# Semantically-aware Mask CycleGAN for Translating Artistic Portraits to
Photo-realistic Visualizations

YIN Zhuohao
Hong Kong University of Science and Technology, Hong Kong
<EMAIL_ADDRESS>
###### Abstract

Image-to-image translation (I2I) is defined as a computer vision task where
the aim is to transfer images in a source domain to a target domain with
minimal loss or alteration of the content representations. Major progress has
been made since I2I was proposed with the invention of a variety of
revolutionary generative models. Among them, GAN-based models perform
exceptionally well as they are mostly tailor-made for specific domains or
tasks. However, few works proposed a tailor-made method for the artistic
domain. In this project, I propose the Semantic-aware Mask CycleGAN
(SMCycleGAN) architecture which can translate artistic portraits to photo-
realistic visualizations. This model can generate realistic human portraits by
feeding the discriminators semantically masked fake samples, thus enforcing
them to make discriminative decisions with partial information so that the
generators can be optimized to synthesize more realistic human portraits
instead of increasing the similarity of other irrelevant components, such as
the background. Experiments have shown that the SMCycleGAN generate images
with significantly increased realism and minimal loss of content
representations.

##  1 Introduction

In 1816, the world’s first camera was invented and cameras later became the
dominant tool for human beings to capture moments in their lives. However,
prior to this revolutionary invention, paintings served as the only feasible
approach to recording visual information. Realistic as some genres of
paintings may seem, such as mannerism and classicism, there still existed a
gap between what was on the canvas and what was before human eyes. This
project seeks to fill this gap, specifically within the domain of artistic
portraits, where the aim is to restore what was presented before the painters
hundreds of years ago. Given a set of art portraits $X$ and a set of realistic
portrait photos $Y$ (Fig. 1), the model is expected to learn a mapping $G$
such that given some $x_{i}$,

$y_{i}=G(x_{i})\in Y.$ (1)

Figure 1: Unpaired data comprised of source set $X=\\{x_{i}\\}_{i=1}^{N}$ and
target set $Y=\\{y_{i}\\}_{i=1}^{N}$.

More broadly, this is a problem of unpaired image-to-image translation [10,
3], where the goal is to translate images in a source domain to a target
domain while preserving the content representations. Since the proposal of
CycleGAN [10], a variety of variants that were built on the cycle-consistent
framework have been proposed [4], tailor-made to solve different problems. The
Augmented CycleGAN (AugCGAN) extended the results of CycleGAN from one-to-one
mappings to many-to-many mappings by adopting latent codes to capture missing
information in the generated images and learning the mappings in the augmented
space[1]. Thus, the AugCGAN is capable of generating diverse images when
different latent codes are sampled. The Unsupervised Image-to-Image
Translation Networks (UNIT) is built on the assumption that any pair of images
in two domains that are intertranslatable has a shared latent code in a shared
latent space. The training process of UNIT is performed in the latent space
after encoding the raw inputs.

Under the unpaired I2I setting, a major drawback of the vanilla CycleGAN [10]
is that the mappings between two domains are learned as a whole for each image
pair. That is, each image is regarded as one distribution and the generators
are optimized to minimize the difference between the distributions in the
source domain and those in the target domain. However, such methods may not
yield satisfactory results when the input images consist of semantically
different objects. For example, art portraits are generally comprised of a
human subject and a background. Naturally, distinct objects tend to belong to
different distributions and it is much more appropriate to view one image as a
mixture of different distributions. Based on this assumption, I propose the
Semantic-aware Mask CycleGAN (SMCycleGAN), which applies semantic segmentation
masks on generated samples masking out the background before they are fed to
the discriminators to differentiate. This enforces the discriminators to make
discriminative decisions based on partial information. Consequently, the
generators can be optimized particularly to synthesize more realistic human
subjects rather than attempting to match the background patterns to increase
realism.

My contributions are as follows. (i) I introduce the Semantic-aware Mask
CycleGAN for translating artistic portraits to photo-realistic visualizations.
(ii) I show that my model can learn the mapping from the artistic domain to
the realistic domain for portraits and generate portraits with high realism.
(iii) I show that the SMCycleGAN achieves better qualitative and quantitative
results than several baseline models.

##  2 Related Work

Cycle Consistency. One of the revolutionary concept proposed by Zhu et al.
[10] was the cycle-consistency loss. Suppose we want to transform images
between domain $X$ and domain $Y$, and we have a translator $G:X\rightarrow Y$
that receives images from domain $X$ and output their corresponding
representation in domain $Y$ and another translator $F:Y\rightarrow X$ that
does the opposite. Consider the translated result $G(x)$ of an image $x\in X$
that is transferred to domain $Y$ through generator $G$. Once we obtain
$G(x)$, an intuitive process is that if we send $G(x)\in Y$ through
$F:Y\rightarrow X$, we should be able to recover the original image $x$. The
cycle-consistency loss depicts the distance between the recovered image
$F(G(x))$ and the original image $x$. Adding the cycle-consistency loss to the
training process encourages the generation of images to be cycle consistent.
While CycleGAN has been shown to yield sensible results on a variety of tasks,
such as interconverting zebras and horses, summer landscapes and winter
landscapes, and Monet paintings and natural images, it remains a generic
method for I2I problems as it learns the mappings between domains regarding
each image as a single distribution instead of mixtures of distributions.
Additionally, CycleGAN fails to utilize any structural or semantic information
embedded in the data, but merely regulates the networks’ behaviors through the
delicately-defined objective functions.

Patch-retrieving Mechanism. The mechanism of retrieving patches of natural
images has been widely adopted for image synthesis tasks [6, 9]. Intuitively,
including patches from real images can significantly increase the realism of
the synthesized images. Several works extended this technique for I2I tasks
[7, 8]. Tomei et al. [7] built their model on vanilla CycleGAN [10] by adding
the patch-retrieval mechanism for transforming Monet paintings to photo-
realistic images. An external memory bank was established, storing patches of
real images. During the training process, each generated image is split into
patches and for each generated patch, one real patch which is the most similar
is retrieved from the memory bank and paired with the generated patch. The
objective included the maximization of the similarity between each generated
patch and its closest real patch in the external memory bank, thus pushing the
generators to produce more realistic images.

Exploiting Semantic Information. Although CycleGAN[10] is proven to be
effective on generating sensible translation results with consistent content
representations as the cycle-consistency loss greatly reduced the space of
total possible mapping functions, the reduction is still inadequate in certain
problem settings so that the training process is highly stochastic and the
performance could not be guaranteed. In other words, there is much problem-
specific information that are failed to be exploited.

To remedy the information waste, Tomei et al. [8] proposed Art2Real, which is
a semantically-aware architecture for translating artworks to photo-realistic
images. During the training process of Art2Real, semantic segmentation is
first performed on the generated samples to obtain segmentation maps. For
instance, a landscape painting will typically be segmented into different
components such as sky, grass, tree, etc. After segmentation, patches are
produced from the segmentation maps in the same way as in [7], but with
predicted class labels. Patches that belong to the same class will again be
paired up with a real patch retrieved from an external memory bank. Different
from the only memory bank where patches of all types of instances are stored
in [7], the revised method adopted a class of memory banks, storing patches of
different semantics separately. For example, a patch sampled from a
segmentation map of sky will search for a patch in the memory bank of sky
patches and find the one with the highest similarity to pair up with. Once all
generated patches are paired up with real ones, the model is optimized to
further increase the similarity between each pair, thus achieving the goal of
synthesizing realistic images of the same content representations.

##  3 Methods

Since the goal of this project is to transfer artistic portraits to photo-
realistic visualizations, my proposed approach is built upon the unpaired
image-to-image translation setting and adopts the same training logic as that
of the CycleGAN [10]. However, the problem of this project differs from most
image-to-image translation problems in that each portraits is comprised of
instances that have distinct semantics, i.e. the background and the human
subject. Whereas, in [7], they studied Monet’s landscape paintings, of which
the composition is much more consistent. Naturally, different semantic
instances tend to have different distributions. Hence, it is only appropriate
to regard each image as a mixture of distributions rather than a single
distribution as a whole. Consequently, it is a must that the component
distributions be learned and mapped separately. Fig. 2 illustrates the
scenarios of correct mappings and mismatched mappings. In cases of mismatched
mappings, even though the mixture of the two distributions are able to match
up, the underlying mapping pattern is erroneous. Thus, my model introduces
segmentation masks and presents the discriminators generated samples with the
backgrounds ruled out, enforcing them to make discriminative decisions only
judging from the human subjects. This mechanism discourages the generators
from a cheating behavior that they try to increase the similarity of the
backgrounds to deceive the discriminators.

Figure 2: Different mapping patterns between two domains. The triangles
represent background pixels and human subjects are represented by squares.
Those in the artistic domain are colored blue and those in the realistic
domain are colored green. The correct mapping pattern is shown on the left and
the mismatched mapping pattern is shown on the right.

###  3.1 Vanilla Adversarial Loss

As the core component of losses of GAN models, the adversarial loss in my
model is applied to two directions of translation. Define $G:X\rightarrow Y$
to be the mapping from artistic portraits to realistic ones, and
$F:Y\rightarrow X$ the opposite. Define $D_{Y}$ as the discriminator to spot
generated realistic portraits and $D_{X}$ the opposite. The adversarial loss
can be expressed as:

$\begin{split}\mathcal{L}_{GAN}(G,D_{Y},X,Y)=&\mathop{\mathbb{E}}_{y\in
Y}[\log{D_{Y}(y)}]\\\ +&\mathop{\mathbb{E}}_{x\in X}[\log{(1-D_{Y}(G(x))}]\\\
\mathcal{L}_{GAN}(F,D_{X},Y,X)=&\mathop{\mathbb{E}}_{x\in
X}[\log{D_{X}(x)}]\\\ +&\mathop{\mathbb{E}}_{y\in
Y}[\log{(1-D_{X}(F(y))}].\end{split}$ (2)

However, the vanilla adversarial loss does not tend to work well when there
are semantically distinct components in the image. To this end, I introduce
the adversarial loss over segmented regions in Section 3.2.

###  3.2 Adversarial Loss over Segmentation Maps

As opposed to Monet’s landscape paintings studied by [10] and [7], the
artistic theme that this project aims to study is portraits, which each has a
clear-cut subject. However, the vanilla adversarial loss does not distinguish
between the subject and the background. Instead, it judges the authenticity of
the images by a general look across the image. The drawback is that the
features in the background of one domain tend to be transferred onto the
subject or the background in the other domain, while neither is desired. As
shown in Figure 3(a) and 3(b), several generated images from the art portraits
have blue and red stripes in the background, as well as on the subjects’
clothes, which are manifest signs of the flag of the United States in the
background of the realistic photos.

(a)

(b)

Figure 3: Undesirable experiment results where patterns of the flag of United
States are transferred onto the generated realistic photos from the art
portraits. The first row shows the original painting. The second row shows the
photo-realistic visualizations of the art portraits. The third row shows the
realistic portraits. The fourth row shows the artistic visualizations of the
realistic portraits.

My proposed solution is to apply different degrees of adversarial loss on
different semantically-segmented regions. Each portrait image, whether
artistic or realistic, is first segmented to produce a mask that softly rules
out the background by assigning low scores to background pixels and high
scores to human body pixels. Denote $M_{x}$ to be the mask matrix for image
$x$ which has the same shape as the input image. A masked image is given by:

$x_{masked}=M_{x}\odot x,$ (3)

where $\odot$ is the element-wise multiplication between M and each color
channel of $x$. The adversarial loss is then applied on the masked image:

$\begin{split}\mathcal{L}_{mGAN}(G,D_{Y},X,Y)=&\mathop{\mathbb{E}}_{y\in
Y}[\log{D_{Y}(M_{y}\odot y)}]\\\ +&\mathop{\mathbb{E}}_{x\in
X}[\log{(1-D_{Y}(M_{G(x)}\odot G(x))}]\\\
\mathcal{L}_{mGAN}(F,D_{X},Y,X)=&\mathop{\mathbb{E}}_{x\in
X}[\log{D_{X}(M_{x}\odot x)}]\\\ +&\mathop{\mathbb{E}}_{y\in
Y}[\log{(1-D_{X}(M_{F(y)}\odot F(y))}].\end{split}$ (4)

Figure 4: After translating an artistic portrait $x\in X$ to a realistic photo
$G(x)\in Y$, the discriminator $D_{Y}$ puts its focus on the subject region of
the generated image, while the background region induces a small or zero
portion of the adversarial loss.

Fig. 4 demonstrates the process of computing the masked adversarial loss for
one artistic portrait $x\in X$. The input first goes through the generator
$G$, before the generated realistic photo is fed into a U-Net [5], which
segments the human body from the background and produces the aforementioned
mask matrix $M_{x}$. $M_{x}$ is a matrix of the same shape as the input image
$x$ and has value $1$ in the purple regions and $0$ in the yellow region in
Fig. 4, where the purple region is called the adversarial loss activation mask
and the yellow region is called the silent region. The discriminator $D_{Y}$
then examines the generated realistic photo with its focus on the purple
active region. Intuitively, this forces the discriminator to only look at the
human body part to decide whether this sample is real or fake, which restrains
the generator’s behavior of learning features in the background to deceive the
discriminator.

###  3.3 Cycle Consistency Loss

Like many existing models tackling the unpaired image-to-image translation
problem, my model also adopts the cycle consistency loss, expressed as:

$\begin{split}\mathcal{L}_{cyc}(G,F)&=\mathop{\mathbb{E}}_{x\in
X}[\begin{Vmatrix}F(G(x)-x)\end{Vmatrix}]\\\ &+\mathop{\mathbb{E}}_{y\in
Y}[\begin{Vmatrix}G(F(y)-y)\end{Vmatrix}].\end{split}$ (5)

The intuition comes from the assumption that an image after translation is
expected to be restored by applying the reverse mapping to the translated
image. The sequence of translations can be expressed as:

$x\in X,x\rightarrow G(x)\rightarrow F(G(x)).$ (6)

In theory, the reconstructed $F(G(x))$ should be identical to the original
$x$. Hence, the difference between the two induces the cycle consistency loss.

###  3.4 Identity Loss

Another technique introduced by Zhu et al. [10] was the identity loss. The
idea is that when the input to a generator is from the target domain to begin
with, the generator is encouraged to be an identity mapping that produces an
identical result to the input. This regularizer has been shown particularly
helpful in photo generation from paintings. Given the context of this project,
it is suitable and critical to include the identity loss in the full
objective.

###  3.5 Full Objective

The full objective of my model is:

$\begin{split}\mathcal{L}(G,F,D_{X},D_{Y})&=\mathcal{L}_{mGAN}(G,D_{Y},X,Y)\\\
&+\mathcal{L}_{mGAN}(F,D_{X},Y,X)\\\ &+\lambda\mathcal{L}_{cyc}(G,F)\\\
&+0.5\lambda\mathcal{L}_{identity}(G,F),\end{split}$ (7)

where $\lambda$ is the importance of different components of the full
objective.

##  4 Experiments

Datasets. Since this project involves training multiple models, multiple
datasets were used to fulfill different tasks. To train the semantic
segmentation model that produces segmentation masks, a subset of the Matting
Human Dataset on Kaggle is utilized. The dataset comes with two folders where
one stores human photographs and the other stores the segmentation masks that
mask out the backgrounds. All images are rescaled to $256\times 256$ pixels
during both training and testing. The size of the training set is 3500. As for
the artworks, they are originally retrieved from Wikiart.org and the Art
Portraits Dataset on Kaggle extracted the portrait category to form a new
dataset. Overly abstract portraits are pruned by hand as it is unlikely to
generate realistic photos from them. During the early stage of development,
the US Senators Dataset on Kaggle was adopted. It is comprised of 100
portraits of the U.S. senators. Since its size is too small and the content is
too domain-specific, it was deprecated and I instead used the human portrait
data in the Matting Human Dataset as data in the realistic domain when
training the SMCycleGAN model.

Training details. The segmentation model I adopted was U-Net proposed by
Ronneberger et al. primarily for biomedical image segmentation [5]. It was
trained for 15 epochs using the Adagrad optimizer.

The SMCycleGAN model was developed on top of the CycleGAN source code. It
inherited most of the training details in [10]. The $\lambda$ that balances
different components of loss functions was set to 10. Additionally, for a more
stable training process, an image pool is built that stores a number of
generated samples in previous iterations and is dynamically updated through
training. However, due to RAM limitation, the size of the pool is restricted
to 3, as opposed to 50 in [10]. During the early epochs of training, the
generated results are far from sensible and the method introduced in 3.2 tend
not to work since the segmentation masks are poorly determined. To remedy
this, a mask-out probability is implemented in a similar practice like the
dropout probability. The probability to use the segmentation mask starts off
to be small numbers like $0.1$, and is increased gradually as training
proceeds and more realistic images are generated. The SMCycleGAN model is
trained for 100 epochs.

Baseline models. To demonstrate that my method indeed brings better
performance, I include two baseline models as competitors. The first is the
vanilla CycleGAN model [10]. It is trained using the same set of data and
training configurations for 100 epochs. Another baseline model is Art2Real
[8], which is an I2I model specialized for transforming artworks to the
realistic domain. The Art2Real checkpoints are downloaded from the authors’
GitHub repository and used directly for testing without further training.

###  4.1 Qualitative Results

As seen in Fig. 5, SMCycleGAN is capable of generating portraits with high
realism by adjusting skin color, smoothing texture, etc. Although the baseline
models are also able to generate sensible outputs, they look less realistic
due to either vagueness or anomalous color composition. In particular, the
vanilla CycleGAN outputs are similar to those of my model. However, as can be
observed in Fig. 5, in some cases there are evident noises either in the
background or on the subject’s clothes. This is presumably caused by the
mismatch of mappings mentioned in Section 3. Thus, qualitative results
indicate that my method improves the vanilla CycleGAN [10] by a certain degree
in that in can produce cleaner and more realistic outputs. More examples are
shown in Fig. 6.

Figure 5: Qualitative results of SMCycleGAN and baseline models. My method can
reduce the degree of artifacts while preserving facial details of the human
subjects as opposed to CycleGAN [10] and Art2Real [8]. Figure 6: More
relatively successful test examples of SMCycleGAN.

The Art2Real [8] framework relies on external image banks that store real
human faces and the training aims to maximize the similarity of the generated
patches and real patches. This may lead to a situation where some facial
features get replaced by or shifted towards those in the image banks. For
example, a subject may be assigned a new nose that belongs to someone else in
the image bank. In those cases, even though the transformed images have higher
realism, the faces would look different from what the subjects really looked
like centuries ago, which contradicts with the original intention of this
project.

###  4.2 Quantitative Evaluation

As GANs does not have an orthodox evaluation metric, it is generally much more
difficult to assess their performance as opposed to models that tackle
conventional vision problems such as classification, segmentation, etc.
However, the Fréchet Inception Distance introduced by Heusel et al. has been
universally adopted for assessment of the quality of the generated images [2].
It measures the distance between two multidimensional Gaussian distributions.
Under the setting of this project, the FID between the generated images and
realistic portraits is computed. The lower FID, the more similar the two sets
are. In other words, the more realistic the generated images look.

Method | Fréchet Inception Distance
---|---
Original paintings | 0.6482
Vanilla CycleGAN [10] | 0.5880
SMCycleGAN | 0.4912
Table 1: Comparison in terms of the Fréchet Inception Distance [2] of
SMCycleGAN and the vanilla CycleGAN [10].

As can be seen in Table 1, the SMCycleGAN model achieved the lowest Fréchet
Inception Distance, meaning that it generates the most similar results to
realistic portrait photos. The numerical improvement indicates that my method
works robustly for translating artistic portraits to realistic photos.

##  5 Conclusion

This project have presented SMCycleGAN, a novel model for translating artistic
portraits to photo-realistic visualizations. The motivation of this project
comes from the wish to restore the ground truth appearances of the subjects
portrayed centuries ago, with the artistic effects ruled out. The model was
trained by adopting a method where semantic information is exploited. In
particular, a much higher penalty is applied on the segmented regions of the
objects of our interest, in my case the human subjects, while the other
objects induce a relatively low level of penalty. This gives the generators
more incentive to refine its performance on the human subjects generation, and
less incentive to shift their attention to match the backgrounds to realistic
scenarios. The model has been shown to produce compelling results.
Additionally, SMCycleGAN achieved more realistic results and lower Fréchet
Inception Distances [2] compared to the vanilla CycleGAN [10] and Art2Real
[8], which are state of the art models for I2I tasks from the artistic domain
to the realistic domain.

Figure 7: Failure cases of my model. Left: SMCycleGAN produces poor results
for African subjects due to the small number of black people in the training
set. Right: SMCycleGAN fails for artworks that are too abstract.

Despite the compelling results, my model has several limitations. First, the
model fails to generate realistic results for Africans or Asians. As seen in
Fig. 7, the test example on the top-left is an African, and the transformation
is poorly done as the generated subject has anomalous red skin. This is
presumably due to the imbalanced ethnicity composition in the training set
where a majority are white people. Similar cases are observed for other
ethnicity such as Asians. Other failures happened when the input portrait is
too abstract or contains a large amount of artifacts. As can been seen in Fig.
7, the other 3 test examples are abstract artworks that do not embed adequate
information for the model to recover the subjects’ appearances. Under such
circumstances, the Art2Real model [8] may outperform SMCycleGAN as it
retrieves real patches from external memory banks and this may compensate for
the lack of information in the source images.

Nevertheless, the SMCycleGAN model proposed in this project provides a novel
way of unsupervised image-to-image translation tasks, which exploits semantic
information embedded in the source images and adjust adversarial loss
accordingly. Further work can be done to explore different problems other than
translating art portraits to photo-realistic images.

Acknowledgements: This project was developed primarily based on this Kaggle
notebook that implemented the vanilla CycleGAN [10] and this Kaggle notebook
that implemented U-Net [5] for human body segmentation.

## References

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  * [2] Martin Heusel, Hubert Ramsauer, Thomas Unterthiner, Bernhard Nessler, and Sepp Hochreiter.  Gans trained by a two time-scale update rule converge to a local nash equilibrium.  Advances in neural information processing systems, 30, 2017.
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  * [4] Yingxue Pang, Jianxin Lin, Tao Qin, and Zhibo Chen.  Image-to-image translation: Methods and applications.  IEEE Transactions on Multimedia, 2021.
  * [5] Olaf Ronneberger, Philipp Fischer, and Thomas Brox.  U-net: Convolutional networks for biomedical image segmentation.  In International Conference on Medical image computing and computer-assisted intervention, pages 234–241. Springer, 2015.
  * [6] Fuwen Tan, Song Feng, and Vicente Ordonez.  Text2scene: Generating compositional scenes from textual descriptions.  In Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition, pages 6710–6719, 2019.
  * [7] Matteo Tomei, Lorenzo Baraldi, Marcella Cornia, and Rita Cucchiara.  What was monet seeing while painting? translating artworks to photo-realistic images.  In Proceedings of the European Conference on Computer Vision (ECCV) Workshops, pages 0–0, 2018.
  * [8] Matteo Tomei, Marcella Cornia, Lorenzo Baraldi, and Rita Cucchiara.  Art2real: Unfolding the reality of artworks via semantically-aware image-to-image translation.  In Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern Recognition, pages 5849–5859, 2019.
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# Matrix power means and new characterizations of operator monotone functions

Trung Hoa Dinh, Cong Trinh Le, The Van Nguyen, Bich Khue Vo  Trung Hoa Dinh,
Department of Mathematics, Troy University, Troy, AL 36082, United States
<EMAIL_ADDRESS>The Van Nguyen, Faculty of Mathematics-Mechanics-Informatics,
VNU University of Science, Vietnam National University, Hanoi, Vietnam
<EMAIL_ADDRESS>Cong Trinh Le, Division of Computational Mathematics
and Engineering, Institute for Computational Science, Ton Duc Thang
University, Ho Chi Minh City, Vietnam
Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh
City, Vietnam
<EMAIL_ADDRESS>Bich Khue Vo, Faculty of Economics and Law, University
of Finance and Marketing, Ho Chi Minh City, Vietnam
<EMAIL_ADDRESS>
###### Abstract.

For positive definite matrices $A$ and $B$, the Kubo-Ando matrix power mean is
defined as

$P_{\mu}(p,A,B)=A^{1/2}\left(\frac{1+(A^{-1/2}BA^{-1/2})^{p}}{2}\right)^{1/p}A^{1/2}\quad(p\geq
0).$

In this paper, for $0\leq p\leq 1\leq q$, we show that if one of the following
inequalities

$\displaystyle f(P_{\mu}(p,A,B))\leq f(P_{\mu}(1,A,B))\leq f(P_{\mu}(q,A,B))$

holds for any positive definite matrices $A$ and $B$, then the function $f$ is
operator monotone on $(0,\infty).$ We also study the inverse problem for non-
Kubo-Ando matrix power means with the powers $1/2$ and $2$. As a consequence,
we establish new charaterizations of operator monotone functions with the non-
Kubo-Ando matrix power means.

###### Key words and phrases:

Matrix power mean, matrix geometric mean, operator monotone functions, inverse
problems for means, Kubo-Ando means

###### 2010 Mathematics Subject Classification:

47A63,47A64,47A56,46E05,15B48

##  1\. Introduction

It is well-known that for $0\leq p\leq 1\leq q$ and for non-negative numbers
$a$ and $b$,

$\sqrt{ab}\leq\left(\frac{a^{p}+b^{p}}{2}\right)^{1/p}\leq\frac{a+b}{2}\leq\left(\frac{a^{q}+b^{q}}{2}\right)^{1/q},$

or,

$\sqrt{ab}\leq\mu(p,a,b)\leq\mu(1,a,b)\leq\mu(q,a,b),$

where $\mu(p,a,b)=\left(\frac{a^{p}+b^{p}}{2}\right)^{1/p}.$ Let $f$ be a
continuous increasingly monotone function on $[0,\infty)$. Then

(1) $f(\sqrt{ab})\leq
f\left(\left(\frac{a^{p}+b^{p}}{2}\right)^{1/p}\right)\leq
f\left(\frac{a+b}{2}\right)\leq
f\left(\left(\frac{a^{q}+b^{q}}{2}\right)^{1/q}\right).$

Suppose that one of inequalities in (1) holds for any non-negative numbers $a$
and $b$. Then, the following question is natural: Is it true that $f$ is
increasingly monotone on $[0,\infty)$?

Mention that if the inequality

$f(\sqrt{ab})\leq f\left(\frac{a+b}{2}\right)$

holds for any $a,b\geq 0$, then $f$ is increasingly monotone on $[0,\infty)$.
In [3] the first author considered the following reverse AGM inequality

(2) $\frac{a+b-|a-b|}{2}\leq\sqrt{ab},$

and show that this inequality also characterizes increasingly monotone
function. Some other characterizations were also obtained by the first author
and his co-authors in [6].

Now, let ${\mathbb{M}\,}_{n}$ be the algebra of $n\times n$ matrices over
$\mathbb{C}$ and $\mathcal{P}_{n}$ denote the cone of positive definite
elements in ${\mathbb{M}\,}_{n}$. Denote by $I_{n}$ the identity matrix of
${\mathbb{M}\,}_{n}$. A continuous function $f$ is said to be operator
monotone on $I\subset\mathbb{R}$ if for any Hermitian matrices $A$ and $B$
with spectra in $I,$

$A\leq B\quad\implies\quad f(A)\leq f(B),$

where $f(A)$ is understood by means of the functional calculus. Operator
monotone functions were firstly introduced by Loewner in 1930 [11]. In 1980,
Kubo and Ando [9] introduced the theory of operator means on the set of
$B(H)^{+}\times B(H)^{+}$, where $B(H)^{+}$ is the set of positive invertible
operators in a Hilbert space $H$. The main result in their paper is the one-
to-one correspondence between operator means $\sigma$ and operator monotone
functions $f$ on $(0,\infty)$ defined by

(3) $A\sigma B=A^{1/2}f(A^{-1/2}BA^{-1/2})A^{1/2}.$

In 1996, Petz [12] introduced the theory of monotone metric in quantum
information theory which was based on operator monotone functions. Therefore,
such functions are important matrix analysis, quantum information and other
areas as well. The authors refer readers to the books of William Donoghue [7]
and Barry Simon [14] for more details about operator monotone functions.

One of the most important means is the geometric mean

$A\sharp B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$

which was firstly defined by Pusz and Woronowicz [13]. The matrix AGM
inequality states that for positive definite matrices $A$ and $B,$

(4) $A\sharp B\leq\frac{A+B}{2}.$

In 2014, Hiai and Ando gave a new characterization of operator monotone
functions using (4). They showed that if

$f(A\sharp B)\leq f\left(\frac{A+B}{2}\right)$

for any positive definite matrices $A$ and $B$, then $f$ is operator monotone
on $(0,\infty)$. In 2015, the first author obtained a new characterization
based on the matrix version of (2):

$f\left(\frac{A+B-|A-B|}{2}\right)\leq f(A\sharp B)$

whenever $A$, $B$ are positive definite matrices. In another papers [5, 6] a
series of new characterizations related to the matrix Heron mean and Powers-
Stormer inequality in quantum hypothesis testing theory [1] were established.

Now, let $p$ be a real number, and $a,b$ be positive. According to the
relation (3) the power mean $\mu(p,a,b)=\big{(}\frac{a^{p}+b^{p}}{2})^{1/p}$
is corresponding to the monotone function $f$ as

$f_{\mu,p}(t)=\left(\frac{1+t^{p}}{2}\right)^{1/p}.$

Then for positive definite matrices $A$ and $B$, the Kubo-Ando matrix power
mean is defined as

$P_{\mu}(p,A,B)=A^{1/2}\left(\frac{1+(A^{-1/2}BA^{-1/2})^{p}}{2}\right)^{1/p}A^{1/2}.$

Therefore, the chain of inequalities (1) for matrices looks like

(5) $\displaystyle f(A\sharp B)$ $\displaystyle\leq
f(A^{1/2}f_{\mu,p}(A^{-1/2}BA^{-1/2})A^{1/2})$ $\displaystyle\leq
f(A^{1/2}f_{\mu,1}(A^{-1/2}BA^{-1/2})A^{1/2})$ $\displaystyle\leq
f(A^{1/2}f_{\mu,q}(A^{-1/2}BA^{-1/2})A^{1/2}).$

Motivated by works mentioned above, in this paper we investigate new
characterizations of operator monotone functions by inequalities in (5). We
show that if one of the inequalities in (5) holds for any positive finite
matrices $A$ and $B$, then the function $f$ is operator monotone on
$(0,\infty).$

The more difficult situation is for the naive matrix extension of the power
means. Let $1/2\leq p\leq 1\leq q$. The function $t^{1/q}$ is operator
concave, while the function $t^{1/p}$ is operator convex. Then we have

(6)
$\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p}\leq\frac{A+B}{2}\leq\left(\frac{A^{q}+B^{q}}{2}\right)^{1/q}$

whenever $A$ and $B$ are positive semidefinite. It is worth noting that the
inequalities in (6) were discussed by Audenaert and Hiai and [2], where they
obtained conditions on $p$ and $q$ for such that (6) holds true. In [4] the
authors also studied a new type of operator convex functions.

In Section 3 we study the inverse problem for the non-Kubo-Ando matrix power
means when $q=1/2$ and $p=2$. As a consequence, we establish a new
characterization of operator monotone functions by inequalities in (6).

##  2\. Kubo-Ando matrix power means and characterizations

In this section we study the problem of characterization of operator monotone
functions using Kubo-Ando matrix power means. We start with the scalar cases.

######  Theorem 1.

Let $f$ be a continuous function on $[0,\infty)$. For $0\leq p\leq 1\leq q$,
suppose that one of the following inequalities holds for all any non-negative
numbers $a\leq b$:

(7) $\displaystyle f(\sqrt{ab})\leq
f\left(\left(\dfrac{a^{p}+b^{p}}{2}\right)^{1/p}\right),$ (8) $\displaystyle
f\left(\left(\dfrac{a^{p}+b^{p}}{2}\right)^{1/p}\right)\leq
f\left(\dfrac{a+b}{2}\right),$ (9) $\displaystyle
f\left(\dfrac{a+b}{2}\right)\leq
f\left(\left(\dfrac{a^{q}+b^{q}}{2}\right)^{1/q}\right).$

Then the function $f$ is increasingly monotone on $[0,\infty)$.

###### Proof.

To prove the theorem, we need to show that $f(x)\leq f(y)$ for any $0\leq
x\leq y$. Since all three inequalities in assumption are homogeneous, it
suffices to prove the theorem in the case when either $x=1$ and $y\geq 1$ or
$x\leq 1$ and $y=1$.

Suppose that the first inequality (7) holds, we have to show that for any
$y\geq 1$ there exist $a,b\geq 0$ such that $1=\sqrt{ab}$ and
$y=\left(\dfrac{a^{p}+b^{p}}{2}\right)^{1/p}$. Or, equivalently, there exists
$a>0$ such that

(10) $y=\left(\dfrac{a^{p}+a^{-p}}{2}\right)^{1/p}.$

Note that the function $f(x)=2^{-1/p}(x^{p}+x^{-p})^{1/p}$ is surjective from
$(0,\infty)$ onto $[1,\infty)$. Therefore, for any $y\geq 1$, there exist
$a>0$ such that the identity (10) holds.

Now, suppose that the second inequality (8) holds, we will show that for any
$x\leq 1$, there exist $a,b\geq 0$ such $1=(a+b)/2$ and
$x=\left(\dfrac{a^{p}+b^{p}}{2}\right)^{1/p}.$ Or, equivalently, there exists
$0\leq a\leq 2$ such that

(11) $x=\left(\dfrac{a^{p}+(2-a)^{p}}{2}\right)^{1/p}.$

The function $g(x)=2^{-1/p}(x^{p}+(2-x)^{p})^{1/p}$ is surjective from $[0,2]$
onto $[2^{1-1/p},1]$. Therefore, for any $\gamma\leq x\leq 1$ where
$\gamma=2^{1-1/p}$, there exists $a>0$ such that the identity (11) holds.
Consequently, if $\gamma y\leq x\leq y$, from the homogeneous property of the
second inequality it implies that

$f(x)\leq f(y).$

If $x<\gamma y$, let $\gamma_{0}\in(\gamma,1)$ and consider the sequence
$\\{\gamma_{0}^{n}\\}_{n\in\mathbb{N}}$. Since $\gamma_{0}^{n}\to 0$ as
$n\to\infty$, there exists $k\in\mathbb{N}$ such that

$0<\gamma\left(\gamma_{0}^{n}y\right)<\gamma_{0}^{n+1}y<x\leq\gamma_{0}^{n}y\leq\gamma_{0}^{n-1}y\leq\dots\leq\gamma_{0}y\leq
y.$

Hence, from the previous argument it implies that

$f(x)\leq f\left(\gamma_{0}^{n}y\right)\leq\dots\leq
f\left(\gamma_{0}y\right)\leq f(y).$

Therefore, $f$ is increasingly monotone on $[0,\infty)$ if the second
inequality holds.

For the case when the last inequality (9) holds, we proceed similarly as the
case of the second inequality, in which we work for $x=1,y\geq 1$, and
$\gamma=2^{1-1/q}>1$. ∎

######  Theorem 2.

Let $0<p\leq 1$, and $f$ be a continuous function on $[0,\infty)$ that
satisfies the following inequality

(12) $\displaystyle f\left(A\sharp B\right)$ $\displaystyle\leq
f\left(A^{1/2}f_{\mu,p}\left(A^{-1/2}BA^{-1/2}\right)A^{1/2}\right),$

for any positive definite matrices $A$ and $B$. Then $f$ is operator monotone
on $(0,\infty)$.

###### Proof.

Suppose that the inequality (12) holds, it suffices to show that for any
$0\leq X\leq Y$, there exist two positive semidefinite matrices $A$ and $B$
such that

$\displaystyle X=A\sharp B,\quad
Y=A^{1/2}f_{\mu,p}\left(A^{-1/2}BA^{-1/2}\right)A^{1/2}.$

Firstly, let us consider the case when $X=I_{n}$. We now show that there exist
positive definite matrices $A_{0}$ and $B_{0}$ such that $A_{0}\sharp
B_{0}=I_{n}$ and

(13)
$Y=A_{0}^{1/2}f_{\mu,p}\left(A_{0}^{-1/2}B_{0}A^{-1/2}\right)A_{0}^{1/2}=A_{0}f_{\mu,p}\left(A_{0}^{-2}\right).$

Since the function
$h(x)=xf_{\mu,p}(x^{-2})=\left(\dfrac{x^{p}+x^{-p}}{2}\right)^{1/p}$ is
surjective from $(0,\infty)$ to $[1,\infty)$, we obtain that for any
$I_{n}\leq Y$ there exist a matrix $A_{0}>0$ satisfying (13). The matrix
$B_{0}$ is equal to $A_{0}^{-1}$.

In general, for $0<X\leq Y$ we have $I_{n}\leq X^{-1/2}YX^{-1/2}$. By the
above arguments, we can find positive semidefinite matrices $A_{0}$ and
$B_{0}$ such that $A_{0}\sharp B_{0}=I_{n}$ and

$X^{-1/2}YX^{-1/2}=A_{0}^{1/2}f_{\mu,p}\left(A_{0}^{-1/2}B_{0}A^{-1/2}\right)A_{0}^{1/2}=P_{\mu}\left(p,A_{0},B_{0}\right).$

Consequently, applying (12) to matrices $A=X^{1/2}A_{0}X^{1/2}$ and
$B=X^{1/2}B_{0}X^{1/2}$, we obtain that $f(X)\leq f(Y)$. In other words, $f$
is operator monotone. ∎

######  Theorem 3.

Let $0<p\leq 1\leq q$, and $f$ be a continuous function on $[0,\infty)$ that
satisfies one of the following inequalities

(14) $\displaystyle f\left(\dfrac{A+B}{2}\right)\leq
f\left(A^{1/2}f_{\mu,q}\left(A^{-1/2}BA^{-1/2}\right)A^{1/2}\right),$ (15)
$\displaystyle
f\left(A^{1/2}f_{\mu,p}\left(A^{-1/2}BA^{-1/2}\right)A^{1/2}\right)\leq
f\left(\dfrac{A+B}{2}\right),$

for any positive definite matrices $A$ and $B$. Then $f$ is operator monotone
on $(0,\infty)$.

To prove Theorem 3, we will need the following lemma.

######  Lemma 4.

Suppose $X$ and $Y$ are positive definite matrices satisfying $X\leq Y<\gamma
X$ (resp. $\gamma X<Y\leq X$) where $\gamma=2^{1-1/q}$ (resp.
$\gamma=2^{1-1/p}$), then there exist positive matrices $A$ and $B$ such that

$X=\frac{A+B}{2}\quad\text{ and }\quad
Y=P_{\mu}\left(q,A,B\right)\,(resp.\,\,Y=P_{\mu}\left(p,A,B\right)),$

where $0<p\leq 1\leq q$.

###### Proof.

We show the lemma when $X\leq Y<\gamma X$, the remaining case can be obtained
similarly. Firstly, let us consider the case when $X=I_{n}$, it suffices to
show that given $I_{n}\leq
Y=U\mathrm{diag}\left(\left\\{\lambda_{i}\left(Y\right)\right\\}\right)U^{*}\leq\gamma
I_{n},$ we can find $A_{0},B_{0}\geq 0$ such that

$I_{n}=\frac{A_{0}+B_{0}}{2}\quad\text{ and }\quad Y=P_{\mu}(q,A_{0},B_{0}).$

Or, equivalently, there exists $0<A_{0}\leq 2I_{n}$ such that
$Y=P_{\mu}\left(q,A_{0},2-A_{0}\right)=\varphi(A_{0}),$ where

$\varphi(x)=x^{1/2}f_{\mu,q}\left(x^{-1/2}(2-x)x^{-1/2}\right)x^{-1/2}=\left(\dfrac{x^{q}+(2-x)^{q}}{2}\right)^{1/q}.$

Note that $\varphi$ is continuous on $[0,2]$ and surjective from $[0,2]$ onto
$[1,\gamma]$. Since $\lambda_{i}(Y)\in[1,\gamma],$ for each $i$, we can choose
$\delta_{i}(Y)\in[0,2]$ such that
$\varphi\left(\delta_{i}(Y)\right)=\lambda_{i}(Y).$ The matrix

$A_{0}:=U\mathrm{diag}\left(\left\\{\delta_{i}(Y)\right\\}\right)U^{*}$

satisfies

$\varphi(A_{0})=U\mathrm{diag}\left(\left\\{\varphi\left(\delta_{i}(Y)\right)\right\\}\right)U^{*}=U\mathrm{diag}\left(\\{\lambda_{i}(Y)\\}\right)U^{*}.$

In general, for $0<X\leq Y<\gamma X$, we have $I_{n}\leq
X^{-1/2}YX^{-1/2}<\gamma I_{n}$. By the above arguments, we can find positive
definite matrices $A_{0}$ and $B_{0}$ such that $(A_{0}+B_{0})/2=I_{n}$ and
$X^{-1/2}YX^{-1/2}=P_{\mu}(q,A_{0},B_{0})$. Now, let $A=X^{1/2}A_{0}X^{1/2}$
and $B=X^{1/2}B_{0}X^{1/2}$, we have $(A+B)/2=X$ and

$Y=X^{1/2}P_{\mu}(q,A_{0},B_{0})X^{1/2}=P_{\mu}(q,A,B),$

which completes the proof of Lemma 4. ∎

We are now ready to prove Theorem 3.

###### Proof of Theorem 3.

First we prove the case when the inequality (14). Let $0\leq X\leq Y$ and
$Y_{0}=X^{-1/2}YX^{-1/2}$, and choose $\gamma_{0}\in(1,2^{1-1/q})$. Consider
the spectral decomposition, $Y_{0}=\sum_{i=1}^{r}\lambda_{i}E_{i}$ with the
eigenvalues $\lambda_{i}$ listed in the decreasing order. Then, there exists a
set of non-ascending integers $\\{m_{i}\,|\,1\leq i\leq r\\}$ such that

$\gamma_{0}^{m_{i}}<\lambda_{i}\leq\gamma_{0}^{m_{i}+1}.$

Let $\ell_{1}<\ell_{2}<\dots<\ell_{t}=r$ be the sequence of indexes such that

$\displaystyle
m_{1}=\dots=m_{\ell_{1}}>m_{\ell_{1}+1}=\dots=m_{\ell_{2}}>m_{\ell_{2}+1}=\dots=m_{\ell_{3}}>\dots>m_{\ell_{t-1}+1}=\dots=m_{\ell_{t}}=m_{r}.$

We have

$\gamma_{0}^{m_{\ell_{t}}}<\lambda_{r}<\dots<\lambda_{\ell_{t-1}+1}\leq\gamma_{0}^{m_{\ell_{t}}+1}\leq\gamma_{0}^{\ell_{t-1}}<\lambda_{\ell_{t-1}}<\dots<\lambda_{\ell_{t-2}+1}\leq\gamma_{0}^{\ell_{t-1}+1}<\dots<\lambda_{1}\leq\gamma_{0}^{\ell_{1}+1}.$

It follows that

$\displaystyle
I<\gamma_{0}I<\gamma_{0}^{2}I<\dots<\gamma_{0}^{m_{\ell_{t}}}I=\gamma_{0}^{m_{\ell_{t}}}\left(E_{1}+E_{2}+\dots+E_{r}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{t-1}+1}E_{\ell_{t-1}+1}+\gamma_{0}^{m_{\ell_{t}}}\left(E_{\ell_{t-1}}+E_{\ell_{t-1}-1}+\dots+E_{1}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{t-1}+1}E_{\ell_{t-1}+1}+\gamma_{0}^{m_{\ell_{t}}+1}\left(E_{\ell_{t-1}}+E_{\ell_{t-1}-1}+\dots+E_{1}\right)$
$\displaystyle\dots$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{t-1}+1}E_{\ell_{t-1}+1}+\gamma_{0}^{m_{\ell_{t-1}}}\left(E_{\ell_{t-1}}+E_{\ell_{t-1}-1}+\dots+E_{1}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{t-2}+1}E_{\ell_{t-2}+1}+\gamma_{0}^{m_{\ell_{t-1}}}\left(E_{\ell_{t-2}}+E_{\ell_{t-2}-1}+\dots+E_{1}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{t-2}+1}E_{\ell_{t-2}+1}+\gamma_{0}^{m_{\ell_{t-1}}+1}\left(E_{\ell_{t-2}}+E_{\ell_{t-2}-1}+\dots+E_{1}\right)$
$\displaystyle\dots$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{2}+1}E_{\ell_{2}+1}+\gamma_{0}^{m_{\ell_{2}}}\left(E_{\ell_{1}}+E_{\ell_{1}-1}+\dots+E_{1}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{2}+1}E_{\ell_{2}+1}+\gamma_{0}^{m_{\ell_{2}}+1}\left(E_{\ell_{1}}+E_{\ell_{1}-1}+\dots+E_{1}\right)$
$\displaystyle\dots$
$\displaystyle\leq\lambda_{r}E_{r}+\dots+\lambda_{\ell_{2}+1}E_{\ell_{2}+1}+\gamma_{0}^{m_{\ell_{1}}}\left(E_{\ell_{1}}+E_{\ell_{1}-1}+\dots+E_{1}\right)$
$\displaystyle\leq\lambda_{r}E_{r}+\dots\lambda_{1}E_{1}=Y_{0}\leq\gamma_{0}^{m_{\ell_{1}}+1}I.$

After multiplying each term of the chain of inequalities on both sides by
$X^{1/2}$, let $Z_{k}$ be the $k$-th expression of the chain, we obtain the
following chain inequalities

$0\leq X=Z_{1}\leq Z_{2}\leq\dots\leq Z_{m-1}\leq Y\leq
Z_{m}=\gamma_{0}^{m_{\ell_{1}}+1}X.$

where $m$ is a positive integer. The previous calculation gives us $Z_{k}\leq
Z_{k+1}\leq\gamma Z_{k}$. Therefore, it follows from Lemma 4 that there exist
positive definite matrices $A$ and $B$ such that

$Z_{k}=(A+B)/2\quad\text{and}\quad Z_{k+1}=P_{\mu}(q,A,B).$

Consequently,

$f(X)\leq f(Z_{1})\leq f(Z_{2})\leq\dots\leq f(Z_{m-1})\leq f(Y).$

In other words, $f$ is operator monotone.

The proof in the case when (15) holds is similar. ∎

##  3\. The inverse problem for non-Kubo-Ando matrix power means

In [2] Audenaert and Hiai determined values of $p$ and $p$ such that the
following inequality holds true

$\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p}\leq\left(\frac{A^{q}+B^{q}}{2}\right)^{1/q}$

whenever $A$, $B$ are positive semidefinite matrices. When $1/2\leq p\leq
1\leq q$, according to the operator convexity of $t^{1/p}$ and operator
concavity of $t^{1/q}$ we have

(16)
$\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p}\leq\frac{A+B}{2}\leq\left(\frac{A^{q}+B^{q}}{2}\right)^{1/q}$

Recently, Lam and Le [10] studied the quantum divergence generated by these
inequalities:

$\Phi(A,B)={\rm
Tr\,}\left(\frac{A+B}{2}-\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p}\right).$

In this section we are going to solve the inverse problem for for
$\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p}$ and
$\left(\frac{A^{q}+B^{q}}{2}\right)^{1/q}$. Namely, suppose that $0\leq X\leq
Y$. Solving the inverse mean problem is to find positive definite matrices $A$
and $B$ such that

(17) $X=\left(\frac{A^{p}+B^{p}}{2}\right)^{1/p},\quad
Y=\left(\frac{A^{q}+B^{q}}{2}\right)^{1/q}.$

If this system has a positive solution, then we may use the result to
characterize operator monotone function.

Unfortunately, inequalities in (16) do not characterize operator monotone
functions, in general.

######  Proposition 5.

For any $q>1,$ there exists a non-monotone operator satisfying

$f\left(\dfrac{A+B}{2}\right)\leq
f\left(\left(\dfrac{A^{q}+B^{q}}{2}\right)^{1/q}\right)$

for all positive definite matrices $A$ and $B$.

###### Proof.

For a fixed number $1<r\leq\min\\{2,q\\}$, we consider the operator
$f(x)=x^{r}$ that is a non-monotone operator. It follows from the operator
convexity of $x^{r}$ and the operator concavity of $x^{r/q}$ that

$\left(\dfrac{A+B}{2}\right)^{r}\leq\dfrac{A^{r}+B^{r}}{2}\leq\left(\dfrac{A^{q}+B^{q}}{2}\right)^{r/q}\leq\left[\left(\dfrac{A^{q}+B^{q}}{2}\right)^{1/q}\right]^{r}.$

Hence, the operator monotone function $f$ satisfies

$f\left(\dfrac{A+B}{2}\right)\leq
f\left(\left(\dfrac{A^{q}+B^{q}}{2}\right)^{1/q}\right),$

which completes the proof of Lemma 5. ∎

Similarly, we hope to have the same conclusion for the first inequality,
namely, for $1/2\leq p\leq 1$, there exists a non-monotone operator satisfying

$f\left(\left(\dfrac{A^{p}+B^{p}}{2}\right)^{1/p}\right)\leq
f\left(\dfrac{A+B}{2}\right).$

However, when $p=1/2$ we could be able to solve the inverse problem and
establish a new characterization of operator monotone functions.

######  Theorem 6.

Let $f$ be a continuous function on $[0,\infty)$ that satisfies the following
inequality

$f\left(\left(\dfrac{A^{1/2}+B^{1/2}}{2}\right)^{2}\right)\leq
f\left(\dfrac{A+B}{2}\right),$

for any positive semidefinite matrices $A$ and $B$. Then $f$ is operator
monotone.

###### Proof.

Firstly, we show that $f(X)\leq f(Y)$ for any positive semidefinite matrices
$X,Y$ with $0\leq X\leq Y\leq 2X$. Indeed, we need to solve the following
system

(18) $\begin{cases}\left(\dfrac{A^{1/2}+B^{1/2}}{2}\right)^{2}=X\\\
\dfrac{A+B}{2}=Y.\end{cases}$

Subtracting the first equation from the second, we obtain

$Y-X=\left(\dfrac{A^{1/2}-B^{1/2}}{2}\right)^{2}.$

Therefore, system (18) is equivalent to

$\begin{cases}\dfrac{A^{1/2}+B^{1/2}}{2}=X^{1/2}\\\
\dfrac{A^{1/2}-B^{1/2}}{2}=(Y-X)^{1/2}.\end{cases}$

The last system has a unique positive solution as

$\displaystyle A=\left(X^{1/2}+(Y-X)^{1/2}\right)^{2},\quad
B=\left(X^{1/2}-(Y-X)^{1/2}\right)^{2}.$

Mention that the condition $0\leq X\leq Y<2X$ guarantees the semidefinite
positivity of $A$ and $B$. Thus, $f(X)\leq f(Y)$.

Now, for any positive semidefinite matrices $0\leq X\leq Y$, we identically
apply the arguments in the proof of Theorem 3 to obtain a positive integer $m$
and positive semidefinite matrices $Z_{1},\dots,Z_{m}$ such that

$0\leq X=Z_{1}\leq Z_{2}\leq Z_{3}\leq\dots\leq Z_{m-1}\leq Y\leq Z_{m}$

with $Z_{k}\leq Z_{k+1}<2Z_{k}$ for all $k=1,2,\dots,m$. Therefore, combining
with the previous arguments, we get

$f(X)=f(Z_{1})\leq f(Z_{2})\leq\dots\leq f\left(Z_{m-1}\right)\leq f(Y).$

∎

From the proof of Theorem 6, under a special condition on $X$ and $Y$, the
inverse problem (17) has a positive solution.

######  Corollary 7.

The following inverse problem has a positive semidefinite solution for
$X^{2}\leq Y^{2}\leq\sqrt{2}X^{2},$

(19) $\frac{A+B}{2}=X,\quad\left(\dfrac{A^{2}+B^{2}}{2}\right)^{1/2}=Y.$

###### Proof.

If we put $A_{0}=A^{2},B_{0}=B^{2}$, then the system (19) is equivalent to the
following

$\frac{A_{0}^{1/2}+B_{0}^{1/2}}{2}=X,\quad\left(\dfrac{A_{0}+B_{0}}{2}\right)^{1/2}=Y,$

or,

(20)
$\left(\frac{A_{0}^{1/2}+B_{0}^{1/2}}{2}\right)^{2}=X^{2},\quad\dfrac{A_{0}+B_{0}}{2}=Y^{2}.$

According to the proof of Theorem 6, the last system has positive semidefinite
solutions when $X^{2}\leq Y^{2}\leq 2X^{2}$. ∎

The inverse problem for general matrix power means is still unsolved.

Open question: Let $1/2\leq p\leq 1\leq q$. Does the following system have
positive solutions

$\begin{cases}\left(\dfrac{A^{p}+B^{p}}{2}\right)^{1/p}=X\\\
\left(\dfrac{A^{q}+B^{q}}{2}\right)^{1/q}=Y\end{cases}$

whenever $0\leq X\leq Y$? This case is just a special case of a general result
obtained by Hiai and Audenaert in [2]. If the answer is positive, then we
immediately have a new characterization: a continuous function $f$ is operator
monotone function on $(0,\infty)$ if and only if for any positive definite
matrices $A$ and $B$ the following inequality

$f\left(\left(\dfrac{A^{p}+B^{p}}{2}\right)^{1/p}\right)\leq
f\left(\left(\dfrac{A^{q}+B^{q}}{2}\right)^{1/q}\right)$

holds true.

### Acknowledgement

We are very grateful to Prof. Mikael de la Salle for useful discussions. The
Van Nguyen was funded by Vingroup Joint Stock Company and supported by the
Domestic Master Scholarship Programme of Vingroup Innovation Foundation
(VINIF), Vingroup Big Data Institute (VINBIGDATA), code
VINIF.2020.ThS.KHTN.05.

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# Cyclic Quantum Teleportation of Two-Qubit Entangled States by using Six-
Qubit Cluster State and Six-Qubit Entangled State

A. Slaoui  0000-0002-5284-3240<EMAIL_ADDRESS>LPHE-Modeling and
Simulation, Faculty of Sciences, Mohammed V University in Rabat, Rabat,
Morocco.  Centre of Physics and Mathematics, CPM, Faculty of Sciences,
Mohammed V University in Rabat, Rabat, Morocco.  M. El Kirdi  LPHE-Modeling
and Simulation, Faculty of Sciences, Mohammed V University in Rabat, Rabat,
Morocco.  R. Ahl Laamara  LPHE-Modeling and Simulation, Faculty of Sciences,
Mohammed V University in Rabat, Rabat, Morocco.  Centre of Physics and
Mathematics, CPM, Faculty of Sciences, Mohammed V University in Rabat, Rabat,
Morocco.  M. Alabdulhafith  Department of Information Technology, College of
Computer and Information Sciences, Princess Nourah bint Abdulrahman
University, P.O. Box 84428, Riyadh 11671, Saudi Arabia.  S. A. Chelloug
Department of Information Technology, College of Computer and Information
Sciences, Princess Nourah bint Abdulrahman University, P.O. Box 84428, Riyadh
11671, Saudi Arabia.  A. A. Abd El-Latif<EMAIL_ADDRESS>EIAS Data
Science Lab, College of Computer and Information Sciences, and Center of
Excellence in Quantum and Intelligent Computing, Prince Sultan University,
Riyadh 11586, Saudi Arabia.  Department of Mathematics and Computer Science,
Faculty of Science, Menoufia University, Shebin El-Koom 32511, Egypt.

###### Abstract

Cyclic quantum teleportation schemes requires at least the existence of three
collaborators acting all as senders and receivers of quantum information, each
one of them has an information to be transmitted to the next neighbour in a
circular manner. Here, new cyclic quantum teleportation scheme is proposed for
perfectly transmitting cyclically three arbitrary unknown two-qubit states
($\alpha$, $\beta$ and $\gamma$) among the three collaborators. In this
scheme, Alice can send to Bob the quantum information contained in her two-
qubit state $\alpha$ and receive from Charlie the quantum information
contained in the two-qubit state in his possession $\gamma$ and similarly, Bob
can transmit to Charlie the quantum information contained in his two-qubit
state $\beta$ through a quantum channel of twelve-qubit state consisting of a
six-qubit cluster state and a six-qubit entangled state by sequentially and
cyclically performing Bell state measurements. Subsequently, each one of the
three participants can afterwards retrieve his own desired two-qubit state
using classical channel and by performing appropriate unitary Pauli operators
and we have shown that our proposed scheme performs efficiently.

Keywords: Cyclic quantum teleportation, Six-qubit cluster state, Six-qubit
entangled state, Bell states measurement.

##  I  Introduction

In the field of secure transmission of confidential information, the quantum
teleportation (QT) is considered as protocol that allows the transfer of
unknown quantum information, not only indecipherable but also imperceptible
between two remote entities without loss of value. This opens the door to
secure communication protocols over long distances and thus offers a reliable
solution for the future quantum internet [1, 2, 3]. The first scientific
scheme that discusses QT as a protocol of secure transmission of quantum
information contained in an unknown quantum state, was first developed by
Bennet $et$ $al$. [4], where they found that quantum information from an
unknown quantum state can be perfectly transmitted from a sender, say Alice,
to a distant receiver, say Bob, via a maximally pre-shared entangled quantum
channel, without physical transfer of the discussed system itself. In a more
convincing way, after the measurement the unknown quantum state of the
discussed physical system disappears from the transmitting entity to appear in
the more distant receiving entity at the same time, without loss of
information, and without taking into account the distance between the two
entities using classical communication and quantum entanglement [7, 5, 6, 8].
Without this recent discovery, which has become in our time the most important
principle of quantum mechanics, today we would not be talking about something
called QT of an unknown quantum state according to the principles of quantum
mechanics at the time, which considered something like this forbidden and
impossible i.e. Heisenberg’s principle of uncertainty. For this reason, QT is
the incipient and direct result of this great discovery and is therefore the
one who can better explain and provide an impressive explanation for the
phenomenon of quantum entanglement and the huge success that we can achieve if
we are lucky enough to have it maximally among quantum entities [9, 10, 11,
12, 13, 14].

Experimentally, several physical systems have been transmitted with
encouraging results [15, 16, 17, 18, 19, 20, 21, 22]. Since QT is essentially
a transmission protocol, it was obvious that most of attention will be focused
on its effectiveness according to the different systems studied and the
possibility of continuing its success in the presence of new constraints, for
example, case of long distance [23] or also the quantum channel pre-shared
between sender and receiver, such as Einstein-Podolsky-Rosen (EPR) state [24],
Greenberger-Horner-Zeilinger (GHZ) state [25, 26], GHZ-like state [27, 28,
29], Werner (W) state [30, 31], W-like state [32] and cluster state [33, 34,
35, 36]. Building on prior research [37], this work investigates faithful
teleportation using an arbitrary shared quantum state (qubit) between Alice
and Bob. The authors introduce a general criterion to determine if this
channel can teleport any $d$-qubit state accurately, regardless of the
specific channel state. This new method offers a significant advantage; it’s
simpler to implement in experiments compared to existing protocols. A novel
d-qudit teleportation protocol is presented in Ref.[38]. This scheme utilizes
the tensor product state of $n$-generalized Bell states as the quantum
channel, allowing for faithful state transmission. Additionally, the
implementation of multi-qubit quantum teleportation and a deterministic secure
communication protocol under the control of multiple agents are reported in
[39]. Practically, parts responsible for QT can be extended to other parts
with more tasks as needed as possible. The extension of QT to multipartite QT
is known as controlled quantum teleportation (CQT) [40], which is an integral
part of quantum teleportation networks [41] and could be the cornerstone of
the future quantum internet. In 2013, an original novel protocol for
bidirectional controlled quantum teleportation (BCQT) was firstly presented by
Zha $et$ $al$. [42], where Alice and Bob decides together to teleport their
quantum states to each other under the supervision of a controller, say
Charlie using five-qubit cluster states. Interesting, Yuan and Zhang [43] have
suggested an asymmetric BQT scheme feasible with present experimental
technologies by using a four-qubit cluster state and a Bell state as a quantum
channel. In addition, Wanbin et al.[44] have theoretically proposed three BQCT
schemes, allowing bidirectional teleportation between two participants with
another as the controller, by utilizing seven-qubit maximally entangled
states. BCQT has been well received by researchers in the field of information
transmission and has been applied using many different entangled pre-shared
quantum channel, as for example GHZ states [45], five-qubit entangled states
[46], five-qubit composite GHZ-Bell states [47], six-qubit entangled states
[48], seven-qubit entangled states [49], eight-qubit entangled state [50] and
multi-qubit entangled states [51]. In 2017, an other novel protocol for cyclic
quantum teleportation (CYQT) was originally proposed by Chen $et$ $al$. [52],
where three collaborators Alice, Bob and Charlie all decides to teleport their
single-qubit quantum states among themselves using a six-qubit entangled
state. Immediately after, Sang [53] proposed a controlled CYQT protocol
introducing a fourth co-operator playing the role of supervisor David,
implying that Alice, Bob and Charlie could exactly share their quantum states
if David gave his consent [54, 55, 56, 57]. Recently, Verma [58] suggested a
new scheme for CYQT, where four co-operators Alice, Bob, Charlie and David can
cyclically teleport their single-qubit state among themselves by using two
G-states.

In our work, an efficient cyclic quantum teleportation of three arbitrary two-
qubit states using a six-qubit cluster state and a six-qubit entangled state
is proposed. In our scheme, Alice has an unknown two-qubit state containing
two qubits $\alpha_{0}$ and $\alpha_{1}$, Bob has a second unknown two-qubit
state containing qubits $\beta_{0}$ and $\beta_{1}$, and the third
collaborator Charlie has also a third unknown two-qubit state containing two
qubits $\gamma_{0}$ and $\gamma_{1}$. The goal of our CYQT scheme is to
teleport the two-qubit state in Alice’s possession to his near neighbour Bob,
and Bob must transmit his two-qubit state to his neighbour Charlie and also
the third co-operator Charlie can teleport his unknown two-qubit state to
Alice. To achieve this, the three parties must first share a quantum channel
containing twelve-qubit distributed among the three collaborators, each of
them has four qubits. Next, each collaborator must to perform Bell state
measurements on his two qubits to be transmitted and his two out of these four
qubits of the quantum channel. Finally, each one of them might to perform
suitable Pauli unitary operators on his resulting state to obtain the original
state and this after, of course, receiving measurement results by classical
channel from the responsible sender.

The outline of this article is organized as follows: In Section 2, we
introduced the cyclic quantum teleportation scheme with a mention of the
construction of the used pre-shared entangled twelve-qubit quantum channel,
the three target states to be teleport for Alice to Bob, from Bob to Charlie
and from Charlie to Alice respectively. Section 3 details the execution of
cyclic quantum teleportation scheme of the three arbitrary two-qubit entangle
entangled state. Section 4 analysis the amout data of classical information,
the success probability and the protocol performance by calculating the
efficiency factor of the proposed CYQT scheme. Concluding remarks are given in
Section 5.

##  II Three arbitrary two-qubit states to be transmitted and selected twelve-
qubit entangled quantum channel

In this section, we aim to represent the quantum channel used, the unknown
selected target states and the CYQT protocol steps. Suppose that there are
three participants, employing twelve-qubit entangled states combining a six-
qubit cluster state and a six-qubit entangled state as a quantum channel.
Alice has an unknown two-qubit state $|{\psi}\rangle_{\alpha}$ to teleport to
Bob, Bob has a second unknown two-qubit state $|{\psi}\rangle_{\beta}$ to
teleport to Charlie, and similarly Charlie has a third unknown two-qubit state
$|{\psi}\rangle_{\gamma}$ to teleport to Alice through a quantum channel. The
quantum channel linking the three participants Alice, Bob and Charlie take the
form

$|{\psi}\rangle_{123456789101112}=|{C_{6}}\rangle_{123456}\otimes|{\psi_{6}}\rangle_{789101112},$
(1)

the six-qubit cluster state $|{C_{6}}\rangle$ used in the quantum channel
above has the form

$|{C_{6}}\rangle=\dfrac{1}{2}\\{|{000000}\rangle+|{000111}\rangle+|{111000}\rangle+|{111111}\rangle\\},$
(2)

and the six-qubit entangled state $|{\psi_{6}}\rangle$ can be described as
follows

$\displaystyle|{\psi_{6}}\rangle=\dfrac{1}{4}\\{|{000000}\rangle$
$\displaystyle+|{000001}\rangle+...+|{000110}\rangle+|{000111}\rangle$
$\displaystyle+|{111000}\rangle+|{111001}\rangle+...+|{111110}\rangle+|{111111}\rangle\\}.$
(3)

###  II.1 Construction of the quantum channel

Figure 1: Quantum circuit illustrating the construction of the twelve-qubit
quantum channel combining a six-qubit cluster state and a six-qubit entangled
state using H-gate and CNOT-gate.

The quantum channel in the simplest case is a twelve-qubit entangled state
shared between the three participants Alice, Bob and Charlie, which can be
realized by applying Hadamard (H) and Controlled-Not (CNOT) quantum gates (see
the Fig.(1)). Firstly, the product state of twelve-qubit state initialized to
$|{0}\rangle$ is used as the input state $|{\phi_{0}}\rangle$ in the quantum
circuit, which can be expressed as

$|{\phi_{0}}\rangle=|{0}\rangle_{1}\otimes|{0}\rangle_{2}\otimes...\otimes|{0}\rangle_{12},$
(4)

then, performing H-gates on qubits (1, 4, 7, 10, 11, 12), the input state
$|{\phi_{0}}\rangle$ is converted to $|{\phi_{1}}\rangle$ as follows

$\displaystyle|{\phi_{1}}\rangle$
$\displaystyle=\dfrac{1}{8}\\{|{000000}\rangle+|{000100}\rangle+|{100000}\rangle+|{100100}\rangle\\}_{123456}\otimes\\{|{000}\rangle+|{100}\rangle\\}_{789}$
$\displaystyle\otimes\\{|{000}\rangle+|{001}\rangle+|{010}\rangle+|{011}\rangle+|{100}\rangle+|{101}\rangle+|{110}\rangle+|{111}\rangle\\}_{101112},$
(5)

after that, several CNOT-gates need to be performed on the following qubit
pairs (1;2), (1;3), (4;5), (4;6), (7;8) and (7;9), qubits 1, 4, 7 control
qubits (2;3), (5;6) and (8;9) respectively. Afterwards, the quantum channel
takes the form (1). Where Alice holds qubits (5,6,7,10), Bob holds qubits
(1,8,9,11) and Charlie holds qubits (2,3,4,12). The twelve-qubit entangled
state can be rewritten as

$\displaystyle|{\psi_{12}}\rangle$
$\displaystyle=\dfrac{1}{8}\\{|{000000}\rangle+|{000111}\rangle+|{111000}\rangle+|{111111}\rangle\\}_{B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}}\otimes\\{|{000}\rangle+|{111}\rangle\\}_{A_{4}B_{1}B_{3}}$
$\displaystyle\otimes\\{|{000}\rangle+|{001}\rangle+|{010}\rangle+|{011}\rangle+|{100}\rangle+|{101}\rangle+|{110}\rangle+|{111}\rangle\\}_{A_{2}B_{2}C_{2}},$
(6)

###  II.2 Description of the proposed CYQT scheme

In this scheme, each one of three participants has an arbitrary unknown two-
qubit state to teleport to the next neighbour. Alice has an unknown two-qubit
state
$|{\psi}\rangle_{\alpha}=a_{0}|{00}\rangle_{\alpha_{0}\alpha_{1}}+a_{1}|{11}\rangle_{\alpha_{0}\alpha_{1}}$
to teleport to Bob, Bob has a second unknown two-qubit state
$|{\psi}\rangle_{\beta}=b_{0}|{00}\rangle_{\beta_{0}\alpha_{1}}+b_{1}|{11}\rangle_{\beta_{0}\beta_{1}}$
to teleport to Charlie, while Charlie has a third unknown two-qubit state
$|{\psi}\rangle_{\gamma}=c_{0}|{00}\rangle_{\gamma_{0}\gamma_{1}}+c_{1}|{11}\rangle_{\gamma_{0}\gamma_{1}}$
to teleport to Alice through the quantum channel discussed above. Where
$a_{0}$, $a_{1}$, $b_{0}$, $b_{1}$, $c_{0}$, $c_{1}$ are arbitrary complex
coefficients with $|a_{0}|^{2}+|a_{1}|^{2}=1$, $|b_{0}|^{2}+|b_{1}|^{2}=1$ and
$|c_{0}|^{2}+|c_{1}|^{2}=1$. The description of our CYQT scheme is shown in
Figure 2. Alice wants to transmit $\alpha_{0}$ and $\alpha_{1}$ to Bob, Bob
wants to transmit his own two qubits $\beta_{0}$ and $\beta_{1}$ to Charlie,
and similarly Charlie wants to transfer the two qubits in its possession
$\gamma_{0}$ and $\gamma_{1}$ to Alice.

Figure 2: Description of the protocol of CYQT, each one of the three observers
Alice, Bob and Charlie has a two-qubit state ($\alpha_{0}$, $\alpha_{1}$),
($\beta_{0}$,$\beta_{1}$) and ($\gamma_{0}$,$\gamma_{1}$) respectively to
teleport among themselves via quantum channel and by performing BSM.

###  II.3 CYQT protocol steps

The three participants Alice, Bob and Charlie, are spatially separated and
entangled by a quantum channel, which means that the quantum information will
be transmitted using the pre-shared quantum channel as well as other classical
communication channels. The steps used to make our CYQT scheme are organized
as follows:

  * •

After combining the three arbitrary two-qubit states to be transmitted with
the twelve-qubit quantum channel, Alice carries out Bell state measurements
(BSM) on qubits $\alpha_{0}$ and $A_{2}$, and will send to Bob the result of
her measurement by classical communication channel.

  * •

Bob carries out BSM on qubits $\beta_{0}$ and $B_{2}$, and will send his BSM
results to Charlie by classical communication channel.

  * •

Charlie carries out BSM on qubits $\gamma_{0}$ and $C_{2}$, and will send to
Alice the result of his measurement by classical communication channel.

  * •

Alice carries out a second BSM on qubits $\alpha_{1}$ and $A_{4}$, and will
send to Bob the result of his last measurement by classical communication
channel.

  * •

Charlie carries out his second BSM on qubits $\gamma_{1}$ and $C_{1}$, and
will send to Alice the result of his measurement by classical communication
channel.

  * •

Bob ends the cycle of Bell state measurements, by carrying out BSM on qubits
$\beta_{1}$ and $B_{4}$, and will send his BSM results to Charlie by classical
communication channel.

Based on the received results, Alice, Bob and Charlie can recover the initial
two-qubit quantum state send by Charlie, Alice and Bob respectively by
applying appropriate unitary transformations ($I$, $X$, $i$$Y$,$Z$).

##  III Cyclic quantum teleportation of three arbitrary two-qubit states among
the three users

To teleport the three arbitrary two-qubit state among the three collaborators
Alice, Bob and Charlie as described in Figure 2, one has to follow the CYQT
protocol steps well detailed in subsection II.3 (see the Fig.(3)). The
combined state of the whole subsystems can be expressed as

$\displaystyle\left|\psi\right>_{\alpha_{0}\alpha_{1}\beta_{0}\beta_{1}\gamma_{0}\gamma_{1}B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}A_{4}B_{1}B_{3}A_{2}B_{2}C_{2}}=\left|\psi\right>_{\alpha}\otimes\left|\psi\right>_{\beta}\otimes\left|\psi\right>_{\gamma}\otimes\left|\psi_{12}\right>,$
(7)

Step 1: To implement the cyclic quantum teleportation protocol, Alice takes
the first step by applying a BSM on her qubits $\alpha_{0}$ and $A_{2}$, the
BSM basis is given by,

$\displaystyle\left|\Phi^{\pm}\right>=\dfrac{1}{\sqrt{2}}\\{\left|00\right>\pm\left|11\right>\\},\hskip
28.45274pt\left|\Psi^{\pm}\right>=\dfrac{1}{\sqrt{2}}\\{\left|01\right>\pm\left|10\right>\\},$
(8)

After measurement, if Alice’s BSM result was
$|{\Phi^{+}}\rangle_{\alpha_{0}A_{2}}$, the state of the remaining sixteen
subsystems suddenly collapsed to

$\displaystyle\dfrac{\sqrt{2}}{4}\\{a_{0}|{0}\rangle_{\alpha_{1}}+a_{1}|{1}\rangle_{\alpha_{1}}\\}\otimes\left|\psi\right>_{\beta}\otimes\left|\psi\right>_{\gamma}\otimes|{C_{6}}\rangle_{B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}}$
$\displaystyle\otimes\\{|{000}\rangle+|{111}\rangle\\}_{A_{4}B_{1}B_{3}}\otimes\\{|{00}\rangle+|{01}\rangle+|{10}\rangle+|{11}\rangle\\}_{B_{2}C2}$
(9)

Step 2: Immediately, Bob realize the second step in this protocol by carrying
out a BSM on his qubits $\beta_{0}$ and $B_{2}$ in the same basis suggested in
$Eq.(8)$. Suppose that Bob’s BSM result was also
$\left|\Phi^{+}\right>_{\beta_{0}B_{2}}$, the state of the remaining fourteen
subsystems suddenly collapsed to

$\displaystyle\dfrac{1}{2}\\{_{0}|{0}\rangle_{\alpha_{1}}+a_{1}|{1}\rangle_{\alpha_{1}}\\}\otimes\\{b_{0}|{0}\rangle_{\beta_{1}}+b_{1}|{1}\rangle_{\beta_{1}}\\}\otimes\left|\psi\right>_{\gamma}\otimes$
$\displaystyle|{C_{6}}\rangle_{B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}}\otimes\\{|{000}\rangle+|{111}\rangle\\}_{A_{4}B_{1}B_{3}}\otimes\\{|{0}\rangle+|{1}\rangle\\}_{C2}$
(10)

Step 3: Basing on the same BSM basis proposed in $Eq.(8)$, the third party
Charlie ends the first cycle of the discussed CYQT by performing a BSM on his
qubits $\gamma_{0}$ and $C_{2}$. If Charlie’s BSM result was
$\left|\Phi^{+}\right>_{\gamma_{0}C_{2}}$, the state of the remaining twelve
subsystems collapsed to

$\displaystyle\dfrac{\sqrt{2}}{2}\\{a_{0}|{0}\rangle_{\alpha_{1}}+$
$\displaystyle
a_{1}|{1}\rangle_{\alpha_{1}}\\}\otimes\\{b_{0}|{0}\rangle_{\beta_{1}}+b_{1}|{1}\rangle_{\beta_{1}}\\}\otimes\\{c_{0}|{0}\rangle_{\gamma_{1}}+c_{1}|{1}\rangle_{\gamma_{1}}\\}\otimes$
$\displaystyle|{C_{6}}\rangle_{B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}}\otimes\\{|{000}\rangle+|{111}\rangle\\}_{A_{4}B_{1}B_{3}}$
(11)

Step 4: In the following steps, the three parties will realize the second
cycle of the discussed CYQT. In this step, Alice carry out a second BSM on her
qubits $\alpha_{1}$ and $A_{4}$. If Alice’s new BSM result was
$\left|\Phi^{+}\right>_{\alpha_{1}A_{4}}$, the state of the remaining ten
subsystems would collapse into

$\displaystyle\\{b_{0}|{0}\rangle_{\beta_{1}}+b_{1}|{1}\rangle_{\beta_{1}}\\}\otimes\\{c_{0}|{0}\rangle_{\gamma_{1}}+c_{1}|{1}\rangle_{\gamma_{1}}\\}\otimes|{C_{6}}\rangle_{B_{4}C_{1}C_{3}C_{4}A_{1}A_{3}}\otimes\\{a_{0}|{00}\rangle+a_{1}|{11}\rangle\\}_{B_{1}B_{3}}$
(12)

at the end of this step, it is clear that Bob received Alice’s two-qubit
state.

Step 5: In this sense, Charlie re-perform a second BSM on his qubits
$\gamma_{1}$ and $C_{4}$. If Charlie’s new BSM result was
$\left|\Phi^{+}\right>_{\gamma_{1}C_{4}}$, the state of the remaining height
qubits collapsed to

$\displaystyle\dfrac{\sqrt{2}}{2}\\{b_{0}|{0}\rangle_{\beta_{1}}+b_{1}|{1}\rangle_{\beta_{1}}\\}\otimes$
$\displaystyle\\{c_{0}|{00000}\rangle+c_{1}|{00011}\rangle+c_{0}|{11100}\rangle+c_{1}|{11111}\rangle\\}_{B_{4}C_{1}C_{3}A_{1}A_{3}}$
$\displaystyle\otimes\\{a_{0}|{00}\rangle+a_{1}|{11}\rangle\\}_{B_{1}B_{3}}$
(13)

Step 6: Finally, Bob ends the second cycle of our proposed CYQT by performing
the last BSM on his qubit $\beta_{1}$ and $B_{4}$. If Bob’s last BSM result
was $|{\Phi^{+}}\rangle_{\beta_{1}B_{4}}$, the remaining six qubits might
collapse to the state below

$\displaystyle\\{b_{0}c_{0}|{0000}\rangle+b_{0}c_{1}|{0011}\rangle+b_{1}c_{0}|{1100}\rangle+b_{1}c_{1}|{1111}\rangle\\}_{C_{1}C_{3}A_{1}A_{3}}\otimes\\{a_{0}|{00}\rangle+a_{1}|{11}\rangle\\}_{B_{1}B_{3}}.$
(14)

This final state can be expressed as

$\displaystyle|{\psi}\rangle_{C_{1}C_{3}A_{1}A_{3}B_{1}B_{3}}=\\{|{\psi^{{}^{\prime}}}\rangle_{\beta}\\}_{C_{1}C_{3}}\otimes\\{|{\psi^{{}^{\prime}}}\rangle_{\gamma}\\}_{A_{1}A_{3}}\otimes\\{|{\psi^{{}^{\prime}}}\rangle_{\alpha}\\}_{B_{1}B_{3}},$
(15)

with

$\displaystyle|{\psi^{{}^{\prime}}}\rangle_{\beta}=\\{b_{0}|{00}\rangle+b_{1}|{11}\rangle\\}$
$\displaystyle|{\psi^{{}^{\prime}}}\rangle_{\gamma}=\\{c_{0}|{00}\rangle+c_{1}|{11}\rangle\\}$
$\displaystyle|{\psi^{{}^{\prime}}}\rangle_{\alpha}=\\{a_{0}|{00}\rangle+a_{1}|{11}\rangle\\},$
(16)

which are exactly the desired three two-qubit arbitrary states sends
respectively by Bob to Charlie, by Charlie to Alice and by Alice to Bob. This
means that receivers will not need to do any conversion to get the original
sent states. This in the case when the BSMs results were respectively
$|{\Phi^{+}}\rangle_{\alpha_{0}A_{2}}$, $|{\Phi^{+}}\rangle_{\beta_{0}B_{2}}$,
$|{\Phi^{+}}\rangle_{\gamma_{0}C_{2}}$,
$|{\Phi^{+}}\rangle_{\alpha_{1}A_{4}}$, $|{\Phi^{+}}\rangle_{\gamma_{1}C_{4}}$
and $|{\Phi^{+}}\rangle_{\beta_{1}B_{4}}$. For other possible BSM outcomes,
receivers will have to apply appropriate transformations before they can
exactly retrieve the original ones. Due to classical communication, receivers
will be able to know exactly which Pauli unitary operators will be necessary
to recover the exactly ones. To be precise, there would be $2^{12}=4096$
possible measurement states (see the Table (1)). Each one of them will have a
certain series of appropriate unitary transformations that must be applied
before the original transmitted information states can be reliably
recuperated. We will not go over all of them, first because they are many, and
secondly, because necessary transformations will be clear and intuitive. But
despite that, and for more clarification, we discussed the first four of them.
In table below, we can find the first four possible BSM result cases and their
necessary conversions. Whatever the result of applied measurements, Alice, Bob
and Charlie will be able to recover the original transmitted information
states and thus we can be sure that the proposed CYQT protocol runs perfectly
in all cases.

Bell State Measurements (BSM) | Pauli Unitary Operators
---|---
Alice |  Bob |  Charlie |  Alice |  Charlie |  Bob |  Alice |  Bob |  Charlie
$|{\Phi^{+}}\rangle_{\alpha_{0}A_{2}}$ |  $|{\Phi^{+}}\rangle_{\beta_{0}B_{2}}$ |  $|{\Phi^{+}}\rangle_{\gamma_{0}C_{2}}$ |  $|{\Phi^{+}}\rangle_{\alpha_{1}A_{4}}$ |  $|{\Phi^{+}}\rangle_{\gamma_{1}C_{4}}$ |  $|{\Phi^{+}}\rangle_{\beta_{1}B_{4}}$ |  $I_{A_{1}}\otimes I_{A_{3}}$ |  $I_{B_{1}}\otimes I_{B_{3}}$ |  $I_{C_{1}}\otimes I_{C_{3}}$
$|{\Phi^{-}}\rangle_{\beta_{1}B_{4}}$ |  $I_{A_{1}}\otimes I_{A_{3}}$ |  $I_{B_{1}}\otimes I_{B_{3}}$ |  $Z_{C_{1}}\otimes I_{C_{3}}$
$|{\Psi^{+}}\rangle_{\beta_{1}B_{4}}$ |  $I_{A_{1}}\otimes I_{A_{3}}$ |  $I_{B_{1}}\otimes I_{B_{3}}$ |  $X_{C_{1}}\otimes X_{C_{3}}$
$|{\Psi^{-}}\rangle_{\beta_{1}B_{4}}$ |  $I_{A_{1}}\otimes I_{A_{3}}$ |  $I_{B_{1}}\otimes I_{B_{3}}$ |  $iY_{C_{1}}\otimes X_{C_{3}}$
Table 1: First four results among ”$4096$” possible BSM outcomes and Pauli
unitary operators need to be applied by the three receivers to retrieve
original states. $i$ is an imaginary unit, $X$, $Y$ and $Z$ are Pauli unitary
operators and $I$ is the identity unitary operator. Figure 3: Quantum circuit
illustrating the scheme of the proposed CYQT protocol of three two-qubit state
$|{\psi}\rangle_{\alpha}$, $|{\psi}\rangle_{\beta}$ and
$|{\psi}\rangle_{\gamma}$ using a twelve-qubit state $|{\psi}\rangle_{12}$
combining a six-qubit cluster state and a six-Bell state. The double lines
decodes the classical wires and $"U"$ decode Pauli unitary operations I, X,
$i$Y and Z. The output states $|{\psi^{{}^{\prime}}}\rangle_{\alpha}$,
$|{\psi^{{}^{\prime}}}\rangle_{\beta}$ and
$|{\psi^{{}^{\prime}}}\rangle_{\gamma}$ represented the recovered states at
the end of the teleportation protocol.

Figure 4: Panel ($a$) presents the quantum circuit fidelity, giving
information about the accuracy of quantum teleportation protocol. Panel ($b$)
shows the Teleportation Density, showing the distribution and characteristics
of quantum states after teleportation.

Using Qiskit, the code (3) prepares the quantum and classical registers. After
defining a Toffoli gate matrix, it builds a quantum circuit that carries out
several quantum operations. These measures correspond to the steps described
above of the CYQT protocol. To transmit the quantum states, Alice, Bob, and
Charlie each execute Bell State Measurements (BSM) on their individual qubits
in the CYQT protocol. These procedures, which involve the use of CNOT gates,
Hadamard gates, and measurements, appear in the circuit. A unitary gate that
reflects the outcomes of the BSMs executed by the three parties is applied at
the circuit’s conclusion. In order to represent the successful teleportation
of the initial states between the parties, the circuit collapses the quantum
state into a final state. the circuit show how the quantum operations transfer
to the steps in the article, where each step is a particular party executing a
BSM and collapsing the quantum state correspondingly. As described in the
attached paper, the circuit functions as an implementation of a quantum
circuit for CYQT, allowing the teleportation of arbitrary quantum states
between multiple parties.

The fidelity of the quantum circuit is shown along the $X$-axis from $0$ to
$900$ in the graph 4(panel ($a$)). The number of shots varies. Data points are
common, appearing approximately every $100$ frames. On the $Y$-axis, the
fidelity is shown as a range between $0.02$ and a little above $0.12$. The
fidelity shows significant variation across different shot counts, with no
clear increasing or negative tendency. Initially, fidelity increases
significantly from zero shots to approximately one hundred. After about $300$
shots, it stabilizes in the upper half of the graph after fluctuating for a
while. The fidelity reaches its maximum at around $700$ shots, or $0.12$. The
fidelity, on the other hand, reaches its minimum approximately 0.04, 200 shots
after the beginning increase. When you get over $300$ shots, fidelity seems to
settle off even more. In the last $600–900$ shots, there is less variation,
but there are still obvious fluctuations. It also means that the performance
of the circuit can vary depending on a number of factors or quantum processes
like decoherence. In general, there is not an obvious increase or decrease
trend, even when fidelity stabilizes at the end, indicating to the detailed
dynamics of the quantum system.

The left side of graph 4(panel ($b$)) shows a three-dimensional grid that
represents the density matrix components along the $z$-axis. This grid
represents Alice’s (the first sender) unknown state. Alice unkown variable was
given a randowm value to represent the density matrix, we could repeat the
process for Bob and Charlie, but with different values. Every block in the
grid represents a matrix element, and the element’s value is indicated by its
height. All elements are non-negative and range from $0$ to $0.25$ on the
$z$-axis scale, with a maximum value of $0.25$. The color blue, doesn’t change
continuously.

##  IV Protocol success probability and efficiency

As we have already shown, Alice, Bob and Charlie each have to apply two Bell
state measurements, i.e. there will be a total of six Bell state measurements.
So, we’ll have 4096 possible Bell state measurements results. The probability
of each result is given by

$\hskip
56.9055ptP_{i}=\left(\dfrac{1}{2^{6}}\right)^{2}=\dfrac{1}{4096};\hskip
28.45274pt\\{i=0,1,...,4095\\}$ (17)

$\bullet$ The total success probability can be expressed as

$P_{total}=\sum_{i=0}^{4095}P_{i}=100\%$ (18)

$\bullet$ The average lack of information $S$ can be expressed as

$S=-\sum_{i=0}^{4095}P_{i}\hskip 7.11317ptlog\hskip 7.11317ptP_{i}=12$ (19)

it means that results have all equal probabilities. To study the performance
of communication protocols [59], we often calculate the efficiency factor
defined by

$e\left(\%\right)=\dfrac{q_{t}}{q_{s}}$ (20)

where $q_{t}$ denotes the number of qubits have been teleportated during the
execution of the protocol (here it was three arbitrary two-qubit state i.e. 6
teleported qubits). And, $q_{s}$ denotes the total number of qubits of the
quantum channel used for teleportation (here we used a combination of a six-
qubit cluster state and a six-qubit entangled state i.e. 12 entangled qubits),
which means that the proposed protocol has an efficiency that can reach
$50\%$.

##  V Concluding remarks

In summary, an original and perfect protocol of cyclic quantum teleportation
by using a combination of six-qubit cluster state and six-qubit entangled
state as quantum channel is presented for the first time. Where, the three
distant parties Alice, Bob and Charlie are not only senders of information but
also receivers of information. Alice can send an arbitrary two-qubit state to
Bob and receive an arbitrary two-qubit state from Charlie and Bob can transmit
an arbitrary two-qubit state to Charlie by performing Bell state measurements
and appropriate unitary Pauli transformations after receiving the Bell state
measurement results sent by others by classical way. In our scheme, the
proposed quantum channel can be constructed by applying Controlled-Not gate
and Hadamard gate on a twelve-qubit states all initialized to $\left|0\right>$
which has been implemented experimentally. During the execution of the
protocol, the three collaborators choose the same basis of Bell states to
perform the necessary measurements on the qubits they have. In fact, each of
the three parties must perform two Bell state measurements on their qubits,
which means that there will be a total of six Bell state measurements, so
there will be 4096 possible output states. All these cases have equal
probabilities and the protocol offers a success probability of up to 100%.
Particular attention was paid to the performance of the proposed protocol by
calculating the information transfer efficiency factor and it was observed
that this efficiency can reach 50%, which is a very important and honourable
percentage. Finally, it is hoped that the results obtained and the proposed
protocol will bring an additional advantage in the world of quantum
communication and data transmission. This proposed protocol offers a novel and
secure method for transmitting quantum information between three distant
parties. It enables not only sending but also receiving information, thereby
enhancing the efficiency of quantum communication. However, generalizing this
approach to other quantum teleportation tasks (like CBCQT) or applying it to
larger systems might pose challenges. Further investigation and analysis are
necessary to determine the protocol’s effectiveness in these scenarios. We
will address this question in a future work.

Data availability: The datasets used and/or analyzed during the current study
available from the corresponding author on request.

Funding statement: Princess Nourah bint Abdulrahman University Researchers
Supporting Project number (PNURSP2024R407), Princess Nourah bint Abdulrahman
University, Riyadh, Saudi Arabia.

Acknowledgements: Princess Nourah bint Abdulrahman University Researchers
Supporting Project number (PNURSP2024R407), Princess Nourah bint Abdulrahman
University, Riyadh, Saudi Arabia. Also, Ahmed A. Abd El-Latif would like to
thank Prince Sultan University for their support.

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