# Quantum Machine Learning and its Supremacy in High Energy Physics

Kapil K. Sharma†
DY Patil International University,
Sect-29, Nigdi Pradhikaran
Akurdi, Pune,
Maharashtra-411044, India
E-mail<EMAIL_ADDRESS>
###### Abstract

This article reveals the future prospects of quantum algorithms in high energy
physics (HEP). Particle identification, knowing their properties and
characteristics is a challenging problem in experimental HEP. The key
technique to solve these problems is pattern recognition, which is an
important application of machine learning and unconditionally used for HEP
problems. To execute pattern recognition task for track and vertex
reconstruction, the particle physics community vastly use statistical machine
learning methods. These methods vary from detector to detector geometry and
magnetic filed used in the experiment. Here in the present introductory
article, we deliver the future possibilities for the lucid application of
quantum computation and quantum machine learning in HEP, rather than focusing
on deep mathematical structures of techniques arise in this domain.

##  I Introduction

The field of high energy physics (HEP) deals with the discovery of varieties
of particles which gives the clue to understand the big bang and origin of
universehep1 ; hep2 . The HEP experimentsehep  demand the high voltage to
operate and need the accelerators for beam collisions. At CERNcern , the Large
Hadron Collider (LHC) is the biggest particle collider in the world, which has
been operated with energy (6.5 TeV/beam) in its second run scheduled in 2015.
However a short run of the accelerator with xenon-xenon collisions has been
performed in 2017. The experimental set-up of LHC is tunnelled underground at
175 meters, it has its huge diameter as 27 kilometres. In LHC the particle
beams are launched in anti directions at very high speed, the beams further
collide at many interaction points available at the periphery of large
accelerator. The interaction points support the bombardment of antiparticle
beams. This bombardment release huge energy with different kind of particles
and varieties of trajectories are developed by particles during this process.
The most important ingredient used to capture the event happened at
interaction point is the detector. Round the periphery of the accelerator,
there are seven detectors (ATLAS, CMS, LHcb, ALICE, TOTEM, LHcf, MoEDAL)
assembled, which are used for different roles in LHClhc1 ; lhc2 . Experimental
part of HEP involve many complexities in terms of designing detectors, high-
end electronics, data acquisition systems and softwaredata . The data gather
in real time at LHC is recorded at a tape and processed through grid
computing, which further can be distributed to many universities and research
centres for particle physics analysis. The changes in designing methodologies
and implementation of various detectors is very crucial and important part.
The detectors play an important role to capture the event and provide the huge
data corresponding to interaction points. The technical journey of detectors
from bubble chamber to semiconductor detectors have long strides. Each
detector has association with front-end electronics equipped with data
acquisition system (DAS)data . The DAS gather the information from detector in
real time during the event happening in accelerator. It also play the role to
avoid the unnecessary background events and to collect only valid events
during the triggering process. For this purpose the triggering may be
implemented at many levels of hardware in real time. Most of the time DAS
suffer from dead time, the time during which no event is captured and there is
also the possibilities of missing the events. The dead time in DAS depends on
many factors such as clock speed of the electronics circuitry, noise, rate of
event happening etc. Because of the dead time, the speed of writing the data
on storage tape is also affected. Once the data is recorded on tape through
grid computing, it can be distributed further to do offline analysis to
extract the information about particle trajectories developed inside the
detectorsdata . These trajectories are important ingredient which have the
hidden information about many characteristics of the particles. During the
analysis of offline data the machine learning come into picture and play the
important rolemch1 ; mch2 . The point where initially two anti beams collide
is called primary vertex, while secondary vertex may also be produced because
of the particle decay inside the LHCtrack1 . The tracks produced inside the
detectors can have complex structure. In particular, the complex structure of
tracks arise because of the magnetic field associated with LHC solenoid. In
this process of colliding of two anti directional beams inside the LHC, there
are always chances for high background noise for which the material of the
detector has the significant impact. The process to determine the particle
characteristics depends on track reconstruction and their fitting, this
process is governed by pattern recognition methods performed on offline data
and no doubt machine learning techniques contribute a lot. There are important
software, which perform these tasks such as GEANT4, ROOT, HERWIG etcsoft1 ;
soft2 ; soft3 . These software use the classical algorithms, which of course
are detector dependent and take many features of detector into account such as
it’s geometry, orientation, diameter, its material etc. Here it is mentioned
that machine learning played an important role to solve the problem of track
reconstruction and fitting in HEP from a decademl1 . To the date, the
techniques of machine learning (supervised, unsupervised) have been
implemented in offline data

Figure 1: The detectors at CERN.

simulation in HEP with many models such as neural network, deep learning,
simulated annealing etcnl1 ; mch2 ; sm1 . These technique are successfully
performed well to discovered the particle Higgs Bosonhiggs , which is a great
example. As high end electronics and fast algorithms is the primary
requirement for HEP experiments. So, to overcome these issues, one can think
towards HEP on quantum computer. Off course quantum hardware is not mature
till date but there is a future hope for quantum processorsdw1 ; dw2 ; dw3 ;
dw4 ; dw5 . This may help to execute algorithms with significant speed and can
lead the scenario towards quantum machine learning algorithms development
which can be utilized in HEPqml1 ; qml2 ; qml3 ; qml4 ; qml5 ; qml6 ; qml7 ;
qml8 ; qml9 ; qml10 . There are landmark quantum algorithms such as (Shor’s,
Love Grover) algorithmssho ; love . These algorithms give the clue to develop
many other quantum algorithms in the domain of machine learning and
optimization used for varieties of tasks. The domain of developing quantum
algorithms and studying the quantum complexityqcom  open the newly emerging
field of quantum machine learning. Recently, there is rapid progress in this
filed, St Loyed et al. have been developed the quantum algorithm to solve the
system of linear equations on the quantum computerle1 , Fernando et al.
developed the semi-infinite programming algorithm which is a step towards
quantum algorithms for optimization problemsle2 ; le3 . D-Wave systems have
been developed quantum annealing based processor, which is an indication for
future solutions of optimization problems on quantum hardwaredw1 ; dw2 ; dw3 ;
dw4 ; dw5 . To the date, there are huge attempts to investigate the properties
of quantum counterpart models such as quantum neural networksn1 ; n2 ; n3 ,
quantum deep learningqml5 ; dp1 , quantum Boltzmann machines, quantum
annealing and others. The area of quantum machine learning can serve better
for many tasks performed in offline data simulation in HEP and set-up a new
domain of research.

The paper is sketched in three sections. In Sect. 2, we discuss the Track
reconstruction and machine learning techniques. Sect 3, is devoted for the
supremacy of quantum machine learning in HEP.

##  II Track reconstruction and pattern recognition

In this section, we introduce the method for track reconstruction and also
give the shadow on machine learning techniques used for the same. The track
reconstruction is the important requirement in HEP which can be divided into
two basic steps as 1) finding the track candidates 2) track fittingdata ;
track1 . The primary requirement of track fitting is that it must be robust
against the error-prone of track finding procedure, it must be fast and
numerically stable. Overall, it is important to mention that the track
reconstruction strictly depends on the type of detectors used in HEP
experiments. Most of the previously used detectors such as bubble chamber,
gaseous chamber etc. are completely obsolete and overtaken by semiconductor
detectors.data  In practical applications semiconductor suffers from radiation
released in the collision of particle beams, hence to overcome this phase, the
research to develop the diamond detectors is very activedd1 ; dd2 . As an
example, the inner detector used in ATLAS use semiconductor technology and has
complicated geometry. The assembling and installation process of detectors
often disturb their geometry over the pre-assumed geometry. This problem is
called misalignment problem in detectors, which is a key element to produce
the track candidatesal1 ; al2 ; al3 . Getting the best track candidates,
contribute to the goodness of algorithm for track reconstruction. Here it is
mentioned that obtaining the track candidates can also be called as feature
extraction and a primary step to reconstruct the track, in this process the
classification is done of all hit points by particles in track detector. Each
class set has all hit points for a single track and each class is called track
candidate. It is important to state that these track candidates many times
carry the noise, in other words, the background hit points. During the track
finding process based on a particular track model, the pattern recognition
plays a significant role, which is the part of machine learning. Before
applying any machine learning techniques, it is always better to reduce the
dimensionality of the data gathered in the experiment, such that, overall the
outcome of the goal must not be affected in terms of better classification and
error reduction. There are countless algorithms for dimensionality
reductiondr1 ; dr2 ; dr3  and these always can be challenged. Any dimensional
reduction algorithm is suitable for one problem but may not fit for another
one. So data dimensionality reduction technique in machine learning is a
highly challenging step and must be performed carefully if required because
the adoption of any bad technique always lead towards wastage of efforts. In
spite of focusing on these methods in detail, here we discuss the method of
track finding, which can be divided into two categories as local and global
methods. In detail for these topics, the reader may refer (Ref:data ). In
continuation of the paper, we proceed the short introduction of these methods
in terms of offline data analysis. The track finding is the crucial part of
track reconstruction and first needs the track modeling. Track modeling takes
into consideration the geometry of the detector, associated magnificent filed,
noise, measurement errors etc. Few important mathematical approximations with
circles, parabola and splines have been used for the samedata , these methods
require the speed of the calculation and also need interpolation or
extrapolation techniques for predictiondata . The next step after track
modeling is the track finding, which definitely uses machine learning
methodspt . During the collision when particles hit the detectors layers and
ionize the detector material than hitting corresponds to a kind of measurement
of particle which is recorded by the sensor assembled in the detector. The set
of measurements recorded by the detector help to find the track candidates and
have the information about the track traced by the particle. There may exist
any situation such that, missing the tracked candidate, or there is no track
candidate and there may be track candidates which do not belong to any track.
For the sake of clarity here we rewrite, track finding methods can be divided
into two categories as local or global methodspt . In local or sequential
methods, the track is reconstructed sequentially by taking a seed. The seed is
a portion of a track got from the measurements done by the detector during the
collisions. Generally, two track modeling approaches have been used in local
methods such as track-road and track-following. In track road method a hollow
cylinder of a desirable diameter is considered around the trajectory, the
points fallen inside the track road are considered for analysis by using the
pattern recognition methods. Often, the track road methods are slower than
track following methods. Track following methods are valid while the track
candidates are easily identified by human senses. Further, in the global
method, the track candidates are supplied to the algorithm at once to produce
the tracks. The order of track candidates do not matter, but the execution
speed of algorithm in this process is low in comparison to local methods. The
computation time is taken by the global method is proportional to the number
of candidates. There are many classical approaches have been used for track
finding such as Hough transformhg1 , Conformal mappingcon1 , Kalman
Filterkalman , Neural networks, Deep learningcon2  etc. Based on the above
discussion we would like to emphasize that machine learning techniques are
highly important for experimental particle physics, which can not be ignored.
So can we think of better situations than existing techniques? Yes, hope so.
In the next section we cover the future perspective of quantum machine
learning techniques in HEP domain.

##  III Neural networks and Track reconstruction

In this section we present classical adaptive method for track findingtr .
Adaptive method are more simpler than tradition methods because they do not
involve the description of detector geometry. Almost all the adaptive methods
use the model of neural network and their various forms. Many forms of neural
networks such as hope filed network, Boltzmann machine have been used for the
same purpose. The basic ingredient of neural network is a neuron and its
firing mechanism, which use an activation function. Let suppose the neural
network is simulated over an Ising system with $N$ spins connected by synaptic
strength $w_{ij}$. A spin may the spin up and spin down configuration, ie.
$s_{i}\in\\{+1,-1\\}$. The dynamics of the neural network may be given by
applying the local updating rule as follows,

$s_{i}=sig(\sum_{j}T_{ij}s_{j})$ (1)

The matrix $T_{ij}$ involve in the above equation has the information about
synaptic strength of whole network. By taking into account the configuration
of spins over the network and the matrix of synaptic strength, the energy
function of the network can be written as follows,

$E(S)=-\frac{1}{2}\sum_{ij}T_{ij}s_{i}s_{j}$ (2)

The simple strategy to obtain the solution of a problem is to encode the
problem in neural network and optimize the energy function. There are many
ways to optimize the energy function, a common way is to setup the Boltzmann
distribution function with introducing the parameter as temperature, it is
given as below,

$P(T)=\frac{e^{-\frac{E}{KT}}}{Tr[e^{-\frac{E}{KT}}]}$ (3)

Where $P(T)$ is the probability of occupying the highest energy state at a
particular temperature. Properties of Boltzmann distribution are very helpful
to develop varieties of algorithms for optimization with heuristic approaches.
By following the basic frame work of neural network here we provide the track
reconstruction encoding method in neural networks developed by Denby and
Peterson independently. In this approach they encode the track segments in
neurons with some assumptions. These assumptions are encoded in terms of
constraints in energy function established over the specified neural network.
The neural network based algorithms have input as a data produced by the
detectors and the goal is to develop the tracks which have the following
constraints, a) The tracks must be smooth b) The tracks must not kink c). The
tracks must not have bifurcation. Debny and Peterson encoded a track segment
formed between to points $(i,j)$ in a neuron named as $s_{ij}$. Let suppose
another track segment connected between the points $(k,l)$ with the neuron
$s_{k,l}$. If the condition $\\{s_{ij}=+1,s_{kl}=+1\\}$, satisfy than the
connections are on, if the condition is followed as
$\\{s_{ij}=-1,s_{kl}=-1\\}$, than the connections are off and no track segment
is formed. After the track reconstruction problem in neural network the energy
function can be written as below,

$E_{ijkl}=-\frac{1}{2}\sum_{ijkl}(T_{ijkl}^{cost}+T_{ijkl}^{constraint})s_{ij}s_{kl}$
(4)

Two track segments may have connections if the condition $(j=k)$ is satisfied.
The smoothness of the segments are encoded in the angle between two track
segments ie $\theta$ with the following eq.,

$T_{ijkl}^{cost}=\delta_{jk}\frac{cos^{m}\theta_{ijl}}{r_{ij}+r_{jl}}$ (5)

Further the bifurcation encoding is done in the following equation,

$T_{ijkl}^{constraint}=-\frac{\alpha}{2}[\delta_{ik}(1-\delta_{jl})-\delta_{jl}(1-\delta_{ik})]-\frac{\beta}{2}[text]$
(6)

with, text$\longrightarrow$“global inhibition”

Where $(\alpha,\beta)$ are Lagrange multipliers. By introducing the Eq.5 and 6
in Eq.4, the final energy function can be written. The final energy function
can be optimized by using any suitable method with by applying some cuts in
the parameters on number of neurons $(N)$, angle parameter $(\theta)$ and
circle with certain radius which put the restrictions on the length of
segments.

##  IV Supremacy of quantum machine learning in HEP

Quantum pattern recognitionqp1 ; qp2 ; qp3  is an important application of
quantum machine learning. It is obvious that there is always research progress
in HEP to develop fast and better algorithms. Can quantum pattern recognition
techniques help in HEP to deal with massive amount of data on the quantum
computer and extract the useful information from the data? Here we mention
that there is recent progress on quantum algorithms in many domains like,
algebraic domain (Hidden subgroup problems)mm1 ; mm2 , semidefinite
programmingle2 , linear differential equationsle22 ; mm22 , finite element
methodsle23  and in pattern recognitionmm3 . There is major developments for
quantum algorithms with black box model and query complexitymm1 . Query
complexity is the quantum equivalent of classical decision tree model. However
the development of quantum algorithms with adiabatic quantum computationad1
model is on slow progress, as this approach does not have any suitable
complexity model to calculate the quality of the algorithm, which is an open
problem. But there are future possibilities for the same, which can contribute
for better quantum algorithm designing. During the track reconstruction
process the problem at many stages can be mapped to suitable optimization
problemsop1 ; op2  which may be solved further by any suitable method. The
research to solve quadratic binary optimization problems subjected to
constrained or unconstrained are on the way by using quantum annealingqd1 ;
qd2 . The quantum annealing exhibit better signatures to handle the problems
on the quantum computer. The quantum algorithms developed based on quantum
strategy (superposition and entanglementen1 ; en2 ) and taking into
consideration the geometrical aspects of detectors may be useful in comparison
to classical algorithms in terms of (time, space) complexity and speed. On the
other hand, as per the literature survey, the algorithmic development in HEP
is less pervasive towards the existence of entanglement during the collision
process inside any detector. Can we also have such algorithms which can catch
the phenomena of entanglement inside the detector if exists, which can help
further to understand the true nature of particles? Pattern recognition in
many forms involving the image processing had been the part of particle
physics community for a long timeim1 . There is literature, number of papers
and Ph.D work in which the community has been solved the problems of particle
physics by using image processing techniquesim2 . It is not worth to mention
that the emergence of quantum image processingqim1 ; qim2  can also contribute
to better improvement for experimental particle physics. However the current
trend deals with neural network and deep learning methods, which overcome to
the difficulty of feature extractionf1  as involved in traditional theory of
pattern recognition, but these techniques involve huge complex structures of
networks which further make the optimization problems very difficult. It is
mentioned that, the development of quantum algorithms and studying quantum
complexities in particle physics domain is really challenging but do not seem
impossible. The rapid progress in quantum domain boost the hope for future
possibilities to solve particle physics problems on quantum computer. We hope
the present article gives the sufficient indication to the HEP and quantum
community to make the development towards the aforementioned directions, which
is almost untouched.

##  V Optimization on analog quantum computer

Recently quantum annealerqa1 ; qa2  has been developed by D-Wave system, which
is a kind of analog quantum computer and suitable for many optimization
problems. Quantum annealing is inspired by simulated annealingsim1 , the basic
difference is that quantum annealing often use magnetic filed as a controlling
parameter, while simulated annealing use temperature. Both the approaches can
be used to simulate spin glass systems, these systems further can be used to
encode NP complete problems. NP complete problems are generally decision
making problems, which produce the answers in yes and no. It is found that
quantum annealing has better advantage in terms of complexity reduction of NP
complete problemsnp1  in comparison to simulated annealing. On the other hand
quantum annealing can also be introduced as as restricted version of adiabetic
quantum computationadia  which do not allow tunneling inside the physical
system. The tunneling mechanism with the framework of quantum annealing has an
extra advantage to reduce the computational complexity. The manufactured
D-Wave quantum computer has a good attempt to solve a subset of optimization
problems involved in mathematics more efficiently than classical computers.
With the promising progress in designing the D-Wave computer, the fundamental
results exhibts that quantum hardware has speedup capability to implement few
quantum algorithms for solving some mathematical problems. The field of
quantum computation is progressing rapidly. Regardless of the results achieved
by D-Wave and its hardware with quantum annealing, fully programmable quantum
computer operate much differently than D-Wave but initial results for
universal quantum computer seems very promising. The method of operation of
these universal quantum computers needs a full architecture incorporating with
many components like storage medium, buses to transport the signal, Quantum
arithmetic and logic unit and quantum softwares.

##  VI Mapping the problem to QUBO

D-Wave quantum annealer have the connectivity of super-conducting qubits on
Chimera graph. The graph has a collection of cells arranged in square fashion,
each cell is consisted of by eight qubits connected in feed forward fashion.
Further all the cells have special connections among each other. The
arrangement of chimera graph is shown below in the Fig.2,

Figure 2: Arrangement of qubits on Chimera graph with nine cells.

Any quantum Ising systems can be mapped to quantum annealer for quadratic
binary optimization (QUBO) problems. The method used for this mapping is
called minor embeddingme . As per the definition of Ising system, it can be
assumed as a set of spins in d-dimensional hyper cubic lattice. But for the
current work, for simplicity we fix the limit over the dimensions and assume
that the spins are scattered in two dimensional space at random in a variable
magnetic filed over each spin. Every spin has interaction with its neighbor
spin modeled with certain coupling strength. Please note that, here we are
dealing with planer Ising system, the scenario of non-planer Ising systems is
different. The classical mathematical expression for planer Ising system in 2
dimensions can be written as below,

$E(J|h)=\sum_{(i=1,j=1)}^{(N,N)}J_{ij}s_{i}s_{j}+\sum_{i}h_{i}s_{i}$ (7)

For quantum Ising system, one has to replace the variables $(s_{i},s_{j})$
with Pauli matrices as $(\sigma_{i},\sigma_{j})$ and the energy function
$E(J|h)$ takes the from of an Hamiltonian operator. Here $J_{ij}$ is the
pairwise coupling strength between $(i^{th},j^{th})$ spins and $(s_{i},s_{j})$
are the spin variables which can have eigenvalues as $(\pm 1)$. The quantity
$(+1)$, denotes spin up and $(-1)$ denotes spin down. In Ising system, every
spin can encode binary data, which is very useful to solve NP problems such as
SAT problem, graph bipartition and many others. D-Wave quantum annealer use
the adiabatic quantum computing strategy in which the system is prepared in
initial ground state with the Hamiltonian $H_{I}$. The initial system is
disturb by an additional Hamiltonian $H_{F}$. The total Hamiltonian of the
system can be written as a convex combination as below,

$H(s)=(1-s)H_{I}+sH_{F}$ (8)

The problem is encoded in $H_{F}$, whose ground state has the solution of the
problem, provide the total Hamiltonian $H(s)$ evolve with adiabatic
conditionad . Before adopting the quantum adiabatic evolutionary plan, it is
advisable to encode the problem first on quadratic polynomial and further it
can be easily mapped to available quantum annealer. Consider the quadratic
polynomial over the three spins $S=(s_{1},s_{2},s_{3})^{T}$ as,

$(as_{1}+bs_{2}+cs_{3})^{2}=(as_{1}+bs_{2}+cs_{3}).(as_{1}+bs_{2}+cs_{3})$ (9)
$=a^{2}(s_{1})^{2}+b^{2}(s_{2})^{2}+c^{2}(s_{3})^{2}+2ab(s_{1}s_{2})+2ac(s_{1}s_{3})+2bc(s_{2}s_{3})$
(10)

manipulating the Eq. (2) in matrix form we obtain,

$\displaystyle=[s_{1},s_{2},s_{3}].\begin{bmatrix}a\\\ b\\\
c\end{bmatrix}.[a,b,c].\begin{bmatrix}s_{1}\\\ s_{2}\\\ s_{3}\end{bmatrix}$
(11) $\displaystyle=[s_{1},s_{2},s_{3}].\begin{bmatrix}a^{2}&ab&ac\\\
ab&b^{2}&bc\\\ ac&bc&c^{2}\end{bmatrix}.\begin{bmatrix}s_{1}\\\ s_{2}\\\
s_{3}\end{bmatrix}=S^{T}QS$ (12)

Here the matrix Q is symmetric, another representation of Q will also give the
same result as obtained in Eq. (3).

$=S^{T}.\begin{bmatrix}a^{2}&2ab&2ac\\\ 0&b^{2}&2bc\\\
0&0&c^{2}\end{bmatrix}.S$ (13)

Let decompose the matrix Q we get,

$\displaystyle=S^{T}.\\{\begin{bmatrix}a^{2}&0&0\\\ 0&b^{2}&0\\\
0&0&c^{2}\end{bmatrix}+\begin{bmatrix}0&2ab&2ac\\\ 0&0&2bc\\\
0&0&0\end{bmatrix}\\}.S$ (14) $\displaystyle=S^{T}.(Q^{diag}+Q^{upper}).S$
(15) $\displaystyle=(S^{T}.Q^{diag}.S)+(S^{T}.Q^{upper}.S)$ (16)

If the vector S is having N number of spins ie.
$S\in\\{s_{1},s_{2},s_{3}.......s_{N}\\}$, than the above expression can be
written in generalized form as below,

$\displaystyle=\sum_{ii}S^{T}Q_{ii}^{diag}.S+\sum_{(i<j)}S^{T}.Q_{ij}^{upper}.S$
(17) $\displaystyle=\sum_{i\leq j}S^{T}.Q_{ij}.S$ (18)

Proceeding for $N$ number of spins the quadratic form of a polynomial can be
written as,

$\displaystyle(as_{1}+bs_{2}+cs_{3}+.........+\gamma s_{N})^{2}$
$\displaystyle=\sum_{ii}S^{T}Q_{ii}^{diag}.S+\sum_{(i<j)}S^{T}.Q_{ij}^{upper}.S$
(19)

Consequently, the quadratic Ising function of spins can be written as below,

$E(J|h)=\sum_{ii}S^{T}Q_{ii}^{diag}.S+\sum_{(i<j)}S^{T}.Q_{ij}^{upper}.S$ (20)

Here the vector $S=\\{s_{i}\\}$ with $s_{i}\in\\{-1,+1\\}$ can be mapped to a
vector $Y=\\{y_{i}\\}$with $y_{i}\in\\{0,1\\}$ by using a simple arithmetic
transformation $y_{i}=\frac{1+s_{i}}{2}$. This transformation is required to
map the quadratic polynomial on available quantum annealer. After
transformation the condition $y_{i}^{2}=y_{i}$ is easily satisfied. The state
$(0)$ represent the spin down and the state $(1)$ represent spin up. Now the
Eq.20 can be written in new variables $y_{i}$ as below,

$\displaystyle E(J|h)=\sum_{i}(2Y-1)^{T}.Q_{ii}.(2Y-1)$ (21)
$\displaystyle+\sum_{i<j}(2Y-1)^{T}.Q_{ij}^{upper}.(2Y-1)$ (22)

In short we can also write as,

$E(J|h)=\sum_{i\leq j}(2Y-1)^{T}.Q_{ij}.(2Y-1)$ (23)

Point to be noted that, the diagonal terms of $Q_{ij}$ has the terms $h_{ii}$
and upper triangular part of the matrix has the coupling strengths $J_{ij}.$
So magnetic filed and coupling strength are absorbed in the equation with
variables $y_{i}\in\\{0,1\\}$. So we can also say that the matrix $(2Y-1)$
encode the binary data with the variables $y_{i}\in\\{0,1\\}$. So it is easy
to re-express the equation explicitly as below,

$E(J|h)=\sum_{ii}X^{T}Q_{ii}X+\sum_{i<j}X^{T}Q_{ij}X$ (24)

Or,

$E(J|h)=\sum_{i\leq j}X^{T}.Q_{ij}.X$ (25)

With the continuation of the above discussion, we can express the future
possibilities of developing quantum algorithms by taking into consideration
the adiabatic computation model. Apart from it the framework of quantum
machine learningqml1 ; qml2  based on quantum annealing may also give new
avenues for solving computational problems in HEP.

##  VII Conclusion

The paper cover the perspective of quantum machine learning in high energy
physics. There are many computational extensive tasks in high energy physics
such track finding, particle identification, unfolding and many others.
Specially track reconstruction is most time consuming task. We have described
as the classical methods for track reconstruction and shown the possibility of
usage of quantum annealing for the same in future. This introductory article
deliver a lucid information towards the application of quantum computation in
high energy physics. This may be helpful for experimental high energy physics
community to produce the efficient quantum algorithms for computational
problems in high energy physics.

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# On the Extremality of the Tensor Product of Quantum Channels

James Miller Simeão Toledo da Silva<EMAIL_ADDRESS>University of São
Paulo, IME, Brazil

###### Abstract

Completely positive and trace preserving (CPT) maps are important for Quantum
Information Theory, because they describe a broad class of of transformations
of quantum states. There are also two other related classes of maps, the
unital completely positive (UCP) maps and the unital completely positive and
trace preserving (UCPT) maps. For these three classes, the maps from a finite
dimensional Hilbert space $X$ to another one $Y$ is a compact convex set and,
as such, it is the convex hull of its extreme points. The extreme points of
these convex sets are yet not well understood. In this article we investigate
the preservation of extremality under the tensor product. We prove that
extremality is preserved for CPT or UCP maps, but for UCPT it is not always
preserved.

##  1 Introduction

Quantum Information Theory is a theory of storage and processing of
information in the form of quantum states. Completely positive and trace
preserving maps [NC10] are used to describe open quantum systems, and is also
a broad class of transformations of quantum states. Being a central object of
Quantum Information Theory, understanding its mathematical structure is
important for the development of the subject. We’ll consider not only CPT
maps, but also related classes of maps, the unital completely positive (UCP)
maps and the unital completely positive and trace preserving (UCPT) maps.

CPT maps from a finite dimensional Hilbert space $X$ to another $Y$ forms a
compact convex set $CPT(X,Y)$. The same is true for $UCP$ or $UCPT$ maps.
Every compact convex set is the convex hull of its extreme points. There are
characterizations of these set of extreme points due to Choi [Cho75] (see
theorem 2.11) and Laudau and Streater[LS93] (see theorem 2.18), but the
extreme points are yet not well understood. An important question is the
stability of extremality under some operations. These include the operations
of composition and tensor product. It is already known that the composition of
two extreme CPT maps need not be extreme [Exc]. But, apparently, it wasn’t yet
answered the question about the extremality of the tensor product of extreme
CPT, UCP or UCPT maps ([nLa23]). In this article we prove that extremality is
preserved by the tensor product in the cases of CPT or UCP maps. For UCPT maps
extremality is preserved if one of the Hilbert spaces has dimension 2, but it
may not be the case for higher dimensions, as counterexamples shows.

For this particular article the maximal Choi ranks of UCPT maps will be
important, as we’ll see in theorem 3.5. In [Ohn10] it was investigated the
maximal Choi ranks of UCPT maps, but they were only computed up to dimension
4. For higher dimensions, there was given a lower bound that the maximal Choi
rank is at least the dimension of the space. Also, for dimensions 3 and 4 were
given single examples of such extreme UCPT maps of maximal rank. In [HMR21]
was constructed families of examples of high rank extreme maps, but the lower
bounds on the maximal ranks were not improved. A better lower bound on the
maximal ranks of extreme UCPT maps may be enough to decide for which
dimensions the tensor product preserves extremality for UCPT maps, but this
wasn’t done yet, as far as I know.

The article is organized as follows. In section 2 we review some concepts of
Linear Algebra, Convex Sets and Quantum Information Theory. Also, we review
the characterization of linear independence as the invertibility of the Gram
matrix, which will be important in the cases of CPT and UCP maps. Then, in
section 3, we apply these results to prove that the tensor product of extreme
CPT or UCP maps is also extreme. Also, we show that if two extreme UCPT maps
have high enough Choi ranks, their tensor product cannot be an extreme point
of UCPT maps. Then we use examples of [Ohn10] of high enough Choi ranks so
that their tensor product is not extreme.

##  2 Preliminaries

###  2.1 Linear Algebra

####  2.1.1 Hilbert spaces

As usual in Quantum Mechanics, the inner product of a Hilbert space is
antilinear in the first argument and linear in the second. Also, we restrict
to complex finite dimensional vector spaces. Therefore, such a Hilbert space
is a finite dimensional complex vector space $X$, together with an inner
product $\langle\;,\;\rangle\colon X\times X\to\mathbb{C}$. The inner product
is a sesquilinear form, which means that the following properties are valid:

  * •

$\forall x,x^{\prime},x^{\prime\prime}\in X$, $\langle
x,x^{\prime}+x^{\prime\prime}\rangle=\langle x,x^{\prime}\rangle+\langle
x,x^{\prime\prime}\rangle$;

  * •

$\forall x,x^{\prime}\in X$, $\forall\lambda\in\mathbb{C}$, $\langle x,\lambda
x^{\prime}\rangle=\lambda\langle x,x^{\prime}\rangle$;

  * •

$\forall x,x^{\prime}\in X$, $\langle x,x^{\prime}\rangle=\overline{\langle
x^{\prime},x\rangle}$.

We denote by $\mathrm{Hom}(X,Y)$ the set of linear maps $A\colon X\to Y$. We
define $\mathrm{End}(X)=\mathrm{Hom}(X,X)$, the set of linear endomorphisms of
$X$. Also, $\mathrm{Hom}(X,Y)$ have the Hilbert-Schmidt inner product, given
by $\langle A,B\rangle=tr(A^{\dagger}B)$, for $A,B\in\mathrm{Hom}(X,Y)$.

A construction that will be important in this article is the tensor product.
The tensor product of vector spaces $X$ and $Y$ is a $\mathbb{C}$-bilinear map
$\otimes\colon X\times Y\to X\otimes Y$ that satisfies the following universal
property: if $B\colon X\times Y\to Z$ is $\mathbb{C}$-bilinear, then there
exists a unique $\mathbb{C}$-linear map $u\colon X\otimes Y\to Z$ such that
$b=u\otimes$.

${X\times Y}$${X\otimes
Y}$${Z}$$\scriptstyle{\otimes}$$\scriptstyle{B}$$\scriptstyle{u}$

Given a basis $x_{1}$,…,$x_{n}$ for $X$ and $y_{1}$,…,$y_{m}$ for $Y$,
$(x_{i}\otimes y_{j})_{\begin{subarray}{c}i=1,...,n\\\
j=1,...,m\end{subarray}}$ is a basis for $X\otimes Y$. We defined the tensor
product of vector spaces, but there is also a tensor product of linear maps.
Let $A\colon X\to X^{\prime}$ and $B\colon Y\to Y^{\prime}$ be
$\mathbb{C}$-linear maps. Their tensor product $A\otimes B\colon X\otimes Y\to
X^{\prime}\otimes Y^{\prime}$ is the unique linear map such that the next
diagram commutes:

${X\times Y}$${X\otimes Y}$${X^{\prime}\times Y^{\prime}}$${X^{\prime}\otimes
Y^{\prime}}$$\scriptstyle{A\times
B}$$\scriptstyle{\otimes}$$\scriptstyle{A\otimes B}$$\scriptstyle{\otimes}$

That is, $A\otimes B$ is computed on elementary tensors as $(A\otimes
B)(x\otimes y)=A(x)\otimes B(y)$, for any $x\in X$ and $y\in Y$. Given basis
$x_{1}$,…,$x_{n}$ for $X$, $x^{\prime}_{1}$,…,$x^{\prime}_{n^{\prime}}$ for
$X^{\prime}$, $y_{1}$,…,$y_{m}$ for $Y$ and
$y^{\prime}_{1}$,…,$y^{\prime}_{m^{\prime}}$ for $Y^{\prime}$, we have the
basis $(x_{i}\otimes y_{j})_{\begin{subarray}{c}i=1,...,n\\\
j=1,...,m\end{subarray}}$ for $X\otimes Y$ and
$(x^{\prime}_{i^{\prime}}\otimes
y^{\prime}_{j^{\prime}})_{\begin{subarray}{c}i^{\prime}=1,...,n^{\prime}\\\
j^{\prime}=1,...,m^{\prime}\end{subarray}}$ for $X^{\prime}\otimes
Y^{\prime}$. We can express the matrix $[A\otimes B]$ of $A\otimes B$ in these
basis. Computing $A\otimes B$ on $x_{i}\otimes y_{j}$, we have $(A\otimes
B)(x_{i}\otimes y_{j})=A(x_{i})\otimes
B(y_{j})=(\sum_{i^{\prime}}[A]_{i^{\prime},i}x^{\prime}_{i^{\prime}})\otimes(\sum_{j^{\prime}}[B]_{j^{\prime},j}y^{\prime}_{j^{\prime}})$.
By bilinearity of $\otimes$, we get $(A\otimes B)(x_{i}\otimes
y_{j})=\sum_{i^{\prime},j^{\prime}}[A]_{i^{\prime},i}[B]_{j^{\prime},j}x^{\prime}_{i^{\prime}}\otimes
y^{\prime}_{j^{\prime}}$. Therefore, $[A\otimes B]$ has elements $[A\otimes
B]_{(i^{\prime},j^{\prime}),(i,j)}=[A]_{i^{\prime},i}[B]_{j^{\prime},j}$.
Since $[A\otimes B]$ has double indices $(i,j)$, to interpret it as a matrix
we have to choose an order. This is done by taking the lexicographic order,
that is, $(i,j)<(i^{\prime},j^{\prime})\iff i<i^{\prime}$ or $i=i^{\prime}$
and $j<j^{\prime}$. With this order, the matrix $[A\otimes B]$ is called the
Kronecker product of $[A]$ and $[B]$, which we denote as $[A]\otimes[B]$.

If $X$ and $Y$ are Hilbert spaces, then their tensor product $X\otimes Y$ is
also a Hilbert space, with inner product given by $\langle x\otimes
y,x^{\prime}\otimes y^{\prime}\rangle=\langle x,x^{\prime}\rangle\langle
y,y^{\prime}\rangle$, for $x,x^{\prime}\in X$ and $y,y^{\prime}\in Y$. We have
a similar property for the Hilbert-Schmidt inner product on
$\mathrm{End}(X\otimes Y)$. Any element of $\mathrm{End}(X\otimes Y)$ is a
linear combination of linear maps of the form $A\otimes B$, for
$A\in\mathrm{End}(X)$ and $B\in\mathrm{End}(Y)$. Therefore, they are
sufficient to characterize the inner product on $\mathrm{End}(X\otimes Y)$.
Given $A,A^{\prime}\in\mathrm{End}(X)$ and $B,B^{\prime}\in\mathrm{End}(Y)$,
we have $\langle A\otimes B,A^{\prime}\otimes B^{\prime}\rangle=tr((A\otimes
B)^{\dagger}(A^{\prime}\otimes B^{\prime}))=tr((A^{\dagger}\otimes
B^{\dagger})(A^{\prime}\otimes B^{\prime}))=tr(A^{\dagger}A^{\prime}\otimes
B^{\dagger}B^{\prime})$. The trace is multiplicative, that is, we have the
property $tr(f\otimes g)=tr(f)tr(g)$. Using this property, we get $\langle
A\otimes B,A^{\prime}\otimes
B^{\prime}\rangle=tr(A^{\dagger}A^{\prime})tr(B^{\dagger}B^{\prime})=\langle
A,A^{\prime}\rangle\langle B,B^{\prime}\rangle$. Therefore

$\langle A\otimes B,A^{\prime}\otimes B^{\prime}\rangle=\langle
A,A^{\prime}\rangle\langle B,B^{\prime}\rangle,$ (1)

for any $A,A^{\prime}\in\mathrm{End}(X)$ and $B,B^{\prime}\in\mathrm{End}(Y)$.

####  2.1.2 Gram matrices

As we’ll see in theorem 2.11, it characterizes extremality of CPT maps in
terms of linear independence. It will be useful to restate linear independence
as the invertibility of the Gram matrix. In this section, we recall some
results regarding Gram matrices. A proof of lemma 2.1 may be found in theorem
7.2.10 of [HJ12].

######  Lemma 2.1.

Let $(V,\langle\;,\;\rangle)$ be an inner product space and $v_{1}$,…, $v_{n}$
be vectors of $V$. Let

$G=\begin{pmatrix}\langle v_{1},v_{1}\rangle&\langle
v_{1},v_{2}\rangle&\cdots&\langle v_{1},v_{n}\rangle\\\ \langle
v_{2},v_{1}\rangle&\langle v_{2},v_{2}\rangle&\cdots&\langle
v_{2},v_{n}\rangle\\\ \vdots&\vdots&\ddots&\vdots\\\ \langle
v_{n},v_{1}\rangle&\langle v_{n},v_{2}\rangle&\cdots&\langle
v_{n},v_{n}\rangle\end{pmatrix}$ (2)

be the Gram matrix of $v_{1}$,…,$v_{n}$. Then $v_{1}$,…, $v_{n}$ are linearly
independent if, and only if, its Gram matrix is invertible.

###### Proof.

First, given vectors $a=(a_{1},...,a_{n})$ and $b=(b_{1},...,b_{n})$, which
we’ll represent as column vectors, lets compute $a^{\dagger}Gb$. We have

$\displaystyle Gb$ $\displaystyle=\begin{pmatrix}\langle
v_{1},v_{1}\rangle&\langle v_{1},v_{2}\rangle&\cdots&\langle
v_{1},v_{n}\rangle\\\ \langle v_{2},v_{1}\rangle&\langle
v_{2},v_{2}\rangle&\cdots&\langle v_{2},v_{n}\rangle\\\
\vdots&\vdots&\ddots&\vdots\\\ \langle v_{n},v_{1}\rangle&\langle
v_{n},v_{2}\rangle&\cdots&\langle
v_{n},v_{n}\rangle\end{pmatrix}\begin{pmatrix}b_{1}\\\ b_{2}\\\ \vdots\\\
b_{n}\end{pmatrix}$ $\displaystyle=\begin{pmatrix}\sum_{j=1}^{n}\langle
v_{1},v_{j}\rangle b_{j}\\\ \sum_{j=1}^{n}\langle v_{2},v_{j}\rangle b_{j}\\\
\vdots\\\ \sum_{j=1}^{n}\langle v_{n},v_{j}\rangle b_{j}\\\ \end{pmatrix}$
$\displaystyle=\begin{pmatrix}\langle v_{1},\sum_{j=1}^{n}b_{j}v_{j}\rangle\\\
\langle v_{2},\sum_{j=1}^{n}b_{j}v_{j}\rangle\\\ \vdots\\\ \langle
v_{n},\sum_{j=1}^{n}b_{j}v_{j}\rangle\\\ \end{pmatrix}.$

Therefore $a^{\dagger}Gb=\sum_{i=1}^{n}\overline{a_{i}}\langle
v_{i},\sum_{j=1}^{n}b_{j}v_{j}\rangle=\langle\sum_{i=1}^{n}a_{i}v_{i},\sum_{j=1}^{n}b_{j}v_{j}\rangle$.
Notice that the inner product is antilinear in the first argument, so $a_{i}$
passes to inside the inner product without the conjugation. The last result
implies that $G$ is a positive matrix, since
$a^{\dagger}Ga=||\sum_{i=1}^{n}a_{i}v_{i}||^{2}$, which is $\geq 0$.
Therefore, $G$ is diagonalizable and its eigenvalues are real and non-
negative.

Now we prove the equivalence of $v_{1}$,…,$v_{n}$ being linearly independent
and $G$ being invertible. By the equation
$a^{\dagger}Ga=||\sum_{i=1}^{n}a_{i}v_{i}||^{2}$, the vectors $v_{1}$,…,
$v_{n}$ are linearly independent if, and only if, $a=0$ is the unique solution
to the equation $a^{\dagger}Ga=0$. Also notice that, since $G$ is
diagonalizable, it is invertible if, and only if, $0$ isn’t an eigenvalue of
$G$. Let $c_{1}$,…, $c_{n}$ be orthonormal eigenvectors of $G$ and
$\lambda_{i}$ the eigenvalue of $c_{i}$. Then $a$ is written as
$a=\sum_{i=1}^{n}\alpha_{i}c_{i}$, for some $\alpha_{i}\in\mathbb{C}$. Then we
have
$a^{\dagger}Ga=a^{\dagger}G\sum_{i=1}^{n}\alpha_{i}c_{i}=\sum_{i=1}^{n}\alpha_{i}a^{\dagger}Gc_{i}=\sum_{i=1}^{n}\alpha_{i}a^{\dagger}\lambda_{i}c_{i}=\sum_{i=1}^{n}\lambda_{i}\alpha_{i}a^{\dagger}c_{i}=\sum_{i=1}^{n}\lambda_{i}\alpha_{i}\overline{\alpha_{i}}=\sum_{i=1}^{n}\lambda_{i}|\alpha_{i}|^{2}$.
If each $\lambda_{i}$ isn’t 0, then for any $a\neq 0$ we have
$\sum_{i=1}^{n}\lambda_{i}|\alpha_{i}|^{2}>0$. Therefore, when each
$\lambda_{i}$ isn’t 0, $a^{\dagger}Ga$ can only be 0 if $a=0$. If for some $j$
we have $\lambda_{j}=0$, then we can pick $\alpha_{j}=1$ and $\alpha_{i}=0$
for $i\neq j$, which gives a vector $a\neq 0$ with $a^{\dagger}Ga=0$.
Therefore, $a=0$ is the unique solution to the equation $a^{\dagger}Ga=0$
precisely when each $\lambda_{i}$ isn’t 0, that is, when $G$ is invertible.
This concludes that $G$ is invertible if, and only if, $v_{1}$,…, $v_{n}$ are
linearly independent. ∎

Gram matrices of tensor products of vectors will be important. For this
reason, we prove the following lemmas:

######  Lemma 2.2.

Let $V,W$ be vectors spaces. Let also $(v_{i})_{i=1,...,n}$ be vectors of $V$
and $(w_{j})_{j=1,...,m}$ be vectors of $W$. Let $G$ be the Gram matrix of
$(v_{i})_{i=1,...,n}$, $G^{\prime}$ be the Gram matrix of
$(w_{j})_{j=1,...,m}$ and $G^{\prime\prime}$ be the Gram matrix of
$(v_{i}\otimes w_{j})_{\begin{subarray}{c}i=1,...,n\\\
j=1,...,m\end{subarray}}$. We regard $(v_{i}\otimes
w_{j})_{\begin{subarray}{c}i=1,...,n\\\ j=1,...,m\end{subarray}}$ as a
sequence ordered in the lexicographic order, that is,
$(i,j)<(i^{\prime},j^{\prime})\iff i<i^{\prime}$ or $i=i^{\prime}$ and
$j<j^{\prime}$. Then $G^{\prime\prime}$ is the Kronecker product of $G$ and
$G^{\prime}$, that is, $G^{\prime\prime}=G\otimes G^{\prime}$. Its matrix
elements are, therefore,
$G^{\prime\prime}_{(i,j),(i^{\prime},j^{\prime})}=G_{i,i^{\prime}}G^{\prime}_{j,j^{\prime}}$.

###### Proof.

The matrix elements of $G$, $G^{\prime}$ and $G^{\prime\prime}$ are, by
definition,

$G_{i,i^{\prime}}=\langle v_{i},v_{i^{\prime}}\rangle,$ (3)
$G^{\prime}_{j,j^{\prime}}=\langle w_{j},w_{j^{\prime}}\rangle,$ (4)
$G^{\prime\prime}_{(i,j),(i^{\prime},j^{\prime})}=\langle v_{i}\otimes
w_{j},v_{i^{\prime}}\otimes w_{j^{\prime}}\rangle.$ (5)

By definition of the inner product of $V\otimes W$, we have
$G^{\prime\prime}_{(i,j),(i^{\prime},j^{\prime})}$ $=$ $\langle v_{i}\otimes
w_{j},v_{i^{\prime}}\otimes w_{j^{\prime}}\rangle$ $=$ $\langle
v_{i},v_{i^{\prime}}\rangle\langle w_{j},w_{j^{\prime}}\rangle$. By (3) and
(4), we have
$G^{\prime\prime}_{(i,j),(i^{\prime},j^{\prime})}=G_{i,i^{\prime}}G^{\prime}_{j,j^{\prime}}$.
Since the indices $(i,j)$ have the lexicographic order, the last equation says
that $G^{\prime\prime}=G\otimes G^{\prime}$. ∎

######  Lemma 2.3.

Let $(v_{i})_{i}$ be a finite sequence in some vector space $V$ and let
$(w_{j})_{j}$ be a finite sequence in some vector space $W$. Then
$(v_{i}\otimes w_{j})_{i,j}$ is linearly independent if, and only if, both
$(v_{i})_{i}$ and $(w_{j})_{j}$ are linearly independent.

###### Proof.

We use the same notation of lemma 2.2. Let $n$ be the size of $(v_{i})_{i}$
and $m$ be the size of $(w_{j})_{j}$. The sequences $(v_{i})_{i}$,
$(w_{j})_{j}$ and $(v_{i}\otimes w_{j})_{i,j}$ have Gram matrices $G$,
$G^{\prime}$ and $G^{\prime\prime}$, respectively. By lemma 2.2, we have
$G^{\prime\prime}=G\otimes G^{\prime}$. Also,
$\det(G^{\prime\prime})=\det(G\otimes
G^{\prime})=\det(G)^{m}\det(G^{\prime})^{n}$, so $G^{\prime\prime}$ is
invertible if, and only if, both $G$ and $G^{\prime}$ are invertible. In this
case, $G^{\prime\prime-1}=G^{-1}\otimes G^{\prime-1}$. Using lemma 2.1 we
conclude that $(v_{i}\otimes w_{j})_{i,j}$ is linearly independent if, and
only if, both $(v_{i})_{i}$ and $(w_{j})_{j}$ are linearly independent. ∎

###  2.2 Convex Sets

Given a vector space $X$, a convex set $C\subseteq X$ is a subset $C$ that is
closed by convex combinations. A convex combination of vectors
$v_{1}$,…,$v_{n}$ is a linear combination $\sum_{i=1}^{n}p_{i}v_{i}$, where
$p_{i}\geq 0$ and $\sum_{i=1}^{n}p_{i}=1$. For any subset $S\subseteq X$, its
convex hull is the set of convex combinations of points of $S$. Therefore, a
subset $C\subseteq X$ is convex if it equals its convex hull. An extreme point
of a convex $C$ is one that can’t be written as a convex combination of
different points of $C$. For example, $[0,1]$ is a convex with 0 and 1 as
extreme points. By a theorem of Krein and Milman, every compact convex set of
a finite dimensional vector space is the convex hull of its extreme points
(theorem 3.3 of [Bar02]).

######  Theorem 2.4.

Let $X$ be a finite dimensional complex vector space and $C\subseteq X$ a
compact convex set. Then $C$ is the convex hull of its extreme points.

Of course, this is not true for general convex sets. For example,
$[0,+\infty)$ is a convex set, but isn’t compact. Also, it has 0 as its unique
extreme point, whose convex hull is $\\{0\\}$, which isn’t $[0,+\infty)$.

###  2.3 CP maps

Every map we’ll consider in this article will be some special kind of
completely positive (CP) map. Let $X$ and $Y$ be finite dimensional Hilbert
spaces. A CP map $\varepsilon\colon X\to Y$ is a linear map
$\varepsilon\colon\mathrm{End}(X)\to\mathrm{End}(Y)$ such that
$\mathrm{id}_{\mathrm{End}(Z)}\otimes\varepsilon\colon\mathrm{End}(Z\otimes
X)\to\mathrm{End}(Z\otimes Y)$ is a positive map, for any finite dimensional
Hilbert space $Z$. A positive map is one that sends positive operators to
positive operators. A positive operator is a diagonalizable operator whose
eigenvalues are all real and non-negative. We’ll denote as $CP(X,Y)$ the set
of CP maps from $X$ to $Y$. Instead of using this definition of complete
positivity, we’ll often use the following characterization. A linear map
$\varepsilon\colon\mathrm{End}(X)\to\mathrm{End}(Y)$ is CP if, and only if, it
is of the form $\varepsilon(A)=\sum_{k}E_{k}AE_{k}^{\dagger}$, for some finite
sequence $(E_{k})_{k}$ of operators $E_{k}\colon X\to Y$. The operators
$E_{k}$ are called operation elements or Kraus operators. Note that the
sequence $(E_{k})_{k}$ that represents a CP map is not unique. The minimum
number of operators $E_{k}$ necessary to represent $\varepsilon$ is called its
Choi rank, which we’ll denote by $CR(\varepsilon)$. The number of indices $k$
in the sequence $(E_{k})_{k}$ equals the Choi rank if, and only if,
$(E_{k})_{k}$ is linearly independent. For a proof of these statements, see
[Cho75] or [Wat18].

There is a duality between CP maps that will be useful to relate CPT and UCP
maps. Since $\mathrm{End}(X)$ and $\mathrm{End}(Y)$ are Hilbert spaces with
the Hilbert-Schmidt inner product, we have a notion of dual map
$\varepsilon^{\dagger}$. It is defined as the unique linear map
$\varepsilon^{\dagger}\colon\mathrm{End}(Y)\to\mathrm{End}(X)$ that satisfies
$\langle\varepsilon(A),B\rangle=\langle A,\varepsilon^{\dagger}(B)\rangle$,
for every $A\in\mathrm{End}(X)$ and $B\in\mathrm{End}(Y)$.

######  Lemma 2.5.

A linear map $\varepsilon\colon\mathrm{End}(X)\to\mathrm{End}(Y)$ is CP if,
and only if, its dual
$\varepsilon^{\dagger}\colon\mathrm{End}(Y)\to\mathrm{End}(X)$ is CP. Also, if
$\varepsilon$ has operation elements $(E_{k})_{k}$, then
$(E_{k}^{\dagger})_{k}$ are operation elements for $\varepsilon^{\dagger}$.

###### Proof.

Suppose that $\varepsilon$ is a completely positive map with operation
elements $(E_{k})_{k}$. For any $A\in\mathrm{End}(X)$ and
$B\in\mathrm{End}(Y)$, we have

$\displaystyle\langle A,\varepsilon^{\dagger}(B)\rangle$
$\displaystyle=\langle\varepsilon(A),B\rangle$
$\displaystyle=\langle\sum_{k}E_{k}AE_{k}^{\dagger},B\rangle$
$\displaystyle=\sum_{k}\langle E_{k}AE_{k}^{\dagger},B\rangle$
$\displaystyle=\sum_{k}tr((E_{k}AE_{k}^{\dagger})^{\dagger}B)$
$\displaystyle=\sum_{k}tr(E_{k}A^{\dagger}E_{k}^{\dagger}B)$
$\displaystyle=\sum_{k}tr(A^{\dagger}E_{k}^{\dagger}BE_{k})$
$\displaystyle=tr(A^{\dagger}\sum_{k}E_{k}^{\dagger}BE_{k})$
$\displaystyle=\langle A,\sum_{k}E_{k}^{\dagger}BE_{k}\rangle.$

Since $\varepsilon^{\dagger}$ is unique, we conclude that
$\varepsilon^{\dagger}(B)=\sum_{k}E_{k}^{\dagger}BE_{k}$. Therefore,
$\varepsilon^{\dagger}$ is completely positive and has $(E_{k}^{\dagger})_{k}$
as operation elements.

If we assume instead that $\varepsilon^{\dagger}$ is completely positive,
using that $\varepsilon^{{\dagger}{\dagger}}=\varepsilon$ we conclude, by the
previous results, that $\varepsilon$ is completely positive. Therefore,
$\varepsilon$ is completely positive if, and only if, $\varepsilon^{\dagger}$
is completely positive. ∎

The tensor product will be a central topic of this article, so we review some
results about it. For any CP maps $\varepsilon\in CP(X,Y)$ and
$\varepsilon^{\prime}\in CP(X^{\prime},Y^{\prime})$ there is the tensor
product $\varepsilon\otimes\varepsilon^{\prime}$. We’ll see that
$\varepsilon\in CP(X\otimes X^{\prime},Y\otimes Y^{\prime})$. Also, given
operation elements for $\varepsilon$ and $\varepsilon^{\prime}$, we can
construct operation elements for $\varepsilon\otimes\varepsilon^{\prime}$ as
in the next lemma.

######  Lemma 2.6.

If $\varepsilon\in CP(X,Y)$ and $\varepsilon^{\prime}\in
CP(X^{\prime},Y^{\prime})$ have operation elements $(E_{k})_{k}$ and
$(F_{l})_{l}$, respectively, then $\varepsilon\otimes\varepsilon^{\prime}$
have $(E_{k}\otimes F_{l})_{k,l}$ as operation elements. In particular,
$\varepsilon\in CP(X\otimes X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

$\varepsilon\otimes\varepsilon^{\prime}$ is characterized as the map that
satisfies $(\varepsilon\otimes\varepsilon^{\prime})(A\otimes
B)=\varepsilon(A)\otimes\varepsilon^{\prime}(B)$, for any
$A\in\mathrm{End}(X)$ and $B\in\mathrm{End}(X^{\prime})$. Therefore
$(\varepsilon\otimes\varepsilon^{\prime})(A\otimes
B)=\varepsilon(A)\otimes\varepsilon^{\prime}(B)=(\sum_{k}E_{k}AE_{k}^{\dagger})\otimes(\sum_{l}F_{l}BF_{l}^{\dagger})=\sum_{k,l}E_{k}AE_{k}^{\dagger}\otimes\sum_{l}F_{l}BF_{l}^{\dagger}\stackrel{{\scriptstyle(*)}}{{=}}\sum_{k,l}(E_{k}\otimes
F_{l})(A\otimes B)(E_{k}^{\dagger}\otimes
F_{l}^{\dagger})=\sum_{k,l}(E_{k}\otimes F_{l})(A\otimes B)(E_{k}\otimes
F_{l})^{\dagger}$. In the equality ($*$) we used the identity
$(f^{\prime}\otimes g^{\prime})(f\otimes g)=f^{\prime}f\otimes g^{\prime}g$.
The identity $(\varepsilon\otimes\varepsilon^{\prime})(A\otimes
B)=\sum_{k,l}(E_{k}\otimes F_{l})(A\otimes B)(E_{k}\otimes F_{l})^{\dagger}$
together with the linearity of $\varepsilon\otimes\varepsilon^{\prime}$
implies that
$(\varepsilon\otimes\varepsilon^{\prime})(C)=\sum_{k,l}(E_{k}\otimes
F_{l})C(E_{k}\otimes F_{l})^{\dagger}$, for any $C\in\mathrm{End}(X\otimes
X^{\prime})$. Therefore $(E_{k}\otimes F_{l})_{k,l}$ are operation elements
for $\varepsilon\otimes\varepsilon^{\prime}$. Since
$\varepsilon\otimes\varepsilon^{\prime}$ is of the form
$(\varepsilon\otimes\varepsilon^{\prime})(C)=\sum_{k,l}(E_{k}\otimes
F_{l})C(E_{k}\otimes F_{l})^{\dagger}$, it is completely positive. ∎

######  Corollary 2.7.

For any CP maps $\varepsilon$ and $\varepsilon^{\prime}$ we have
$CR(\varepsilon\otimes\varepsilon^{\prime})=CR(\varepsilon)CR(\varepsilon^{\prime})$.

###### Proof.

Let $(E_{k})_{k}$ be linearly independent operation elements for $\varepsilon$
and let $(F_{l})_{l}$ be linearly independent operation elements for
$\varepsilon^{\prime}$. By lemmas 2.3 and 2.6 we know that $(E_{k}\otimes
F_{l})_{k,l}$ are linearly independent operation elements for
$\varepsilon\otimes\varepsilon^{\prime}$. Since the operations elements are
linearly independent, the sequences $(E_{k})_{k}$ and $(F_{l})_{l}$ have
$CR(\varepsilon)$ and $CR(\varepsilon^{\prime})$ operators, respectively.
Therefore $(E_{k}\otimes F_{l})_{k,l}$ is a sequence with
$CR(\varepsilon)CR(\varepsilon^{\prime})$ operators. But $(E_{k}\otimes
F_{l})_{k,l}$ is linearly independent, so it must have
$CR(\varepsilon\otimes\varepsilon^{\prime})$ operators. Therefore
$CR(\varepsilon\otimes\varepsilon^{\prime})=CR(\varepsilon)CR(\varepsilon^{\prime})$.
∎

For later use, we’ll show that
$(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}=\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}$.

######  Lemma 2.8.

Let $\varepsilon$ and $\varepsilon^{\prime}$ be completely positive maps. We
have that
$(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}=\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}$.

###### Proof.

Lets write that $\varepsilon\in CP(X,Y)$ and $\varepsilon^{\prime}\in
CP(X^{\prime},Y^{\prime})$. Both
$(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}$ and
$\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}$ are elements of
$CP(Y\otimes Y^{\prime},X\otimes X^{\prime})$. Their domain is
$\mathrm{End}(Y\otimes Y^{\prime})$, which is spanned by the maps $A\otimes
A^{\prime}$, for $A\in\mathrm{End}(Y)$ and
$A^{\prime}\in\mathrm{End}(Y^{\prime})$. Similarly for the codomain
$\mathrm{End}(X\otimes X^{\prime})$, it is spanned by the maps $B\otimes
B^{\prime}$ for $B\in\mathrm{End}(X)$ and
$B^{\prime}\in\mathrm{End}(X^{\prime})$. By linearity, we’ll only need to
consider these $A\otimes A^{\prime}$ and $B\otimes B^{\prime}$. By definition
of the dual, $(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}$ is the
unique linear map that satisfies
$\langle(\varepsilon\otimes\varepsilon^{\prime})(A\otimes A^{\prime}),B\otimes
B^{\prime}\rangle=\langle A\otimes
A^{\prime},(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}(B\otimes
B^{\prime})\rangle$. We have that

$\displaystyle\langle(\varepsilon\otimes\varepsilon^{\prime})(A\otimes
A^{\prime}),B\otimes B^{\prime}\rangle$
$\displaystyle=\langle\varepsilon(A)\otimes\varepsilon^{\prime}(A^{\prime}),B\otimes
B^{\prime}\rangle$
$\displaystyle=\langle\varepsilon(A),B\rangle\langle\varepsilon^{\prime}(A^{\prime}),B^{\prime}\rangle$
$\displaystyle=\langle A,\varepsilon^{\dagger}(B)\rangle\langle
A^{\prime},\varepsilon^{\dagger}(B^{\prime})\rangle$ $\displaystyle=\langle
A\otimes
A^{\prime},\varepsilon(B)\otimes\varepsilon^{\prime}(B^{\prime})\rangle$
$\displaystyle=\langle A\otimes
A^{\prime},(\varepsilon\otimes\varepsilon^{\prime})(B\otimes
B^{\prime})\rangle.$

By uniqueness it follows that
$(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}=\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}$.
∎

####  2.3.1 CPT maps

A CPT map $\varepsilon\colon X\to Y$ is an $\varepsilon\in CP(X,Y)$ which is
also trace preserving. Being trace preserving means that
$tr(\varepsilon(A))=trA$, for every $A\in\mathrm{End}(X)$. This is equivalent
to the equation $\sum_{k}E_{k}^{\dagger}E_{k}=id_{X}$. Lets denote by
$CPT(X,Y)$ the set of CPT maps $\varepsilon\colon X\to Y$.

As the two next propositions shows, $CPT(X,Y)$ is convex and compact.

######  Proposition 2.9.

$CPT(X,Y)$ is a convex set.

###### Proof.

Let $p_{1},...,p_{n}$ be numbers satisfying $p_{i}\geq 0$ and
$\sum_{i=1}^{n}p_{i}=1$. Also, let $\varepsilon_{1}$,…,$\varepsilon_{n}$ be
CPT maps in $CPT(X,Y)$. Let $(E_{i,k_{i}})_{k_{i}}$ be operation elements for
$\varepsilon_{i}$, so that
$\varepsilon_{i}(A)=\sum_{k_{i}}E_{i,k_{i}}AE_{i,k_{i}}^{\dagger}$. Then the
convex combination of $\varepsilon_{i}$ is given by
$(\sum_{i=1}^{n}p_{i}\varepsilon_{i})(A)$ $=$
$\sum_{i=1}^{n}p_{i}\sum_{k_{i}}E_{i,k_{i}}AE_{i,k_{i}}^{\dagger}$ $=$
$\sum_{i=1}^{n}\sum_{k_{i}}$ $(\sqrt{p_{i}}E_{i,k_{i}})A$
$(\sqrt{p_{i}}E_{i,k_{i}})^{\dagger}$. This shows that
$\sum_{i=1}^{n}p_{i}\varepsilon_{i}$ is completely positive and has operation
elements $(\sqrt{p_{i}}E_{i,k_{i}})_{i,k_{i}}$. Also, it is trace preserving,
since $\sum_{i,k_{i}}(\sqrt{p_{i}}E_{i,k_{i}})^{\dagger}$
$(\sqrt{p_{i}}E_{i,k_{i}})$ $=$
$\sum_{i}p_{i}\sum_{k_{i}}E_{i,k_{i}}^{\dagger}$ $E_{i,k_{i}}$ $=$
$\sum_{i}p_{i}id_{X}=id_{X}$. ∎

For the compacity we use the Choi-Jamiołkowski isomorphism [Wat18]. There is a
well known basis dependent version of this isomorphism, but here we use a
basis independent one. Given an orthonormal basis $x_{1}$, …, $x_{n}$ of $X$,
let $x^{1}$,…,$x^{n}$ denote the dual basis for $X^{*}$. That is, $x^{i}\in
X^{*}$ and $x^{i}(x_{j})=\delta_{i,j}$. Let $\Omega=\sum_{i=1}^{n}x^{i}\otimes
x_{i}\in X^{*}\otimes X$. The Choi-Jamiołkowski isomorphism is the linear
isomorphism $CJ\colon$ $\mathrm{Hom}(\mathrm{End}(X),\mathrm{End}(Y))$ $\to$
$\mathrm{End}(X^{*}\otimes Y)$ given by
$CJ(\varepsilon)=(id_{\mathrm{End}(X^{*})}\otimes\varepsilon)(|\Omega\rangle\langle\Omega|)$
(in Dirac notation). When restricted to $CPT(X,Y)$, it gives a bijection
$CJ\colon CPT(X,Y)\to\\{A\in\mathrm{End}(X^{*}\otimes Y)\mid A\text{ is
positive, }tr_{Y}A=id_{X^{*}}\\}$. The well known basis dependent version
differs by replacing $x^{i}$ with $x_{i}$ and $X^{*}$ with $X$ in the previous
definitions.

######  Proposition 2.10.

$CPT(X,Y)$ is compact, in the norm topology.

###### Proof.

By the Choi-Jamiołkowski isomorphism, $CPT(X,Y)$ is homeomorphic to
$CJ(X,Y)\coloneqq\\{A\in\mathrm{End}(X^{*}\otimes Y)\mid A\text{ is positive,
}tr_{Y}A=id_{X^{*}}\\}$. Since for finite dimensions every norm is equivalent,
it suffices to show that $CJ(X,Y)$ is bounded and closed, with respect to any
norm on $CJ(X,Y)$. We show this for the operator norm.

First we show that $CJ(X,Y)$ is bounded. For any $A\in CJ(X,Y)$ we have
$tr_{Y}A=id_{X^{*}}$, therefore
$trA=tr_{X^{*}}tr_{Y}A=tr_{X^{*}}id_{X^{*}}=\dim X^{*}=\dim X$. Since $A$ is
also positive, it is diagonalizable and has non negative eigenvalues. Let
$Sp(A)$ be the spectrum of $A$, that is, its set of eigenvalues. Then
$||A||=max_{\lambda\in
Sp(A)}|\lambda|\stackrel{{\scriptstyle(*)}}{{=}}max_{\lambda\in
Sp(A)}\lambda=max(Sp(A))$. In $(*)$ we’ve used that $A$ has non negative
eigenvalues. Since $trA$ is the sum of the eigenvalues of $A$, we have
$\lambda\leq trA\leq dimX$, for any $\lambda\in Sp(A)$. Therefore $||A||\leq
dimX$. This concludes that $CJ(X,Y)$ is bounded.

Now lets show that $CJ(X,Y)$ is closed. Let $(A_{n})_{n\in\mathbb{N}}$ be a
sequence in $CJ(X,Y)$ that converges to some operator
$A\in\mathrm{End}(X^{*}\otimes Y)$. We can easily show that $A$ is positive.
Let $v\in X^{*}\otimes Y$, then $\langle v,A(v)\rangle=\langle
v,\lim_{n}A_{n}(v)\rangle$. Since the inner product is continuous in both its
entries, then $\langle v,\lim_{n}A_{n}(v)\rangle=\lim_{n}\langle
v,A_{n}(v)\rangle$. Since each $A_{n}$ is positive, we have $\langle
v,A_{n}(v)\rangle\geq 0$, so $\lim_{n}\langle v,$ $A_{n}(v)\rangle$ $\geq$
$0$. Therefore $\langle v,A(v)\rangle\geq 0$, for any $v\in X^{*}\otimes Y$,
which implies that $A$ is positive. Also, since $tr_{Y}$ is linear, it is
continuous. Therefore, $tr_{Y}A=tr_{Y}(\lim_{n}A_{n})=\lim_{n}tr_{Y}A_{n}$.
Since $tr_{Y}A_{n}=id_{X^{*}}$, we have
$\lim_{n}tr_{Y}A_{n}=\lim_{n}id_{X^{*}}=id_{X^{*}}$. Therefore,
$tr_{Y}A=id_{X}$ and we conclude that $A\in CJ(X,Y)$. This proves that
$CJ(X,Y)$ is closed.

Since $CJ(X,Y)$ is bounded and closed, it is compact. Using the Choi-
Jamiołkowski isomorphism we conclude that $CPT(X,Y)$ is also compact.

∎

Being a compact convex set, by theorem 2.4 we know that $CPT(X,Y)$ is the
convex hull of its extreme points.

In [Wat18] is given the following characterization of extremality of a CPT map
(theorem 2.31 of [Wat18]):

######  Theorem 2.11.

Let $\varepsilon\in CPT(X,Y)$ be a CPT map with linearly independent operation
elements $(E_{k})_{k}$. Then $\varepsilon$ is an extreme point of $CPT(X,Y)$
if, and only if, the family $(E_{k}^{\dagger}E_{k^{\prime}})_{k,k^{\prime}}$
is linearly independent.

Notice that we can always find operation elements for $\varepsilon$ which are
linearly independent. This can be achieved by applying the Choi-Jamiołkowski
isomorphism to $\varepsilon$ and computing an orthonormal basis that
diagonalizes $CJ(\varepsilon)$. Then the $E_{k}$’s can be constructed from the
eigenvectors with non zero eigenvalue.

In lemma 2.6 we’ve shown that $CP$ maps are closed under the tensor product.
For completeness, lets show the same result for CPT maps.

######  Lemma 2.12.

If $\varepsilon\in CPT(X,Y)$ and $\varepsilon^{\prime}\in
CPT(X^{\prime},Y^{\prime})$ then $\varepsilon\otimes\varepsilon^{\prime}\in
CPT(X\otimes X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

By lemma 2.6 we already know that $\varepsilon\otimes\varepsilon^{\prime}\in
CP(X\otimes X^{\prime},Y\otimes Y^{\prime})$. It remains to check that
$\varepsilon\otimes\varepsilon^{\prime}$ is trace preserving. Lets use the
same notation of lemma 2.6. Being trace preserving means that
$\sum_{k,l}(E_{k}\otimes F_{l})^{\dagger}(E_{k}\otimes F_{l})=id_{X\otimes
X^{\prime}}$. We prove it as follows: $\sum_{k,l}(E_{k}\otimes
F_{l})^{\dagger}(E_{k}\otimes F_{l})=\sum_{k,l}(E_{k}^{\dagger}\otimes
F_{l}^{\dagger})(E_{k}\otimes F_{l})=\sum_{k,l}E_{k}^{\dagger}E_{k}\otimes
F_{l}^{\dagger}F_{l}=(\sum_{k}E_{k}^{\dagger}E_{k})\otimes(\sum_{l}F_{l}^{\dagger}F_{l})=id_{X}\otimes
id_{X^{\prime}}=id_{X\otimes X^{\prime}}$. ∎

####  2.3.2 UCP maps

Unital completely positive (UCP) maps are CP maps $\varepsilon\in CP(X,Y)$
that preserve the identity, that is, $\varepsilon(id_{X})=id_{Y}$. We denote
the set of UCP maps from $X$ to $Y$ by $UCP(X,Y)$. If $\varepsilon$ has
operation elements $(E_{k})_{k}$, being unital is equivalent to the equation
$\sum_{k}E_{k}E_{k}^{\dagger}=id_{Y}$. We can see that this equation is
similar to $\sum_{k}E_{k}^{\dagger}E_{k}=id_{X}$, which is equivalent to being
trace preserving. Such similarity can be seen as a consequence of the fact
that UCP maps are dual to CPT maps.

######  Proposition 2.13.

A linear map $\varepsilon\colon\mathrm{End}(X)\to\mathrm{End}(Y)$ is CPT if,
and only if, $\varepsilon^{\dagger}\colon\mathrm{End}(Y)\to\mathrm{End}(X)$ is
UCP. In particular, we have an anti-linear bijection ${\dagger}\colon
CPT(X,Y)\to UCP(Y,X)$.

###### Proof.

Assume that $\varepsilon$ is completely positive with operation elements
$(E_{k})_{k}$. By lemma 2.5 we know that $\varepsilon^{\dagger}$ is also
completely positive and has $(E_{k}^{\dagger})_{k}$ as operation elements.
Then $\varepsilon$ is trace preserving if
$\sum_{k}E_{k}^{\dagger}E_{k}=id_{X}$, and $\varepsilon^{\dagger}$ is unital
if $\sum_{k}E_{k}^{\dagger}(E_{k}^{\dagger})^{\dagger}=id_{X}$. Both are the
same equation, therefore $\varepsilon$ is trace preserving if, and only if,
$\varepsilon^{\dagger}$ is unital. ∎

We can use the duality between $CPT$ and $UCP$ maps to transfer results
between them.

######  Proposition 2.14.

UCP(X,Y) is compact convex, in the norm topology.

###### Proof.

By proposition 2.13, we have an anti-linear bijection ${\dagger}\colon
CPT(X,Y)\to UCP(Y,X)$, for any finite dimensional Hilbert spaces $X$ and $Y$.
Since it is anti-linear, it is also continuous. Then $UCP(X,Y)$ is the image
of the compact $CPT(Y,X)$ by the continuous function ${\dagger}$, and
therefore $UCP(X,Y)$ is compact. Also, if $\varepsilon_{i}\in UCP(X,Y)$ and
$(p_{i})_{i}$ is a discrete probability distribution, then
$\sum_{i}p_{i}\varepsilon_{i}=(\sum_{i}p_{i}\varepsilon_{i})^{{\dagger}{\dagger}}=(\sum_{i}p_{i}(\varepsilon_{i})^{\dagger})^{\dagger}$.
Sincd $CPT(Y,X)$ is convex, we know that
$\sum_{i}p_{i}(\varepsilon_{i})^{\dagger}\in CPT(Y,X)$, and therefore
$(\sum_{i}p_{i}(\varepsilon_{i})^{\dagger})^{\dagger}\in UCP(X,Y)$. That is,
$\sum_{i}p_{i}\varepsilon_{i}\in UCP(X,Y)$, which implies that $UCP(X,Y)$ is
convex. ∎

The anti-linear bijection ${\dagger}\colon CPT(X,Y)\to UCP(Y,X)$ stablishes a
bijection between the extreme points of $CPT(X,Y)$ and the extreme points of
$UCP(Y,X)$. Using this duality, we obtain a characterization of extreme points
for $UCP(X,Y)$ that is similar to theorem 2.11.

######  Theorem 2.15.

Let $\varepsilon\in UCP(X,Y)$ be a UCP map with linearly independent operation
elements $(E_{k})_{k}$. Then it is an extreme point of $UCP(X,Y)$ if, and only
if, $(E_{k}E_{l}^{\dagger})_{k,l}$ is linearly independent.

###### Proof.

$\varepsilon$ is an extreme point of $UCP(X,Y)$ if, and only if,
$\varepsilon^{\dagger}$ is an extreme point of $CPT(Y,X)$. By lemma 2.5,
$(E_{k}^{\dagger})_{k}$ are operation elements for $\varepsilon$. Since
$(E_{k})_{k}$ is linearly independent, it follows that $(E_{k}^{\dagger})_{k}$
is also linearly independent. Then, by theorem 2.11, $\varepsilon^{\dagger}$
is an extreme point of $CPT(Y,X)$ if, and only if,
$(E_{k}^{{\dagger}{\dagger}}E_{l}^{\dagger})_{k,l}$ is linearly independent.
But $E_{k}^{{\dagger}{\dagger}}=E_{k}$, so $\varepsilon$ is an extreme point
of $UCP(X,Y)$ if, and only if, $(E_{k}E_{l}^{\dagger})_{k,l}$ is linearly
independent. ∎

Finally, lets prove that UCP maps are closed under the tensor product.

######  Lemma 2.16.

If $\varepsilon\in UCP(X,Y)$ and $\varepsilon^{\prime}\in
UCP(X^{\prime},Y^{\prime})$ then $\varepsilon\otimes\varepsilon^{\prime}\in
UCP(X\otimes X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

We can prove this result by using the duality between UCP and CPT maps and
lemma 2.12. Using lemma 2.8 we have that
$\varepsilon\otimes\varepsilon^{\prime}=(\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}})^{\dagger}$.
We know that $\varepsilon\in UCP(X,Y)$ and $\varepsilon^{\prime}\in
UCP(X^{\prime},Y^{\prime})$, then proposition 2.13 implies that
$\varepsilon^{\dagger}\in CPT(Y,X)$ and $\varepsilon^{\prime{\dagger}}\in
CPT(Y^{\prime},X^{\prime})$. By lemma 2.12 we know that
$\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}\in CPT(Y\otimes
Y^{\prime},X\otimes X^{\prime})$. Using proposition 2.13 again we conclude
that $(\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}})^{\dagger}\in
UCP(X\otimes X^{\prime},Y\otimes Y^{\prime})$. ∎

####  2.3.3 UCPT maps

An UCPT map is one that is both unital and trace preserving, that is,
$UCPT(X,$ $Y)$ $=$ $UCP(X,Y)\cap CPT(X,Y)$. In this case, we restrict
attention to endomorphisms ($X=Y$), since $X$ and $Y$ have necessarily the
same dimension. In fact, let $\varepsilon\in UCPT(X,Y)$, then
$\varepsilon(id_{X})=id_{Y}$, because it is unital. But it is also trace
preserving, so $tr(\varepsilon(id_{X}))=tr(id_{X})$, so
$tr(id_{Y})=tr(id_{X})$. But $tr(id_{X})=\dim X$, so $\dim X=\dim Y$. We’ll
denote $UCPT(X,X)$ as $UCPT(X)$.

######  Proposition 2.17.

$UCPT(X)$ is compact convex, in the norm topology.

###### Proof.

Since both $UCP(X)$ and $CPT(X)$ are convex sets, its intersection $UCPT(X)$
is also convex. Since, both $UCP(X)$ and $CPT(X)$ are bounded sets, $UCPT(X)$
is also bounded. And since both $UCP(X)$ and $CPT(X)$ are closed sets,
$UCPT(X)$ is also closed. Therefore $UCPT(X)$ is closed and bounded, so it is
compact. ∎

As $UCPT(X)$ is compact convex, it is the convex hull of its extreme points.
The following characterization of the extreme points of $UCPT(X)$ was proven
is [LS93]. It can also be found in theorem 4.21 of [Wat18].

######  Theorem 2.18.

Let $\varepsilon\in UCPT(X)$ be an UCPT map with linearly independent
operation elements $(E_{k})_{k}$. Then it is an extreme point of $UCPT(X)$ if,
and only if, $(E_{k}^{\dagger}E_{l}\oplus E_{l}E_{k}^{\dagger})_{k,l}$ is
linearly independent as vectors of $\mathrm{End}(X)\oplus\mathrm{End}(X)$.

As done for the other types of CP maps, lets prove that UCPT is closed under
the tensor product.

######  Lemma 2.19.

If $\varepsilon\in UCPT(X,Y)$ and $\varepsilon^{\prime}\in
UCPT(X^{\prime},Y^{\prime})$ then $\varepsilon\otimes\varepsilon^{\prime}\in
UCPT(X\otimes X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

By lemmas 2.12 and 2.16 we know that
$\varepsilon\otimes\varepsilon^{\prime}\in CPT(X\otimes X^{\prime},Y\otimes
Y^{\prime})$ and $\varepsilon\otimes\varepsilon^{\prime}\in UCP(X\otimes
X^{\prime},Y\otimes Y^{\prime})$. Since $UCPT(X\otimes X^{\prime},Y\otimes
Y^{\prime})=CPT(X\otimes X^{\prime},Y\otimes Y^{\prime})\cap UCP(X\otimes
X^{\prime},Y\otimes Y^{\prime})$, this concludes the proof. ∎

##  3 Main results

###  3.1 Extremality is preserved for CPT maps

Now we prove that the tensor product preserves extramality for CPT maps.

######  Theorem 3.1.

If $\varepsilon$ is an extreme point of $CPT(X,Y)$ and $\varepsilon^{\prime}$
is an extreme point of $CPT(X^{\prime},Y^{\prime})$, then
$\varepsilon\otimes\varepsilon^{\prime}$ is an extreme point of $CPT(X\otimes
X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

We always use the Hilbert-Schmidt inner product. Let $(E_{k})_{k}$ be linearly
independent operation elements for $\varepsilon$, and let $(F_{l})_{l}$ be
linearly independent operation elements for $\varepsilon^{\prime}$. By lemma
2.6, $(E_{k}\otimes F_{l})_{k,l}$ are operation elements for
$\varepsilon\otimes\varepsilon^{\prime}$. Also, lemma 2.3 implies that
$(E_{k}\otimes F_{l})_{k,l}$ are linearly independent. Then, by theorem 2.11,
$\varepsilon\otimes\varepsilon^{\prime}$ is extreme if, and only if,
$((E_{k}\otimes F_{l})^{\dagger}(E_{k^{\prime}}\otimes
F_{l^{\prime}}))_{k,k^{\prime},l,l^{\prime}}$ is linearly independent (as
vectors of $\mathrm{End}(X\otimes Y)$). By lemma 2.1, linear independence is
equivalent to the invertibility of the Gram matrix. To prove the invertibility
of the Gram matrix, we’ll show that it is the Kronecker product of two
invertible matrices. Let $G$, $G^{\prime}$ and $G^{\prime\prime}$ be the Gram
matrices of $(E_{k}^{\dagger}E_{k^{\prime}})_{k,k^{\prime}}$,
$(F_{l}^{\dagger}F_{l^{\prime}})_{l,l^{\prime}}$ and $((E_{k}\otimes
F_{l})^{\dagger}(E_{k^{\prime}}\otimes
F_{l^{\prime}}))_{k,k^{\prime},l,l^{\prime}}$, respectively. Its matrix
elements are

$G_{(k,k^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime})}=\langle
E_{k}^{\dagger}E_{k^{\prime}},E_{k^{\prime\prime}}^{\dagger}E_{k^{\prime\prime\prime}}\rangle,$
(6)
$G^{\prime}_{(l,l^{\prime}),(l^{\prime\prime},l^{\prime\prime\prime})}=\langle
F_{l}^{\dagger}F_{l^{\prime}},F_{l^{\prime\prime}}^{\dagger}F_{l^{\prime\prime\prime}}\rangle,$
(7)
$G^{\prime\prime}_{(k,k^{\prime},l,l^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime},l^{\prime\prime},l^{\prime\prime\prime})}=\langle(E_{k}\otimes
F_{l})^{\dagger}(E_{k^{\prime}}\otimes
F_{l^{\prime}}),(E_{k^{\prime\prime}}\otimes
F_{l^{\prime\prime}})^{\dagger}(E_{k^{\prime\prime\prime}}\otimes
F_{l^{\prime\prime\prime}})\rangle.$ (8)

Since $\varepsilon$ and $\varepsilon^{\prime}$ are extreme CPT maps, and since
$(E_{k})_{k}$ and $(F_{l})_{l}$ are linearly independent operation elements,
by theorem 2.11 we know that $(E_{k}^{\dagger}E_{k^{\prime}})_{k,k^{\prime}}$
and $(F_{l}^{\dagger}F_{l^{\prime}})_{l,l^{\prime}}$ are both linearly
independent. By lemma 2.1, it implies that $G$ and $G^{\prime}$ are invertible
matrices. We can rewrite $G^{\prime\prime}$ in terms of $G$ and $G^{\prime}$
as follows:

$\displaystyle
G^{\prime\prime}_{(k,k^{\prime},l,l^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime},l^{\prime\prime},l^{\prime\prime\prime})}$
$\displaystyle=\langle(E_{k}\otimes F_{l})^{\dagger}(E_{k^{\prime}}\otimes
F_{l^{\prime}}),(E_{k^{\prime\prime}}\otimes
F_{l^{\prime\prime}})^{\dagger}(E_{k^{\prime\prime\prime}}\otimes
F_{l^{\prime\prime\prime}})\rangle$
$\displaystyle=\langle(E_{k}^{\dagger}\otimes
F_{l}^{\dagger})(E_{k^{\prime}}\otimes
F_{l^{\prime}}),(E_{k^{\prime\prime}}^{\dagger}\otimes
F_{l^{\prime\prime}}^{\dagger})(E_{k^{\prime\prime\prime}}\otimes
F_{l^{\prime\prime\prime}})\rangle$ $\displaystyle=\langle
E_{k}^{\dagger}E_{k^{\prime}}\otimes
F_{l}^{\dagger}F_{l^{\prime}},E_{k^{\prime\prime}}^{\dagger}E_{k^{\prime\prime\prime}}\otimes
F_{l^{\prime\prime}}^{\dagger}F_{l^{\prime\prime\prime}}\rangle$
$\displaystyle\stackrel{{\scriptstyle(*)}}{{=}}\langle
E_{k}^{\dagger}E_{k^{\prime}},E_{k^{\prime\prime}}^{\dagger}E_{k^{\prime\prime\prime}}\rangle\langle
F_{l}^{\dagger}F_{l^{\prime}},F_{l^{\prime\prime}}^{\dagger}F_{l^{\prime\prime\prime}}\rangle$
$\displaystyle=G_{(k,k^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime})}G^{\prime}_{(l,l^{\prime}),(l^{\prime\prime},l^{\prime\prime\prime})}.$

In equality (*) we’ve used equation (1). Given any order of $(k,k^{\prime})$
and $(l,l^{\prime})$, we can give the lexicographic order to
$(k,k^{\prime},l,l^{\prime})$. In this way, the identity
$G^{\prime\prime}_{(k,k^{\prime},l,l^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime},l^{\prime\prime},l^{\prime\prime\prime})}=G_{(k,k^{\prime}),(k^{\prime\prime},k^{\prime\prime\prime})}G^{\prime}_{(l,l^{\prime}),(l^{\prime\prime},l^{\prime\prime\prime})}$
says that $G^{\prime\prime}=G\otimes G^{\prime}$, that is, $G^{\prime\prime}$
is the Kronecker product of $G$ and $G^{\prime}$. Since $G$ and $G^{\prime}$
are invertible, then $G^{\prime\prime}$ is also invertible, with inverse
$G^{\prime\prime-1}=G^{-1}\otimes G^{\prime-1}$. Then lemma 2.1 implies that
$((E_{k}\otimes F_{l})^{\dagger}(E_{k^{\prime}}\otimes
F_{l^{\prime}}))_{k,k^{\prime},l,l^{\prime}}$ is linearly independent.
Therefore, by theorem 2.11, $\varepsilon\otimes\varepsilon^{\prime}$ is an
extreme CPT map.

∎

###  3.2 Extremality is preserved for UCP maps

The result for UCP maps can be obtained by using the duality between CPT and
UCP maps.

######  Theorem 3.2.

If $\varepsilon$ is an extreme point of $UCP(X,Y)$ and $\varepsilon^{\prime}$
is an extreme point of $UCP(X^{\prime},Y^{\prime})$, then
$\varepsilon\otimes\varepsilon^{\prime}$ is an extreme point of $UCP(X\otimes
X^{\prime},Y\otimes Y^{\prime})$.

###### Proof.

Since $\varepsilon\in UCP(X,Y)$ and $\varepsilon^{\prime}\in
UCP(X^{\prime},Y^{\prime})$, we know that $\varepsilon^{\dagger}\in CPT(Y,X)$
and $\varepsilon^{\prime{\dagger}}\in CPT(Y^{\prime},X^{\prime})$. Also,
$\varepsilon$ and $\varepsilon^{\prime}$ are extreme points of $UCP(X,Y)$ and
$UCP(X^{\prime},Y^{\prime})$, so $\varepsilon^{\dagger}$ and
$\varepsilon^{\prime{\dagger}}$ are extreme points of $CPT(Y,X)$ and
$CPT(Y^{\prime},X^{\prime})$. By theorem 3.1, we know that
$\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}$ is an extreme
point of $CPT(Y\otimes Y^{\prime},X\otimes X^{\prime})$. By lemma 2.8 we know
that
$\varepsilon^{\dagger}\otimes\varepsilon^{\prime{\dagger}}=(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}$,
so $(\varepsilon\otimes\varepsilon^{\prime})^{\dagger}$ is extreme. Using
again the duality between CPT and UCP maps we conclude that
$\varepsilon\otimes\varepsilon^{\prime}$ is an extreme point of $UCP(X\otimes
X^{\prime},Y\otimes Y^{\prime})$. ∎

###  3.3 The case of UCPT maps

The tensor product of extreme UCPT maps may not be extreme. But it is always
extreme if one of the two maps is over a Hilbert space of dimension 2. This is
a consequence of the next proposition and the fact that, for dimension 2,
extreme maps are unitary.

######  Proposition 3.3.

Let $\varepsilon\in UCPT(X)$ and $\mathcal{U}\in UCPT(Y)$ be a unitary map,
that is, $\mathcal{U}(A)=UAU^{\dagger}$ for some unitary operator
$U\in\mathrm{End}(Y)$. We assume that $\dim Y>0$. Then $\varepsilon$ is an
extreme point of $UCPT(X)$ if, and only if, $\varepsilon\otimes\mathcal{U}$ is
an extreme point of $UCPT(X\otimes Y)$.

###### Proof.

Let $(E_{k})_{k}$ be linearly independent operation elements for
$\varepsilon$. $\mathcal{U}$ has already $(U)$ as linearly independent
operation elements, since there is only one operator. By lemmas 2.3 and 2.6,
$\varepsilon\otimes\mathcal{U}$ has linearly independent operation elements
$(E_{k}\otimes U)$. By theorem 2.18, $\varepsilon\otimes\mathcal{U}$ is an
extreme point of $UCPT(X\otimes Y)$ if, and only if, $((E_{k}\otimes
U)^{\dagger}(E_{l}\otimes U)\oplus(E_{l}\otimes U)(E_{k}\otimes
U)^{\dagger})_{k,l}$ is linearly independent. We have that

$\displaystyle(E_{k}\otimes U)^{\dagger}(E_{l}\otimes U)\oplus(E_{l}\otimes
U)(E_{k}\otimes U)^{\dagger}$ $\displaystyle=(E_{k}^{\dagger}\otimes
U^{\dagger})(E_{l}\otimes U)\oplus(E_{l}\otimes U)$
$\displaystyle\quad\;(E_{k}^{\dagger}\otimes U^{\dagger})$
$\displaystyle=(E_{k}^{\dagger}E_{l}\otimes
U^{\dagger}U)\oplus(UU^{\dagger}\otimes E_{l}E_{k}^{\dagger})$
$\displaystyle=(E_{k}^{\dagger}E_{l}\otimes id_{Y})\oplus(id_{Y}\otimes
E_{l}E_{k}^{\dagger}).$

By lemma 2.1, we can check $((E_{k}^{\dagger}E_{l}\otimes
id_{Y})\oplus(id_{Y}\otimes E_{l}E_{k}^{\dagger}))_{k,l}$ is linearly
independent or not by analysing its Gram matrix. Let $G$ be the Gram matrix of
$(E_{k}^{\dagger}E_{l}\oplus E_{l}E_{k}^{\dagger})_{k,l}$ and $G^{\prime}$ be
the Gram matrix of $((E_{k}^{\dagger}E_{l}\otimes id_{Y})\oplus(id_{Y}\otimes
E_{l}E_{k}^{\dagger}))_{k,l}$. The matrix elements of $G$ are

$\displaystyle G_{(k,l),(k^{\prime},l^{\prime})}$ $\displaystyle=\langle
E_{k}^{\dagger}E_{l}\oplus
E_{l}E_{k}^{\dagger},E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\oplus
E_{l^{\prime}}E_{k^{\prime}}^{\dagger}\rangle$ $\displaystyle=\langle
E_{k}^{\dagger}E_{l},E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\rangle+\langle
E_{l}E_{k}^{\dagger},E_{l^{\prime}}E_{k^{\prime}}^{\dagger}\rangle.$

The matrix elements of $G^{\prime}$ are

$\displaystyle G^{\prime}_{(k,l),(k^{\prime},l^{\prime})}$
$\displaystyle=\langle(E_{k}^{\dagger}E_{l}\otimes id_{Y})\oplus(id_{Y}\otimes
E_{l}E_{k}^{\dagger}),(E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\otimes
id_{Y})\oplus(id_{Y}\otimes E_{l^{\prime}}E_{k^{\prime}}^{\dagger})\rangle$
$\displaystyle=\langle E_{k}^{\dagger}E_{l}\otimes
id_{Y},E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\otimes id_{Y}\rangle+\langle
id_{Y}\otimes E_{l}E_{k}^{\dagger},id_{Y}\otimes
E_{l^{\prime}}E_{k^{\prime}}^{\dagger}\rangle$ $\displaystyle=\langle
E_{k}^{\dagger}E_{l},E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\rangle\langle
id_{Y},id_{Y}\rangle+\langle id_{Y},id_{Y}\rangle\langle
E_{l}E_{k}^{\dagger},E_{l^{\prime}}E_{k^{\prime}}^{\dagger}\rangle$
$\displaystyle=\dim Y(\langle
E_{k}^{\dagger}E_{l},E_{k^{\prime}}^{\dagger}E_{l^{\prime}}\rangle+\langle
E_{l}E_{k}^{\dagger},E_{l^{\prime}}E_{k^{\prime}}^{\dagger}\rangle)$
$\displaystyle=\dim Y\;G_{(k,l),(k^{\prime},l^{\prime})}.$

That is, $G^{\prime}=\dim Y\;G$. Therefore $G$ is invertible if, and only if,
$G^{\prime}$ is invertible. By lemma 2.1, this says that
$(E_{k}^{\dagger}E_{l}\oplus E_{l}E_{k}^{\dagger})_{k,l}$ is linearly
independent if, and only if, $((E_{k}^{\dagger}E_{l}\otimes
id_{Y})\oplus(id_{Y}\otimes E_{l}E_{k}^{\dagger}))_{k,l}$ is linearly
independent. Then, by theorem 2.18, $\varepsilon$ is an extreme point of
$UCPT(X)$ if, and only if, $\varepsilon\otimes\mathcal{U}$ is an extreme point
of $UCPT(X\otimes Y)$. ∎

By theorem 4.23 of [Wat18], for $\dim X=2$, the extreme points of $UCPT(X)$
are the unitary maps. Combining this result with proposition 3.3 we obtain the
next theorem.

######  Theorem 3.4.

Let $\varepsilon$ be an extreme point of $UCPT(X)$ and $\varepsilon^{\prime}$
be an extreme point of $UCPT(Y)$. If $\dim X=2$ or $\dim Y=2$, then
$\varepsilon\otimes\varepsilon^{\prime}$ is an extreme point of $UCPT(X\otimes
Y)$.

For higher dimensions the extremality may not be preserved. One reason for
this can be seen if we analyze theorem 2.18. Given $\varepsilon\in UCPT(X)$
with $CR(\varepsilon)$ linearly independent operation elements $(E_{k})_{k}$,
it is an extreme point of the UCPT maps if $(E_{k}^{\dagger}E_{l}\oplus
E_{l}E_{k}^{\dagger})_{k,l}$ is linearly independent as vectors of
$\mathrm{End}(X)\oplus\mathrm{End}(X)$. The sequence
$(E_{k}^{\dagger}E_{l}\oplus E_{l}E_{k}^{\dagger})_{k,l}$ has
$CR(\varepsilon)^{2}$ elements. For it to be linearly independent,
$CR(\varepsilon)^{2}$ must be at most the dimension of
$\mathrm{End}(X)\oplus\mathrm{End}(X)$, which is $2(\dim X)^{2}$. Therefore,
if $\varepsilon$ is extreme then $CR(\varepsilon)\leq\sqrt{2}\dim X$.
Actually, a slightly better upper bound is given in [Ohn10], which says that
$CR(\varepsilon)\leq\sqrt{2(\dim X)^{2}-1}$. If we take two extreme UCPT maps
of high enough Choi ranks, their tensor product can have a Choi rank that
exceeds the previous upper bound, so the tensor product can’t be extreme. This
is proven in the next theorem.

######  Theorem 3.5.

Let $\varepsilon$ be an extreme point of $UCPT(X)$ and $\varepsilon^{\prime}$
be an extreme point of $UCPT(Y)$. In particular, their Choi ranks satisfy
$CR(\varepsilon)\leq\sqrt{2}\dim X$ and
$CR(\varepsilon^{\prime})\leq\sqrt{2}\dim Y$. If they further satisfy that
$\sqrt[4]{2}\dim X<CR(\varepsilon)$ and $\sqrt[4]{2}\dim
Y<CR(\varepsilon^{\prime})$, then $\sqrt{2}\dim(X\otimes
Y)<CR(\varepsilon\otimes\varepsilon^{\prime})$, so
$\varepsilon\otimes\varepsilon^{\prime}$ isn’t an extreme point of
$UCPT(X\otimes Y)$.

###### Proof.

By corollary 2.7, we know that
$CR(\varepsilon\otimes\varepsilon^{\prime})=CR(\varepsilon)$
$CR(\varepsilon^{\prime})$. If $\sqrt[4]{2}\dim X$ $<$ $CR(\varepsilon)$ and
$\sqrt[4]{2}\dim Y<CR(\varepsilon^{\prime})$, multiplying both inequalities we
get $\sqrt[4]{2}\dim X\sqrt[4]{2}\dim
Y<CR(\varepsilon)CR(\varepsilon^{\prime})$, so $\sqrt{2}\dim(X\otimes
Y)<CR(\varepsilon\otimes\varepsilon^{\prime})$. As seen previously, this
implies that $\varepsilon\otimes\varepsilon^{\prime}$ isn’t an extreme point
of $UCPT(X\otimes Y)$. ∎

By theorem 3.5, to prove that the tensor product of extreme UCPT maps may not
be extreme, it is sufficient to find extreme maps of high enough Choi rank.
According to theorems 2.5 and 2.6 of [Ohn10], there is an extreme map
$\varepsilon_{3}\in UCPT(\mathbb{C}^{3})$ with $CR(\varepsilon_{3})=4$ and an
extreme map of $\varepsilon_{4}\in UCPT(\mathbb{C}^{4})$ with
$CR(\varepsilon_{4})=5$. More precisely, the operation elements (in the
standard basis) for $\varepsilon_{3}$ are

$E_{0}=\frac{1}{2}\begin{pmatrix}1&0&0\\\ 0&0&0\\\ 0&0&0\end{pmatrix},$ (9)
$E_{1}=\frac{1}{2}\begin{pmatrix}0&0&0\\\ 1&0&0\\\ 0&\sqrt{2}&0\end{pmatrix},$
(10) $E_{2}=\frac{1}{2}\begin{pmatrix}0&\sqrt{2}&0\\\ 0&0&\sqrt{3}\\\
0&0&0\end{pmatrix},$ (11) $E_{3}=\frac{1}{2}\begin{pmatrix}0&0&1\\\ 0&0&0\\\
\sqrt{2}&0&0\end{pmatrix}.$ (12)

The operation elements for $\varepsilon_{4}$ are

$F_{0}=\frac{1}{2}\begin{pmatrix}0&0&0&0\\\ 0&0&1&0\\\ 1&0&0&0\\\
0&0&0&0\end{pmatrix},$ (13) $F_{1}=\frac{1}{2}\begin{pmatrix}0&0&0&0\\\
0&0&0&0\\\ 0&0&0&\sqrt{2}\\\ 0&\sqrt{2}&0&0\end{pmatrix},$ (14)
$F_{2}=\frac{1}{2}\begin{pmatrix}0&0&\sqrt{3}&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
\sqrt{2}&0&0&0\end{pmatrix},$ (15) $F_{3}=\frac{1}{2}\begin{pmatrix}0&1&0&0\\\
0&0&0&\sqrt{2}\\\ 0&0&0&0\\\ 0&0&0&0\end{pmatrix},$ (16)
$F_{4}=\frac{1}{2}\begin{pmatrix}0&0&0&0\\\ 1&0&0&0\\\ 0&1&0&0\\\
0&0&0&0\end{pmatrix}.$ (17)

Notice that the above matrices are the complex conjugate of the ones defined
in [Ohn10]. This occurs because in [Ohn10] a UCPT map is written as
$\varepsilon(A)=\sum_{k}E_{k}^{\dagger}AE_{k}$, while in this article the
convention is to write $\varepsilon(A)=\sum_{k}E_{k}AE_{k}^{\dagger}$, which
is the same convention of [NC10, Wat18].

Both $\varepsilon_{3}$ and $\varepsilon_{4}$ have maximal Choi ranks. For both
$n=3$ and $n=4$ we have $CR(\varepsilon_{n})>\sqrt[4]{2}\,n$. Then, by theorem
3.5, we know that $\varepsilon_{n}\otimes\varepsilon_{m}$ is not an extreme
UCPT map, for $n,m\in\\{3,4\\}$. It remains to show if we can apply theorem
3.5 to obtain counterexamples for even higher dimensions.

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