\right)^{2}}$
$\displaystyle
v_{i}=M(t_{i}^{\prime})\Bigg{/}\sqrt{1+\left(\sum\limits_{s=0}^{d-1}t_{i}^{\prime
2s}\right)^{2}}$ (47)

concludes the proof. ∎

##  Appendix D Proof of proposition 3

######  Proof.

In this proof, we use the two lemmas of the Appendix C. To ensure clarity, we
will first explain the case of qubits and then generalize it for qudits ,
i.e., a system of general dimension $d$.

Case $d=2$ and $N=2$
We begin with the case $N=2$ and then proceed by induction. Let’s consider
$U\in\mathcal{C}_{2}(U(2))$. Based on the Lemma 2, we know that we can compose
$U$ with local gates so that we can assume without loos of generality that the
coefficient of $U$ are all non-zero. We can then write $U$ in block form with
each block being non-zero:

$U=\left(\begin{matrix}x_{1}&x_{2}&x_{3}&x_{4}\\\
y_{1}&y_{2}&y_{3}&y_{4}\end{matrix}\right)=\left(\begin{matrix}z_{1}&t_{1}\\\
z_{2}&t_{2}\\\ z_{3}&t_{3}\\\ z_{4}&t_{4}\end{matrix}\right).$ (48)

Since $\ket{00}=\left(\begin{matrix}1&0&0&0\end{matrix}\right)^{T}$ is a
product-state, $U\ket{00}=\left(\begin{matrix}x_{1}\\\
y_{1}\end{matrix}\right)$ is also a product-state. This implies that the
blocks $x_{1}$ and $y_{1}$ are proportional to each other, and similarly,
$x_{2}\propto y_{2}$, $x_{3}\propto y_{3}$ and $x_{4}\propto y_{4}$. By the
same procedure with
$\ket{0}\otimes(\ket{0}+\ket{1})=\left(\begin{matrix}1&1&0&0\end{matrix}\right)^{T}$
and
$(\ket{0}+\ket{1})\otimes\ket{0}=\left(\begin{matrix}1&0&1&0\end{matrix}\right)^{T}$
which are also product-states, we observe that $x_{1}+x_{2}\propto
y_{1}+y_{2}$ and $x_{1}+x_{3}\propto y_{1}+y_{3}$.

We have seen that $U$ preserve the product “bra”-states (see the discussion
after prop. 2). Taking the hermitian conjugate thus shows that
$U^{-1}=U^{\dagger}$ also preserve the product-state structure. This implies
that $U$ cannot map an entangled state to a product one, otherwise, $U^{-1}$
would send a product-state to an entangled one. So,
$\ket{01}+\ket{10}=\left(\begin{matrix}0&1&1&0\end{matrix}\right)^{T}$ which
is not a product-state, is mapped to an entangled one. This implies
$x_{2}+x_{3}\not\propto y_{2}+y_{3}$. Since all the vectors involved are non-
zero we have $x_{2}\not\propto x_{3}$ otherwise it would imply
$x_{2}+x_{3}\propto y_{2}+y_{3}$. In complete analogy, we have
$x_{1}\not\propto x_{4}$. We get the same type of equation by considering the
expressions $\bra{\psi}U$, for analogously chosen $\bra{\psi}$: $z_{1}\propto
t_{1}$, $z_{2}\propto t_{2}$, $z_{3}\propto t_{3}$, $z_{4}\propto t_{4}$,
$z_{1}+z_{2}\propto t_{1}+t_{2}$… as well as $z_{1}\not\propto z_{4}$ and
$z_{2}\not\propto z_{3}$.

Without loss of generality, we can assume that we have $x_{1}\not\propto
x_{2}$ and $x_{3}\not\propto x_{4}$. Indeed the previously shown
proportionality relations imply that we have either $(x_{1}\not\propto
x_{2}~{}\&~{}x_{4}\not\propto x_{3})$ or $(x_{1}\not\propto
x_{3}~{}\&~{}x_{4}\not\propto x_{2})$. So, if necessary, we can multiply $U$
by the swapping gate $P=\left(\begin{matrix}1&0&0&0\\\ 0&0&1&0\\\ 0&1&0&0\\\
0&0&0&1\end{matrix}\right)$, whose effect is to exchange the second and third
columns of $U$, to have $x_{1}\not\propto x_{2}$ and $x_{3}\not\propto x_{4}$.
The Lemma 1 then implies proportionality relation between larger blocks
$\left(\begin{matrix}x_{1}&x_{2}\end{matrix}\right)\propto\left(\begin{matrix}y_{1}&y_{2}\end{matrix}\right)$
and
$\left(\begin{matrix}x_{3}&x_{4}\end{matrix}\right)\propto\left(\begin{matrix}y_{3}&y_{4}\end{matrix}\right)$.

Now, we want to apply the same procedure to the rows. We repeat the same
process as before, but this time we cannot use a swapping gate. This is not a
problem since, thanks to the proportionality relations between larger blocks,
we know that $z_{1}\propto z_{3}$. This implies $z_{1}\not\propto z_{2}$ and
$z_{3}\not\propto z_{4}$, and we can apply the Lemma 1 again to get
$\left(\begin{matrix}z_{1}\\\
z_{2}\end{matrix}\right)\propto\left(\begin{matrix}t_{1}\\\
t_{2}\end{matrix}\right)$ and $\left(\begin{matrix}z_{3}\\\
z_{4}\end{matrix}\right)\propto\left(\begin{matrix}t_{3}\\\
t_{4}\end{matrix}\right)$.

The proportionality relation obtained for the bigger blocks implies that $U$
can be written in the form of a tensor product: $U=V\otimes W$. Furthermore,
since $U$ is unitary, we have
$UU^{\dagger}=\mathds{1}=(VV^{\dagger})\otimes(WW^{\dagger})$. This implies
that $VV^{\dagger}\propto WW^{\dagger}\propto\mathds{1}$, indicating that with
the appropriate normalization factor, $V$ and $W$ can be chosen to be unitary.
Since we used the possibility of composition with a swapping gate to reach
this form, we have obtained the result of the theorem for the case $N=2$ and
$d=2$.

Case $d=2$ and $N>2$
The case with $N$ qubits is similar. We will show that $U$ can be factored on
the first qubit and then the proof follows by induction. We apply the same
procedure by writing $U$

$U=\left(\begin{matrix}A&B\\\
C&D\end{matrix}\right)=\left(\begin{matrix}x_{1}&\cdots&x_{2^{N}}\\\
y_{1}&\cdots&y_{2^{N}}\end{matrix}\right)=\left(\begin{matrix}z_{1}&t_{1}\\\
\vdots&\vdots\\\ z_{2^{N}}&t_{2^{N}}\end{matrix}\right),$ (49)

where $A$, $B$, $C$, $D$ are square matrices of size $2^{N-1}$, the $x_{i}$,
$y_{i}$ are length $2^{N-1}$ column vectors and $z_{i}$ and $t_{i}$ are length
$2^{N-1}$ row vectors. Thanks to Lemma 2, we can still assume without loss of
generality all the coefficients of $U$ are non zero, and thus that $x_{i}$,
$y_{i}$, $z_{i}$ and $t_{i}$ are all non-zero blocks.

In the following discussion, instead of referring to each column or row with
an index $i$ running from $1$ to $2^{N}$, we will label them using the binary
string associated to the corresponding basis state $\ket{i_{1},\cdots,i_{N}}$.
For instance, we will speak of the column labeled as $000\cdots 0100$, which
would be the fourth column. We start by examining the columns, more
specifically only the upper half, i.e., the $x_{i}$’s. By using the same
methodology as before, we know that the following holds:

  1. A)

If the strings $i$ and $j$ labeling two columns differ at exactly one position
and if $x_{i}\not\propto x_{j}$, then
$\left(\begin{matrix}x_{i}&x_{j}\end{matrix}\right)\propto\left(\begin{matrix}y_{i}&y_{j}\end{matrix}\right)$.

Indeed under these hypothesis the sum of basis states $\ket{i}+\ket{j}$ is a
product-state, which then implies that $x_{i}+x_{j}\propto y_{i}+y_{j}$. We
can apply lemma 1 which allows to conclude that
$\left(\begin{matrix}x_{i}&x_{j}\end{matrix}\right)\propto\left(\begin{matrix}y_{i}&y_{j}\end{matrix}\right)$.

  2. B)

If the strings $i$ and $j$ of two columns differ at two positions or more,
then we know that $x_{i}\not\propto x_{j}$.

To see why this is true, suppose the strings differ at two positions. The sum
$\ket{i}+\ket{j}$ is an entangled state and its image
$\left(\begin{matrix}x_{i}+x_{j}\\\ y_{i}+y_{j}\end{matrix}\right)$ is thus
also entangled. Now assume by contradiction that $x_{i}\propto x_{j}$. Then
$\left(\begin{matrix}x_{i}\\\ x_{j}\end{matrix}\right)$ is a product-state,
which can be written as $\left(\begin{matrix}x_{i}\\\
x_{j}\end{matrix}\right)=\ket{\psi_{1}}\otimes\cdots\otimes\ket{\psi_{N}}$.
Consequently, $x_{i}$ must be the vector
$a\ket{\psi_{2}}\otimes\cdots\otimes\ket{\psi_{N}}$. All the proportionality
relation then imply that
$x_{j}=b\ket{\psi_{2}}\otimes\cdots\otimes\ket{\psi_{N}}$,
$y_{i}=c\ket{\psi_{2}}\otimes\cdots\otimes\ket{\psi_{N}}$ and
$y_{j}=d\ket{\psi_{2}}\otimes\cdots\otimes\ket{\psi_{N}}$. Meaning that
$\left(\begin{matrix}x_{i}+x_{j}\\\
y_{i}+y_{j}\end{matrix}\right)=(a+b,c+d)^{T}\otimes\ket{\psi_{2}}\otimes\cdots\otimes\ket{\psi_{N}}$,
which is a contradiction since it is supposed to be an entangled state.

Now we begin the analysis with $x_{0\cdots 0}$. Using the point B) from above
we know that, $x_{0\cdots 01}\not\propto x_{10\cdots 0}$ and thus $x_{0\cdots
0}$ has to be non-collinear to either $x_{0\cdots 01}$ or $x_{10\cdots 0}$. If
necessary, we can multiply $U$ by a swapping gate that exchanges qubit $1$ and
qubit $N$. This permutes the columns $0\cdots 01$ and $10\cdots 0$ so that, we
can assume that $x_{0\cdots 0}\not\propto x_{0\cdots 01}$. Similarly, we can
assume $x_{0\cdots 0}\not\propto x_{0\cdots 010}$ by using the same trick of
multiply $U$ by the swapping gate that exchange qubit $1$ and qubit $N-1$.
This would not affect the relation $x_{0\cdots 0}\not\propto x_{0\cdots 01}$
we just shown since the corresponding columns are invariant under this
exchange of qubits. The argument continues in the same fashion: we have
$x_{0\cdots 0}\not\propto x_{i}$, where the string $i$ has exactly one ‘$1$’
that is not at the first position. The point A) tells us that the
proportionality coefficients $\lambda_{i}$ (such that
$x_{i}=\lambda_{i}y_{i}$) are all the same for $i=0\cdots 0$ or $i$ has
exactly one ‘$1$’ that is not at the first position. We will continue with the
same methodology propagating the fact that $\lambda_{i}$ is a constant for all
string $i$ starting with ‘$0$’. Indeed, $x_{0\cdots 01}\not\propto x_{0\cdots
010}$ thus $x_{0\cdots 011}$ is non-proportional to at least $x_{0\cdots 01}$
or $x_{0\cdots 010}$ and we can apply the point A), to get $\lambda_{0\cdots
011}=\lambda_{0\cdots 0}$. This extension applies to all $x_{i}$ with $i$
having two ‘$1$’ (not on the first qubit), then three, and so on, until
$x_{01\cdots 1}$. In the end, we get that $A\propto C$.

We now apply the same reasoning for rows of $U$, except we do not have the
flexibility of multiplying by some swapping gates. Nonetheless, this is not
needed, because $z_{0\cdots 0}$ is non-proportional to at least $z_{0\cdots
01}$ or $z_{10\cdots 0}$ and we have just shown that $z_{0\cdots 0}\propto
z_{10\cdots 0}$. Thus $A\propto B$. The same thing works for the lower half:
$C\propto D$. The proportionality relation between the large block $A$, $B$,
$C$ and $D$ finally show that $B\propto D$.

All these four proportionality relations between large blocks ($A\propto B$,
$B\propto D$, $A\propto C$ and $C\propto D$) imply that $U$ can be written in
the form of a tensor product: $U=V\otimes W$. Here $V$ is a $2\times 2$
matrix, and $W$ is $2^{N-1}\times 2^{N-1}$. Furthermore, $U$ is unitary, which
means that we can chose $V$ and $W$ to be unitary as well. $W$ will also
stabilize the set of product-states on $N-1$ qubit, thus we can apply the
induction hypothesis to $W$. We used the possibility to compose $U$ with swap
gates to arrive at this form but since for a local gate $u$ and a permutation
gate $P$, $PuP^{\dagger}$ is local, we can “move” the permutation gate to the
left to arrive at the result of the theorem in the case $d=2$.

Case $d\geq 1$ and $N\geq 1$
For the case of $N$ qudits, the proof works in a similar way. We start by
applying the Lemma 2 and multiply $U$ by local gates, so that we write

$U=\left(\begin{matrix}A_{1,1}&\cdots&A_{1,d}\\\ \vdots&&\vdots\\\
A_{d,1}&\cdots&A_{d,d}\end{matrix}\right)=\left(\begin{matrix}\alpha_{1,1}&\cdots&\alpha_{1,d^{N}}\\\
\vdots&&\vdots\\\ \alpha_{d,1}&\cdots&\alpha_{d,d^{N}}\end{matrix}\right),$
(50)

with all $\alpha$-blocks being non-zero column vectors of length $d$. Acting
on the right of equation (50) by a basis state, it becomes clear that for all
fixed $j$, $\alpha_{1,j},\alpha_{2,j},\dots,\alpha_{d,j}$ are proportional to
each other. Let’s denote the proportionality coefficients as $\lambda_{i,j}$,
i.e., $\alpha_{i,j}=\lambda_{i,j}\alpha_{1,j}$. Now, what we have to show is
that $\lambda_{i,j}=\lambda_{i,0\cdots 0}$ for all $i=1,\dots,d$ and string
$j$ which start with a ‘$0$’.

By using remarks A) and B) listed previously and generalized on qudits
(replace $x_{j}$ by $\alpha_{1,j}$), we know that
$\lambda_{i,j}=\lambda_{i,0\cdots 0}$, for either $j=0\cdots 01$ or
$j=10\cdots 0$. As before, we can multiply $U$ by the swapping gate permuting
qudit $1$ and $N$, such that we get $\lambda_{i,0\cdots 01}=\lambda_{i,0\cdots
0}$. Like in the qubit case we show that the swapping freedom allows us to
assume $\alpha_{1,0\cdots 0}\not\propto\alpha_{1,j}$ whenever $j$ is a string
composed of only zeros and one $1$ which is not at the beginning. For such
string $j$, we have that $\lambda_{i,j}=\lambda_{i,0\cdots 0}$.

For the column $0\cdots 02$, we know $\alpha_{1,0\cdots 02}$ cannot be
proportional to $\alpha_{1,0\cdots 0}$ and $\alpha_{1,0\cdots 01}$ at the same
time (since both are non-collinear). Thus by point A), we have
$\lambda_{i,0\cdots 02}=\lambda_{i,0\cdots 0}$ or $\lambda_{i,0\cdots 01}$.
But since we already showed $\lambda_{i,0\cdots 0}=\lambda_{i,0\cdots 01}$ we
indeed manage to obtain $\lambda_{i,0\cdots 02}=\lambda_{i,0\cdots 0}$. The
same holds ($\lambda_{i,j}=\lambda_{i,0\cdots 0}$) for all string $j$ only
made of ‘$0$’ and one non-zero index which is not at the first place.

We can then gradually extend this equality of the proportionality coefficients
($\lambda_{i,j}=\lambda_{i,0\cdots 0}$) to all columns with their label $j$
starting with ‘$0$’. In this case, we reason like for the qubit case: we
compare the column $j$ we are interested in, with two other columns $k$ and
$l$ for which we already know that
$\lambda_{i,k}=\lambda_{i,l}=\lambda_{i,0\cdots 0}$. We can choose the strings
$l$ and $k$ such that: $k$ differ from $j$ at one position, $l$ differ form
$j$ at one position and $k$ differ form $l$ at two positions. Application of
point B) then shows that $\alpha_{1,k}\not\propto\alpha_{1,l}$ and thus
$\alpha_{1,j}$ is not proportional to $\alpha_{1,k}$ or $\alpha_{1,l}$. The
point A) allows then to conclude that $\lambda_{i,j}=\lambda_{i,0\cdots 0}$.

The last equality means that the $d^{N-1}\times d^{N-1}$ block $A_{i,1}$ (
$i=1,\dots,d$) are all proportional to each other: $A_{1,1}\propto
A_{2,1}\propto\cdots\propto A_{d,1}$. We now continue the analysis on the
rows. This time multiplying $U$ by swapping gates is not necessary, like in
the case of qubits, because we already know that the row $0\cdots 0$ and
$10\cdots 0$ are proportional. Thus, this procedures show that all the square
blocks on the same line are proportional to each other, that is, for every
$i$, $A_{i,1}\propto\cdots\propto A_{i,d}$. This, with the proportionality
relation we showed for the $A_{i,1}$’s block, shows that in fact all the
square blocks are proportional to each other, regardless of their positions.
This means $U=V\otimes W$ where $V$ (of size $d\times d$) and $W$ (of size
$d^{N-1}\times d^{N-1}$) can be chosen to be unitary. Additionally, since $W$
stabilizes product-states on $N-1$ qudits, we can conclude the proof by
induction.

∎

##  Appendix E Proof of theorem 3

######  Proof.

Since the projector $P_{0,\cdots,0}$ is of rank $2^{n-1}$, we will have at
least $2^{n-1}$ zeros as eigenvalues. To show that the states defined by the
equation (7) are indeed eigenvectors with their respective eigenvalues, we
proceed with the following computation

$\displaystyle
P_{\theta_{1},\cdots,\theta_{n}}\ket{\psi_{j_{1},\cdots,j_{n}}}$
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{\operatorname{Re}}{2}\Bigg{[}\sum\limits_{k_{1},\cdots,k_{n}}(ij_{1})^{k_{1}}\cdots(ij_{n})^{k_{n}}A_{\theta_{1}}\ket{k_{1}}\cdots
A_{\theta_{n}}\ket{k_{n}}\Bigg{]}$ (51a)
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{\operatorname{Re}}{2}\Bigg{[}\prod\limits_{l=1}^{n}(\cos\theta_{l}\ket{0}+\sin\theta_{l}\ket{1}-ij_{l}\cos\theta_{l}\ket{1}+ij_{l}\sin\theta_{l}\ket{0}\Bigg{]}$
(51b)
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{\operatorname{Re}}{2}\Bigg{[}\prod\limits_{l=1}^{n}(e^{ij_{l}\theta_{l}}\ket{0}-ij_{l}e^{ij_{l}\theta_{l}}\ket{1})\Bigg{]}$
(51c)
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{\operatorname{Re}}{2}\Bigg{[}e^{i(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}\sum\limits_{k_{1},\cdots,k_{n}}(-ij_{1})^{k_{1}}\cdots(-ij_{n})^{k_{n}}\ket{k_{1}\cdots
k_{n}}\Bigg{]}$ (51d)
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{1}{4}\Bigg{[}e^{i(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}\sum\limits_{k_{1},\cdots,k_{n}}(-ij_{1})^{k_{1}}\cdots(-ij_{n})^{k_{n}}\ket{k_{1}\cdots
k_{n}}$ (51e)
$\displaystyle~{}~{}~{}~{}~{}~{}+e^{-i(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}\sum\limits_{k_{1},\cdots,k_{n}}(ij_{1})^{k_{1}}\cdots(ij_{n})^{k_{n}}\ket{k_{1}\cdots
k_{n}}\Bigg{]}$ (51f)
$\displaystyle=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{1}{4}\Big{[}\sum\limits_{k_{1},\cdots,k_{n}=0,1}\Big{(}e^{i(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}(-ij_{1})^{k_{1}}\cdots(-ij_{n})^{k_{n}}$
(51g)
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}+e^{-i(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}(ij_{1})^{k_{1}}\cdots(ij_{n})^{k_{n}}\Big{)}\ket{k_{1}\cdots
k_{n}}\Bigg{]}.$ (51h)

We can examine the terms within the last summation symbol and divide our
analysis into two cases based on the parity of the number of indices $k_{l}$
that have the value $1$.

  * •

For an even number of $k_{l}$’s taking the value $1$, the term in the sum is
$(ij_{1})^{k_{1}}\cdots(ij_{n})^{k_{n}}2\cos(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})\ket{k_{1}\cdots
k_{n}}$. Summing over all such terms yields $2\cos(j_{1}\theta
1+\cdots+j_{n}\theta_{n})\ket{\psi_{j_{1},\cdots,j_{n}}}$.

  * •

For an odd number of $k_{l}$’s taking the value $1$, the term in the sum
becomes
$(-ij_{1})^{k_{1}}\cdots(-ij_{n})^{k_{n}}2i\sin(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})\ket{k_{1}\cdots
k_{n}}$. These terms are all eigenvectors of
$\sigma_{Z}\otimes\cdots\otimes\sigma_{Z}$ and we denote their sum as
$\ket{\varphi}$.

Therefore, we have

$P_{\theta_{1},\cdots,\theta_{n}}\ket{\psi_{j_{1},\cdots,j_{n}}}=\frac{1}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\frac{\cos(j_{1}\theta_{1}+\cdots+j_{n}\theta_{n})}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}+\ket{\varphi}.$
(52)

Since $\ket{\varphi}$ is a $-1$-eigenvector of
$\sigma_{Z}\otimes\cdots\otimes\sigma_{Z}$, by applying $P_{0,\cdots,0}$ on
the left-hand side of equation (52) will yield zero. Finally, we conclude that

$P_{0,\cdots,0}P_{\theta_{1},\cdots,\theta_{n}}\ket{\psi_{j_{1},\cdots,j_{n}}}=\frac{1+\cos(j_{1}\theta
1+\cdots+j_{n}\theta_{n})}{2}\ket{\psi_{j_{1},\cdots,j_{n}}}.$ (53)

∎

##  Appendix F 3 qubit stabilization pattern analysis

In this section of the appendix, we describe in details the analysis of the
stabilization involving $3$ qubits. Following the definitions of IV.2.1, we
seek minimal and unique stabilization. To achieve this, we will need either
$2$ or $3$ stabilizers as defined in equation (IV.2.1). A operator, denoted as
$\mathcal{O}=A_{1}\otimes\cdots\otimes A_{n}$, where
$A_{i}\in\\{A_{\theta,\phi},\mathds{1}_{2}\\}$, cannot have a unique
eigenvalue of +1. Consequently, a single operator alone cannot uniquely
stabilize a state. On the other hand, the investigation of stabilization with
two operators $\mathcal{O}_{1}$ and $\mathcal{O}_{2}$ reveals that
stabilization involving four or more operators is never minimal. Therefore, in
the following two subsections, we examine the cases of stabilization using
either $2$ or $3$ operators.

###  F.1 Stabilization with $2$ operators

All non-equivalent stabilization patterns with $2$ operators on $3$ qubits are
as follows. We analyze these patterns using the methods described in section
IV.2.1. Our focus extends beyond the states stabilized by these operators; we
also examine the dimensions of the corresponding $+1$ eigenspaces, as a
function of the various parameters. This leads to a two-operators
compatibility condition: if multiple operators stabilize a state, then, for
every pair of operators, the shared $+1$ eigenspace must have a minimum
dimension of $1$. The results are presented in the table 2.

No. | Operators | Dimension | Condition | Stabilizer state
---|---|---|---|---
$1$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\ A_{\theta}\otimes A_{\phi}\otimes A_{\omega}\end{array}$ | $4$ | $\theta=\phi=\omega=0~{}[2\pi]$ | Non unique
$2$ |  $(\theta=0~{}\&~{}\phi=\omega)$ or $(\theta=\pi~{}\&~{}\phi=\pi-\omega)$ and permutations | Non unique
$1$ | Only one of the equation $\theta\pm\phi\pm\omega=0~{}[2\pi]$ | $\ket{\psi_{\pm 1,\pm 1,\pm 1}}$
$0$ | Other | None
$2$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\ A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\end{array}$ | $4$ | $\theta=\phi=0,\pi~{}[2\pi]$ | Non unique
$2$ | $\theta=\pm\phi$ | Non unique
$0$ | Other | None
$3$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\ A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\end{array}$ | $2$ | $\theta=0,\pi~{}[2\pi]$ | Non unique
$0$ | Other | None
$4$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\ A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\end{array}$ | $4$ | $\theta=\phi=0,\pi~{}[2\pi]$ | Non unique
$2$ | $\theta=\pm\phi$ | Non unique
$0$ | Other | None
$5$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\ A_{\theta}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}$ | $2$ | $\theta=0,\pi~{}[2\pi]$ | Non unique
$0$ | Other | None
$6$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\ A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\end{array}$ | $2$ | $\theta=0,\pi~{}[2\pi]$ | Non unique
$0$ | Other | None
$7$ | $\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\ \mathds{1}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}$ | $2$ | None | Non unique
$8$ |  $\begin{array}[]{ll}\sigma_{Z}\otimes\mathds{1}\otimes\mathds{1}\\\ A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\end{array}$ | $4$ | $\theta=0~{}[2\pi]$ | Non unique
$0$ | Other | None
$9$ | $\begin{array}[]{ll}\sigma_{Z}\otimes\mathds{1}\otimes\mathds{1}\\\ \mathds{1}\otimes\sigma_{Z}\otimes\mathds{1}\end{array}$ | $2$ | None | Non unique
Table 2:  Different ways two binary operators, constructed as tensor product
of $\mathds{1}$ and $A_{\theta}$ can stabilize subspaces of various dimension.
The first column give the expression of the operators, the second give the
list of possible dimensions the stabilized subspace can have. The third, shows
constraints on the set of free parameters that define the stabilizing
operators, see equation (6). The last column shows if a state is uniquely
stabilized or not.

From such analysis, we can observe the following facts

  * •

The only way to have unique stabilization with two operators, is when no
tensor factor of the stabilizers are the identity operator $\mathds{1}_{2}$
(correspond to the case number $1$). The corresponding state is
$\ket{\psi_{\pm 1,\pm 1,\pm 1}}$ which is a Pauli stabilizer state, as shown
section IV.1.

  * •

In all situations when two parameters $\theta$ and $\varphi$ are needed to
define the stabilizers (case number $2$ and $4$), the condition so that there
is a two dimensional stabilized subspace is $\theta=\pm\phi~{}[2\pi]$ and
$\theta,\phi\neq 0,\pi$. This gives rise to a first instance of the two-
operators compatibility rule.

  * •

In a similar way, when only one parameter $\theta$ is needed to define the
stabilizer (situation number $3$, $5$, $6$ and $8$), the condition so that
there is a two dimensional stabilized subspace is $\theta=0,\pi$. This
correspond to another version of the two-operators compatibility rule.

  * •

There is two situations (number $7$ and $9$) where the stabilized subspace is
of dimension $2$ space of dimension and no parameters are needed to define the
stabilizers.

  * •

Since we see that there is no situations, where the dimension of the
stabilized subspace is $3$, it follows that we only need up to $3$ operators
to do a unique stabilization on $3$ qubits.

###  F.2 Stabilization with 3 operators

All non-equivalent stabilization patterns are as follows. Here we have already
discarded the situations where it is easy to see that unique stabilization is
not possible, for example, due to too many identity operators used to
construct the stabilizing operators. An example such trivial situation, is
when the operators are $\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}$,
$A_{\theta}\otimes A_{\phi}\otimes\mathds{1}$ and $A_{\psi,\alpha}\otimes
A_{\kappa,\beta}\otimes\mathds{1}$. In this case the last operators are all
the identity $\mathds{1}$, and thus either, there are no states stabilized or
the dimension of the stabilized subspace is at least $2$. This kind of
situation cannot uniquely stabilize a state and can safely be ignored.

  1. 1.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta_{1}}\otimes A_{\theta_{2}}\otimes A_{\theta_{3}}\\\
A_{\phi_{1},\alpha}\otimes A_{\phi_{2},\beta}\otimes
A_{\phi_{3},\gamma}\end{array}\right.$. The stabilized states are equivalent
to $\ket{0}(\ket{00}+\ket{11})$ or $\ket{\psi_{\pm 1,\pm 1,\pm 1}}$ (See
details below).

  2. 2.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes A_{\omega}\\\ A_{\psi,\alpha}\otimes
A_{\kappa,\beta}\otimes\mathds{1}\end{array}\right.$. The stabilized states
are equivalent to $\ket{0}(\ket{00}+\ket{11})$ (see details below).

  3. 3.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes A_{\omega}\\\
A_{\psi,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. The
stabilization is non-unique.

  4. 4.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\ A_{\omega,\alpha}\otimes
A_{\psi,\beta}\otimes\mathds{1}\end{array}\right.$. The stabilized states are
equivalent to $\ket{0}(\ket{00}+\ket{11})$.

  5. 5.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
A_{\omega,\alpha}\otimes\mathds{1}\otimes A_{\psi}\end{array}\right.$. The
stabilized states are equivalent to $\ket{000}$ or
$\ket{000}+\ket{011}+\ket{101}+\ket{110}$.

  6. 6.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
A_{\omega,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. Here
$\theta=\pm\phi$, $\omega=0,\pi$ and the stabilizes states are equivalent to
$\ket{000}$.

  7. 7.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
\mathds{1}\otimes\mathds{1}\otimes A_{\omega}\end{array}\right.$. Here
$\theta=\pm\phi$, $\omega=0,\pi$ and the stabilized states are equivalent to
$\ket{0}(\ket{00}+\ket{11})$.

  8. 8.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\\\
A_{\phi,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. The
stabilization is non-unique.

  9. 9.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}\\\
A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\\\ \mathds{1}\otimes
A_{\phi}\otimes\mathds{1}\end{array}\right.$. Here $\theta,\phi=0,\pi$ and the
stabilized states are equivalent to $\ket{000}$.

  10. 10.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\ A_{\omega,\alpha}\otimes
A_{\psi,\beta}\otimes\mathds{1}\end{array}\right.$. The stabilization is non-
unique.

  11. 11.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
A_{\omega,\alpha}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}\right.$. The
stabilization is non-unique.

  12. 12.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
A_{\omega,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. The
stabilization is non-unique.

  13. 13.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes A_{\phi}\otimes\mathds{1}\\\
\mathds{1}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}\right.$. Here
$\theta=\pm\phi$ and the stabilized states are equivalent to
$(\ket{00}+\ket{11})\ket{0}$.

  14. 14.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\sigma_{Z}\\\ \mathds{1}\otimes
A_{\phi}\otimes A_{\omega}\end{array}\right.$. The stabilization is non-
unique.

  15. 15.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\sigma_{Z}\\\
A_{\phi,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. Here
$\theta,\phi=0,\pi$ and the stabilized states are equivalent to $\ket{000}$.

  16. 16.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\sigma_{Z}\\\
\mathds{1}\otimes\mathds{1}\otimes A_{\phi}\end{array}\right.$. Here
$\theta,\phi=0,\pi$ and the stabilized states are equivalent to $\ket{000}$.

  17. 17.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\\\
A_{\phi,\alpha}\otimes\mathds{1}\otimes\mathds{1}\end{array}\right.$. The
stabilization is non-unique.

  18. 18.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\\\ \mathds{1}\otimes
A_{\phi}\otimes\mathds{1}\end{array}\right.$. The stabilization is non-unique.

  19. 19.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
A_{\theta}\otimes\mathds{1}\otimes\mathds{1}\\\
\mathds{1}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}\right.$. Here
$\theta=0,\pi$ and the stabilized states are equivalent to
$\ket{\psi}=\ket{000}$.

  20. 20.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\sigma_{Z}\otimes\mathds{1}\\\
\mathds{1}\otimes\mathds{1}\otimes\sigma_{Z}\\\
\mathds{1}\otimes\mathds{1}\otimes A_{\theta}\end{array}\right.$. The
stabilization is non-unique.

  21. 21.

$\left\\{\begin{array}[]{ll}\sigma_{Z}\otimes\mathds{1}\otimes\mathds{1}\\\
\mathds{1}\otimes\sigma_{Z}\otimes\mathds{1}\\\
\mathds{1}\otimes\mathds{1}\otimes\sigma_{Z}\end{array}\right.$. The
stabilized states are equivalent to $\ket{\psi}=\ket{000}$.

Since certain cases cannot be obtained by directly applying the methods
described in Section IV.2.1, we will now elaborate on some of the more complex
sub-cases.

For the case 1, we can use the study of stabilization with two operators, and
more specifically the case $1$ of table 2. By applying the obtained results to
the different pairs of operators we get that $\theta_{1}=0$ and
$\theta_{2}=\theta_{3}$ or $\theta_{1}=\pi$ and $\theta_{2}=\pi-\theta_{3}$,
or any permutations thereof (the same holds for $\phi_{1}$, $\phi_{2}$,
$\phi_{3}$). We can thus divide the problem into different sub-cases and use a
numerical solver to get the complete set of solutions, which are set forth in
the table 3.

Operators | Conditions | stabilizer state
---|---|---
$\sigma_{Z}\otimes A_{\varphi,0}\otimes A_{\varphi,0}$ | $\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,\beta}$ | $\varphi,\omega,\beta\neq 0,\pi$ | $\ket{101}-\ket{110}$
$\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,-\beta}$ | $\ket{000}+\ket{011}$
$-\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,\pi-\beta}$ | $\ket{101}-\ket{110}$
$-\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,\beta-\pi}$ | $\ket{000}+\ket{011}$
$-\sigma_{Z}\otimes A_{\varphi,0}\otimes A_{\pi-\varphi,0}$ | $-\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,\beta}$ | $\ket{101}+\ket{110}$
$-\sigma_{Z}\otimes A_{\omega,\beta}\otimes A_{\omega,-\beta}$ | $\ket{000}-\ket{011}$
$\sigma_{Z}\otimes A_{\varphi,0}\otimes A_{\varphi,0}$ | $A_{\omega,0}\otimes\sigma_{Z}\otimes A_{\omega,0}$ | $\varphi,\omega\neq 0,\pi$ | $\ket{000}+\ket{011}+\ket{101}-\ket{110}$
$A_{\omega,0}\otimes\sigma_{Z}\otimes A_{\omega,\pi}$ | $\ket{000}+\ket{011}-\ket{101}+\ket{110}$
$A_{\omega,0}\otimes-\sigma_{Z}\otimes A_{\pi-\omega,0}$ | $\ket{000}+\ket{011}-\ket{101}+\ket{110}$
$A_{\omega,0}\otimes-\sigma_{Z}\otimes A_{\pi-\omega,\pi}$ | $\ket{000}-\ket{011}-\ket{101}-\ket{110}$
$-\sigma_{Z}\otimes A_{\varphi,0}\otimes A_{\pi-\varphi,0}$ | $A_{\omega,0}\otimes-\sigma_{Z}\otimes A_{\pi-\omega,0}$ | $\ket{000}-\ket{011}-\ket{101}-\ket{110}$
$A_{\omega,0}\otimes-\sigma_{Z}\otimes A_{\pi-\omega,\pi}$ | $\ket{000}-\ket{011}+\ket{101}+\ket{110}$
Table 3:  Different ways three binary operators
$\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}$,
$A_{\theta_{1},\varphi_{1}}\otimes A_{\theta_{2},\varphi_{2}}\otimes
A_{\theta_{3},\varphi_{3}}$ and $A_{\theta_{4},\varphi_{4}}\otimes
A_{\theta_{5},\varphi_{5}}\otimes A_{\theta_{6},\varphi_{6}}$ can stabilize a
unique state on three qubits. The first two columns give the expression of the
second and third operators. This table shows all possibilities up to the
permutation of the qubits.

The analysis of the case 2 is conducted as follows. We apply the $2$-operators
compatibility condition to the first two stabilizers, which allows us split
the study into two cases. In the first case, the stabilizing operators are
$\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}$, $A_{\theta}\otimes
A_{\theta}\otimes\sigma_{Z}$ and $A_{\omega,\alpha}\otimes
A_{\omega,\pm\alpha}\otimes\mathds{1}$. For the $+\alpha$ choice, the
stabilization can be treated with the already presented tools. For the choice
$-\alpha$, we can verify that $(\ket{00}+\ket{11})\ket{0}$ is stabilized for
all values of $\alpha$. This means that the stabilization is either non-
unique, or exclusively fixes this particular state. Ultimately, we conclude
that whatever the sign in front of $\alpha$, a uniquely stabilized state is
equivalent to $(\ket{00}+\ket{11})\ket{0}$.

For the second case, the operators are
$\sigma_{Z}\otimes\sigma_{Z}\otimes\sigma_{Z}$,
$A_{\theta}\otimes\sigma_{Z}\otimes A_{\theta}$ and $A_{\omega,\alpha}\otimes
A_{\omega}\otimes\mathds{1}$. We can apply local rotations to get a equivalent
stabilization given by $A_{\theta}\otimes A_{\theta}\otimes A_{\theta}$,
$A_{-\theta}\otimes A_{\theta}\otimes A_{-\theta}$ and
$A_{\omega,\pi/2}\otimes A_{\omega,\pi/2}\otimes\mathds{1}$. Applying the
determinant method (see section IV.2.1) reveals that stabilization is only
possible when certain conditions are met. Indeed the determinant
$\det(M^{\dagger}M)=0$ can be factored into multiple terms. Asking that the
determinant vanish imply one of the following condition

  1. 1.

$\theta=0,\pi/2,\pi$

  2. 2.

$\omega=\pi/2$

  3. 3.

$25\cos(2\theta-2\omega)-4\cos(4\theta-2\omega)+25\cos(2(\theta+\omega))-4\cos(4\theta+2\omega)+146\cos(2\theta)+8\cos(4\theta)+54\cos(2\omega)-250=0$

We need to discard the cases $\theta=0,\pi/2,\pi$ (case 1), since it would
result in the first two operators being the same. By further analysis, using
trigonometric expansion and substituting $\cos(\theta)^{2}\mapsto x$ and
$\cos(\omega)^{2}\mapsto y$, it can be shown that the complex term (case 3) is
zero only when $\theta=0,\pi$ and $\omega=0,\pi$, which we should also
exclude. Therefore, the only remaining case is $\omega=\pi/2$ (case 2). To
analyze this case, we reverse the local rotation and find that the state
$\ket{000}+\ket{011}+\ket{101}+\ket{110}$ is uniquely stabilized if
$\theta\neq 0,\pi/2,\pi$.