# Towards Two Bloch Sphere Representation of Pure Two Qubit States and
Unitaries.

Stanislav Filatov and Marcis Auzinsh
University of Latvia, Department of Physics
Raina boulevard 19, LV-1586, Riga, Latvia
E-mail<EMAIL_ADDRESS>
###### Abstract

We extend Bloch Sphere formalism to pure two qubit systems. Combining insights
from Geometric Algebra and analysis of entanglement in different conjugate
bases we identify Two Bloch Sphere geometry that is suitable for representing
maximally entangled states. It turns out that relative direction of coordinate
axes of the two Bloch Spheres may be used to describe the states. Moreover,
coordinate axes of one Bloch sphere should be rignt-handed and of the other
one - left-handed. We describe and depict separable and maximally entangled
states as well as entangling and non-entangling rotations. We also offer
graphical representation of workings of a CNOT gate for different inputs.
Finally we provide a way to also represent partially entangled states and
describe entanglement measure related to the surface area of the sphere
enclosing the state representation.

##  1 Introduction

The Bloch sphere’s historical roots extend to the early 20th century, driven
by the contributions of physicist Felix Bloch. His work in nuclear magnetic
resonance (NMR), which delves into the behavior of atomic nuclei in magnetic
fields, led to the development of the Bloch sphere as a visualization tool for
comprehending quantum properties, particularly those associated with quantum
spins in the presence of magnetic fields. In 1946, Bloch published a seminal
paper titled “Nuclear Induction” [1], which played a crucial role in advancing
our knowledge of how quantum systems interact with magnetic fields. This work
laid the groundwork for fundamental concepts in quantum physics, including
nuclear magnetic resonance and spin coherence.

Mathematically, the Bloch sphere representation relies on the use of complex
vector spaces to visually depict quantum states. This approach provides a
geometric framework for understanding quantum systems while preserving
essential information related to the amplitude and phase of these states [2].
In this representation, every vector within the Bloch sphere corresponds to a
specific quantum state (Figure 1). Pure, coherent states extend precisely to
the surface of the Bloch sphere with a length of 1, while mixed, incoherent
states fall short of reaching the sphere’s surface with a length less than 1.
The length of each representation’s vector on the Bloch sphere provides
insight into the degree of coherence within the quantum state. At the center
of the sphere resides the maximally mixed state with length 0, characterized
by a density matrix proportional to the identity matrix.

$x$$y$$z$$\left|0\right>$$\left|1\right>$$\left|\psi\right>$ Figure 1:
Arbitrary State Vector on Bloch Sphere

In addition to serving as a mathematical representation, the Bloch sphere
finds practical applications in various physical fields. In the field of
quantum optics, it facilitates the understanding of light polarization states,
enabling insights into phenomena such as the polarization of single photons[3,
4]. In nuclear and atomic physics, the Bloch sphere serves as a crucial tool
for describing atomic and nuclear spins. For instance, in nuclear magnetic
resonance (NMR) experiments, it is employed to visualize the behavior of
nuclear spins under the influence of magnetic fields, aiding in the analysis
of spin dynamics[5, 6, 7]. In condensed matter physics, the Bloch sphere
provides valuable insights into electron spin states in materials. It can be
used to illustrate the behavior of electron spins in magnetic materials,
enhancing our understanding of their magnetic properties[8, 9]

In the realm of quantum information, the Bloch sphere serves as a valuable
tool for simplifying the visualization of quantum states, particularly in the
context of qubits. Here, the concept of a qubit is abstracted away from its
physical substrate, allowing us to focus solely on its logical and
information-theoretical quantum mechanical properties[2, 10]. Various
implementations of qubits exist, ranging from superconducting circuits to
trapped ions and quantum dots[10, 11], but the Bloch sphere transcends the
differences, emphasizing the fundamental quantum nature of these entities.
Within the domain of quantum information processing, the Bloch sphere aids in
understanding the workings of quantum gates and operations, which are
fundamental components of quantum circuits[12]. This visualization simplifies
the design and implementation of quantum algorithms and protocols, and it also
highlights the challenges posed by decoherence, as it becomes evident how
quantum states evolve and deviate from their ideal representations over
time[11].

This geometric representation simplifies the visualization of complex quantum
states, allowing for an intuitive understanding of concepts related to quantum
coherence and mixed states[13]. It provides a bridge between the abstract
mathematics of quantum mechanics and the geometric visualization, enhancing
our grasp of quantum phenomena without imposing judgments on their nature or
significance.

$x$$y$$z$ (a) $\ket{M}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(b) $\ket{M}\otimes\ket{N}$

Figure 2: (a) Arbitrary Single Qubit state. (b) Atribtrary Separable Two Qubit
state

Extending the BS formalism to two qubits is not only logical but also highly
desirable. This extension would significantly enhance our ability to analyze
experiments in the Foundations of Quantum Physics and comprehend the workings
of Quantum Circuits. Two-qubit systems represent the simplest scenarios where
entanglement, a quintessentially quantum phenomenon, may emerge.

In the context of quantum information, entanglement is pivotal for quantum
communication protocols, ensuring secure information transmission through
quantum key distribution [14, 15]. Moving to quantum computation, entanglement
becomes a crucial resource for quantum gates, enabling parallel information
processing and the potential to solve specific problems exponentially faster
than classical computers [16].

In the foundations of quantum mechanics, entanglement challenges classical
intuitions by highlighting the non-local nature of quantum correlations. This,
in turn, sparks debates about the fundamental principles of the quantum world
[17, 18]. Moreover, advancements in the interplay between entanglement and
gravity have opened up new avenues for understanding the profound connections
between quantum mechanics and the nature of spacetime [19].

Entanglement, however, is what makes the two BS representation difficult to
obtain. While separable states are easily visualized (Figure 2), entanglement
introduces complexity without a corresponding structural representation. The
single-qubit reduced density matrix of any maximally entangled state is
proportional to the identity. Simply depicting Bloch sphere with a dot in the
middle for every qubit of entangled pair results in the same representation
for all maximally entangled states — two spheres with dots in the middle.
Which is not informative at all. While one can visualize clouds of different
color around the two Bloch spheres to signify entanglement and its different
kinds, this approach still fails to reveal the distinct geometries of the
underlying states.

There have been attempts at creating thorus geometry visualizations [20] or
even sphere geometries [21], an important milestone in creating the two BS
representation. We haven’t seen any, though, that would use only two Bloch
Spheres and connect straightforward representation of separable states with
representation of entanglement. Moreover, there has been an extensive research
into the geometry and symmetries of the SU(N) and specifically SU(4) groups
[22, 23, 24, 25]. In particular, [22] present an extension of the Bloch sphere
to two qubits involving a four-sphere at each point of which is a fiber
consisting of two two-spheres of SU(2).

At the core of our work lies the uncovering of the structure that describes
maximally entangled states within two Bloch Sphere formalism aided by the
insights from Geometric Algebra. The structure is the relative direction of
the Bloch Sphere’s coordinate axes, one being right-handed, the other one
being left-handed. Insights from Geometric Algebra are association of each of
the 15 two-qubit pauli matrices with the 15 planes of rotation formed by the
coordinate axes of two Bloch Spheres. Combining everything together we get
representations of separable and maximally entangled states as well as
dynamics of all the two-qubit unitaries. We navigate the space of two qubit
states with a step of $\pi/2$. Not only these rotations are the most
widespread in quantum computation but, importantly, they allow to obtain a set
of visually orthogonal representations (Stabilizer set of two-qubit states)
that can further be merged to create representation of an arbitrary pure state
and rotation.

In this work, we specifically focus on pure two-qubit states, so unless
otherwise stated, the term ”state” refers to pure states. If not specified
otherwise, ”entanglement” denotes maximal entanglement, and an ”entangled
state” refers to a Maximally Entangled State (MES), while Partially Entangled
States (PES) will be explicitly identified or referred to as such to make a
clear distinction from MES.

The subsequent sections are organized as follows: Chapter 2 provides a brief
overview of the relevant concepts in Geometric Algebra; Chapter 3 focuses on
representing two-qubit entanglement; Chapter 4 elucidates the visual dynamics
of a two-qubit system, detailing the process of inferring eigenmatrices from
graphical representations and performing various Unitary Operations without
relying on matrix formalism; Chapter 5 demonstrates usefulness of this
approach for analysis of logical operations and as an example presents a step-
by-step breakdown of the workings of the CNOT gate for three different inputs;
Chapter 6 addresses the representation of arbitrary two-qubit pure states and
rotations and provides a way to measure entanglement within the state
graphically; the graphical representation of the entanglement and
disentanglement dynamics can be found in Figure 14 of that chapter; In Chapter
7 we discuss possible questions regarding the model and Chapter 8 serves as
the conclusion of this work.

##  2 Geometric Algebra of two qubits

Geometric algebra offers tools for visually interpreting the matrix formalism
of Quantum Mechanics. It provides the structure to represent four-dimensional
complex Hilbert space of two qubit system in six-dimensional real Euclidean
space. Visualizing what is going on in three dimensions is simple, but it gets
more complex when trying to capture even a facet of what’s unfolding in six
dimensions. GA serves as a bridge between abstract mathematics and embodied
intuition by employing concepts such as “bivector,” “plane of rotation,” and
“handedness”. At the same time, GA remains an exceptionally versatile system,
offering a unified language for diverse areas of mathematics, physics, and
engineering that traditionally employ field-specific formalisms [26, 27].
Therefore, we will take some time to introduce the essential concepts within
this system.

Several sources [28, 29, 30, 31, 32] offer valuable insights into GA, but our
primary reference is the work by Havel and Doran [33]. The introduction of
their work explains the fundamental principles of GA, while its core employs
GA to create a Bloch-sphere-like GA model for a two-qubit system in a six-
dimensional Real Euclidean space. This work has been the main inspiration and
guide in our explorations. Although the authors do not explicitly address a
representation involving two Bloch Spheres, their graph model and the clear
connection between GA concepts and standard quantum mechanical formalism were
crucial in shaping the representations that will be developed later in this
work. For detailed mathematical information and formal proofs, we refer the
reader to this specific work.

They note, among other things, that just as in the case of two-dimensional
complex Hilbert space of a single qubit where Lie algebra isomorphism so(3)
$\approx$ su(2) allows to represent states of a qubit in a three-dimensional
real Euclidean vector space, in the case of four-dimensional complex Hilbert
space of two qubits, the Lie algebra isomorphisms spin(6) $\approx$ so(6)
$\approx$ su(4) allow to represent states of two-qubit system in a six-
dimensional real Euclidean vector space. More precisely, the respective
Geometric algebras of real Euclidean three- and six- dimensional vector spaces
are suitable to represent single- and two- qubit states and operators.

The fundamental component of geometric algebra (GA) is the bivector concept.
For a rigorous account of GA concepts we refer the reader to the literature on
GA [28, 29, 30, 31, 32, 33]. We will give an intuitive description of concepts
relevant to our work. Bivectors result from the wedge product (notated as
$a\land b$), involving two vectors $a$ and $b$. Similar to how vectors define
an oriented line, bivectors define an oriented plane. They also specify a
rotation within that plane, with the rotation direction determined by the
order of the vectors in the wedge product. For example, $a\land b$ signifies a
rotation from vector $a$ to vector $b$. In three-dimensional space, each
rotation plane possesses a single perpendicular vector, which can be termed an
axis of rotation. In higher-dimensional spaces, multiple normals exist for
each rotation plane, rendering the concept of an “axis of rotation” ill-
defined for rotations in spaces exceeding three dimensions. Conversely, the
notion of a “plane of rotation” remains well-defined in spaces of any
dimensionality. Bivectors are anticommutative $a\land b=-(b\land a)$ and
$-(b\land a)=-b\land a=b\land-a$. One could interpret the anticommutativity
relations as follows: To reverse a rotation from $a$ to $b$ we need to make a
rotation from $b$ to $a$.

We employ the concept of plane of rotation to visualize rotations of the
statevector on a Bloch sphere. Moreover, we use passive rotations instead of
active ones. Namely, the arrow representing the statevector remains fixed and
the coordinate axes are rotated instead. While such an approach arguably
doesn’t add more clarity when representing the separable states, it will be
seen later that it is essential in representing the entangled states.

Following [33] we associate each generator of rotations, which has the form of
a tensor product of two elements of the set
$\\{I;\sigma_{x};\sigma_{y};\sigma_{z}\\}$ with a plane of rotation formed by
two out of six axes of the two Bloch spheres
$\\{x_{1};y_{1};z_{1};x_{2};y_{2};z_{2}\\}$. See equations below that list all
the correspondences. The correspondences for local rotations (i.e. rotations
in plane formed by the axes of the same Bloch Sphere) are seen in eq. 1 and
correspondences for double-Pauli rotations (rotations in plane formed by the
axes of different Bloch Spheres) in eq. 2.

We refrain from the term “non-local” when speaking about the plane or
rotation. Instead to describe those planes we use term “double-Pauli” to
emphasize the nature of the rotation when expressed in matrix formalism.
However, if we interpret wedge products of the form $x_{1}\land y_{2}$ as
planes and forget the matrix description of the rotation these planes may be
characterized as “non-local” to reflect the fact that they are formed by
coordinate axes of different Bloch Spheres. Furthermore because those planes
are formed by axes of different BS they won’t be visible in the two-Bloch-
sphere visualization unlike the planes formed by two axes of the same Bloch
sphere.

local rotations

$\begin{split}I\otimes\sigma_{x}&\leftrightarrow y_{2}\land z_{2}\\\
I\otimes\sigma_{y}&\leftrightarrow z_{2}\land x_{2}\\\
I\otimes\sigma_{z}&\leftrightarrow x_{2}\land y_{2}\\\
\end{split}\quad\begin{split}\sigma_{x}\otimes I&\leftrightarrow y_{1}\land
z_{1}\\\ \sigma_{y}\otimes I&\leftrightarrow z_{1}\land x_{1}\\\
\sigma_{z}\otimes I&\leftrightarrow x_{1}\land y_{1}\\\ \end{split}$ (1)

The local rotations of separable states are seen in Figure 3(g). The order of
axes in the wedge product determines the directionality of the plane they
define and the direction of rotation of that plane on the picture. Once the
internal coordinates of each BS are explicitly added to the picture there is
considerable freedom in how to represent the states because rigid rotations of
the whole BS (coordinates and the statevector) don’t change the state
represented. Rather they amount to a change of point of view from which we are
looking at the BS.

Convention we have chosen is following: we always represent two BS in such a
way that the statevector arrows are pointing in the same absolute direction;
the coordinates of the first BS are always fixed. These rules make it easier
to obtain and visually recognize the entangled states later. Due to such
notational choice local rotations in planes that belong to the first BS
require not only rotation of the plane relative to the statevector, but also a
rigid rotation of both BS so that the two aforementioned rules are satisfied
(see Figure 3(g) (c) through (f)). Rotations depicted in (c) to (d) are
physically the same rotations as (e) to (f). (f) is representing the final
state in a convention we have chosen: the coordinates of the first BS should
be always drawn in the same way and both statevector arrows must face in the
same absolute direction.. A move from (d) to (f) is essentially a change of
point of view on both BS.

Eigentorations, rotations that don’t change the state of the system are
rotations that happen in the plane that is perpendicular to the statevector
arrow. In other words, any rotation of the plane perpendicular to the
statevector arrow may be considered a change of point of view on the BS.
Strictly speaking, the direction of the axes perpendicular to the statevector
arrow cannot be well defined, only the bivector formed by them can be -
without language of GA it would be difficult to formulate such an observation.
We will see later that such ambiguity in where those coordinate axes should be
drawn is an important ingredient in the mechanism of rotation between
separable and entangled states.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\uparrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$z_{2}$$y_{2}$

(b) $\ket{\uparrow}\otimes\ket{\uparrow+i\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(c) $\ket{\uparrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes I)}]{y_{1}\land z_{1}}$

$x_{1}$$z_{1}$$y_{1}$

$x_{2}$$y_{2}$$z_{2}$

(d) $\ket{\uparrow+i\downarrow}\otimes\ket{\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(e) $\ket{\uparrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes I)}]{y_{1}\land z_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$z_{2}$$y_{2}$

(f) $\ket{\uparrow+i\downarrow}\otimes\ket{\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(g) $\ket{\uparrow\uparrow}$ Figure 3: Local rotations of separable states.
(a) to (b) rotation of plane defined by $y\land z$ of the second BS. (c) to
(d) rotation of plane defined by $y\land z$ of the first BS. (e) to (f) same
rotation as in (c) to (d) depicted in the notation we will be using. (g) to
(h) eigenrotaion that doesn’t change the state.

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{z})}]{x_{2}\land y_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(h) $\ket{\uparrow\uparrow}$

double-Pauli rotations

$\begin{split}\sigma_{x}\otimes\sigma_{x}&\leftrightarrow x_{1}\land x_{2}\\\
\sigma_{x}\otimes\sigma_{y}&\leftrightarrow x_{1}\land y_{2}\\\
\sigma_{x}\otimes\sigma_{z}&\leftrightarrow x_{1}\land z_{2}\\\
\end{split}\quad\begin{split}\sigma_{y}\otimes\sigma_{x}&\leftrightarrow
y_{1}\land x_{2}\\\ \sigma_{y}\otimes\sigma_{y}&\leftrightarrow y_{1}\land
y_{2}\\\ \sigma_{y}\otimes\sigma_{z}&\leftrightarrow y_{1}\land z_{2}\\\
\end{split}\quad\begin{split}\sigma_{z}\otimes\sigma_{x}&\leftrightarrow
z_{1}\land x_{2}\\\ \sigma_{z}\otimes\sigma_{y}&\leftrightarrow z_{1}\land
y_{2}\\\ \sigma_{z}\otimes\sigma_{z}&\leftrightarrow z_{1}\land z_{2}\\\
\end{split}$ (2)

Furthermore, it can be shown using Cartan decomposition [33, 34] that any
rotation in our six-dimensional space (in other words, any unitary
transformation) may be written as a combination of rotations in six local
planes and rotations in only three double-Pauli planes. Which leaves us with 9
relevant planes of rotation. The most symmetric choice of three double-Pauli
planes of rotation would be $\\{x_{1}\land x_{2};y_{1}\land y_{2};z_{1}\land
z_{2}\\}$. In fact, such decomposition may be done using the formalism of
Pauli Matrices (see Chapter 4 in [25]).

##  3 Representing Two-Qubit Entanglement

Creating a representation for a two-qubit system requires careful
consideration, especially when dealing with entanglement. The structure needs
to reflect that an entangled state doesn’t provide information about
individual qubits but rather conveys information about both qubits as a
unified system. Moreover, since entangled states are non-separable, there
aren’t specific directions in which the qubits face. While we can make
statements about the relative direction of the qubits, like “qubits always
face in opposite directions for the state $\ket{\Psi^{-}}$,” this holds true
for some entangled states and not for others. Despite adhering to certain
symmetries, translating these symmetries into a clear representation in a two
Bloch Sphere framework is not straightforward. Thus, graphically illustrating
two-partite entanglement becomes a challenge.

Surprisingly, the key element for constructing the representation of entangled
states did not originate from GA. Instead, it emerged through the process of
rewriting the four Bell basis states in all three conjugate bases.
Disregarding normalization constant and having
$\uparrow_{x}=(\uparrow+\downarrow)_{z}$;
$\downarrow_{x}=(\uparrow-\downarrow)_{z}$; $\uparrow_{y}=(\uparrow+\
i\downarrow)_{z}$; $\uparrow_{y}=(\uparrow-\ i\downarrow)_{z}$ one obtains:

${\begin{array}[]{cc}\begin{aligned}
\Psi^{-}&=(\uparrow\downarrow-\downarrow\uparrow)_{z}\\\
&=(\uparrow\downarrow-\downarrow\uparrow)_{x}\\\
&=(\uparrow\downarrow-\downarrow\uparrow)_{y}\end{aligned}&\begin{aligned}
\Psi^{+}&=(\uparrow\downarrow+\downarrow\uparrow)_{z}\\\
&=(\uparrow\uparrow-\downarrow\downarrow)_{x}\\\
&=(\uparrow\uparrow-\downarrow\downarrow)_{y}\end{aligned}\\\ \begin{aligned}
\Phi^{-}&=(\uparrow\uparrow-\downarrow\downarrow)_{z}\\\
&=(\uparrow\downarrow+\downarrow\uparrow)_{x}\\\
&=(\uparrow\uparrow+\downarrow\downarrow)_{y}\end{aligned}&\begin{aligned}
\Phi^{+}&=(\uparrow\uparrow+\downarrow\downarrow)_{z}\\\
&=(\uparrow\uparrow+\downarrow\downarrow)_{x}\\\
&=(\uparrow\downarrow+\downarrow\uparrow)_{y}\end{aligned}\end{array}}$ (3)

Now, recalling that the reduced density matrix for each of the qubits in a
maximally entangled state is Identity and therefore corresponds to the dot in
the middle of the Bloch sphere, let us try to use the information provided by
different conjugate basis expansions to create a representation of the four
Bell basis states. Let us use the following rule: whenever we see the arrows
in the description of a state being aligned, e.g.
$(\uparrow\uparrow+\downarrow\downarrow)$, we look at the label and draw the
corresponding coordinate axes of the two Bloch spheres as facing in the same
direction. And whenever they are anti-aligned in the description - we draw the
corresponding axes to be anti-aligned. Going through all the labels ($x,y,z$)
we draw all the axes on both Bloch spheres. We put a dot in the middle of the
Bloch sphere instead of an arrow of the statevector. If we follow these simple
rules we get the following visual representations (Figure 4).

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\Psi^{-}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(b) $\Psi^{+}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(c) $\Phi^{-}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(d) $\Phi^{+}$

Figure 4: Bell states depicted using two Bloch spheres. A pair of Bloch
Spheres, with differing handedness in their coordinate axes, represents each
state. Odd number of coordinate axes is always inverted on the second BS. For
example, $\Phi^{-}$ has the following representations (up to a global phase
and dropping the normalization constant):
$(z)\ket{\uparrow\uparrow}-\ket{\downarrow\downarrow}$;
$(x)\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}$;
$(y)\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow}$. Therefore the Bloch
spheres will have $z_{1}$ and $z_{2}$; $y_{1}$ and $y_{2}$ aligned and $x_{1}$
and $x_{2}$ anti-aligned. Similarly for the rest of Bell states.

Let us note that the four representations reflect exactly the four ways in
which the right-handed coordinates may be transformed into left-handed using
only inversion of axes: there are three ways of inverting a single axis ($x,y$
or $z$) and there is one way to invert all three axes simultaneously. If one
inverts two axes the coordinates will stay right-handed and such an inversion
would be equivalent to some rotation. Also, it is impossible to go from right-
to left-handed coordinates by rotation of axes only.

The next important thing to note about this representation is that the choice
of the orientation of axes on the first Bloch sphere is arbitrary, and the
rules of drawing the axes specify only the aligned/anti-aligned relationships
between the axes, not their absolute orientation. Hence is the next difference
from the separable state representation: in the representation of entangled
states only the relative directions of coordinate axes matter. This property
together with a dot in the centre of the Bloch spheres reflects precisely the
fact that looking at a single qubit of an entangled pair doesn’t reveal
anything about the entangled state.

The next question one may ask is “How to depict other maximally entangled
states?” This is the moment when the concept of plane of rotation and all the
relations between GA and standard Matrix representation of quantum mechanics
come into play. Every maximally entangled state may be reached by single-qubit
local rotations from any other entangled state [36]. From our representation
it becomes visually obvious why - because any relative configuration of
coordinate axes may be arrived at through rotations on a single sphere. That
is why we can use the correspondences from eq. 1 and just rotate the relevant
planes of relevant Bell state to obtain all the maximally entangled state
representations. A great benefit of GA isomorphism is that even in left-handed
coordinates and even without a statevector arrow to rotate (!) rotations are
very simple. One just needs to rotate the plane of rotation formed by the
relevant axes in the direction specified by the order of axes in the wedge
product. Further as an example we use $\Psi^{-}$ state to obtain other
maximally entangled state. For clarity of representation we drop the
normalization constants. In eq. 4 we apply $\pi/2$ rotation to the state
$\Psi^{-}$ using the $I\otimes\sigma_{x}$ generator of rotation and obtain the
state
$\ket{\uparrow\uparrow-i\uparrow\downarrow+i\downarrow\uparrow-\downarrow\downarrow}$.
In fig. 5(a) we show a graphical representation of this rotation.

$\displaystyle Rot(\theta)=e^{i(I\otimes\sigma_{x})\frac{\theta}{2}}$
$\displaystyle=\cos{\frac{\theta}{2}}\ (I\otimes
I)+i\sin{\frac{\theta}{2}}(I\otimes\sigma_{x})\Rightarrow$ (4) $\displaystyle
Rot(\frac{\pi}{2})=e^{i(I\otimes\sigma_{x})\frac{\pi}{4}}$
$\displaystyle=\cos{\frac{\pi}{4}}\ (I\otimes
I)+i\sin{\frac{\pi}{4}}(I\otimes\sigma_{x})\Rightarrow$ $\displaystyle
Rot(\frac{\pi}{2})\ket{\Psi^{-}}$ $\displaystyle=$
$\displaystyle\frac{(I\otimes I+iI\otimes\sigma_{x})}{\sqrt{2}}\ket{\Psi^{-}}$
$\displaystyle=\frac{1}{2}\left[\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{pmatrix}+i\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\
0&0&0&1\\\ 0&0&1&0\end{pmatrix}\right]\cdot\begin{pmatrix}0\\\ 1\\\ -1\\\
0\end{pmatrix}$ $\displaystyle=\frac{1}{2}\begin{pmatrix}1&i&0&0\\\ i&1&0&0\\\
0&0&1&i\\\ 0&0&i&1\end{pmatrix}\cdot\begin{pmatrix}0\\\ 1\\\ -1\\\
0\end{pmatrix}=\frac{i}{2}\begin{pmatrix}1\\\ -i\\\ i\\\ -1\end{pmatrix}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\Psi^{-}=\ket{\uparrow\downarrow-\downarrow\uparrow}$ Figure 5: In order
to graphically apply $\pi/2$ rotation using $I\otimes\sigma_{x}$ to the
$\Psi^{-}$ state we take the plane that corresponds to such a rotation (see
eq. 1), in this case it’s $y_{2}\land z_{2}$, and rotate this plane by $\pi/2$
in the direction indicated by the letter order in the wedge product. In this
case the direction may be described as “from $y_{2}$ to $z_{2}$”.

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(b)
$\ket{\uparrow\uparrow-i\uparrow\downarrow+\downarrow\uparrow+i\downarrow\downarrow}$

Following this procedure we can generate representations of all maximally
entangled states. See appendices for representations of 24 maximally entangled
states generated by $\pi/2$ rotations. Incidentally, one could examine
stabilizer group for a given maximally entangled state and use it to obtain
our graphical representation. We focus on $\pi/2$ rotations because they
produce a set of visually orthogonal representations on the Bloch Spheres
(Stabilizer set) that can later be combined with each other to obtain
representation of any state. As opposed to $\pi$ rotations which produce
mathematically orthogonal vectors in Hilbert Space, but create too few
representations for unambiguous interpolation. The set of separable states may
be generated in a similar way. It consists of 36 states. The simplest way to
visualize it is to note that there are 6 visually orthogonal directions which
an arrow of the BS might be facing (along or opposite each of the three
coordinate axes). Squaring this number to account for two BS and therefore two
qubits - gives us 36.

The differences in the number of states are reflected in the different
symmetries of representations. For the separable states absolute direction of
the single arrow matters while for entangled states relative direction of
coordinate axes is important. In the language of GA this observation may be
expressed as follows. For separable states absolute direction of two vectors
are important while for entangled states relative directions of two trivectors
are important.

For every separable state there are 3 planes - two of them local - whose
rotation doesn’t result in the change of state (eigenplanes); 8 planes - 4
local, 4 double-Pauli - that result in separable states only and 4 planes -
all double-Pauli - that connect the state to entangled states. For every
entangled state the 15 planes of rotation are distributed, respectively as 3
eigenplanes - all double-Pauli; 6 planes that result in entangled states only
- all local; 6 planes that connect to separable states - all double-Pauli.

One could say that it is easier to stay inside the set of separable states or
that it is easier to get out of the set of entangled states, hence - the
difference in the size of sets (36 vs 24). The total number 60 confirms that
we have obtained the whole set of Stabilizer states for two qubits [35].

##  4 Exploring Two Qubit Stabilizer Set Dynamics Graphically

After examining the structure of bipartite entanglement, our focus now shifts
to exploring the capabilities within our framework. Within this framework, we
can visually perform rotations without resorting to matrix formalism, and we
can deduce the eigenmatrices of a given state just by looking at its
representataion. To systematize this information, we’ve established a set of
visual rules. While these rules may appear somewhat extensive, one should keep
in mind that they accommodate the intricacies of rotations in six-dimensional
space. Indeed, after a bit of practice one realizes that behind the words lay
straightforward intuitive moves. Importantly, the articulation of these rules
not only aids in manual application but also lays the groundwork for their
computational implementation. This coding aspect enables the creation of
visualizations that are even more efficient than manual rule application.
These rules unveil interesting symmetries within two-qubit dynamics, providing
an intuitive understanding of such systems.

We have already mentioned rule for performing visual rotations in local
planes. We just take the coordinate axes corresponding to a local rotation
(eq. 1) and rotate the plane defined by the coordinate axes pair in the wedge
product. This rule works both in entangled mode and separable mode and
accounts for 6 out of 15 planes planes of rotation (Figures 3(g), 5(a)).

###  4.1 Eigenrotations

Before speaking about double-Pauli rotations, let’s speak about visual rules
for guessing the eigenplanes (planes of rotation that correspond to the matrix
that doesn’t change the state up to an overall phase). There are always three
eigenmatrices for every separable and entangled state. Two local and one
double-Pauli for separable states and three double-Pauli for entangled.

For separable states eigenmatrices will be a) tensor product of eigenmatrices
of each individual qubit; b,c) tensor product of each individual qubit’s
eigenmatrix with Identity. In other words, planes that are perpendicular to
the direction of statevector on the Bloch spheres will be eigenplanes. For
example, for the state
$\ket{\uparrow\leftarrow}=\ket{\uparrow\uparrow-\uparrow\downarrow}$ the arrow
of the first qubit is facing in $z$ direction and the arrow of the second
qubit is facing in $-x$ direction. Therefore the eigenmatrices will be
$\\{\sigma_{z}\otimes\sigma_{x}$; $\sigma_{z}\otimes I$;
$I\otimes\sigma_{x}\\}$, which translates into GA language as the eigenplanes
are $\\{z_{1}\land x_{2}$; $x_{1}\land y_{1}$; $y_{2}\land z_{2}\\}$. The
alignment or anti-alignment along the axis doesn’t play a role in determining
the eigenplane, only the axis itself which is logical because if it is an
eigenrotation it doesn’t matter whether it’s performed clockwise or
counterclockwise - it won’t change the state.

For maximally entangled states everything is about the relative direction of
coordinate axes, which also determines the eigenplanes. Here aligned axes
determine the eigenplanes. For example, the Bell states (Figure 4) all have
$x_{1}$ aligned with $x_{2}$, $y_{1}$ with $y_{2}$, $z_{1}$ with $z_{2}$,
therefore all of them will have the same sets of eigenplanes and
eigenmatrices: $\\{x_{1}\land x_{2}$; $y_{1}\land y_{2}$; $z_{1}\land
z_{2}\\}$ and $\\{\sigma_{x}\otimes\sigma_{x}$; $\sigma_{y}\otimes\sigma_{y}$;
$\sigma_{z}\otimes\sigma_{z}\\}$. State
$\ket{\uparrow\uparrow-i\uparrow\downarrow+\downarrow\uparrow+i\downarrow\downarrow}$
which we’ve been describing in Figure 5(a) will have the following sets of
eigneplanes and eigenmatrices: $\\{z_{1}\land y_{2}$; $x_{1}\land x_{2}$;
$y_{1}\land z_{2}\\}$ and $\\{\sigma_{z}\otimes\sigma_{y}$;
$\sigma_{x}\otimes\sigma_{x}$; $\sigma_{y}\otimes\sigma_{z}\\}$.

###  4.2 Rules for Double-Pauli Rotations.

Rules for performing rotations in double-Pauli planes are somewhat more
intricate, but most of the times they are dealing with axis inversion.
Essentially there are three kinds of dynamics that are obtainable with
rotations generated by tensor product of two Pauli matrices: movement from
Entanglement to Separability; movement from Separability to Entanglement and
movement from Separability to Separability.

####  4.2.1 Entanglement - Separability

Rule “E-S”: Take the matrix for rotation and the wedge product corresponding
to that rotation. The axes of the wedge product belong to different BS, and
are orthogonal to each other. If translated to originate from the same point
they are forming an imaginary plane. Second axis in the wedge product inverts
and state arrows appear on both BS in direction opposite to the directionality
of imaginary plane.

We call it an imaginary plane because this plane is formed by axes of
different BS. Directionality of this plane can be settled by applying right-
hand-rule or taking a cross product of the two vectors in the order specified
by the wedge product. The vector resulting from the cross product is the
directionality of the plane. In case rotation matrix and therefore the
resulting wedge product has a minus sign in it, use negative direction of one
of the axes in the wedge product and apply the rules as usual.

Now let us apply this rule to two rotations illustrated in Figure 6(c). First
of all (a to b) we will be rotating state
$\Psi^{-}=\ket{\uparrow\downarrow-\downarrow\uparrow}$ in the plane
$x_{1}\land y_{2}$ by an angle of $\pi/2$ to obtain state
$\ket{\uparrow\downarrow}$. The vectors $x_{1}$ and $y_{2}$ forming a wedge
product are drawn as thicker arrows. If we imagine that they originate from
the same point (say, origin of the first BS) they will be defining a plane
that is perpendicular to $z_{1}$ axis. The directionality of this plane may be
determined by taking cross product of vectors $x_{1}$ and $y_{2}$ or applying
a right-hand rule. If we did so, we would see that the plane is facing down or
in the direction opposite to $z_{1}$ axis. Let’s call this direction $-z_{1}$.
According to the rule the statevector arrows should appear in the direction
opposite of the imaginary plane direction. Hence in this case the statevector
arrows will appear in the $z_{1}$ direction, in other words they will be
facing up (Figure 6(c) (b)). Also according to the rule the second vector of
the wedge product $x_{1}\land y_{2}$, will invert. That is why $y_{2}$ got
inverted moving from (a) to (b).

As a result we have the state $\ket{\uparrow\downarrow}$ (note that the second
statevector is facing in the $-z_{2}$ direction). If we were to rotate the
second BS rigidly with all the coordinates and statevectors around the $y_{2}$
axis we would obtain a more conventional representation of the same state.

Now let us look at a more intricate rotation in the same Figure 6(c) (c) to
(d). Here we are rotating Maximally entangled state
$\ket{\uparrow\uparrow+\uparrow\downarrow-i\downarrow\uparrow+i\downarrow\downarrow}$
to obtain a separable state $\ket{\uparrow-i\downarrow}\otimes\ket{\uparrow}$.
Having a minus sign in the exponent of the Pauli Matrix description of
rotation we are free to choose the bivector corresponding to it: $-x_{1}\land
x_{2}$ or $x_{1}\land-x_{2}$. This choice does not affect the final state. We
decided to choose the former. The plane of rotation is now $-x_{1}\land
x_{2}$. As in previous case we have used thicker arrows to explicitly show the
vectors forming the wedge product: those are vector $-x_{1}$ on the first BS
and vector $x_{2}$ on the second. If we imagine them originating from the same
point (say, origin of the first BS) the wedge product $-x_{1}\land x_{2}$ is
forming a plane that is facing in the $y_{1}$ direction. Let’s call this
direction Right. According to the rule the statevector arrows will appear in
the direction opposite. Therefore in (d) both statevector arrows are facing in
the direction Left or the direction opposite to $y_{1}$. The second vector of
the wedge product, in this case $x_{2}$, inverts.

Note that if we were to interpret the wedge product as $x_{1}\land-x_{2}$
nothing would change in the analysis because they define the same plane with
same directionality and inversion of $-x_{2}$ is equivalent to inversion of
$x_{2}$. Also one should remember that it is important to pay attention to the
axes along which the statevectors are pointing, not their absolute direction
to read out the final separable state. In (d) the first statevector is facing
in $-y_{1}$ direction and the second statevector is facing in $z_{2}$
direction. Hence the final state:
$\ket{\uparrow-i\downarrow}\otimes\ket{\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\Psi^{-}=\ket{\uparrow\downarrow-\downarrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{y})}]{x_{1}\land
y_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(b) $\ket{\uparrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(c)
$\ket{\uparrow\uparrow+\uparrow\downarrow-i\downarrow\uparrow+i\downarrow\downarrow}$
Figure 6: (a) to (b):
$e^{i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{y})}\ket{\uparrow\downarrow-\downarrow\uparrow}=\ket{\uparrow\downarrow}$
Imaginary plane formed by the vectors $x_{1}$ and $y_{2}$ through the wedge
product $x_{1}\land y_{2}$ is facing down on this picture. Therefore the
orange statevectors appear in the opposite direction - up. The second vector
in the wedge product ($y_{2}$) inverts.
(c) to (d):
$e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}\ket{\uparrow\uparrow+\uparrow\downarrow-i\downarrow\uparrow+i\downarrow\downarrow}=\ket{\uparrow\uparrow-i\downarrow\uparrow}=\ket{\uparrow-i\downarrow}\otimes\ket{\uparrow}$
Imaginary plane formed by the vectors $-x_{1}$ and $x_{2}$ through the wedge
product $-x_{1}\land x_{2}$ is facing in the $y_{1}$ direction -right, so both
statevector arrows will appear in the opposite direction - left. Second vector
of the wedge product $-x_{1}\land x_{2}$ inverts.

$\xrightarrow[e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}]{-x_{1}\land
x_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(d)
$\ket{\uparrow\uparrow-i\downarrow\uparrow}=\ket{\uparrow-i\downarrow}\otimes\ket{\uparrow}$

####  4.2.2 Separability - Entanglement.

Rule “S-E”: First draw the coordinate axes of the first BS in a usual way. Now
draw the statevector arrows so that they are pointing in the same absolute
direction. Next, draw the axes of the second BS so that they are right-handed,
the axes of the wedge product corresponding to rotation are perpendicular to
each other and form imaginary plane whose directionality is opposite to the
directionality of the statevector arrows. Now invert the second axis of the
wedge product and statevector arrows are pulled into the center of BS.

Let us now turn to Figure 7(c) for examples. In (a) to (b) we will be rotating
the state $\ket{\uparrow\uparrow}$ in the plane $x_{1}\land y_{2}$ to obtain
the state $\ket{\Phi^{-}}$. Following the rule we draw the two BS with
statevectors. Note how the $x_{2}$ and $y_{2}$ axes are rotated with respect
to $x_{1}$ and $y_{1}$. This is done to comply with the …the axes of the wedge
product corresponding to rotation are perpendicular to each other and form
imaginary plane whose directionality is opposite to the directionality of the
statevector arrows part of the rule. The vectors forming the wedge product
$x_{1}$ and $y_{2}$ are drawn as thicker arrows. If we were to translate them
so that they originate from the same point, say the origin of the first BS,
they would form an (imaginary) plane. It is facing in the direction $-z_{1}$
or Down, which is opposite to the direction statevector arrows are pointing.
We have complied with all the requirements of the rule. Now we just pull the
statevectors into the center of BS drawing a dot instead and invert the second
vector of the wedge product, $y_{2}$. We have obtained a representation for
the state $\ket{\uparrow\uparrow-\downarrow\downarrow}$.

Now let us turn to a more intricate (c) - (d) case in Figure 7(c). Here we
start with the state
$\ket{\uparrow+i\downarrow}\otimes\ket{\uparrow-\downarrow}$. Using this
information according to the Rule “S-E” we can draw the coordinates of the
first BS and both statevectors. The first statevector corresponds to the
single-qubit state $\ket{\uparrow+i\downarrow}$ meaning it is facing the
$y_{1}$ direction. We draw the first statevector in the appropriate direction
and immediately draw the second statevector arrow in the empty second BS in
such a way that it faces the same absolute direction. Now we can draw one
coordinate axis of the second BS. We know that the second qubit is in the
state $\ket{\uparrow-\downarrow}$ meaning it is facing in the $-x_{2}$
direction. Hence we draw $x_{2}$ axis as pointing opposite to statevector
direction.

To draw the $z_{2}$ and $y_{2}$ axes we should look at the rotation bivector.
In this case it is $z_{1}\land-y_{2}$. Therefore we have to draw $-y_{2}$ axis
in such a way that were it to originate from the same origin with $z_{1}$ the
plane formed by the wedge product $z_{1}\land-y_{2}$ would face in the
direction opposite to the statevector arrows. This could be done in only one
way: $-y_{2}$ (thicker arrow) should face into the page and therefore $y_{2}$
should face out of the page. In other words $y_{2}$ should face in the same
absolute direction as $x_{1}$. Having $x_{2}$ and $y_{2}$ axes in place there
is only one choice for the $z_{2}$ axis to keep the coordinates of the second
BS right-handed.

Now, the most difficult part past us, we pull the statevectors inside the
Bloch Spheres and invert the second axis of the wedge product, $-y_{2}$. That
of course implies that $y_{2}$ is also inverted. If we check appendices where
all 24 stabilizer maximally entangled states are depicted and find the one
labelled (g), we will see that this state corresponds to
$\ket{\uparrow\uparrow-i\downarrow\downarrow}$.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\uparrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{y})}]{x_{1}\land
y_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(b) $\ket{\Phi^{-}}=\ket{\uparrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(c) $\ket{\uparrow+i\downarrow}\otimes\ket{\uparrow-\downarrow}$ Figure 7: (a)
to (b):
$e^{i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{y})}\ket{\uparrow\downarrow}=\ket{\uparrow\uparrow-\downarrow\downarrow}$
The coordinates are drawn in such a way that both orange statevectors are
facing the same direction and imaginary plane formed by the wedge product
mutually orthogonal vectors $x_{1}$ and $y_{2}$ is facing the opposite
direction. The second vector of the wedge product ($y_{2}$) inverts and
statevectors are pulled into the center of the BS.
(c) to (d):
$e^{-i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{y})}\ket{\uparrow+i\downarrow}\otimes\ket{\uparrow-\downarrow}=e^{-i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{y})}\ket{\uparrow\uparrow-\uparrow\downarrow+i\downarrow\uparrow-i\downarrow\downarrow}=\ket{\uparrow\uparrow-i\downarrow\downarrow}$
This time we have chosen the second vector of the wedge product to bear the
minus sign. Plane formed by the vectors of the wedge product is directed in
the opposite direction from the orange statevector arrows; $z_{1}$ and
$-y_{2}$ are orthogonal. $-y_{2}$ inverts, arrows are again pulled into the
center.

$\xrightarrow[e^{-i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{y})}]{z_{1}\land-
y_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(d) $\ket{\uparrow\uparrow-i\downarrow\downarrow}$

####  4.2.3 Separability - Separability.

Rule “S-S”: When the rotation matrix is a tensor product of two Pauli
matrices, but the state is such that one of those Pauli matrices is an
eigenmatrix, double-Pauli rotations act like local rotation. We can substitute
the eigenmatrix by I multiplied by an eigenvalue.

For example
$(\sigma_{x}\otimes\sigma_{z})\ket{\uparrow\downarrow}=-(\sigma_{x}\otimes
I)\ket{\uparrow\downarrow}$ Thus the eigenvalue ($\pm 1$) determines the
direction of rotation of the other qubit. We could also formulate this rule in
terms of axes of the wedge product. If one axis of the rotation wedge product
aligns with the state vector arrow, only rotate the other qubit around the
remaining axis of the wedge product. Reverse the rotation direction if the
state vector anti-aligns with the former axis of the wedge product.

Let us take a look at the Figure 8(c). (a) to (b): We are trying to rotate the
state $\ket{\uparrow\uparrow}$ in the plane defined by a wedge product
$z_{1}\land x_{2}$. The statevector of the first BS is aligned with $z_{1}$,
in other words $\sigma_{z}$ from the tensor $\sigma_{z}\otimes\sigma_{x}$ is
an eigenrotation for the state $\ket{\uparrow}$ with an eigenvalue $+1$.
Therefore we are rotating the second qubit around the remaining axis of the
wedge product, $x_{2}$. That is equivalent to a rotation using the wedge
product $y_{2}\land z_{2}$. As a result we obtain the state
$\ket{\uparrow}\otimes\ket{\uparrow+i\downarrow}$.

In case the initial state is $\ket{\downarrow\uparrow}$ then rotation using
the same Paulis and the same wedge product $z_{1}\land x_{2}$ ((c) to (d))
results in the reversal of direction of rotation for the second qubit. Let us
see how it happens. The statevector of the first BS is anti-aligned with the
first vector in the wedge product. In other words the eigenvalue of the
$\sigma_{z}$ acting on state $\ket{\downarrow}$ is $-1$. Therefore the
rotation around the relaining axis of the wedge product will be reversed. That
is equivalent to a rotation of the second BS via the wedge product
$-y_{2}\land z_{2}$ resulting in the state
$\ket{\uparrow}\otimes\ket{\uparrow-i\downarrow}$.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\uparrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{x})}]{z_{1}\land
x_{2}\ \sim\ y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(b) $\ket{\uparrow}\otimes\ket{\uparrow+i\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(c) $\ket{\downarrow\uparrow}$ Figure 8: (a) to (b):
$e^{i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{x})}\ket{\uparrow\uparrow}$
becomes equivalent to
$e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}\ket{\uparrow\uparrow}$ and therefore
to a rotation of the plane $y_{2}\land z_{2}$.
(c) to (d):
$e^{i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{x})}\ket{\downarrow\uparrow}$
becomes equivalent to
$e^{-i\frac{\pi}{4}(I\otimes\sigma_{x})}\ket{\downarrow\uparrow}$ and
therefore to a rotation of the plane $-y_{2}\land z_{2}$. The direction of
rotation has reversed. We do not stick to the convention of representation for
more vivid difference between two initial state representations.

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{z}\otimes\sigma_{x})}]{z_{1}\land
x_{2}\ \sim\ -y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(d) $\ket{\downarrow}\otimes\ket{\uparrow-i\downarrow}$

###  4.3 Summary

The aforementioned rules combined with rules for local rotation and restricted
to $\pi/2$ angles of rotation allow to navigate the whole stabilizer set of
two-qubit states starting from a single state of the set. We have seen how
double-pauli rotations can entangle, disentangle and be single-qubit rotators.
In the latter case the direction of rotation of the qubit is conditioned on
the state of the qubit that doesn’t get rotated. Such behaviour reminds of the
action of CNOT gate. In the next chapter we will see explicitly how such
dynamics combined with local rotations produces CNOT gate.

##  5 Representation of CNOT Gate

The CNOT gate holds significant importance in quantum computation due to its
entangling nature. Understanding its role is essential for quantum computing
practitioners, especially in quantum algorithms. This chapter delves into
visualizing the CNOT gate’s operations within our established framework.
Observing the CNOT gate in action offers valuable insights, explicitly
revealing why it entangles specific states and leaves others unaltered. The
dynamics of this entanglement systematically unfold from the visual rules
discussed in the preceding chapter.

We will show how CNOT gate acts on three different input states. First where
the control qubit is such that the target qubit doesn’t flip
$CNOT\ket{\uparrow\uparrow}=\ket{\uparrow\uparrow}$; second when the control
qubit is such that the target qubit gets flipped
$CNOT\ket{\downarrow\uparrow}=\ket{\downarrow\downarrow}$; third - a
superposition of the two previous cases which results in entanglement
$CNOT(\ket{\uparrow\uparrow}+\ket{\downarrow\uparrow})=\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow}$.
Being Unitary, CNOT may be expressed in terms of a combination of rotations of
some of the 15 rotation planes mentioned earlier.

$CNOT=e^{-i\tfrac{\pi}{4}}e^{i\tfrac{\pi}{4}\sigma_{y}\otimes
I}e^{i\tfrac{\pi}{4}\sigma_{x}\otimes
I}e^{i\tfrac{\pi}{4}I\otimes\sigma_{x}}e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}e^{-i\tfrac{\pi}{4}\sigma_{y}\otimes
I}$ (5)

We can see from the Figures 9, 10, 11 that the key to such rich dynamics is
the entangling interaction $e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}$.
In the three cases it is doing three different things: rotates along the plane
$-y_{2}\land z_{2}$, rotates in the opposite direction of that plane
($y_{2}\land z_{2}$) and finally creates entanglement through rotation in the
plane $x_{1}\land x_{2}$ when conditions are right for it. It is interesting
how in the first two cases (Figures 9, 10) the dependence on the state of the
first qubit becomes apparent. The control qubit obtains its controlling
function through becoming an eigenvector of the first sigma matrix in
$e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}$ rotation. In case the
control qubit is $\ket{\uparrow}$ the eigenvalue is $1$ and if it is
$\ket{\downarrow}$ the eigenvalue is $-1$ which reverses the direction of
rotation of the plane of the second qubit. Note how in case of non-entangling
interactions the local rotation just after
$e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}$ undoes the rotation in case
when control qubit is $\ket{\uparrow}$ and continues the rotation in case when
control qubit is $\ket{\downarrow}$. Hence in case the control qubit is
$\ket{\uparrow}$ the two rotations are $-\pi/2$ and $\pi/2$ rotations in the
plane $y_{2}\land z_{2}$ and in case the control qubit is $\ket{\downarrow}$
the rotations are $\pi/2$ and $\pi/2$ in the same plane. As a result in the
first case the target qubit remains intact and in the second - it gets
flipped.

Speaking of the entangling case (Figure 11), note how the first local rotation
prepares the right configuration for the
$e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}$ rotation to work in
entangling mode. How in the (b) subfigure of Figure 11 $-x_{1}$ and $x_{2}$
form an imaginary plane that is directed opposite to the statevector arrows
and how in the (c) subfigure $x_{2}$ axis inverts and how the next three local
rotations are gradually navigating the state towards
$\ket{\uparrow\uparrow+\downarrow\downarrow}$.

There are many other interesting cases that may be analyzed graphically. For
example one can see explicitly why for states lying along the x axes the
control-target qubit relation reverses and the second qubit becomes control.
The first rotation $-z_{1}\land x_{1}$ keeps the second qubit aligned along
the $x_{2}$ axis, therefore it will be an eigenstate of the second
$\sigma_{x}$ matrix in $e^{-i\tfrac{\pi}{4}\sigma_{x}\otimes\sigma_{x}}$,
hence alignment or anti-alignment with the $x_{2}$ axis will determine the
flip or no flip of the first qubit.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\uparrow\uparrow}$

$\xleftarrow[e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}\ \sim\
e^{-i\frac{\pi}{4}(I\otimes\sigma_{x})}]{-x_{1}\land x_{2}\ \sim\ -y_{2}\land
z_{2}}$

$\xrightarrow[e^{-i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{-z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(b) $\ket{\uparrow+\downarrow}\otimes\ket{\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(c) $\ket{\uparrow+\downarrow}\otimes\ket{\uparrow-i\downarrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(d) $\ket{\uparrow+\downarrow}\otimes\ket{\uparrow}$

$\xleftarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes I)}]{y_{1}\land z_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(e) $\ket{\uparrow+\downarrow}\otimes\ket{\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(f) $\ket{\uparrow\uparrow}$

Figure 9: Step-by-step representation of CNOT acting on the state
$\ket{\uparrow\uparrow}$. (a) to (b): this rotation changes the state of the
control qubit to $\ket{\rightarrow}$. (b) to (c) is a rotation by
$e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}$ which in this case (see
rules) is the same as rotation by $e^{-i\frac{\pi}{4}(I\otimes\sigma_{x})}$ or
the plane $-y_{2}\land z_{2}=z_{2}\land y_{2}$ so only the second BS rotates.
State of the target qubit has changed. (c) to (d)
$e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}$ or the plane $y_{2}\land z_{2}$ which
undoes the rotation of the previous step so the state of the target qubit is
back to initial. (d) to (e) is an eigenrotation. (e) to (f) rotates the state
of the control qubit to its initial state.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\downarrow\uparrow}$

$\xleftarrow[e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}\ \sim\
e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{-x_{1}\land x_{2}\ \sim\ y_{2}\land
z_{2}}$

$\xrightarrow[e^{-i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{-z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(b) $\ket{\uparrow-\downarrow}\otimes\ket{\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(c) $\ket{\uparrow-\downarrow}\otimes\ket{\uparrow+i\downarrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(d) $\ket{\uparrow-\downarrow}\otimes\ket{\downarrow}$

$\xleftarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes I)}]{y_{1}\land z_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(e) $\ket{\uparrow-\downarrow}\otimes\ket{\downarrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(f) $\ket{\downarrow\downarrow}$

Figure 10: Step-by-step representation of CNOT acting on the state
$\ket{\downarrow\uparrow}$. (a) to (b): this rotation changes the state of the
control qubit to $\ket{\leftarrow}$. (b) to (c) is a rotation by
$e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}$ which in this case (see
rules) is the same as rotation by $e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}$ or
the plane $y_{2}\land z_{2}$ so only the second BS rotates. The absence of
minus sign due to negative eigenvalue is an important difference from the
previous case. State of the target qubit has changed. (c) to (d)
$e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}$ or the plane $y_{2}\land z_{2}$ which
continues the rotation of the previous step so the state of the target qubit
is flipped compared to beginning. (d) to (e) is an eigenrotation. (e) to (f)
rotates the state of the control qubit to its initial state.

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(a) $\ket{\uparrow+\downarrow}\otimes\ket{\uparrow}$

$\xleftarrow[e^{-i\frac{\pi}{4}(\sigma_{x}\otimes\sigma_{x})}]{-x_{1}\land
x_{2}}$

$\xrightarrow[e^{-i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{-z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(b) $\ket{\downarrow\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

(c) $\ket{\uparrow\downarrow+i\downarrow\uparrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(I\otimes\sigma_{x})}]{y_{2}\land z_{2}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(d)
$\ket{\uparrow\uparrow-i\uparrow\downarrow+\downarrow\uparrow+i\downarrow\downarrow}$

$\xleftarrow[e^{i\frac{\pi}{4}(\sigma_{x}\otimes I)}]{y_{1}\land z_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$y_{2}$$x_{2}$

(e)
$\ket{\uparrow\uparrow-\uparrow\downarrow+\downarrow\uparrow+\downarrow\downarrow}$

$\xrightarrow[e^{i\frac{\pi}{4}(\sigma_{y}\otimes I)}]{z_{1}\land x_{1}}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(f) $\ket{\uparrow\uparrow+\downarrow\downarrow}$

Figure 11: Step-by-step representation of CNOT acting on the state
$\ket{\uparrow+\downarrow}\otimes\ket{\uparrow}$. (a) to (b): this rotation
changes the state of the control qubit. (b) to (c) is entangling interaction
(see rules): $x_{2}$ inverts and the statevector arrows are drawn into the
center of BS; (c) through (f) rotations of local planes in entangled mode that
finally arrive to the state $\ket{\uparrow\uparrow+\downarrow\downarrow}$

##  6 Rotations by Arbitrary Angles, Partial Entanglement and Measures of
Entanglement

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

Figure 12: Example of a state of partial entanglement. It has been generated
by a $\pi/4$ rotation of state $\ket{\uparrow\uparrow}$ in the plane
$y_{1}\land x_{2}$ or in matrix formalism by a rotation
$e^{i\frac{\pi}{8}(\sigma_{y}\otimes\sigma_{x})}$. Note how $x_{2}$ axis is
partially inverted. How the inner sphere represents the separable part of the
state and the outer shell - entangled part.

So far we have depicted $\pi/2$ rotations in all of the planes. What about
arbitrary angle rotations? Being able to generalize the representation to
rotations by arbitrary angles we would cover all the two-qubit Hilbert space
and, hence, complete the Two Bloch Sphere Representation of Pure Two Qubit
States. In this section we develop a method to represent Partially Entangled
States (PES) shown in Figure 12. Furthermore, we discuss measures of
entanglement and relate them to the graphical representation. Finally we show
dynamics of a full rotation in a double-Pauli plane $y_{1}\land x_{2}$ that
goes through Separable, Partially Entangled and Maximally entangled states in
Figure 14.

Generalization to arbitrary angles is rather straightforward for local
rotations. We just rotate the relevant coordinate axes of the wedge product by
the relevant angle. This generalization works for local rotations of both
separable and entangled states. Let us pause here and discuss alternative
representations of a single qubit that is half way between $\ket{\uparrow}$
and $\ket{\leftarrow}$ that is
$\ket{k}=\frac{\sqrt{2+\sqrt{2}}}{2}\ket{\uparrow}-\frac{\sqrt{2-\sqrt{2}}}{2}\ket{\downarrow}$.

The three possible representations are shown in Figure 13 (a) through (c). In
(a) $\ket{k}$ is shown in the usual way: as a statevector arrow tilted by
$\pi/4$ angle. In (b) - as a combination of two BS with statevectors pointing
along $z$ and $-x$ directions. The lengths of the statevectors are
$\sqrt{2}/2$. Note that the combined surface area of these BS is equal to the
surface area of the original BS. (c) is a way to merge the two BS from (b) in
such a way that the size of the resulting BS is the same as in (a) and the
statevectors are kept on the coordinate axes.

In this case the length of the outer statevector arrow aligned with $z$ axis
in (c) doesn’t seem to match its length inside a smaller BS in (b).
Nevertheless, if we define surface area enclosing the vector as surface area
at the tip of the vector minus surface area at the origin of the vector, this
quantity remains equal to that in (b). In other words, in all three
representations (a), (b) and (c) the surface area enclosing the statevector is
the same. One can see that this works for less trivial case when there are
projections of the statevector on all three coordinate axes. We will see
usefulness of such attention to the surface area when we speak about
entanglement measures later in this chapter.

$x$$y$$z$

(a)

$x$$y$$z$

$x$$y$$z$

(b)

$x$$y$$z$

(c)

$x$$y$$z$

(d)

$x$$y$$z$

$x$$y$$z$

(e)

$x$$y$$z$

(f)

Figure 13: Alternative representations of a single qubit from the two qubit
Stabilizer state rotated by $\pi/4$ angle in local plane - (a) through (c);
double Pauli plane - (d) through (f).

Similar approach could be taken for double-Pauli rotations by arbitrary angle.
Partially entangled states, for example, could be represented as a separable
state rotated in double-Pauli plane by a less than $\pi/2$ angle. Hence we can
take two representations - one for separable state, another for entangled
state - and combine them using the angle of rotation for scaling.

This is done in Figure 13 (d) - (f) where three alternative representations of
a single qubit from a partially entangled pair are shown. We take two BS with
squared radii summing to one (e), or equivalently, with surface areas adding
up to that of a standard BS. Then, we combine them analogously to (c). There
are multiple methods to combine them, but the two most straightforward
approaches are depicted in (d) and (f). They differ in which of the two BS is
situated inside the compound BS and which touches the surface. Note, that the
size of the Spheres in (e) need not be the same and will generally vary
depending on how far the state is from being separable.

Now let us use this technique to represent the partially entangled state
$\ket{P}=(0.924\ket{\uparrow\uparrow}-0.383\ket{\downarrow\downarrow})=\frac{\sqrt{2+\sqrt{2}}}{2}\ket{\uparrow\uparrow}-\frac{\sqrt{2-\sqrt{2}}}{2}\ket{\downarrow\downarrow}$.
It is generated by applying $\pi/4$ rotation in the plane $y_{1}\land x_{2}$
to the state $\ket{\uparrow\uparrow}$. Given that same rotation by twice the
angle ($\pi/2$) results in the state
$\ket{\uparrow\uparrow}-\ket{\downarrow\downarrow}$ we will say that $\ket{P}$
is half-way between entanglement and separability.

Figure 12 provides a representation of $\ket{P}$. The separable part is
represented by the inner sphere enclosing the statevector arrow and the
entangled part is represented by the outer shell lacking the statevector
arrow. Note that although it might not be obvious from the picture, the ends
of coordinate axes reach the surface of the outer shell. To represent
entanglement we need the outer shell of the second BS to have inverted
coordinates. In this particular case $x_{2}$ should be inverted. Therefore in
the Figure 12 $x_{2}$ axis reaches the edge of the inner shell in usual right-
handed way and then becomes inverted and reaches the edge of the outer shell
in the left-handed way. In the picture we don’t draw in negative parts of the
axes, but logically negative part of the $x_{2}$ axis also inverts in the same
way.

This representation maintains a connection with standard purity measures, such
as $Tr(\rho^{2})$, serving as a separability measure in our context. $\rho$ in
this case becomes the reduced density matrix $\rho_{A}$ of one of the
particles of the pair; $Tr(\rho_{A}^{2})=Tr(\rho_{B}^{2})$. Additionally, the
representation hints at a relation between the separability/entanglement
measure and the surface area of the Bloch Sphere section enfolding
separable/entangled state. The conventional purity measure, the Trace of the
Square of the density matrix, is related to the square of the statevector
arrow $r^{2}$:

$\centering Tr(\rho^{2})=\frac{1+r^{2}}{2}\@add@centering$ (6)

$r^{2}$ is proportional to the surface area of the sub-sphere containing the
statevector arrow. Consequently, $Tr(\rho_{A}^{2})$ as a separability measure
is directly related to the surface area of the sub-sphere representing
separable part of the PES. A measure of the entanglement in a given state
therefore may be obtained by subtracting the surface area of the inner sub-
sphere from the surface area of the whole BS. Note that this is the “surface
area enclosing the state” we’ve defined discussing Figure 13. Hence we can say
that entanglement in a PES is proportional to the surface area of the sub-
sphere enclosing the entangled part of the PES. Or equivalently, proportional
to the surface area of the BS without a statevector in Figure 13 (e) which
represents the entangled part of the PES.

Other measure of entanglement, Concurrence [37, 38, 39] may be related to the
graphical representation too. Being proportional to square root of
separability measure $Tr(\rho_{A}^{2})$ for pure states

$\centering
C=\sqrt{2(1-Tr(\rho^{2})}=\sqrt{2(1-\frac{1+r^{2}}{2})}=\sqrt{1-r^{2}}=\sqrt{\tilde{r}^{2}}=\tilde{r}\@add@centering$
(7)

Concurrence may be associated with $\tilde{r}$ which is the radius of the
Bloch sphere that represents entangled part of the partially entangled state,
i.e. BS without a statevector in Figure 13 (e). Hence Concurrence for a
Partially entangled state $\ket{P}$ represented in Figure 12 that we called
being exactly half way between separable and entangled state is $\sqrt{2}/2$,
not $1/2$. But given graphical interpretation we understand why. In this case
$r=\tilde{r}$ nevertheless, meaning that the Bloch Spheres representing
separable and entangled part are of equal size. And $\sqrt{2}/2$ is an
artifact of the fact that when we rotate a vector by $\pi/4$, its projection
doesn’t halve. Moreover, we understand that one can relate Concurrence and
Separability through length of radius and surface area. Concurrence being
length of radius of the BS representing entangled part of the PES and
Separability being proportional the surface area of the BS representing the
separable part of the PES.

In fact, being able to represent arbitrary separable state, arbitrary MES and
one PES is sufficient to be able to represent arbitrary PES. We can make
Schmidt decomposition of the PES arriving at
$\ket{P}=\alpha\ket{a_{0}}\ket{b_{0}}+\beta\ket{a_{1}}\ket{b_{1}}$. This makes
obvious between which separable and MES states the PES lies:
$\ket{a_{0}}\ket{b_{0}}$ and $\ket{a_{0}}\ket{b_{0}}+\ket{a_{1}}\ket{b_{1}}$.
We just take the representations of those two states, which we know, scale and
combine them accordingly. This observation implies that rules and
representations shown in this work are sufficient to create representations of
arbitrary pure two-qubit state. Having representations of arbitrary states we
can simulate the action of a unitary on a given state by creating
representations of all the intermediate states on the trajectory of the action
of the unitary. Moreover, having representation of a single separable state,
single MES and being able to redefine the coordinate axes is enough to create
representation of an arbitrary pure state too.

As a side note, having the rules, being able to redefine the coordinates and
perform Cartan decomposition should be enough to perform arbitrary rotations
on arbitrary state. Redefinition of coordinates can always give us stabilizer
state. Even for a PES, which may be disassembled into a separable and a ME
parts and redefinition of coordinates for each of those parts gives the
stabilizers. In the newly redefined basis we can always use Cartan
decomposition to reduce any unitary to a combination of rotations in some of
the 15 planes. Then we can apply the rules.

Now let us examine the Figure 14 where a full rotation of the state
$\ket{\uparrow\uparrow}$ in the plane $y_{1}\land x_{2}$ is represented. The
Figure should be read column-by-column up to down. The rotation transitions
through the states $\ket{\uparrow\uparrow}$,
$\ket{\uparrow\uparrow}-\ket{\downarrow\downarrow}$,
$\ket{\downarrow\downarrow}$,
$\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow}$, back to
$\ket{\uparrow\uparrow}$.

During this double-Pauli rotation, the statevector arrows are drawn into the
center of the Bloch Spheres before flipping to the opposite side.
Additionally, the coordinate axes undergo inversion: first, the $x_{2}$ axis
flips, followed by a reversal, and then the $y_{2}$ axis undergoes a similar
inversion. These flips align with the previously discussed rules, where we
utilized the flexibility in representing separable state coordinate axes to
enhance the visual dynamics. Note how double-Pauli rotations are composed of
inversions rather than rotations. And how such representation utilizes the
whole volume of Bloch Spheres to represent states and rotations.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$x_{2}$$z_{2}$

Figure 14: Full rotation of state $\ket{\uparrow\uparrow}$ in the plane
$y_{1}\land x_{2}$, in matrix formalism rotation generator is
$\sigma_{y}\otimes\sigma_{x}$. The figure should be read column by column,
from top to bottom. We gradually move through states $\ket{\uparrow\uparrow}$,
$\ket{\uparrow\uparrow-\downarrow\downarrow}$, $\ket{\downarrow\downarrow}$,
$\ket{\uparrow\uparrow+\downarrow\downarrow}$ back to $\ket{\uparrow\uparrow}$

##  7 Discussion

The model presented here has a lot of structure which we have only started to
explore. We see this paper as an invitation for the members of community to
join the exploration. We’ve decided to organize this section in the form of
Q&A to highlight some questions that might arise regarding the model.

Although in the previous section it is shown how a representation of an
arbitrary pure two-qubit state and, in some sense, arbitrary unitary may be
obtained if one is able to redefine the coordinates, exploring manual rule
application alone in different contexts may also be valuable.

Q1. Does the model work only for 15 rotations or for any arbitrary rotation?

As we mentioned at the end of Chapter 2 any arbitrary two-qubit unitary may be
expressed as a product of some of the 15 rotations using Cartan decomposition
[33, 25, 41, 40]. In fact, not all of the 15 planes are necessary and lesser
number suffices.

Q2. How does one rotate when the initial set is not from a stabilizer set,
i.e. if the statevector arrows don’t align with coordinate axes?

Decompose the state into pairs of BS of smaller size similar to how it is done
in 13 so that the statevector arrows align with the coordinate axes, perform
rotations using rules, recompose the Spheres. We have tried it on some simple
examples and it works. Second option: redefine coordinates so that the
statevector arrows are aligned with the coordinate axes, transform the
rotation planes to conform to new description, use the rules as usual.

Q3. What about PES states - do rules still hold?

PES, just like non-stabilizer states in the question above may be disassembled
into pairs of smaller BS which can be rotated using the rules and then
reassembled. Again, we have tried a couple of simpler cases and they work.
Even for the case when a double-Pauli rotation applied to a PES doesn’t lead
to MES or separable state, but moves towards another PES. That happens because
application of the same rule to MES part of the PES reduces its entanglement,
but when applied on the separable part of the PES - increases its
entanglement. As a result we move from PES to PES.

Q4. Can we do some kind of visual calculations using the representations and
rules?

Some visual calculations might be done using the representations. We can
estimate the amount of entanglement in a PES; guess the eigenrotations and
stabilizers of the state, tell what planes of rotation/Pauli matrices are
needed to entangle/disentangle the state. It would be great if someone found a
way to connect representation in a direct and straightforward way to its
numeric representation in the vector or density matrix form.

Q5. How can we be sure that these rules work? They are just stated as fact.

We have developed the rules for moving between separable and ME stabilizer
states after we have obtained the representations of stabilizer MES. In
principle one does not need the rules to create representation of an arbitrary
state - representation of single separable and single ME state combined with
all the possible basis changes are sufficient to create representations. But
these rules seem to obey the structure of the states in question. Every
rotation from separable to ME state involves coordinate handedness change in
one of the BS - rules reflect that; they also reflect the dynamics of
statevector arrows being pulled into and outside the center of BS. Those are
two main features that any set of rules for given representation must
incorporate and rules listed in Chapter 4 do so. But we do agree that tying
those rules to some fundamental considerations would be important, especially
for the possible extension of the model to more qubits where the number of
state combinations grows beyond what is possible to check manually.

Q6. Are these representations and rules unique?

No. There are alternative equivalent representations. We have chosen a
convention in which the coordinate axes of the first BS are always fixed for
the ease of manual manipulation of the representations. But MES
representations are equivalent up to simultaneous rotation of both BS that
doesn’t change the relative axis orientation. Separable states are defined up
to an arbitrary rotation of each BS axes around the statevector arrow. In
fact, if we were to represent separable states as faithfully as possible we
shouldn’t draw the two BS axes perpendicular to the statevector at all.
Instead we should only depict the plane defined by those axes. This ambiguity
is actually very important for the functioning of the rules.

As a result when we combine the separable and MES BS representations to create
a PES representation there is a lot of freedom in the way they get combined.
In Figure 14 we have chosen the representation so that the inner and outer BS
seem to have a common set of coordinate axes, but it need not be the case.
Generally the Spheres may be merged in ways that the description of dynamics
would change - different axes would invert for example. So the rules and
dynamics of the process depend on the convention one chooses to represent the
states. What does not change in the description of the process of entanglement
are the BS dynamics of handedness and statevector arrows mentioned in previous
question. But even here there is some flexibility and one could create a model
where during a full double-Pauli rotation the handedness of both BS
coordinates would change from Right-Right to RL, to LL, to LR back to RR.

Considering alternative representations we think it would be an interesting
challenge to create a representation that would combine the different rules of
action of double-Pauli planes on separable states into one.

##  8 Conclusion

We have shown a way to graphically represent two-qubit pure states and
unitaries using two Bloch Spheres. This approach applies the concept of Plane
of Rotation from Geometric Algebra to the context of two Bloch Spheres.
Separable states are represented as a pair of Bloch Spheres with statevector
arrows; coordinate axes handedness is the same; the pair of coordinates of the
plane perpendicular to the statevector are defined up to an arbitrary rotation
of the pair. Maximally entangled states are represented as a pair of Bloch
Spheres with a dot in the middle instead of statevector arrows; the handedness
of the coordinate axes of the Bloch Spheres is opposite; relative directions
of coordinate axes of two BS determine the exact maximally entangled state the
BS pair represents.

We have provided graphical representations of separable and maximally
entangled states. We used a method from a paper by Havel and Doran [33] to
translate 15 matrix generators of rotation into 15 planes of rotation. With
this, we derived rules for performing graphical rotations in each of these
planes by a $\pi/2$ angle. Using these rules we can navigate the set of two-
qubit stabilizer states. Next we have shown through an example of CNOT gate
how quantum gates may be graphically implemented on two Bloch Spheres.
Furthermore, we have generalized the representations to include arbitrary
angle rotations in both local and double-Pauli planes and represent any pure
two-qubit state. This is done by interpolation between representations of two
relevant stabilizer states by combining and properly scaling them.

While the model technically permits the representation of arbitrary pure two-
qubit states and unitaries, the level of effort and computation varies
depending on the specific representation required. Rotations of any stabilizer
state in any of the 15 planes by any angle can be visually performed without
the need to compute Pauli Matrices for rotation, offering a straightforward
approach. However, for now, rotations of arbitrary states in arbitrary planes
by arbitrary angles typically demand a higher level of sophistication.
Alongside rule application, tasks may include coordinate redefinition, Cartan
decomposition, Schmidt decomposition, and decomposition of BS into multiple
smaller BS.

When working on this model we were guided by the symmetries of the uncovered
representations. Some ambiguities in representation eventually turned out to
be important for the functioning of the model. Some ambiguities were resolved
by choosing a suitable and consistent convention. Our main criterion for
selecting the convention was its ease of manual manipulation of the
representations, without relying on simulation. Those developing a
computational application of this model might choose different conventions to
emphasize different aspects of two-qubit dynamics. Indeed we would be very
happy to see a simulation developed for this model.

Representation provided in this work allows for more intuitive manipulation of
quantum states - one that does not even require computation of rotation
matrices. Furthermore, it allows to see the properties of the state, like
eigenmatrices of its rotation or even degree of entanglement. This model
utilizes the entire volume of the Bloch Sphere for pure two-qubit state
representation. Simultaneously, if one examines the representation of a single
qubit from an entangled pair, it appears as a maximally mixed single qubit
state.

Three important takeaways for the foundations of QM in our opinion are: a) the
idea that double-Pauli entangling unitaries are rotations in the plane formed
by axes of different Bloch Spheres like $z_{1}\land x_{2}$ and that such a
rotation leads to a change of handedness in coordiante axes of one of the BS,
hence b) entanglement is linked to an act of inverting BS inside-out and c) to
relations of handedness of particles’ internal degrees of freedom or
particles’ coordinate frames.

##  9 Acknowledgements

We would like to acknowledge the help of Quantum Computing Stack Exchange
users Adam Zalcman in understanding more exotic maximally entangled states;
Rammus and Norbert Schuch for pointing out that Schmidt Decomposition of a PES
allows one to find separable and MES state between which the PES lies;
3blue1brown visualizations of 4D rotations; Vaxjo 2023 conference where this
work has been first presented and especially Andrei Khrennikov and John Small;
Navin Khaneja; Francesco Buscemi and members of his group at the University of
Nagoya; anonymous Reviewers.

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$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\Psi^{-}=\ket{\uparrow\downarrow-\downarrow\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(b) $\Psi^{+}=\ket{\uparrow\downarrow+\downarrow\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(c)
$\ket{\uparrow\uparrow-\uparrow\downarrow-i\downarrow\uparrow-i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(d)
$\ket{\uparrow\uparrow-\uparrow\downarrow+i\downarrow\uparrow+i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(e)
$\ket{\uparrow\uparrow-i\uparrow\downarrow-\downarrow\uparrow-i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(f)
$\ket{\uparrow\uparrow-i\uparrow\downarrow+\downarrow\uparrow+i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(g) $\Phi^{-}=\ket{\uparrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(h) $\Phi^{+}=\ket{\uparrow\uparrow+\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(i)
$\ket{\uparrow\uparrow+\uparrow\downarrow+i\downarrow\uparrow-i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(j)
$\ket{\uparrow\uparrow+\uparrow\downarrow-i\downarrow\uparrow+i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(k)
$\ket{\uparrow\uparrow+i\uparrow\downarrow-\downarrow\uparrow+i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(l)
$\ket{\uparrow\uparrow+i\uparrow\downarrow+\downarrow\uparrow-i\downarrow\downarrow}$

Figure 15: Representation of the maximally entangled states from two qubit
Stabilizer set. States are divided into four groups three each to indicate
more symmetries shared within each group.

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(a) $\ket{\uparrow\downarrow-i\downarrow\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(b) $\ket{\uparrow\downarrow+i\downarrow\uparrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(c)
$\ket{\uparrow\uparrow-\uparrow\downarrow+\downarrow\uparrow+\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(d)
$\ket{\uparrow\uparrow-\uparrow\downarrow-\downarrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(e)
$\ket{\uparrow\uparrow-i\uparrow\downarrow-i\downarrow\uparrow+\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(f)
$\ket{\uparrow\uparrow-i\uparrow\downarrow+i\downarrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(g) $\ket{\uparrow\uparrow-i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$x_{2}$$y_{2}$$z_{2}$

(h) $\ket{\uparrow\uparrow+i\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(i)
$\ket{\uparrow\uparrow+\uparrow\downarrow-\downarrow\uparrow+\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$y_{2}$$z_{2}$$x_{2}$

(j)
$\ket{\uparrow\uparrow+\uparrow\downarrow+\downarrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(k)
$\ket{\uparrow\uparrow+i\uparrow\downarrow-i\downarrow\uparrow-\downarrow\downarrow}$

$x_{1}$$y_{1}$$z_{1}$

$z_{2}$$x_{2}$$y_{2}$

(l)
$\ket{\uparrow\uparrow+i\uparrow\downarrow+i\downarrow\uparrow+\downarrow\downarrow}$

Figure 16: (continued) Representation of the maximally entangled states from
two qubit Stabilizer set. States are divided into four groups three each to
indicate more symmetries shared within each group.