# Prediction of fully metallic $\sigma$-bonded boron framework induced high
superconductivity
above 100 K in thermodynamically stable Sr2B5 at 40 GPa

Xin Yang  Key Laboratory of Material Simulation Methods & Software of Ministry
of Education, College of Physics, Jilin University, Changchun 130012, China
State Key Laboratory for Superhard Materials, College of Physics, Jilin
University, Changchun 130012, China  Wenbo Zhao  Key Laboratory of Material
Simulation Methods & Software of Ministry of Education, College of Physics,
Jilin University, Changchun 130012, China  State Key Laboratory for Superhard
Materials, College of Physics, Jilin University, Changchun 130012, China
International Center of Future Science, Jilin University, Changchun 130012,
China  Liang Ma  Key Laboratory of Material Simulation Methods & Software of
Ministry of Education, College of Physics, Jilin University, Changchun 130012,
China  State Key Laboratory for Superhard Materials, College of Physics, Jilin
University, Changchun 130012, China  International Center of Future Science,
Jilin University, Changchun 130012, China  Wencheng Lu  Key Laboratory of
Material Simulation Methods & Software of Ministry of Education, College of
Physics, Jilin University, Changchun 130012, China  Xin Zhong  Key Laboratory
of Material Simulation Methods & Software of Ministry of Education, College of
Physics, Jilin University, Changchun 130012, China  Yu Xie<EMAIL_ADDRESS>Key Laboratory of Material Simulation Methods & Software of Ministry of
Education, College of Physics, Jilin University, Changchun 130012, China  Key
Laboratory of Physics and Technology for Advanced Batteries of Ministry of
Education, College of Physics, Jilin University, Changchun 130012, China
Hanyu Liu<EMAIL_ADDRESS>Key Laboratory of Material Simulation Methods &
Software of Ministry of Education, College of Physics, Jilin University,
Changchun 130012, China  State Key Laboratory for Superhard Materials, College
of Physics, Jilin University, Changchun 130012, China  International Center of
Future Science, Jilin University, Changchun 130012, China  Key Laboratory of
Physics and Technology for Advanced Batteries of Ministry of Education,
College of Physics, Jilin University, Changchun 130012, China  Yanming Ma
<EMAIL_ADDRESS>Key Laboratory of Material Simulation Methods & Software of
Ministry of Education, College of Physics, Jilin University, Changchun 130012,
China  State Key Laboratory for Superhard Materials, College of Physics, Jilin
University, Changchun 130012, China  International Center of Future Science,
Jilin University, Changchun 130012, China

###### Abstract

Metal borides have been considered as potential high-temperature
superconductors since the discovery of record-holding 39 K superconductivity
in bulk MgB2. In this work, we identified a superconducting yet
thermodynamically stable $F$$\overline{4}$3$m$ Sr2B5 at 40 GPa with a unique
covalent $sp^{3}$-hybridized boron framework through extensive first-
principles structure searches. Remarkably, solving the anisotropic Migdal-
Eliashberg equations resulted in a high superconducting critical temperature
($T_{\mathrm{c}}$) around 100 K, exceeding the boiling point (77 K) of liquid
nitrogen. Our in-depth analysis revealed that the high-temperature
superconductivity mainly originates from the strong coupling between the
metalized $\sigma$-bonded electronic bands and $E$ phonon modes of boron
atoms. Moreover, anharmonic phonon simulations suggest that
$F$$\overline{4}$3$m$ Sr2B5 might be recovered to ambient pressure. Our
current findings provide a prototype structure with a full $\sigma$-bonded
boron framework for the design of high-$T_{\mathrm{c}}$ superconducting
borides that may expand to a broader variety of lightweight compounds.

The quest for high-temperature superconductors has been the long-sought target
since the discovery of superconductivity in solid mercury in 1911 Onnes
(1911). According to the Bardeen-Cooper-Schrieffer (BCS) theory of
superconductivity, light-element compounds are considered as promising
candidates for high superconducting critical temperature ($T_{\mathrm{c}}$)
superconductors since they possess high Debye temperature, which is
proportional to the $T_{\mathrm{c}}$ Suhl et al. (1959). Recently, the theory-
orientated experiments have established near-room-temperature
superconductivity in a class of hydrides at high pressure, such as H3S, LaH10,
and CaH6 Drozdov et al. (2015); Peng et al. (2017); Drozdov et al. (2019);
Somayazulu et al. (2019); Ma et al. (2022); Li et al. (2014); Liu et al.
(2017); Wang et al. (2012a). Nevertheless, the pressure required to realize
these superconducting hydrides is exceptionally high and challenging to
achieve. Therefore, it is of significant importance to search for alternative
high-$T_{\mathrm{c}}$ superconductors at moderate pressures and even at
ambient pressure for practical applications.

At low pressure, MgB2 is the BCS superconductor with the highest
$T_{\mathrm{c}}$ of 39 K owing to its high Debye temperature and strong
electron-phonon coupling (EPC), primarily originating from the metalized
$\sigma$-bonding bands of in-plane $sp^{2}$-hybridized boron atoms Nagamatsu
et al. (2001); Kortus et al. (2001). Thus, metal borides characterized by
metallic $\sigma$ bonds have attracted considerable attention in searching for
high-$T_{\mathrm{c}}$ superconductors. Besides MgB2, a class of MB2, MB6, and
MB12 (M represents transition metals) have also been experimentally identified
to be superconducting with $T_{\mathrm{c}}$ below 20 K at ambient pressure
Fisk et al. (1971); Schneider et al. (1987); Lortz et al. (2006); Akopov et
al. (2017). Recently, Pei et al. discovered that $\alpha$-MoB2, isostructural
with MgB2, exhibits the second-highest $T_{\mathrm{c}}$ of 32 K at 100 GPa
among all known borides Pei et al. (2023). However, as the main contribution
at the Fermi level of these borides is not derived from $\sigma$ bonds of
boron atoms, none of them have achieved a $T_{\mathrm{c}}$ surpassing that of
MgB2 yet.

In addition to experiments, much effort has also been devoted to theoretical
investigations to design borides with high $T_{\mathrm{c}}$. On one hand,
thermodynamically stable boride superconductors, such as BeB6, CaB, and SrB
characterized by deformed $sp^{2}$-hybridization Wu et al. (2016); Cui et al.
(2022); Shah and Kolmogorov (2013), exhibit lower $T_{\mathrm{c}}$ compared to
MgB2 due to the weaker metallic $\sigma$ bonds, which is similar to the
experimental observation. On the other hand, a series of MB2 monolayer and MB4
trilayers, such as AlB2, MgB4, and InB4 Zhao et al. (2019, 2020); Wang et al.
(2021a), have been predicted to exhibit high $T_{\mathrm{c}}$ due to the
strong metallic $\sigma$ bonds originating from $sp^{2}$-hybridized boron
atoms, while their producibility is unclear since their thermodynamic
stability remains unknown. Similarly, various metastable $sp^{3}$-hybridized
borocarbides and boronitrides have also been predicted to exhibit higher
$T_{\mathrm{c}}$ than MgB2 Geng et al. (2023); Gai et al. (2022); Ding et al.
(2022); Zhu et al. (2020); Zhang et al. (2022); Wang et al. (2021b), although
none of them has been experimentally synthesized to date. Thus, it is
encouraging to design metal borides with favorable thermodynamic stability and
strong metallic $sp^{2}$/$sp^{3}$ $\sigma$-bonding characteristics that may
lead to synthesizable high-temperature superconductors with a $T_{\mathrm{c}}$
higher than MgB2.

In this work, we performed a comprehensive first-principles structure searches
for the alkaline-earth (AE) metal borides (MxBy, M = Mg, Ca, Sr, Ba, $x$ =
1-3, $y$ = 1-8) at pressures below 100 GPa via the swarm intelligence-based
CALYPSO method Wang et al. (2010, 2012b). Besides the reproduced already known
phases Esfahani et al. (2017); Shah and Kolmogorov (2013); Kolmogorov et al.
(2012); Cui et al. (2022), we predicted several new thermodynamically stable
borides under high pressures. Among them, the $F$$\overline{4}$3$m$ Sr2B5
consists of a unique fully $sp^{3}$-hybridized $\sigma$-bonded boron
framework, where the density of states at Fermi level is mainly contributed by
$\sigma$-bond. This results in a high $T_{\mathrm{c}}$ above 100 K, not only
surpassing the boiling temperature (77 K) of liquid nitrogen but also setting
a new record for the highest $T_{\mathrm{c}}$ among all known
thermodynamically stable boride superconductors.

Figure 1: (Color online) Phase stabilities of various AE borides MxBy (M = Mg,
Ca, Sr, Ba, $x$ = 1-3, $y$ = 1-8) (a) Convex hull diagrams for MxBy at 50 GPa.
(b) Pressure-composition phase diagram of newly-predicted Mg3B8, Ca2B5, Ca3B5,
Sr2B5, and BaB.  Figure 2: (Color online) The side view of the structure of
$F$$\overline{4}$3$m$ Sr2B5 at 40 GPa. The large and small spheres represent
the Sr and B atoms, respectively.

Our main structure searching results of AE-metal borides MxBy below 100 GPa
are presented in the convex hull and phase diagrams of Fig. 1 and Fig. S1 SI .
Considering the important contribution of zero-point energy (ZPE) to the free
energy of compounds containing light elements Hu et al. (2013); Peng et al.
(2012), the formation energies of the metal borides were calculated with the
inclusion of ZPE. In addition to the known AE-metal borides Esfahani et al.
(2017); Shah and Kolmogorov (2013); Kolmogorov et al. (2012); Cui et al.
(2022), we uncovered several new thermodynamically stable MxBy compounds,
including Mg3B8, Ca2B5, Ca3B5, Sr2B5, and BaB with $C$2/$m$, $P$63/$mmc$,
$P$63/$mmc$, $F$$\overline{4}$3$m$, and $R$$\overline{3}$$m$ structure,
respectively, where Sr2B5 further transforms into $P$63/$mmc$ phase at 44 GPa
(The detailed structural information is listed in Table S1 SI ). The absence
of imaginary frequency in the phonon spectra indicates that these borides are
also dynamically stable (Fig. S2 SI ), confirming the robustness of the
stability. By examining the structures, we found that, except for
$F$$\overline{4}$3$m$-Sr2B5, all the MxBy compounds adopt layered-like
morphology with alternatively stacked boron and metal atomic layers (Fig. S3
SI ). The boron atoms should be mainly $sp^{2}$-hybridized in these MxBy
compounds like other layered metal borides Wu et al. (2016); Cui et al.
(2022); Shah and Kolmogorov (2013). On the other hand,
$F$$\overline{4}$3$m$-Sr2B5 exhibits a unique three-dimensional boron
framework as presented in Fig. 2. It consists of two nonequivalent boron
atoms, namely BD and BT, occupying 4c (0.25, 0.25, 0.25) and 16e (0.59, 0.59,
0.41) Wyckoff position, respectively, where both of them are four coordinated.
Specifically, BD is surrounded by four BT atoms, which forms a regular
tetrahedron B5 unit similar to C atoms in diamond, indicating BD is
$sp^{3}$-hybridized. BT has three shorter BT-BT bonds and one slightly longer
BT-BD bond and forms an irregular tetrahedron B${}_{5}^{{}^{\prime}}$ unit, in
which BT atoms also form a regular tetrahedron B4 unit by itself. This is
close to the motif of C atoms in T-carbon Sheng et al. (2011); Zhang et al.
(2017), implying BT exhibits resembling anisotropic $sp^{3}$-hybridization due
to the unevenly distributed bond lengths. Further electron localization
function (ELF) and crystal orbital Hamiltonian population (COHP) calculations
confirm the strong covalent nature between B atoms (Fig. S4 SI ). Meanwhile,
Sr-B is ionically bonded since electrons are transferred from Sr to B.
Remarkably, the boron sublattice of $F$$\overline{4}$3$m$-Sr2B5 can be viewed
as isostructure to diamond, where one of the nonequivalent atomic position is
occupied by B4 unit instead of B atom. Thus, to the best of our knowledge,
$F$$\overline{4}$3$m$-Sr2B5 should be the first metal boride exhibits fully
$sp^{3}$-hybridized $\sigma$-bonded boron framework that might be beneficial
to the superconductivity.

Figure 3: (Color online) (a) Band structure and DOS of $F$$\overline{4}$3$m$
Sr2B5 at 40 GPa. (b) The projected density of states (PDOS) of boron atoms at
the Fermi level in MgB2, Mg3B8, Ca2B5, Ca3B5, $F$$\overline{4}$3$m$ Sr2B5,
$P$63/$mmc$ Sr2B5, and BaB at 0, 50 ,30, 40, 40, 50, and 80 GPa, respectively.

We next examined the electronic properties of the AE-metal borides by modeling
the band structures and density of states (DOS) as shown in Fig. 3 and Fig. S5
SI . Clearly, the newly predicted metal borides are all metallic. Based on the
boron hybridization scheme, these borides exhibit different band structure
characteristics. The layered-like MxBy compounds have both $\sigma$ and $\pi$
bands crossing the Fermi level and $F$$\overline{4}$3$m$-Sr2B5 only has
$\sigma$ conducting bands, analogous to that of MgB2 and hole-doped diamond,
respectively. Besides the metal-rich Ca3B5 and BaB, boron dominates the
contribution to the DOS at the Fermi level (NEf) due to the charge transfer
from metal to boron atoms, as deciphered in Fig. 3b. In the case of metal-rich
borides, the amount of electrons provided by the metal atoms exceeds the needs
of fully shelled boron atoms. Thus, NEf is dominated by metal atoms.
Importantly, among these metal borides, $F$$\overline{4}$3$m$ Sr2B5 exhibits
not only the highest total NEf but also the highest boron contribution, which
is almost twice as much as that of MgB2. This is because the Fermi level is
located right at the peak position of the DOS, corresponding to a van Hove
singularity along the K-$\Gamma$ direction. The projected DOS further reveals
that the electrons in the vicinity of the Fermi level are coming from BT,
where BD merely contributes to the NEf because it is nearly fully shelled due
to the ideal $sp^{3}$-hybridization. The high NEf together with the solely
$\sigma$ characteristic of boron suggest $F$$\overline{4}$3$m$ Sr2B5 should be
a potential high-temperature superconductor if $\sigma$ electronic bands
couple strongly to the corresponding phonon modes. We also noticed that Ca2B5
has a boron contributed NEf higher than that of MgB2. However, it is a mixture
of $\sigma$ and $\pi$ electrons. The superconductivity of Ca2B5 might not be
able to surpass that of MgB2.

To investigate the potential superconductivity in these AE-metal borides,
electron-phonon coupling simulations were performed. Encouragingly, the
estimated EPC parameters, $\lambda$, of $F$$\overline{4}$3$m$ Sr2B5 is as high
as 3.32 at 40 GPa. This value not only is nearly 3-4 times higher than that of
MgB2 but also higher than that of any $sp^{3}$-hybridized borocarbides studied
so far and close to that of high-$T_{\mathrm{c}}$ clathrate superhyrides. As
depicted in Fig. 4(a), the large coupling strength is derived primarily from
the $T_{2}$ and E modes over the whole Brillouin zone, which corresponded to
the stretching of Sr-B bonds (2-7 THz) and rocking vibrations of B4 units
(8-10 THz). The integral ($\lambda$($\omega$)) of the frequency divided
Eliashberg phonon spectral function ($\alpha^{2}F(\omega)/\omega$) suggests
these two modes contribute almost equally to the overall $\lambda$. This
phenomenon is different from MgB2 and metal borocarbides but similar to
$\alpha$-MoB2 as metal atom shows pronounced contribution to NEf. Since the
electron-phonon coupling is quite strong in $F$$\overline{4}$3$m$ Sr2B5
($\lambda$ > 1.5), the $T_{\mathrm{c}}$ values and superconducting energy gaps
($\Delta$) were evaluated through the direct numerical solutions of the
anisotropic Migdal-Eliashberg equations Eliashberg (1960); Sanna et al.
(2018); Giustino et al. (2007) using the calculated logarithmic average
frequency ($\omega_{\mathrm{log}}$) and typical Coulomb potential parameters
($\mu$*) of 0.1, 0.13, and 0.16 . The estimated single $\sigma$ gap is about
22 meV (Fig. S8 SI ) and $T_{\mathrm{c}}$ value is 105 K using $\mu$* = 0.13
for $F$$\overline{4}$3$m$ Sr2B5 at 40 GPa (Fig. 4), tripling that of MgB2 and
validating the observation of $\lambda$ (those for $\mu$* = 0.1, 0.16 are also
provided in Fig.4 and Table S3 SI ). This predicted $T_{\mathrm{c}}$ certainly
exceeds the boiling point of liquid-nitrogen (77 K), setting a new
superconducting record among all known thermodynamically stable metal borides
to the best of our knowledge.

Given the low mass of B, previous studies have demonstrated that anharmonicity
played an important role in determining the superconductivity of MgB2,
especially for the $E_{2g}$ phonons Liu et al. (2001); Yildirim et al. (2001);
Lazzeri et al. (2003). We then carried out additional EPC calculations to
investigate the occurrence of anharmonic effects using the stochastic self-
consistent harmonic approximation Errea et al. (2014). As shown in Fig. 4, the
anharmonic correction only hardens and broadens the $B_{4}$ rocking E modes,
while other phonon modes barely change, similar to that of MgB2. The average
phonon frequency of E modes increases from 9.3 THz to 11.6 THz and the width
of E modes is increased by 1 THz. Meanwhile, E modes become the major
contributor to the anharmonic Eliashberg function, accounting for 71$\%$ of
total $\lambda^{\mathrm{anh}}$, while the contribution of $T_{2}$ modes
reduces to 25$\%$, indicating the superconductivity of $F$$\overline{4}$3$m$
Sr2B5 should be mainly determined by the coupling between $\sigma$ bands and E
phonons. This results in a reduction of the anharmonic EPC parameter
$\lambda^{\mathrm{anh}}$ to 2.02, 40$\%$ smaller than the harmonic one. In
contrast, the anharmonicity enhances the logarithmic average frequency
$\omega_{\mathrm{log}}$ by 34$\%$ from 354 K to 477 K. Consequently, the
anharmonic $T_{\mathrm{c}}^{\mathrm{anh}}$ value is dropped by 10$\%$ to 95 K,
which still is quite high.

As for other newly predicted borides, only Ca2B5, $P$63/$mmc$ Sr2B5, and BaB
exhibit superconductivity, with calculated $T_{\mathrm{c}}$ values of 3.6 K,
6.4 K, and 3.9 K at 30, 50, and 80 GPa, respectively. The details of
superconducting properties are shown in Table S4 SI .

Figure 4: (Color online) (a) Phonon dispersion relations, Eliashberg spectral
function, (b) the electron-phonon coupling parameter $\lambda$, the
logarithmic average frequency $\omega_{\mathrm{log}}$ and the superconducting
temperature $T_{\mathrm{c}}$ of $F$$\overline{4}$3$m$ Sr2B5 by performing
harmonic and anharmonic calculations at 40 GPa.

The high-temperature superconductivity of $F$$\overline{4}$3$m$ Sr2B5 drives
us to further evaluate its synthesizability. As shown in Fig. S10 SI , the
pressure-temperature phase diagram of Sr2B5 was constructed up to 100 GPa and
2000 K via harmonic free-energy calculations since most of the metal borides
are synthesized at high temperatures Friedrich et al. (2011). Clearly, once
temperature effects are considered, the stable range of $F$$\overline{4}$3$m$
Sr2B5 increases from 38-44 GPa at 0 K to 34-65 GPa at 2000 K, suggesting high
temperature is good for the stabilization of $F$$\overline{4}$3$m$ Sr2B5.
Thus, $F$$\overline{4}$3$m$ Sr2B5 could be synthesized at moderate pressure by
the large-volume press (LVP) since its stable range lies within the operating
pressure and temperature range of LVP Ishii et al. (2019). Compared to
high-$T_{\mathrm{c}}$ BCS hydride superconductors, which can only be realized
by the diamond anvil cell with micrometer-sized samples, the production of
$F$$\overline{4}$3$m$ Sr2B5 shows a certain advantage as LVP can provide much
larger millimeter-sized samples for structure characterization and properties
measurements. Moreover, we also found $F$$\overline{4}$3$m$ Sr2B5 is
dynamically stable down to ambient pressure with anharmonic corrections (Fig.
S11 SI ), implying it might be able to retain to 0 GPa with the proper
quenching method. The $T_{\mathrm{c}}^{\mathrm{anh}}$ of $F$$\overline{4}$3$m$
Sr2B5 at 0 GPa is predicted to be even higher around 115 K (Fig. S12 SI ).

As part of efforts to develop high-temperature BCS superconductors, we
designed a promising thermodynamically stable $F$$\overline{4}$3$m$ Sr2B5 with
record-high $T_{\mathrm{c}}$ at moderate synthetic pressure ($\sim$ 40 GPa).
We note this is the first fully metallic $sp^{3}$-hybridized boron framework
example of metal borides, and we emphasize that the established B-B $\sigma$
bonding is a key to achieving high-$T_{\mathrm{c}}$ at low pressures. For
instance, if BD atom is also replaced by the strongly coupled $B_{4}$ unit in
$F$$\overline{4}$3$m$ Sr2B5, the $T_{\mathrm{c}}$ might be even higher.
Further investigation of other metal borides and lightweight compounds
consisting of similar $\sigma$-bonding frameworks is expected which may lead
to the discovery of potential high-$T_{\mathrm{c}}$ BCS superconductor (> 77
K) at ambient pressure.

In conclusion, our extensive first-principles structure searches for AE-metal
(Mg, Ca, Sr, Ba) borides have revealed the appearance of stable Mg3B8, Ca2B5,
Ca3B5, Sr2B5, and BaB below 100 GPa. Among them, $F$$\overline{4}$3$m$ Sr2B5
consisting of fully metallic $sp^{3}$-hybridized B-B $\sigma$ bonds exhibits
potential high-$T_{\mathrm{c}}$ of 105 K at 40 GPa, with the possibility
retaining to ambient pressure, that originates from the strong coupling
between $\sigma$ electronic bands and rocking vibrations of $B_{4}$ units.
This unique boron framework resembles the diamond structure, serving as a
design guidance for future investigation. Our results provide a useful
structural prototype for designing high-temperature superconductors at low
pressures and will stimulate further experimental synthesis and theoretical
predictions for a large variety of metallic $sp^{3}$-hybridized lightweight
compounds.

This work is supported by by National Key Research and Development Program of
China (Grant No. 2022YFA1402304), the National Natural Science Foundation of
China (Grant no. 12022408, 12374008, 1202290458, 52288102, 52090024, 12074138,
12374007), the Interdisciplinary Integration and Innovation Project of JLU,
Fundamental Research Funds for the Central Universities, Program for Jilin
University Science and Technology Innovative Research Team (2021TD–05), Jilin
Province Science and Technology Development Program (Grant No.
YDZJ202102CXJD016), the Strategic Priority Research Program of Chinese Academy
of Sciences (Grant No. XDB33000000) and computing facilities at the High-
Performance Computing Centre of Jilin University.

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# General quantum Chinos games

Daniel Centeno and Germán Sierra
Instituto de Física Teórica UAM-CSIC, Universidad Autónoma de Madrid,
Cantoblanco, Madrid, Spain

###### Abstract

The Chinos game is a non-cooperative game between players who try to guess the
total sum of coins drawn collectively. Semiclassical and quantum versions of
this game were proposed by F. Guinea and M. A. Martín-Delgado, in J. Phys. A:
Math. Gen. 36 L197 (2003), where the coins are replaced by a boson whose
number occupancy is the aim of player’s guesses. Here, we propose other
versions of the Chinos game using a hard-core boson, one qubit and two qubits.
In the latter case, using entangled states the second player has a stable
winning strategy that becomes symmetric for non-entangled states. Finally, we
use the IBM Quantum Experience to compute the basic quantities involved in the
two-qubit version of the game.

##  I Introduction

Game theory is a field that has fascinated mathematicians since the early 20th
century. The idea is to be able to interpret complex problems in many
different fields as a game of one or more players who, using logical
reasoning, try to optimize their strategies to obtain the highest possible
profit. The pioneers who formalized game theory were John von Neumann and
Oskar Morgenstern Neumann , followed soon later by John Nash Nash . A
fundamental concept is that of the Nash equilibrium, in which it is assumed
that every player knows and chooses their best possible strategy knowing the
strategies of the other players. This implies that the Nash equilibrium
situation is one in which no player would profit by changing her/his strategy
if the other players maintain theirs. Most of multiplayer games tend to reach
Nash equilibrium after a certain number of iterations. This area of study
appeared with the aim of better understanding the economy, but it quickly
extended to biology biologia , politics politica , computation and computer
science computacion  (see review games ).

In recent decades, physicists have begun to introduce features of the quantum
world into game theory in order to gain advantages over classical strategies
Meyers -K18 . There are several reasons why quantifying games can be
interesting. The first is simply because of the number of applications game
theory has had in different fields. Moreover, its probabilistic nature makes
one want to extend it to quantum probabilities. Another reason is the
connection between game theory and quantum information theory. In fact, in the
games themselves, players transmit information to other players and, since our
world is quantum, it can be interpreted as quantum information Eisert .

There are several examples of well-known quantum game models. One is the
prisoner’s dilemma where, by exploiting the peculiarities of quantum
behaviour, both players can escape from the dilemma Eisert . Another case is
the PQ game (flipping or not flipping a coin a certain number of times each
player), where it has been shown that if the first player can use quantum
strategies (superposition of both options) she/he will always win no matter
what action the second player takes Meyers . Not only have quantum games been
proposed, but they are already being used to model, for example, human
decision-making behaviour Y08 -Y20 .

This paper deals with the well-known (in Spain) Chinos game, which
traditionally consists of a group of people who hide a certain number of coins
in their hands. The aim of each of them is to guess the total number of coins
hidden by all of them. The Chinos game is a variant of the Morra game that is
played with fingers instead of coins. It dates back thousands of years to
ancient Roman and Greek times morra . This simple game shows a variety of
behavioural patterns that have been used to model financial markets and
information transmission Pastor .

This is a non-cooperative game, in which each player will seek to maximize
their chances of victory and minimize those of the other players. For this
reason, throughout the work we will always look for this situation in the
analysis of the possible strategies. In other words, the Nash equilibrium of
the game model will be pursued. We must stress the importance of entanglement
in quantum games when looking for the Nash equilibrium entrelazamiento ;
entrelazamiento2 ; entrelazamiento3 . It is worth mentioning that some quantum
games like the prisioner’s dilemma and the PQ game have been already
implemented using IBM quantum computers IBM ; IBM2 .

The organization of the paper is as follows. In section II we define in an
abstract manner the Chinos game. In section III we review a classical Chinos
game. In section IV we review the semiclassical model of the Chinos game
proposed in GMD  and we propose two new semiclassical models. In section V we
define a one-qubit game using unitary transformations. In section VI we
introduce a two-qubit game that incorporates entanglement. In section VII we
implement the latter game using an IBM quantum computer, and finally, in
section VIII we state our conclusions and prospects.

##  II Abstract Chinos game for two players

Let Alice and Bob have access to a set of objects ${\cal C}$. At the start of
the game, Alice chooses one object from ${\cal C}$, denoted $c_{A}$, and Bob
chooses an object $c_{B}\in{\cal C}$. Both players are unaware of the other
player’s choice. Alice and Bob then send $c_{A}$ and $c_{B}$ to a device that
assigns them an object $g_{AB}$ of a set ${\cal G}$. The goal of the players
is to guess $g_{AB}$. To do so there is a set ${\cal\tilde{C}}$ where Alice
and Bob selects the objects $\tilde{c}_{A}$ and $\tilde{c}_{B}$ respectively,
and send them to the previous device that assigns them the objects $g_{A}$ and
$g_{B}$ in ${\cal G}$. The protocol is shown in Fig. 1.

Figure 1: Protocol of the Chinos game

In order to quantify how good the players’ guesses are, we shall use a
distance $d(g,g^{\prime})$ between objects $g,g^{\prime}\in{\cal G}$, that is,
${\cal G}$ is a metric space. The player whose guess has the shortest distance
to the joint object will win. This means that both players will try to
minimize the distances $d(g_{A},g_{AB})$ and $d(g_{B},g_{AB})$. In the Chinos
game, Bob’s guess must be different from Alice’s. We shall call this the
restriction rule. This is implemented by imposing a minimum distance between
both guesses, that is $d(g_{A},g_{B})\geq d_{0}>0$. To check this condition,
Alice sends her guess $g_{A}$ to Bob, who calculates the distance with his
guess $g_{B}$. Finally, we shall suppose that Alice and Bob are intelligent
agents. This implies that Alice’s choice of $c_{A}$ can determine a set of
choices of $\tilde{c}_{A}$ whose associated guesses $g_{A}$ may have some
chance of winning. The same criteria apply to Bob. The intelligence of the
players is done by a map ${\cal I}:{\cal C}\rightarrow 2^{\cal\tilde{G}}$,
where $2^{\cal\tilde{G}}$ is the set of all partitions of ${\cal\tilde{G}}$.

In summary, the ingredients of the game are

  * •

Player’s choices: ${\cal C}$ and ${\cal\tilde{C}}$.

  * •

Player’s guesses: ${\cal G}$.

  * •

Device: choices $\rightarrow$ guesses.

  * •

Payoffs: distances.

  * •

Restriction rule: minimal distance.

  * •

Intelligence rule.

A consequence of the triangle inequality of the metric and the restriction
placed of Bob’s guesses is

$d_{0}\leq d(g_{A},g_{B})\leq d(g_{A},g_{AB})+d(g_{B},g_{AB})\,.$ (1)

This inequality implies that if the distance of Alice’s conjecture to $g_{AB}$
is less than $d_{0}/2$ then Bob’s guess will be more than $d_{0}/2$ and
therefore he will lose the game.

We are going to present below several concrete realizations of the Chinos
game. These can be classified into classical games or quantum games according
to the classical or quantum nature of the space ${\cal G}$.

##  III Classical Chinos game

The game is defined by two parameters: $N_{p}$ that is the number of players
and $N_{c}$ that is the number of coins that each player holds in her/his
hand. In each round, each player draws from 0 to $N_{c}$ coins and guesses the
total number of coins that they all draw, with the restriction that the result
predicted by the previous players cannot be repeated GMD .

We shall consider below the game with only two players, $N_{p}=2$, and one
coin, $N_{c}=1$. The best strategy for Alice is to choose randomly $c_{A}$
coins and to guess always $g_{A}=1$, so as not to reveal information to Bob
GMD . This is also based on the fact that with four possible tosses the most
likely value of the sum is 1. Bob’s best strategy is to choose $c_{B}$ at
random and make his attempt $g_{B}$ in an “intelligent” way. This means that,
if Bob chooses $c_{B}$=0, then he must exclude the option $g_{B}$=2 and if he
chooses $c_{B}$=1, he must exclude $g_{B}$=0. Table 1 shows all the possible
options assuming that each player follows her/his best strategy.

$c_{A}$ | $c_{B}$ | $g_{AB}$ | $g_{A}$ | $g_{B}$ | Winner
---|---|---|---|---|---
0 | 0 | 0 | 1 | 0 | B
0 | 1 | 1 | 1 | 2 | A
1 | 0 | 1 | 1 | 0 | A
1 | 1 | 2 | 1 | 2 | B
Table 1: All possible options for the Chinos game with $N_{p}=2,N_{c}=1$.

It is clear that each player will win half of the time and therefore their
winning probabilities are equal, that is

$P_{A}=P_{B}=\frac{1}{2}.$ (2)

Moreover, their strategies are stable. In reference GMD  it was shown that (2)
also holds for two players and a generic number of coins $N_{c}$. This was
called a classical symmetry between the players. These authors also proposed
in GMD  a semiclassical and quantum version of the Chinos game to test whether
the classical symmetry (2) is broken by quantum fluctuations.

From the formal viewpoint introduced in section II, this game involves the
following sets:

${\cal C}=\\{0,1\\},\quad{\cal\tilde{C}}={\cal G}=\\{0,1,2\\}\,.$ (3)

where ${\cal C}$ contains the number of coins, $c_{A}$ and $c_{B}$, of Alice
and Bob, ${\cal G}$ contains their guesses $g_{A}$ and $g_{B}$ for the total
number of coins $g_{AB}$. The device operates as follows

$g_{A}=\tilde{c}_{A},\quad g_{B}=\tilde{c}_{B},\quad g_{AB}=c_{A}+c_{B}\,.$
(4)

The guesses $g_{A},g_{B}$ are identical to the choices
$\tilde{c}_{A},\tilde{c}_{B}$, but not in the games we shall consider below.
The metric in ${\cal G}$ is given by

$d(g,g^{\prime})=|g-g^{\prime}|,\qquad g,g^{\prime}\in{\cal G}\,.$ (5)

and the minimal distance is set to $d_{0}=1$. The restriction rule amounts to

$\displaystyle{\rm if}\;g_{A}=0$ $\displaystyle\rightarrow g_{B}=1,2\;,$ (6)
$\displaystyle{\rm if}\;g_{A}=1$ $\displaystyle\rightarrow g_{B}=0,2\;,$
$\displaystyle{\rm if}\;g_{A}=2$ $\displaystyle\rightarrow g_{B}=0,1\;,$

and the intelligent rule to

$\displaystyle{\rm if}\;c_{A/B}=0$
$\displaystyle\rightarrow\tilde{c}_{A/B}=g_{A/B}=0,1\;,$ (7)
$\displaystyle{\rm if}\;c_{A/B}=1$
$\displaystyle\rightarrow\tilde{c}_{A/B}=g_{A/B}=1,2\;.$

##  IV Semiclassical models

A semi-classical version of the Chinos game consists of replacing coins with a
quantum degree of freedom and measuring an observable on it. The value taken
by that observable is the goal of the players’ guesses GMD . We shall first
review the proposal of reference GMD  that uses a single boson degree of
freedom and later on we shall propose two related games.

Let $b$ and $b^{\dagger}$ be the bosonic creation and annihilation operators
satisfying the canonical commutation rule [$b,b^{\dagger}$] = 1. They generate
the bosonic states in the standard way: $b|0\rangle$ = 0,
$b^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle$ where
$|n\rangle=(b^{\dagger})^{n}|0\rangle/\sqrt{n!}$.

We will study the case of two players. Each of them can act on the bosonic
state $|0\rangle$ with one of the following operators

$O_{1}=I,O_{2}=\frac{I+b^{\dagger}}{\sqrt{2}},O_{3}=\frac{I-b^{\dagger}}{\sqrt{2}},O_{4}=b^{\dagger}.$
(8)

This is the quantum analogue of drawing a coin. Let us denote by $O_{i}^{A}$
the operator chosen by Alice and by $O_{j}^{B}$ the operator chosen by Bob.
The joint state produced by both players is given by

$|\Psi_{i,j}\rangle=N^{-1/2}_{i,j}O_{i}^{A}O_{j}^{B}|0\rangle=\sum_{n=0}^{2}c_{i,j}(n)|n\rangle,$
(9)

where $N_{i,j}$ is a normalization constant. The operators $O_{1}$ and
$O_{4}$, are equivalent to the classical choices of 0 and 1 coins
respectively. In this case the state (9) is simply $|0\rangle,|1\rangle$ or
$|2\rangle$, that brings us back to the classical game. Allowing Alice and Bob
to use also $O_{2}$ and $O_{3}$ generates linear superpositions in (9), that
leads to probabilistic outcomes. They will be analyzed in terms of the
probability of finding the state $|n\rangle$ in (9)

$p_{i,j}(n)=|\langle n|\Psi_{i,j}\rangle|^{2}=c_{i,j}^{2}(n).$ (10)

Suppose the both players follow the classical strategy, i.e. choose randomly
among the operators $O_{i}$, $i=1,2,3,4$. The probability of Alice guessing
$n$ under the choice $O_{i}^{A}$ is given by the average over Bob’s choices,

$\langle p_{i}(n)\rangle=\frac{1}{4}\sum_{j=1}^{4}p_{i,j}(n).$ (11)

The results are given in Table 2.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$ | $O_{4}^{A}$
---|---|---|---|---
$\langle p(0)\rangle$ | 1/2 | 41/168 | 41/168 | 0
$\langle p(1)\rangle$ | 1/2 | 59/168 | 59/168 | 5/12
$\langle p(2)\rangle$ | 0 | 68/168 | 68/168 | 7/12
Table 2: Average odds of getting 0, 1 or 2 for Alice.

Hence if Alice draws $O_{1}^{A}$ her best guess is 0 (or 1) and on drawing
$O_{2}^{A}$, $O_{3}^{A}$ and $O_{4}^{A}$ her best guess is 2. Therefore, the
total probability of winning for Alice is

$P_{A}=\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{68}{168}+\frac{1}{4}\cdot\frac{7}{12}=\frac{53}{112}<\frac{1}{2}.$
(12)

After many rounds Alice realizes that she is loosing the game and decides to
change her strategy by randomly choosing between the operators $O_{1}^{A}$ and
$O_{4}^{A}$. Her probability of winning changes to

$P_{A}=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{7}{12}=\frac{13}{24}>\frac{1}{2}.$
(13)

Bob then notices this imbalance and decides to choose randomly between the
operators $O_{1}^{B}$ and $O_{4}^{B}$, which reproduces the classical game
where the classical symmetry (2) is restored GMD .

The formalization of the game is as follows. Alice and Bob choose $i,j\in{\cal
C}$ where

${\cal C}=\\{1,2,3,4\\},$ (14)

and send them to the device that constructs the state (9) belonging to the
Hilbert space expanded by the boson states with 0, 1 and 2 occupancies,

${\cal G}={\rm Span}\\{|0\rangle,|1\rangle,|2\rangle\\}\ .$ (15)

Each player’s guess of these occupancies, $n$, is mapped by device into the
basis of ${\cal G}$ as

$n\in{\cal\tilde{C}}=\\{0,1,2\\}\rightarrow|n\rangle\in{\cal G}\,.$ (16)

To define a distance between two states $|g\rangle,|g^{\prime}\rangle\in{\cal
G}$, we use the trace distance between the corresponding density matrices,
$\rho_{g}=|g\rangle\langle g|$ and
$\rho_{g^{\prime}}=|g^{\prime}\rangle\langle g^{\prime}|$ nielsen

$d(|g\rangle,|g^{\prime}\rangle)=\frac{1}{2}{\rm
tr}|\rho_{g}-\rho_{g^{\prime}}|=\sqrt{1-|\langle g|g^{\prime}\rangle|^{2}}\,.$
(17)

The probability (10) is related to this distance as

$d(|n\rangle,|\Psi_{i,j}\rangle)=\sqrt{1-p_{i,j}(n)}.$ (18)

Hence minimizing the distance of the guess $|n\rangle$ to the state
$|\Psi_{i,j}\rangle$ is equivalent to maximizing the probability $p_{i,j}(n)$.
The latter quantity is equal to the square of the fidelity between these
states.

###  IV.1 Hard-core boson

We shall next propose a model where the boson operator $b^{\dagger}$, used
previously, is replaced by a hard-core boson, that is, an operator satisfying
the condition:

$(b^{\dagger})^{2}=0.$ (19)

This forces the elimination of the operator $O_{4}$ as an option since its use
by Alice and Bob would lead to a null move. This also implies that the
classical version of the chinos game is not included in the new version.

The study of strategies is based again on the probabilities of obtaining 0 and
1 for each possible move. They are given in Table 3.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$
---|---|---|---
$O_{1}^{B}$ |  |  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = 1/2
---
$p(1)$ = 1/2
|  $p(0)$ = 1/2
---
$p(1)$ =1/2
$O_{2}^{B}$ |  |  $p(0)$ =1/2
---
$p(1)$ = 1/2
|  $p(0)$ = 1/5
---
$p(1)$ = 4/5
|  $p(0)$ = 1
---
$p(1)$ = 0
$O_{3}^{B}$ |  |  $p(0)$ =1/2
---
$p(1)$ = 1/2
|  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = 1/5
---
$p(1)$ = 4/5
Table 3: Odds of getting 0 or 1 for all possible moves in the game with hard-
core bosons.

Following the classical strategy, the players will draw at random among the
operators $O_{i}$, $i=1,2,3$. Averaging over Bob’s choices, the probabilities
of Alice getting 0 and 1 are given in Table 4.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$
---|---|---|---
$\langle p(0)\rangle$ | 2/3 | 17/30 | 17/30
$\langle p(1)\rangle$ | 1/3 | 13/30 | 13/30
Table 4: Average odds of getting 0, 1 or 2 for Alice.

It is clear that Alice’s best guess is 0, regardless her choice of operator,
with a winning probability

$P_{A}=\frac{1}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\frac{17}{30}=\frac{3}{5}>\frac{1}{2}.$
(20)

To reverse this outcome Bob will choose randomly between the $O_{2}$ and
$O_{3}$ that seem more favorable from Table 4. However, Alice’s winning
probability is still higher than a half,

$\frac{1}{2}<P_{A}=\frac{17}{30}<\frac{3}{5}.$ (21)

Hence, Alice has achieved a winning and stable strategy that breaks the
symmetry of the classical game.

To improve the chances of Bob we shall use the following operators

$O_{1}=I,O_{2}=cI+sb^{\dagger},O_{3}=cI-sb^{\dagger},$ (22)

where $c=\cos(\theta),s=\sin(\theta)$ with $\theta\in(0,\pi/2)$. The previous
case corresponds to $\theta=\pi/4$. The values $\theta=0,\pi/2$ are excluded
because in these cases $O_{2}=\pm O_{3}$.

The probabilities of obtaining 0 or 1 in the boson occupation are shown in
Table 5.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$
---|---|---|---
$O_{1}^{B}$ |  |  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = $c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = $c^{2}$
---
$p(1)$ =$s^{2}$
$O_{2}^{B}$ |  |  $p(0)$ =$c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = $\frac{c^{2}}{c^{2}+4s^{2}}$
---
$p(1)$ = $\frac{4s^{2}}{c^{2}+4s^{2}}$
|  $p(0)$ = 1
---
$p(1)$ = 0
$O_{3}^{B}$ |  |  $p(0)$ =$c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = $\frac{c^{2}}{c^{2}+4s^{2}}$
---
$p(1)$ = $\frac{4s^{2}}{c^{2}+4s^{2}}$
Table 5: Odds of getting 0 or 1 for all possible moves as a function of
$\theta$.

Applying the classical strategies, the probabilities for Alice averaged over
Bob’s choices are given in Table 6 and plotted in Fig. 2.

| $O_{1}^{A}$ | $O_{2,3}^{A}$
---|---|---
$\langle p(0)\rangle$ | $\frac{1}{3}(1+2c^{2})$ | $\frac{2c^{2}+4s^{2}+c^{4}+4c^{2}s^{2}}{3(c^{2}+4s^{2})}$
$\langle p(1)\rangle$ | $\frac{2}{3}s^{2}$ | $\frac{s^{2}c^{2}+4s^{4}+4s^{2}}{3(c^{2}+4s^{2})}$
Table 6:  Average odds of getting 0 or 1 for Alice. Figure 2: Plot of the
probabilities of Table 6

For every Alice’s strategy, the probabilities $\langle p(n)\rangle(n=0,1)$
cross at $1/2$ at the angles $\theta_{1}<\theta_{2}$ given by (the probability
of $O_{2}^{A}$ and $O_{3}^{A}$ are the same)

$\displaystyle\langle p(0)\rangle_{O_{2}^{A}}$ $\displaystyle=$
$\displaystyle\langle
p(1)\rangle_{O_{2}^{A}}=\frac{1}{2}\Longrightarrow\theta_{1}=0.9155..$ (23)
$\displaystyle\langle p(0)\rangle_{O_{1}^{A}}$ $\displaystyle=$
$\displaystyle\langle
p(1)\rangle_{O_{1}^{A}}=\frac{1}{2}\Longrightarrow\theta_{2}=1.0472..$

Consequently the strategies followed by Alice depend on the value of $\theta$
as follows

  * •

0 $<\theta<\theta_{1}$: Alice will guess 0 for all $O_{i}^{A}$.

  * •

$\theta_{1}<\theta<\theta_{2}$ : Alice will guess 0 for $O_{1}^{A}$, and 1 for
$O_{2}^{A}$ or $O_{3}^{A}$.

  * •

$\theta_{2}<\theta<\pi/2$ : Alice will guess 1 for all $O_{i}^{A}$.

Averaging the corresponding probabilities yields a value of $P_{A}$ larger
than $1/2$ for all $\theta^{\prime}s$, so reproducing the previous case where
$\theta=\pi/4$. An analysis of the strategies that Bob may adopt does not
change the situation even at the values of $\theta_{1}$ and $\theta_{2}$. We
conclude that Alice always has a winning and stable strategy in this game.

The sets ${\cal C},{\cal\tilde{C}}$ and the space ${\cal G}$ involved in the
hard-core boson game are simply the truncation of those used in the boson
game, namely

$\displaystyle{\cal C}$ $\displaystyle=$
$\displaystyle\\{1,2,3\\},\;{\cal\tilde{C}}=\\{0,1\\},\;{\cal G}={\rm
Span}\\{|0\rangle,|1\rangle\\},$ (24)

while the distance for ${\cal G}$ is the same as in(18).

##  V One-qubit games

In previous games, the players used the operators $O_{i}$ that are not
unitary, which makes it difficult to carry out experimentally. In this section
and the next one we are going to propose two games that solve this difficulty.
In the first, game we shall replace the hard-core boson by a qubit initialized
in the state $|0\rangle$ on which the players act choosing between the
following unitary transformations

$O_{1}=I,\quad O_{2}=e^{-\frac{i\theta}{2}\sigma^{y}},\quad
O_{3}=e^{\frac{i\theta}{2}\sigma^{y}}\,,$ (25)

with $\theta\in(0,\pi/2)$. The operators $O_{2}$ and $O_{3}$ are rotations of
angles $\theta$ and $-\theta$ around the $y$-axis. The state constructed by
the device with the information provided by Alice and Bob is

$|\Psi_{i,j}\rangle=O^{A}_{i}O^{B}_{j}|0\rangle$ (26)

This state is already normalized since the operators (25) are unitary.

The probabilities of obtaining the states $|0\rangle$ or $|1\rangle$ are shown
in Table 7. Notice the similarities with Table 5.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$
---|---|---|---
$O_{1}^{B}$ |  |  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = $c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = $c^{2}$
---
$p(1)$ =$s^{2}$
$O_{2}^{B}$ |  |  $p(0)$ =$c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = $c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = 1
---
$p(1)$ = 0
$O_{3}^{B}$ |  |  $p(0)$ =$c^{2}$
---
$p(1)$ = $s^{2}$
|  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = $c^{2}$
---
$p(1)$ = $s^{2}$
Table 7: $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$.

Applying the classical strategies, the probabilities for Alice averaged over
Bob’s choices are given in Table 8 and plotted in Fig. 3.

| $O_{1}^{A}$ | $O_{2,3}^{A}$
---|---|---
$\langle p(0)\rangle$ | $\frac{1}{3}(1+2c^{2})$ | $\frac{1}{3}(2-3c^{2}+4c^{4})$
$\langle p(1)\rangle$ | $\frac{2}{3}s^{2}$ | $\frac{1}{3}(1+3c^{2}-4c^{4})$
Table 8:  Average odds for Alice. Figure 3: Plot of probabilities given in
Table 8.

As in the previous game there are three regions where Alice’s strategy is
winning and stable. However, at their boundaries, Bob is able to balance the
game, restoring the classical symmetry. The values of $\theta$ at the
boundaries are given by

$\displaystyle\langle p(0)\rangle_{O_{2,3}^{A}}$ $\displaystyle=$
$\displaystyle\langle
p(1)\rangle_{O_{2,3}^{A}}=\frac{1}{2}\Longrightarrow\theta_{1}^{\prime}=\frac{\pi}{2},$
(27) $\displaystyle\langle p(0)\rangle_{O_{1,2,3}^{A}}$ $\displaystyle=$
$\displaystyle\langle
p(1)\rangle_{O_{1,2,3}^{A}}=\frac{1}{2}\Longrightarrow\theta_{2}^{\prime}=\frac{2\pi}{3}.$

The strategies followed by Alice are the following:

  * •

0 $\leq\theta<\theta_{1}^{\prime}$ : Alice will guess 0 regardless of her
choice of operator between $O_{1}^{A}$, $O_{2}^{A}$ and $O_{3}^{A}$.

  * •

$\theta_{1}^{\prime}<\theta<\theta_{2}^{\prime}$ :Alice will guess 0 if she
chooses $O_{1}^{A}$ and 1 if she chooses $O_{2}^{A}$ or $O_{3}^{A}$.

  * •

$\theta_{2}^{\prime}<\theta<\pi/2$ : Alice will guess 1 if she chooses
$O_{1}^{A}$ and 0 if she chooses $O_{2}^{A}$ or $O_{3}^{A}$.

At the boundary points we have the following situations:

  * •

$\theta=\theta_{1}^{\prime}$: Bob notices that choosing randomly between
$O_{2}^{B}$ and $O_{3}^{B}$ the game becomes symmetric for all Alice’s draws.
Table 9 shows the corresponding probabilities. Alice has therefore a winning
but unstable strategy as Bob can balance it.

| $O_{1}^{A}$ | $O_{2}^{A}$ | $O_{3}^{A}$
---|---|---|---
$O_{2}^{B}$ |  |  $p(0)$ = 1/2
---
$p(1)$ = 1/2
|  $p(0)$ = 0
---
$p(1)$ = 1
|  $p(0)$ = 1
---
$p(1)$ = 0
$O_{3}^{B}$ |  |  $p(0)$ =1/2
---
$p(1)$ = 1/2
|  $p(0)$ = 1
---
$p(1)$ = 0
|  $p(0)$ = 0
---
$p(1)$ = 1
Table 9: Values given in Table 7 for $\theta^{\prime}_{1}$ and Bob’s choices
$O_{2}^{B}$ and $O_{3}^{B}$

  * •

$\theta=\theta_{2}^{\prime}$ : At this point all the curves in Fig. 3
intersect at the value of 1/2. Bob does not even have to change his strategy
to make the game symmetrical again. Moreover, Alice can do nothing to prevent
this. This is observed by evaluating the probabilities shown in Table 7 at
$\theta_{2}^{\prime}$.

We conclude that in the one-qubit Chinos game with unitary operators, Alice
will have a winning and stable strategy for any angle $\theta\in(0,\pi)$
except at the values $\theta_{1}^{\prime}$ and $\theta_{2}^{\prime}$. In the
former, Bob can choose a strategy that balances the game, and in the latter,
the game is symmetrized without the need for Bob to change his strategy. This
result is essentially the same as the one obtained using a hard-core boson,
where Alice has always a winning stable strategy.

In this one-qubit game, the sets ${\cal C},{\cal\tilde{C}}$ and the space
${\cal G}$ are the same as those of the hard-core boson given in eq.(24).

##  VI Two-qubits games

In the previous quantum games the players’ guesses where mapped by the device
onto the states belonging to the orthonormal basis of the Hilbert space ${\cal
G}$. Along the lines of reference GMD , we shall next propose games where this
condition is not imposed. This is implemented as follows. Alice and Bob will
guess the states $|g^{A}\rangle$ and $|g^{B}\rangle$ respectively, with the
condition that Bob’s state should be orthogonal to Alice’s state, that is

$\langle g^{A}|g^{B}\rangle=0.$ (28)

In view of eq.(17), this implies that the distance between the players’s guess
is exactly 1, i.e. $d_{0}=1$. This condition is the quantum version of the
classical rule where players cannot repeat the guesses of the previous ones.
Alice and Bob payoffs evaluate how successful each player was in predicting
the state generated jointly $|\Psi_{a,b}\rangle$,

$f^{A}=|\langle g^{A}|\Psi_{a,b}\rangle|^{2},\quad f^{B}=|\langle
g^{B}|\Psi_{a,b}\rangle|^{2}\,.$ (29)

Eq.(17) relates these quantities to the distance of the players’ guess to the
joint state. The winner of the game is the one with the highest payoff.

We shall consider a two qubit system on the computational basis
$|i_{1},i_{0}\rangle$ with $i_{0},i_{1}=0,1$. The operators that generates the
joint state $|\Psi_{a,b}\rangle$ will be those that produce the Bell states
acting on $|0,0\rangle$. The Bell states form an orthonormal basis of
maximally entangled states of a two-qubit system given by

$|\phi^{\pm}\rangle=\frac{|00\rangle\pm|11\rangle}{\sqrt{2}},\quad|\psi^{\pm}\rangle=\frac{|10\rangle\pm|11\rangle}{\sqrt{2}}\,.$
(30)

They can be constructed as

$\begin{split}|\phi^{+}\rangle=Bell(0,0)|00\rangle,\quad|\phi^{-}\rangle=Bell(0,1)|00\rangle,\\\
|\psi^{+}\rangle=Bell(1,0)|00\rangle,\quad|\psi^{-}\rangle=Bell(1,1)|00\rangle.\end{split}$
(31)

with

$Bell(i_{1},i_{0})=CNOT(id\otimes H)(X^{i_{1}}\otimes X^{i_{0}})\,,$ (32)

and where $H$ is the Hadamard gate, $X$ the NOT gate and $CNOT$ the gate

$CNOT|i_{1},i_{0}\rangle=|i_{1}\oplus i_{0},i_{0}\rangle,$ (33)

where $i_{0}$ is the control qubit and $i_{1}$ is the target qubit.

We shall use in what follows the notation

$\begin{split}O_{0}=Bell(0,0),\quad&O_{1}=Bell(0,1),\\\
O_{2}=Bell(1,0),\quad&O_{3}=Bell(1,1).\end{split}$ (34)

Using these operators Alice and Bob will generate the joint state

$|\psi_{a_{0},b_{0}}\rangle=O^{A}_{a_{0}}O^{B}_{b_{0}}|00\rangle\,,$ (35)

that is normalized because the operators (34) are unitary.

An interesting property of all the joint states (35) is that they are
maximally entangled. This can be proved computing the reduced density matrix
of the qubit 0

$\rho_{1}=\frac{1}{2}\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\,.\quad$ (36)

Alice’s and Bob’s guesses are chosen as

$\displaystyle|g^{A}_{a_{1},a_{2}}\rangle$ $\displaystyle=$ $\displaystyle
O^{A}_{a_{1}}O^{A}_{a_{2}}|00\rangle;\quad a_{1},a_{2}=0,1,2,3\,,$ (37)
$\displaystyle|g^{B}_{b_{1},b_{2}}\rangle$ $\displaystyle=$ $\displaystyle
O^{B}_{b_{1}}O^{B}_{b_{2}}|00\rangle;\quad b_{1},b_{2}=0,1,2,3\,,$

that should be orthogonal to satisfy (28),

$\langle g^{A}_{a_{1},a_{2}}|g^{B}_{b_{1},b_{2}}\rangle=0.$ (38)

The payoffs of the players are given by eq.(29) with $a,b$ replaced by
$a_{0},b_{0}$.

It is convenient to define a metric in the space expanded by the operators
$O_{i}O_{j}$ acting on $|00\rangle$ GMD ,

$\begin{split}G_{(i_{1},j_{1}),(i_{2},j_{2})}&=\langle
00|O^{\dagger}_{j_{1}}O^{\dagger}_{i_{1}}O_{i_{2}}O_{j_{2}}|00\rangle\,,\end{split}$
(39)

where $i_{1},j_{1},i_{2},j_{2}=0,1,2,3$. In terms of this metric the condition
eq. (38) becomes

$G_{(a_{1},a_{2}),(b_{1},b_{2})}=0,$ (40)

and the payoffs of each player,

$\begin{split}f^{A}_{(a_{1},a_{2}),(a_{0},b_{0})}=|G_{(a_{1},a_{2}),(a_{0},b_{0})}|^{2},\\\
f^{B}_{(b_{1},b_{2}),(a_{0},b_{0})}=|G_{(b_{1},b_{2}),(a_{0},b_{0})}|^{2}.\end{split}$
(41)

$G$ | 00 | 22 | 12 | 30 | 13 | 31 | 01 | 23 | 02 | 20 | 10 | 32 | 11 | 33 | 03 | 21
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
00 | 1 | 1 | $1$ | $1$ | $0$ | $0$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
22 | 1 | 1 | $1$ | $1$ | $0$ | $0$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 | $1$ | $1$ | 1 | 1 | 0 | 0 | $0$ | $0$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
30 | $1$ | $1$ | 1 | 1 | 0 | 0 | $0$ | $0$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 | $0$ | $0$ | 0 | 0 | 1 | 1 | $-1$ | $-1$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
31 | $0$ | $0$ | 0 | 0 | 1 | 1 | $-1$ | $-1$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
01 | 0 | 0 | $0$ | $0$ | $-1$ | $-1$ | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
23 | 0 | 0 | $0$ | $0$ | $-1$ | $-1$ | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
02 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | $1$ | $1$ | $0$ | $0$ | 0 | 0
20 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | $1$ | $1$ | $0$ | $0$ | 0 | 0
10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $1$ | $1$ | 1 | 1 | 0 | 0 | $0$ | $0$
32 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $1$ | $1$ | 1 | 1 | 0 | 0 | $0$ | $0$
11 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $0$ | $0$ | 0 | 0 | 1 | 1 | $-1$ | $-1$
33 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $0$ | $0$ | 0 | 0 | 1 | 1 | $-1$ | $-1$
03 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $0$ | $0$ | $-1$ | $-1$ | 1 | 1
21 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $0$ | $0$ | $-1$ | $-1$ | 1 | 1

Table 10: Metric $G_{(i_{1},j_{1}),(i_{2},j_{2})}$ defined in eq. (39) using
the operators of eq. (34) organized in the sets (42).

The entries of the metric (39) are given in Table 10 and has a block diagonal
structure in terms of the following subsets,

$\begin{split}Set_{1}=\\{00,22,12,30\\},&\quad Set_{2}=\\{13,31,01,23\\},\\\
Set_{3}=\\{02,20,10,32\\},&\quad Set_{4}=\\{11,33,03,21\\}.\end{split}$ (42)

Since these sets are mutually orthogonal, if Alice chooses a state belonging
to one set, then Bob has to choose a state belonging to a different set to
satisfy condition (40).

With this metric, $f^{A}$ and $f^{B}$ can be either 0 or 1, that is Alice and
Bob can either win or lose. In particular, for any of Alice’s choices, she can
win with a certain Bob’s choice of $b_{0}$, i.e.

$\forall
a_{0},a_{1},a_{2},\exists\,b_{0}\;/\;f^{A}_{(a_{1},a_{2}),(a_{0},b_{0})}=1\ .$
(43)

The situation of Bob is different. His choice of $b_{0}$ reduces the possible
winning guesses to the following sets

$\displaystyle{\rm if}\;b_{0}$ $\displaystyle=$ $\displaystyle 0\;{\rm
or}\;2\Longrightarrow(b_{1},b_{2})\in Set_{1}\;{\rm or}\;Set_{3}\,,$ (44)
$\displaystyle{\rm if}\;b_{0}$ $\displaystyle=$ $\displaystyle 1\;{\rm
or}\;3\Longrightarrow(b_{1},b_{2})\in Set_{2}\;{\rm or}\;Set_{4}$

as can be verified in Table 10.

Let us consider an example. Suppose that Alice chooses at random among the
four sets of eq. (42) and that Bob chooses $b_{0}=0$. We shall consider the
following cases:

  * •

Alice selects $(a_{1},a_{2})\in Set_{1}$ then Bob, applying eq.(44) and the
orthogonality rule (40), will choose $(b_{1},b_{2})\in Set_{3}$. The payoffs
(41) for $a_{0}=0,1,2,3$ and $b_{0}=0$ are given by

$f^{A}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$Set_{1}$ | 1 | 0 | 0 | 1

$f^{B}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$Set_{3}$ | 0 | 1 | 1 | 0

  * •

Alice selects $(a_{1},a_{2})\in Set_{2}$ then Bob choosing $(b_{1},b_{2})\in
Set_{1}\cup Set_{3}$. Their payoffs are

$f^{A}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$Set_{2}$ | 0 | 0 | 0 | 0

$f^{B}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$Set_{1}$ | 1 | 0 | 0 | 1
$Set_{3}$ | 0 | 1 | 1 | 0

  * •

Alice selects $(a_{1},a_{2})\in Set_{3}$. This case is similar to the choice
of $Set_{1}$.

  * •

Alice selects $(a_{1},a_{2})\in Set_{4}$. This case is similar to the choice
of $Set_{2}$.

The average payoffs of Alice and Bob are given by

$\displaystyle\langle f^{A}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{2}{4}+0\right)=\frac{1}{4}\,,$ (45)
$\displaystyle\langle f^{B}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{2}{4}+\frac{2}{4}\right)=\frac{1}{2}\,,$

that gives the normalized probabilities

$P_{A}=\frac{1}{3},\quad P_{B}=\frac{2}{3}.$ (46)

In view of this result, Alice will try to reverse the situation, but it will
not be possible since, according to eq. (43), given $a_{0}$, any set can make
her win depending on the value of $b_{0}$. We conclude that in this Chinos
game Bob has a winning and stable strategy. This is in contrast with with game
using bosons GMD  where Alice has a winning and stable strategy.

###  VI.1 States with different entanglement

We shall next extend the previous model replacing the CNOT gate with a CU gate
given by

$CU(|i_{1}\rangle\otimes|i_{0}\rangle)=(U(\theta)^{i_{0}}|i_{1}\rangle)\otimes|i_{0}\rangle,$
(47)

where

$U(\theta)=e^{-i\frac{\theta}{2}X}\,,$ (48)

is a rotation around the $x$-axis of angle $\theta$. The rotated Bell
operators will be defined as

$RBell(i_{1},i_{0})=CU(id\otimes H)(X^{i_{1}}\otimes X^{i_{0}}).$ (49)

Obviously, for $\theta=0$ we recover the states and operators considered
previously.

The operators (49) will be denoted as

$\begin{split}O_{0}^{\prime}=RBell(0,0);&\quad O_{1}^{\prime}=RBell(0,1);\\\
O_{2}^{\prime}=RBell(1,0);&\quad O_{3}^{\prime}=RBell(1,1).\end{split}$ (50)

and using them Alice and Bob will generate the states

$|\psi^{\prime}_{a_{0},b_{0}}\rangle=O^{{}^{\prime}A}_{a_{0}}O^{{}^{\prime}B}_{b_{0}}|00\rangle\,.$
(51)

$G^{\prime}$ | 00 | 22 | 12 | 30 | 13 | 31 | 01 | 23 | 02 | 20 | 10 | 32 | 11 | 33 | 03 | 21
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
00 | 1 | 1 | $c$ | $c$ | $s$ | $s$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
22 | 1 | 1 | $c$ | $c$ | $s$ | $s$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 | $c$ | $c$ | 1 | 1 | 0 | 0 | $s$ | $s$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
30 | $c$ | $c$ | 1 | 1 | 0 | 0 | $s$ | $s$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 | $-s$ | $-s$ | 0 | 0 | 1 | 1 | $-c$ | $-c$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
31 | $-s$ | $-s$ | 0 | 0 | 1 | 1 | $-c$ | $-c$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
01 | 0 | 0 | $-s$ | $-s$ | $-c$ | $-c$ | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
23 | 0 | 0 | $-s$ | $-s$ | $-c$ | $-c$ | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
02 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | $c$ | $c$ | $s$ | $s$ | 0 | 0
20 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | $c$ | $c$ | $s$ | $s$ | 0 | 0
10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $c$ | $c$ | 1 | 1 | 0 | 0 | $s$ | $s$
32 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $c$ | $c$ | 1 | 1 | 0 | 0 | $s$ | $s$
11 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $-s$ | $-s$ | 0 | 0 | 1 | 1 | $-c$ | $-c$
33 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $-s$ | $-s$ | 0 | 0 | 1 | 1 | $-c$ | $-c$
03 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $-s$ | $-s$ | $-c$ | $-c$ | 1 | 1
21 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | $-s$ | $-s$ | $-c$ | $-c$ | 1 | 1

Table 11: Metric $G^{\prime}_{(i_{1},j_{1}),(i_{2},j_{2})}$ as a function of
the parameter $\theta$ where $c=\cos\frac{\theta}{2}$ and
$s=i\sin\frac{\theta}{2}$.

The entanglement of the states (51) depends on the value of $\theta$. The
density matrix obtained by tracing over the qubit 0 reads

$\rho^{\prime}_{1}=\frac{1}{2}\begin{pmatrix}1&k\\\ k^{*}&1\end{pmatrix},\quad
k=\pm i\sin(\theta/2)\ ,$ (52)

and has eigenvalues $\lambda_{\pm}=\frac{1}{2}(1\pm\sin(\theta/2))$ that are
invariant under the replacement $\theta\rightarrow 2\pi-\theta$. This symmetry
allows us to restrict ourselves to the interval $\theta\in[0,\pi]$. The
maximal entanglement corresponds to $\theta=0$, and the minimum to
$\theta=\pi$ where the states (51) become separable.

The metric $G^{\prime}_{(i_{1},j_{1}),(i_{2},j_{2})}$ is defined by

$\begin{split}G^{\prime}_{(i_{1},j_{1}),(i_{2},j_{2})}&=\langle
00|O^{\prime\dagger}_{j_{1}}O^{\prime\dagger}_{i_{1}}O^{\prime}_{i_{2}}O^{\prime}_{j_{2}}|00\rangle,\end{split}$
(53)

and its entries are given in Table 11, which coincide with those in Table 10
for $\theta=0$. When $\theta\neq 0$ these entries $G^{\prime}$ can be
organized into the following pairs

$\displaystyle P_{1}$ $\displaystyle=$
$\displaystyle\\{00,22\\},P_{2}=\\{12,30\\},$ $\displaystyle P_{3}$
$\displaystyle=$ $\displaystyle\\{13,31\\},P_{4}=\\{01,23\\},$ (54)
$\displaystyle P_{5}$ $\displaystyle=$
$\displaystyle\\{02,20\\},P_{6}=\\{10,32\\},$ $\displaystyle P_{7}$
$\displaystyle=$ $\displaystyle\\{11,33\\},P_{8}=\\{03,21\\}\,,$

that are related to the sets (42) as

$\displaystyle Set_{1}$ $\displaystyle=$ $\displaystyle P_{1}\cup P_{2},\quad
Set_{2}=P_{3}\cup P_{4}\,,$ (55) $\displaystyle Set_{3}$ $\displaystyle=$
$\displaystyle P_{5}\cup P_{6},\quad Set_{4}=P_{7}\cup P_{8}\,.$

We shall next analyze the different strategies of Alice and Bob. Suppose that
Alice chooses at random among the pairs given in eq. (54) and that Bob chooses
$b_{0}=0$. This leads to the following cases:

  * •

Alice selects $(a_{1},a_{2})\in P_{1}$ then Bob, applying the orthogonality
rule (40), will choose $(b_{1},b_{2})\in P_{4,5,6,7,8}$. The payoffs (41) for
$a_{0}=0,1,2,3$ and $b_{0}=0$ are given by

$f^{A}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$P_{1}$ | 1 | 0 | 0 | $c^{2}$

$f^{B}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$P_{4}$ | 0 | 0 | 0 | $|s|^{2}$
$P_{5}$ | 0 | $c^{2}$ | 1 | 0
$P_{6}$ | 0 | 1 | $c^{2}$ | 0
$P_{7}$ | 0 | 0 | $|s|^{2}$ | 0
$P_{8}$ | 0 | $|s|^{2}$ | 0 | 0

  * •

Alice selects $(a_{1},a_{2})\in P_{3}$ then Bob, applying the orthogonality
rule (40), will choose $(b_{1},b_{2})\in P_{2,5,6,7,8}$. The payoffs (41) are
given by

$f^{A}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$P_{3}$ | $|s|^{2}$ | 0 | 0 | $0$

$f^{B}$ | 00 | 10 | 20 | 30
---|---|---|---|---
$P_{2}$ | $c^{2}$ | 0 | 0 | $1$
$P_{5}$ | 0 | $c^{2}$ | 1 | 0
$P_{6}$ | 0 | 1 | $c^{2}$ | 0
$P_{7}$ | 0 | 0 | $|s|^{2}$ | 0
$P_{8}$ | 0 | $|s|^{2}$ | 0 | 0

  * •

The cases where Alice selects $P_{2,4,5,6,7,8}$ are evenly distributed among
the previous ones.

Taking into account that $1+c^{2}\geq|s|^{2}$, the best strategy for Bob is to
select the pairs $P_{5}$ and $P_{6}$ with equal probability. In this case, the
average payoffs of Alice and Bob are given by

$\displaystyle\langle f^{A}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{1+c^{2}}{4}+\frac{|s|^{2}}{4}\right)=\frac{1}{4}\,,$
(56) $\displaystyle\langle f^{B}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\frac{1+c^{2}}{4}+\frac{1+c^{2}}{4}\right)=\frac{1+c^{2}}{4}\,,$

that yields the normalized probabilities

$P_{A}=\frac{1}{2+c^{2}},\qquad P_{B}=\frac{1+c^{2}}{2+c^{2}}\,.$ (57)

For $\theta=0$, one recovers eq. (46), while for $\theta=\pi$, one gets
$P_{A}=P_{B}=1/2$. Fig. 4 plots the values of (57), that shows. The highest
the entanglement of the states (51) the highest probability has Bob for
winning over Alice. Indeed the ratio of their probabilities can be related to
the purity of the states, ${\rm tr}({\rho^{\prime}_{1}}^{2})$,

$\frac{P_{B}}{P_{A}}=3-2{\rm tr}({\rho^{\prime}_{1}}^{2})\ .$ (58)

Figure 4: Normalized probabilities (57) as a function of $\theta\in[0,\pi]$

###  VI.2 The order matters

In the quantum games based on the boson, hard-core boson and one-qubit, the
operators $O^{A}$ and $O^{B}$, that generate the joint state, commute.
However, in the two-qubit game, the operators (34) or (50) do not. This
implies that the order of their action on the initial state may lead to
different outcomes. We shall next analyze the case $\theta=0$. The joint state
is now created by Alice acting first and followed by Bob

$|\widetilde{\psi}_{a_{0},b_{0}}\rangle=O^{B}_{b_{0}}O^{A}_{a_{0}}|00\rangle\,.$
(59)

The definition of the guess states (37) remains the same. This implies that
the overlaps between the guess states and the joint state yield a matrix
$\tilde{G}$ that is the partial transposed of the matrix $G$, that is

$\widetilde{G}_{(i,j),(a_{0},b_{0})}=G_{(i,j),(b_{0},a_{0})}\,.$ (60)

Hence, the payoffs on the Alice-first game, are given by those of the Bob-
first game as

$\begin{split}\tilde{f}^{A}_{(a_{1},a_{2}),(a_{0},b_{0})}=f^{A}_{(a_{1},a_{2}),(b_{0},a_{0})},\\\
\tilde{f}^{B}_{(b_{1},b_{2}),(a_{0},b_{0})}=f^{B}_{(b_{1},b_{2}),(b_{0},a_{0})}.\end{split}$
(61)

In this new game, one can expect that Alice will have some advantage because
her choice of $a_{0}$ will restrict possible winning guesses. Nevertheless,
this does not happens because Alice’s guess is done before Bob’s one, and then
she will reveal information about the value of $a_{0}$ to Bob. This leads in
turn to a symmetric game. Let us illustrate this result with an example.
Suppose that Alice chooses $a_{0}=0$. Replacing $b_{0}$ in eq.(44) by $a_{0}$,
one finds that the winning sets are $Set_{1}$ or $Set_{3}$. So, she will
choose at random among both. However, Bob, knowing this information, will
notice that $a_{0}$ is equal to 0 or 2\. Then, he will choose the set which is
not chosen by Alice among the pair of sets.

$\tilde{f}^{A}$ | 00 | 01 | 02 | 03
---|---|---|---|---
$Set_{1}$ | 1 | 0 | 0 | 1

,  $\tilde{f}^{B}$ 00 01 02 03 $Set_{3}$ 0 1 1 0

This scenario ends with each one winning half of the times on average. The
same probabilities will appear for every value of $a_{0}$, so the normalized
probabilities are

$P_{A}=P_{B}=\frac{1}{2}\ .$ (62)

The sets ${\cal C},{\cal\tilde{C}}$ and the space ${\cal G}$ involved in the
two-qubit game are given by

$\displaystyle{\cal C}$ $\displaystyle=$
$\displaystyle\\{0,1\\},\;{\cal\tilde{C}}=\\{(0,0),(0,1),(1,0),(1,1)\\},\;$
(63) $\displaystyle{\cal G}$ $\displaystyle=$ $\displaystyle{\rm
Span}\\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\\},$

where the distance in ${\cal G}$ is defined by (17).

##  VII Simulation on an IBM quantum computer

The previous games are completely characterized by the metric $G$ given in
eq.(39), and its extended version $G^{\prime}$ given in (53). Their values are
given in Tables 10 and 11 respectively. In this section, we shall present the
results obtained using an IBM quantum computer to find the matrix $G$ that
corresponds to the case with $\theta=0$. We have also studied the case
$\theta=\pi$, but it will not be presented here.

To compute the matrix $G$ we apply two $O_{i}$-operators and two
$O^{\dagger}_{j}$-operators on the state $|00\rangle$. The probability of
measuring the state $|00\rangle$ provides $|G|^{2}$. Some entries of $G$ have
phases but they are not relevant to the strategies of the game. Nevertheless,
we have included them in the results.

The circuit to compute $G_{(2,2),(3,0)}$ is given by

We used the qubits $q0$ and $q1$ of the quantum computer ibmq-Manila whose
topology is

The number of shots was set to its maximum, 8192, obtaining the results
plotted in Fig. 5.

Figure 5:  Histogram of the probabilities obtained to measure
$|G_{(2,2),(3,0)}|^{2}$ after 8192 shots

The experimental value of $0.964$ is very closed to the theoretical value of 1
given in Table 10. This is of course due to the noise and decoherence in this
computer. We have determined experimentally all the entries of $|G|$ and
collect them in Table 12, including the signs. The theoretical values equal to
1 have an averaged error of 2-3%, while the null values have an average error
of 10-24%. This implies that the orthogonality condition (38) has to a more
relaxed, say to $|\langle g_{A}|g_{B}\rangle|\lessapprox 0.25$, if the game is
to be close to the theoretical one.

$G_{\rm exp}$ | 00 | 22 | 12 | 30 | 13 | 31 | 01 | 23 | 02 | 20 | 10 | 32 | 11 | 33 | 03 | 21
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
00 | 0,97 | 0,98 | 0,97 | 0,97 | 0,24 | 0,24 | 0,20 | 0,24 | 0,20 | 0,20 | 0,22 | 0,20 | 0,10 | 0,10 | 0,17 | 0,10
22 | 0,98 | 0,97 | 0,97 | 0,98 | 0,24 | 0,24 | 0,24 | 0,24 | 0,20 | 0,20 | 0,20 | 0,22 | 0,14 | 0,10 | 0,10 | 0,10
12 | 0,97 | 0,97 | 0,97 | 0,97 | 0,24 | 0,24 | 0,24 | 0,24 | 0,20 | 0,22 | 0,20 | 0,20 | 0,10 | 0,10 | 0,14 | 0,14
30 | 0,97 | 0,97 | 0,98 | 0,98 | 0,24 | 0,24 | 0,22 | 0,24 | 0,20 | 0,20 | 0,24 | 0,22 | 0,10 | 0,10 | 0,14 | 0,10
13 | 0,26 | 0,24 | 0,24 | 0,22 | 0,97 | 0,97 | -0,97 | -0,97 | 0,10 | 0,10 | 0,10 | 0,10 | 0,20 | 0,20 | 0,22 | 0,20
31 | 0,24 | 0,24 | 0,22 | 0,24 | 0,95 | 0,97 | -0,97 | -0,97 | 0,17 | 0,10 | 0,10 | 0,10 | 0,22 | 0,20 | 0,22 | 0,20
01 | 0,26 | 0,24 | 0,24 | 0,22 | -0,97 | -0,97 | 0,96 | 0,98 | 0,10 | 0,14 | 0,10 | 0,10 | 0,20 | 0,22 | 0,22 | 0,20
23 | 0,24 | 0,24 | 0,24 | 0,24 | -0,96 | -0,95 | 0,97 | 0,98 | 0,14 | 0,10 | 0,10 | 0,14 | 0,20 | 0,20 | 0,20 | 0,22
02 | 0,22 | 0,20 | 0,22 | 0,20 | 0,14 | 0,14 | 0,10 | 0,10 | 0,98 | 0,97 | 0,97 | 0,97 | 0,26 | 0,24 | 0,24 | 0,24
20 | 0,22 | 0,24 | 0,24 | 0,22 | 0,14 | 0,10 | 0,10 | 0,10 | 0,97 | 0,97 | 0,97 | 0,97 | 0,24 | 0,24 | 0,26 | 0,24
10 | 0,20 | 0,20 | 0,22 | 0,20 | 0,10 | 0,14 | 0,10 | 0,10 | 0,97 | 0,97 | 0,97 | 0,97 | 0,22 | 0,24 | 0,28 | 0,24
32 | 0,20 | 0,20 | 0,22 | 0,20 | 0,10 | 0,10 | 0,10 | 0,10 | 0,95 | 0,97 | 0,95 | 0,97 | 0,24 | 0,22 | 0,24 | 0,24
11 | 0,10 | 0,10 | 0,10 | 0,14 | 0,22 | 0,20 | 0,20 | 0,20 | 0,24 | 0,24 | 0,24 | 0,24 | 0,98 | 0,97 | -0,95 | -0,97
33 | 0,10 | 0,10 | 0,10 | 0,10 | 0,20 | 0,20 | 0,20 | 0,22 | 0,24 | 0,22 | 0,26 | 0,22 | 0,97 | 0,97 | -0,97 | -0,97
03 | 0,10 | 0,10 | 0,10 | 0,10 | 0,20 | 0,20 | 0,20 | 0,20 | 0,24 | 0,24 | 0,26 | 0,22 | -0,97 | -0,97 | 0,95 | 0,97
21 | 0,10 | 0,10 | 0,10 | 0,10 | 0,20 | 0,22 | 0,20 | 0,20 | 0,26 | 0,24 | 0,24 | 0,24 | -0,97 | -0,96 | 0,96 | 0,97

Table 12: Experimental values of $G_{(i_{1},j_{1}),(i_{2},j_{2})}$ measured
obtained on ibmq-Manila.

##  VIII Conclusions and prospects

We have presented in this work a general formalism of the Chinos game between
two players that includes the quantum games introduced in reference GMD  using
a boson and extend them to a hard-core boson, a one-qubit system, and a two-
qubits system.

A general Chinos game is defined by: i) two sets, ${\cal C},\tilde{\cal C}$,
containing the players’ choices and guesses, ii) a device that maps the
previous data to a metric space ${\cal G}$ whose distance is used to compute
the players’ payoffs, iii) a restriction rule on the second player, and iv) an
intelligence rule to optimize the players’ guesses. In the games we have
considered, the sets ${\cal C},\tilde{\cal C}$ are a discrete and finite
collection of integers. In the classical Chinos game the space ${\cal G}$ is
also discrete and finite, but in the quantum Chinos games ${\cal G}$ is a
Hilbert space. We have employed the trace norm of density matrices to define
the payoffs and the restriction rule, which for pure states is in direct
correspondence with the fidelity. However, one can consider mixed states in
which case the two criteria, i.e. distance or fidelity, can lead to different
results.

The game based on two-qubits shows an interesting interplay between
entanglement and the success probabilities’ of the players. Moreover, the game
is sensitive to the way the joint state is created. In the one and two qubit
games, the device can be implemented on a quantum computer, which has allowed
us to simulated the basic matrix involved in the game using an IBM quantum
computer.

The rules of the Chinos game make it similar to those considered in the
quantum decision theory whose mathematics is based on the theory of quantum
measurement Y08 -Y20 . These classes of games can also be categorized as
abstract economics because the player’s strategies are not independent of each
other K18 . Finally, we would like to notice that the formalism presented here
can be extended in a natural way to more than two players. It will be
interesting to explore its application to quantum communication protocols.

Acknowledgements.-

GS acknowledges financial support through the Spanish MINECO grant PGC2018-
095862-B-C21, the Comunidad de Madrid grant No. S2018/TCS-4342, the Centro de
Excelencia Severo Ochoa Program SEV-2016-0597 and the CSIC Research Platform
on Quantum Technologies PTI-001.

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