# Gould Belt members in X-ray RAVE: Cross-matching RAVE stars with 3XMM point
sources

B. Coşkunoğlu 11affiliationmark<EMAIL_ADDRESS>O. Plevne
22affiliationmark:  M. T. Özkan 11affiliationmark:  33affiliationmark:

###### Abstract

In this paper the results of matching the RAdial Velocity Experiment (RAVE), a
spectroscopic Southern hemisphere survey (9 $<I_{DENIS}<$ 12), and XMM-Newton
Serendipitous Source Catalogue (3XMM) are presented. The latest data releases
of RAVE and XMM were matched and a X-ray RAVE catalogue of 1071 stars was
obtained. Then the catalogue was checked for possible Gould Belt (GB) members.
We obtained a subsample of 10 stars that meet the GB membership criteria. This
subsample and GB member candidates were tested photometrically and
kinematically. Among the members there are two BY Dra type variables, an
NGC2451 open cluster member, a high proper motion star. The rest are regular
main sequence stars. The members have very low velocity dispersions which lead
us to think that the members belong in a single structure. We also found out
that a kinematical GB membership test might be possible to derive given a
large enough GB member sample as they fit in a narrow interval in space
velocity diagrams.

00footnotetext: Istanbul University, Faculty of Science, Department of
Astronomy and Space Sciences, 34119 University, Istanbul, Turkey
00footnotetext: Istanbul University, Graduate School of Science and
Engineering, Department of Astronomy and Space Sciences, 34116, Beyazıt,
Istanbul, Turkey
00footnotetext: TUG (TÜBİTAK National Observatory), Akdeniz University Campus,
07058, Antalya, Turkey

##  1 Introduction

Herschel (1847) observed that a fraction of bright stars in the southern
hemisphere of the sky to be a part of a structure other than Milky Way with a
tilt of $\approx$ 20∘ to the Galactic equator. 32 years later Gould (1879)
also observed it and determined its poles’ and nodes’ Galactic coordinates.
Therefore, this structure is named after him, namely the Gould Belt (hereafter
GB). The GB is relatively flat: its length, width and height semiaxes are 350,
250 and 50 pc, respectively. Its ascending node is at $l$=285∘ and the Sun is
200 pc off the center in the $l$=130∘ direction (Guillout et al., 1998). The
GB consists of X-ray active young stars with distances up to 600 pc (Olano,
1982; Fresneau et al., 1996; Poppel, 1997; Guillout et al., 1998). The age
interval varies greatly in the literature: most authors state that it is
between 20-30 and 70-80 Myr (Ogorodnikov, 1965; Olano, 1982; Poppel, 1997;
Guillout et al., 1998; Bekki, 2009), however, Frogel & Stothers (1977)’s
review suggests a wider interval: 30-220 Myr. Using Strömgren photometry Torra
et al. (2000) determined stars with ages ¡ 90 Myr belong in the GB. Bobylev
(2014) wrote an extensive review regarding the GB, summarizing past studies
about GB’s formation, structure, age, kinematics, mass. The GB is important
because it is the closest star-gas complex to the Sun. These complexes are
regions where star formation is active and they are observed in Milky Way
(Efremov, 1998), as well as other galaxies (Efremov & Elmegreen, 1998).

The GB is not defined very precisely. There are several criteria that are used
to define GB membership, which will be discussed in detail in Chapter 2. This
causes controversial opinions to form about its formation. Bobylev (2014)
states that the GB is a separate structure from the Milky Way, whereas Bekki
(2009) says that a dark matter clump collided with a gas cloud to trigger GB’s
formation. Comerón (2001) observed a GB-like structure in M83 which is
approximately the same size with the GB, is separate from the spiral structure
of the galaxy, lies at a similar distance from the galactic center and has the
same order of age ($\log$ age/yr=7). This means GB-like structures are not
unique to our Galaxy and, at the very least, can exist in other spiral
galaxies as well. Therefore, analysing and better defining the GB can help us
understand the evolution of other galaxies. The aim of study is to determine
GB members and their properties that are in both RAVE and 3XMM catalogues and
to contribute towards defining the GB better.

Having spectral information about stars in multiple regions of the
electromagnetic spectrum is useful for many purposes in astronomy. We have
cross-matched the RAdial Velocity Experiment (RAVE)’s catalogue with XMM-
Newton Serendipitous Source Catalogue (3XMM) in order to obtain both infrared
and X-ray data of the stars in common in the surveys in question. The
resulting X-ray RAVE (XRAVE) catalogue was used to determine GB members. For
the purposes of this study XMM-Newton data provides the X-ray luminosity of
the stars, whereas the RAVE data provides the radial velocities, atmospheric
parameters and distances of the stars in the matched catalogue. However, since
the distances provided in RAVE have relatively large errors, the accurate
distances of GAIA DR2, which includes almost all RAVE stars, were used (Gaia
Collaboration et al., 2018). The atmospheric parameters of RAVE were utilised
in determining ages using the method of Jørgensen & Lindegren (2005).

The data and method are presented in Section 2, the properties of GB members
are given in Section 3 and the conclusion and summary are in Section 4.

##  2 Data & Method

###  2.1 Data

Fig. 1 : 92 XRAVE Stage 1 GB member candidates. Blue dots represent the
candidates, gray dots represent XRAVE stars outside the GB region, the red
solid line is the Galactic latitude obtained from Eq. 2, whereas the dashed
red lines represent the thickness of the GB, i.e. GB’s upper and lower edges.

A part of the data used in creating the catalogue was taken from RAVE, which
is a spectroscopic stellar survey based in the Southern hemisphere. RAVE’s
general aim is to make use of the Sun’s position in the Galaxy and study its
formation and evolution. As one can deduce from its name RAVE specializes in
determining stellar radial velocities for stars in the thin and thick disks
and halo. Its observations are performed with a 1.2 m Schmidt telescope in
Anglo-Australian Observatory. Radial velocities are calculated using medium
resolution spectra (R = 7500) in the Ca-triplet region (8400-8750 Å)
(Steinmetz et al., 2006). Combining radial velocity with proper motion and
distance one can obtain the kinematic components of a star. In the study we
used RAVE’s most recent data release, i.e. RAVE Data Release 5 (RAVE DR 5)
(Kunder et al., 2017). RAVE DR5 has 520781 spectra of 457588 unique stars
Kunder et al. (2017) and their radial velocities, atmospheric parameters,
distances along with other parameters.

Another part of the data used in the study was taken from European Space
Agency’s XMM-Newton satellite’s Serendipitous Source Catalogue’s Data Release
8 (3XMM DR8) (Rosen et al., 2016). XMM-Newton was launched in 1999. It has
three X-ray telescopes, an X-ray detector with three CCDs, an optical monitor
and two reflection grating spectrometers on board. Each telescope has 58
cylindrical, nested Wolter Type-1 mirrors which are 600 mm in length and 306
to 700 mm in diameter. European Photon Imaging Camera (EPIC), i.e. the X-ray
detector has two Metal Oxide Semi-conductor (MOS) cameras and a p-n junction
(PN) camera. The Optical Monitor (OM) helps XMM-Newton perform multi-
wavelength observations simultaneously: in X-rays and optical and ultraviolet
regions of the electromagnetic spectrum. 3XMM DR8 has 775153 detections of
531454 unique sources and provides X-ray fluxes in nine different bands as
well as other parameters.

We also used GAIA DR2 so that we could have higher astrometric accuracy.
Launched in 2013, GAIA is the successor of RAVE and will observe stars
brighter than 20m. It aims to create the most accurate map of Milky Way. GAIA
provides precise astrometric data, up to a few micro arc seconds of accuracy
(Gaia Collaboration et al., 2016).

Upon cross-matching the catalogues of both data releases and removing
duplicates by keeping the highest signal-to-noise ratio observations and
discarding others an XRAVE subsample of 1071 unique stars was obtained. Stars
in RAVE DR5 and 3XMM DR8 catalogues were matched using the angular separation
formula for equatorial coordinates (Eq. 1):

$\theta=cos^{-1}[sin(\delta_{1})sin(\delta_{2})+cos(\delta_{1})cos(\delta_{2})cos(\alpha_{1}-\alpha_{2})],$
(1)

where $\delta_{1}$ and $\delta_{2}$ are the declination of RAVE and XMM stars
and $\alpha_{1}$ and $\alpha_{2}$ are the right ascension of RAVE and XMM
stars, respectively. We accepted matches that yielded an angular separation of
$\leq$ 1” because of the high astrometic accuracy of RAVE (Kunder et al.,
2017) and the increased astrometric accuracy of 3XMM (Rosen et al., 2016). We
checked our resulting catalogue using Vizier
(111http://vizier.u-strasbg.fr/viz-bin/VizieR)’s X-match tool and obtained the
same sample.

###  2.2 Identification of GB members

In order for a star to be a GB member it has to meet certain criteria for
Galactic coordinates, distance, X-ray luminosity and age. These criteria are
as follows:

  * •

Galactic coordinates should fulfill Eq. 2,

  * •

Distances should be less than 600 pc,

  * •

X-ray luminosities in 0.2-12 keV energy band should be between 1028 and 1031
${\rm erg\,s^{-1}}$,

  * •

Ages should be between 10 and 100 Myr.

In order to identify the stars that meet the first criterion we used the GB
description given by Guillout et al. (1998) and shown in their Fig. 4. Their
description can be best represented with a fourth degree polynomial that
associates Galactic latitude with Galactic longitude (Eq. 2).

$\displaystyle b$ $\displaystyle=5.50\times 10^{-8}\times l^{4}-8.19\times
10^{-5}\times l^{3}$ (2) $\displaystyle+4.25\times 10^{-2}\times
l^{2}-8.96\times l+634,$

where $l$ and $b$ are Galactic longitude and latitude, respectively. A fourth
degree polynomial was used because it most accurately describes GB’s position
in Galactic coordinates. Higher degree polynomials yield almost the same
amount of stars with very insignificant coefficients in the order of 10-10,
which have no physical meaning, whereas lower degrees distort and flatten the
shape of GB. Considering the GB has a thickness of $\approx$ 20∘ (Comerón &
Torra, 1994) in latitude, we accepted stars within $\pm$ 10∘ of the latitude
provided by Eq. 2 for the longitude in question. The GB we obtained from Eq. 2
is given in Fig. 1. 92 out of 1071 XRAVE stars fall within the dashed red
lines in Fig. 1, i.e. they are within the GB sky area. We refer to these stars
as Stage 1 GB candidates, since they passed one criterion out of four.

After sorting out Galactic positions of GB candidates, we moved on to
distances. Since the distances given in RAVE DR5 have average relative errors
of greater than 40 per cent, we used GAIA DR2’s parallaxes. The average
relative error of parallaxes in GAIA is $\approx$ 5 per cent. Therefore, we
adopted GAIA parallaxes in order to determine distances and X-ray luminosities
using those distances. One star in XRAVE had no match in GAIA, therefore it
was excluded, reducing our Stage 1 subsample to 91 stars. After removing three
stars with zero parallax and 32 stars with distances larger than 600 pc we
ended up with 56 Stage 2 XRAVE stars that meet both the position and distance
criteria for GB membership (Fig. 2).

Fig. 2 : Distance histogram of 91 Stage 1 GB candidates. The red dashed line
represents the distance limit for GB, namely 600 pc. The stars in green
histograms, whose distances are below the limit, i.e. stars that remain to the
left of the dashed line are the Stage 2 candidates, stars to the right of the
dashed line in blue histograms remain as Stage 1 candidates.

In the next step we determined X-ray luminosities of our Stage 2 candidates.
Guillout et al. (1998); Stelzer & Neuhäuser (2000) stated that $L_{X}$ shows a
broad distribution between 1028 and 1031 erg.s-1 for young stars which is the
interval we adopted for our GB candidates. After calculating the distances and
using them to compute luminosities along with the fluxes from the 0.2-12 keV
band. The 0.2-12 keV band, which is the widest energy band in XMM catalogue,
was adopted in order to obtain matches from all types of stars. The
luminosities were computed as follows:

$L_{X}=4\pi d^{2}F_{0.2-12},$ (3)

where $d$ is the distance, $F_{0.2-12}$ is the X-ray flux of the star in the
0.2-12 keV energy band. Afterwards, we looked for the stars with X-ray
luminosities in the 1028 and 1031 erg.s-1 interval (Guillout et al., 1998). We
obtained a sample of 49 Stage 3 stars meeting all three criteria, i.e.
Galactic position, distance and X-ray luminosity. The X-ray luminosity
distribution of 56 Stage 2 candidates are shown in Fig. 3.

Fig. 3 : X-ray luminosity distribution of 56 Stage 2 GB candidates. The dashed
red line to the left and right represent X-ray luminosities of 1028 and 1031
erg.s-1, respectively. The stars in purple histograms that are between the
lines are the 49 Stage 3 candidates, the stars in green histograms remain as
Stage 2 candidates.

###  2.3 Calculating ages of GB stars

The last step towards determining GB members is calculating ages. Since
neither RAVE and XMM catalogue nor GAIA provide ages we used the Bayesian
method of Jørgensen & Lindegren (2005) in order to derive ages. This method
includes probability density functions (PDF) that are obtained from
theoretical model parameters and observational parameters (Pont & Eyer, 2004;
Jørgensen & Lindegren, 2005). Pont & Eyer (2004) stated that the Solar
neighborhood mass, age and metallicity prior distributions are assumed to be
independent from each other, hence they are taken into account separately.
Jørgensen & Lindegren (2005) assumed that observational parameter errors are
independent and that they show a normal distribution. According to these
assumptions the likelihood function is defined as:

${\cal
L}(\tau,\zeta,m)=\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi}\sigma_{i}}\exp{[-\chi^{2}/2]}.$
(4)

$\chi^{2}$ is given as

$\chi^{2}=\sum_{i=1}^{n}\left(\frac{q_{i}^{obs}-q_{i}(\tau,\zeta,m)}{\sigma_{i}}\right),$
(5)

where $q$ represents the model atmospheric parameters taken from RAVE DR5,
i.e. $T_{eff}$, $\log g$ and $[Fe/H]$, whereas $\tau$, $\zeta$ and $m$ are the
theoretical model parameters which represent age, metallicity, and mass,
respectively. $n$ is the number of objects and $\sigma$ represent parameter
errors. Then, the final PDF is obtained by applying a Bayesian correction
given below:

$f(\tau,\zeta,m)\varpropto f_{0}(\tau,\zeta,m)\times{\cal L}(\tau,\zeta,m)$
(6)

where ${f_{0}}$ is the initial pdf and it is given as

$f_{0}=\psi(\tau)\varphi(\zeta|\tau)\xi(m|\zeta,\tau).$ (7)

In this equation $\psi(\tau)$, $\varphi(\zeta)$ and $\xi(m)$ are the star
formation rate (SFR), the metallicity distribution (MDF) and the initial mass
function (IMF), respectively. SFR and MDF are assumed to be constant since
$\tau$, $\zeta$ and $m$ are independent from each other. However, since there
are more low mass stars than high mass stars in the Galaxy, the IMF is
expressed as a power law (Jørgensen & Lindegren, 2005). We then inserted Eq.
(7) into Eq. (6) and integrated the resulting equation and obtained the
$G(\tau)$ function (Eq. 8) which shows the PDF calculated from a comparison
between observational and theoretical parameters.

$G(\tau)\varpropto\int\int{\cal L}(\tau,\zeta,m)\xi(m)dmd\zeta$ (8)

We integrated Eq. 8 over mass and metallicity for each age on isochrones.
Taking into account the maximum value of the $G$ function we determined the
age for the star in question. Due to the complex and non-Gaussian nature of
Bayesian age estimation method, overall uncertainties on estimated star ages
vary between 20 per cent and 50 per cent.

We applied this method to Stage 3 candidates using the PARSEC (Bressan et al.,
2012) stellar evolution models. Since GB members are young stars with ages
between 30 and 80 Myr (Fresneau et al., 1996; Guillout et al., 1998), we
adopted 10 Myr to be the lower limit to eliminate Classical T Tauri Stars
(Bobylev, 2014) and 100 Myr (Wichmann et al., 2003) to be the upper limit for
this study to make up for errors. The age distribution of XRAVE is shown in
Fig. 4.

Fig. 4 : Age distribution of 49 Stage 3 GB candidates. The dashed red line to
the left represents 10 Myr, whereas the one to the right stands for 100 Myr,
which are the adopted lower and upper limits for GB members for this study,
respectively. 10 stars between the lines that are in the red histograms are
the members, and the ones in purple histograms that are not between the lines
remain Stage 3 candidates.

As seen from the figure, there are 10 GB members. There is also a star with
age 3 Myr that did not meet the lower limit of GB membership criterion even
within error bars. After checking the literature for this star we have found
that it is an Algol type eclipsing binary (Avvakumova et al., 2013), which
means it would throw off RAVE measurements yielding such an unpredictable and
possibly incorrect age. This star’s age was calculated as $\approx$ 3 Myr and
is the only star below the age limit as seen in Fig. 4.

Fig. 5 shows GB candidates for all stages: Stage 1, Stage 2, Stage 3 and
member. 35 Stage 1 candidates which only passed the Galactic position
criterion are represented by blue circles, 7 Stage 2 candidates which passed
the Galactic position and distance criteria are denoted by green circles, 39
Stage 3 candidates which passed Galactic position, distance and X-ray
luminosity criteria are shown with purple circles and finally 10 members who
meet all criteria are given in big red circles.

Fig. 5 : Galactic distribution of all 91 GB member candidates. Blue dots
represent Stage 1 GB candidates, green dots stand for Stage 2, whereas purple
dots are Stage 3 and finally, the red ones represent GB members.

##  3 Properties of GB members

In this section we discuss the 10 stars, which meet all four aforementioned GB
membership criteria, i.e. GB members. Their Galactic coordinates, RAVE and XMM
catalogue identifiers, distances, X-ray luminosities in the 0.2-12 keV energy
band and ages and the errors for these three parameters are given in Table 1.

Table 1 : Galactic positions, catalogue identifiers, distances, distance errors, X-ray luminosities in 0.2-12 keV energy band, luminosity errors, logarithmic ages and age errors of GB members. $l$ | $b$ | RAVEID | XMMID | $d$ | $\sigma_{d}$ | $L_{0.2-12}$ | $\sigma_{L_{0.2-12}}$ | $\log(age)$ | $\sigma_{\log(age)}$
---|---|---|---|---|---|---|---|---|---
∘ | ∘ |  |  | (pc) | (pc) | $\times 10^{30}{\rm erg\,s^{-1}}$ | $\times 10^{30}{\rm erg\,s^{-1}}$ | (yr) | (yr)
183.3054 | -22.9945 | J043225.7+130648 | 3XMM J043225.7+130645 | 46.8 | 0.1 | 0.45 | 0.02 | 7.40 | 0.06
195.3055 | -29.2149 | J043636.6+004331 | 3XMM J043636.5+004330 | 141.2 | 0.7 | 0.45 | 0.07 | 7.39 | 0.05
212.0724 | -25.9166 | J051633.7-104740 | 3XMM J051633.7-104735 | 133.1 | 0.5 | 0.27 | 0.11 | 7.99 | 0.75
214.8583 | -26.4542 | J051848.1-131732 | 3XMM J051848.0-131731 | 465.5 | 7.0 | 2.22 | 0.29 | 7.87 | 1.34
239.9732 | -21.4176 | J061515.4-325403 | 3XMM J061515.6-325403 | 151.0 | 0.7 | 0.02 | 0.01 | 7.08 | 0.02
253.9847 | -19.4802 | J064646.5-444036 | 3XMM J064646.6-444036 | 440.0 | 113.8 | 2.27 | 0.93 | 7.89 | 0.11
252.4563 | -6.3732 | J074700.5-375037 | 3XMM J074700.3-375036 | 178.2 | 7.1 | 0.07 | 0.05 | 7.90 | 0.26
264.8783 | -7.7405 | J081554.9-490555 | 3XMM J081554.9-490554 | 363.5 | 2.9 | 8.26 | 0.15 | 7.38 | 0.01
270.6851 | -6.1511 | J084526.9-525202 | 3XMM J084527.0-525201 | 150.3 | 0.5 | 0.71 | 0.02 | 7.66 | 0.01
282.1352 | 7.5193 | J103851.9-495522 | 3XMM J103851.7-495521 | 597.4 | 10.7 | 1.25 | 0.29 | 7.06 | 0.01

###  3.1 Evolutionary status of GB members

The Hertzsprung-Russell (HR) diagram for the entire XRAVE catalogue (1071
stars) is given Fig. 6. The bands used in the axes are from 2MASS (Cutri et
al., 2003) photometry because that is readily provided by RAVE. The de-
reddening procedure of $J$ and $H$ bands are as follows: we used Schlegel et
al. (1998)’s maps and obtained $E_{\infty}(B-V)$. Then we reduced
$E_{\infty}(B-V)$ to $E_{d}(B-V)$ by taking into account distances obtained
from the parallaxes provided by GAIA DR2 as follows:

$E_{d}(B-V)=E_{\infty}(B-V)\Biggl{[}1-\exp\Biggl{(}\frac{-\mid
d~{}\sin(b)\mid}{H}\Biggr{)}\Biggr{]},$ (9)

where $b$ is the Galactic latitude and $d$ is the distance of the star,
whereas $H$ is the scaleheight for the interstellar dust which is 125 pc
(Marshall et al., 2006). $E_{\infty}(B-V)$ stands for the maximum theoretical
colour excess at the Galaxy’s edge for that direction and $E_{d}(B-V)$ is the
reduced absorption adjusted for distance in that direction. After obtaining
$E_{d}(B-V)$ we calculated the total reddening for $J$ and $H$ bands using
Fiorucci & Munari (2003)’s equations:

$\displaystyle A_{J}=0.885\times E_{d}(B-V),$ $\displaystyle A_{H}=0.565\times
E_{d}(B-V),$ (10)

where $A_{J}$ and $A_{H}$ are the distance adjusted absorptions in $J$ and $H$
bands in the direction of the star in question. The absorptions were used in
Pogson’s equation to determine absolute magnitudes:

$\displaystyle m_{J}-M_{J}=5\log(d/pc)-5+A_{J},$ $\displaystyle
m_{H}-M_{H}=5\log(d/pc)-5+A_{H}.$ (11)

After de-reddening the 2MASS photometric bands and calculating absolute
magnitudes for all XRAVE stars, we plotted the HR diagram (Fig. 6). There are
several giants as seen in the figure, mostly confirmed by their RAVE surface
gravities. The main sequence can be clearly observed, and the number of stars
that belong there comprise the majority of the sample. This is expected
because of RAVE’s observation criteria as it focuses mainly on relatively
bright main sequence stars with F or G spectral types. The turnoff point is
also apparent in the figure. The 10 GB members are also mostly main sequence
as well with a few giants.

Fig. 6 : HR diagram for the entire XRAVE sample. Grey circles are non GB
stars, blue circles denote Stage 1 candidates, green circles are for Stage 2,
whereas purple circles represent Stage 3 and big red circles are the GB
members. The x-axis is the de-reddened $(J-H)_{0}$ colour, the y-axis is the
absolute $J$ magnitude.

###  3.2 Special GB members

There are interesting objects amongst the GB members. We provide their
atmospheric parameters, i.e. effective temperature, surface gravity and
metallicity along with their $J$, $H$ and $K_{S}$ magnitudes in Table 2. The
”Remarks” column provides information about a star if it has any peculiar
features. Two of the members are BY Dra variables, which are chromospherically
active late spectral type main sequence stars. Their effective temperatures
and surface gravities confirm this result. One member is a high proper motion
star. Another is a member of the open cluster NGC2451, which has an age of
30-50 Myr (Platais et al., 2001; Hünsch et al., 2003), which is in agreement
with our estimate of 80 Myr. This result lends credence to our ages. However,
it is unexpected to have a single member of an open cluster in the GB,
therefore, we checked XRAVE for other NGC2451 members. We looked for its
members in WEBDA 222https://www.univie.ac.at/webda/navigation.html and found
136 of them (Williams, 1967). Among those 136 stars only the one in the GB is
in XRAVE, and seven are in RAVE DR5. This means the other six stars in RAVE
DR5 were either not observed by XMM-Newton which is likely because the GB
member is at the edge of the open cluster or are not X-ray active. As a
result, we could not check if other NGC2451 members belong in the GB or not.
However, according to Elias et al. (2009)’s spatial criteria method more than
half of the members of NGC2451 are part of the GB. This suggests that we could
only find a single member of NGC2451 in the GB, because we did not have
membership criteria related data for its other members. Two of the GB members
are well past their turn off points and are giants, more specifically, one of
them is a subgiant and the other is a giant. Their surface temperatures
suggest that they are G-type stars. Taking this into account along with their
young ages, it does not seem possible. The reason for this discrepancy might
be faulty atmospheric parameter determination due to the RAVE pipeline as it
is adjusted for main sequence stars. A literature check has not yielded any
decisive results, neither for the two stars to be giants, nor against it.

Table 2 : Atmospheric parameters and 2MASS photometry bands’ magnitudes of GB members. $l$ | $b$ | $T_{eff}$ | $\log\,g$ | $[Fe/H]$ | $M_{J}$ | $(J-H)_{0}$ | $(H-K_{s})_{0}$ | Remarks
---|---|---|---|---|---|---|---|---
$(^{\circ})$ | $(^{\circ})$ | (K) | $(cm\,s^{-2})$ | (dex) |  |  |  |
183.3054 | -22.9945 | 3943 | 4.97 | -0.31 | 4.58 | 0.53 | -0.03 | BY Dra variable
195.3055 | -29.2149 | 4000 | 5.00 | 0.00 | 4.43 | 0.57 | 0.13 |
212.0724 | -25.9166 | 5030 | 4.88 | 0.00 | 4.26 | 0.36 | 0.08 | high proper motion
214.8583 | -26.4542 | 7262 | 3.95 | 0.00 | 1.80 | 0.19 | 0.06 |
239.9732 | -21.4176 | 4932 | 4.38 | 0.39 | 3.84 | 0.47 | 0.08 |
253.9847 | -19.4802 | 5000 | 4.50 | 0.00 | 2.08 | 0.48 | 0.12 |
252.4563 | -6.3732 | 5025 | 4.90 | 0.10 | 3.64 | 0.33 | -0.14 | member of NGC 2451
264.8783 | -7.7405 | 5678 | 1.25 | 0.00 | 1.94 | 0.33 | -0.06 |
270.6851 | -6.1511 | 4000 | 4.99 | 0.00 | 4.31 | 0.46 | -0.09 | BY Dra variable
282.1352 | 7.5193 | 4925 | 2.03 | -0.61 | -0.81 | 0.40 | -0.03 |

##  4 Conclusion and Summary

In this study we cross matched RAVE DR5 and 3XMM DR8 and obtained a resulting
XRAVE catalogue of 1071 stars. We looked for GB members in XRAVE. According to
Olano (1982); Poppel (1997); Guillout et al. (1998); Torra et al. (2000) a
member of GB has to meet four criteria as discussed in Section 2: a certain
Galactic position, distance less than 600 pc, X-ray luminosity between 1028
and 1031 ${\rm erg\,s^{-1}}$ and age between 10 and 100 Myr. We applied those
criteria to XRAVE and obtained a sample of 10 stars that meet all of them,
i.e. that are GB members.

In order to test the GB membership criteria with another method we approached
the GB candidate sample of 91 stars kinematically. $U$, $V$ and $W$ Galactic
velocity components were obtained using distances and radial velocities in
conjunction with radial and tangential proper motions in Johnson & Soderblom
(1987)’s algorithms and transformation matrices. A detailed explanation of the
calculations is given by Coşkunoǧlu et al. (2011, 2012). After obtaining total
space velocities Galactic differential rotation corrections were obtained and
performed for the $U$ and $V$ components using Mihalas & Binney (1981)’s
procedure. The $W$ component is not affected by Galactic differential rotation
(Mihalas & Binney, 1981). Then we performed the Local Standard of Rest (LSR)
correction on the Galactic velocity components as provided by Coşkunoǧlu et
al. (2011). LSR correction is required to determine velocities accurately as
it describes the Sun’s movement with respect to nearby stars (Coşkunoǧlu et
al., 2011). Finally, the errors of Galactic velocity components were
calculated Johnson & Soderblom (1987)’s algorithm by propagating the errors of
distances, radial velocities and radial and tangential proper motions. The
median space velocity components and median total space velocity and their
respective errors along with the velocity dispersions for each component and
total velocity are given in Table 3 for Stages 1, 2 and 3 and GB members. The
values in the table are in km s-1.

Table 3 : Median space velocities and their respective errors and dispersions of 91 Stage 1, 56 Stage 2, 49 Stage 3 candidates and 10 members. All velocities are given in km s-1. | $\tilde{U_{LSR}}$ | $\tilde{V_{LSR}}$ | $\tilde{W_{LSR}}$ | $\tilde{S_{LSR}}$ | $\tilde{U_{Err}}$ | $\tilde{V_{Err}}$ | $\tilde{W_{Err}}$ | $\tilde{S_{Err}}$ | $\sigma_{U_{LSR}}$ | $\sigma_{V_{LSR}}$ | $\sigma_{W_{LSR}}$ | $\sigma_{S_{LSR}}$
---|---|---|---|---|---|---|---|---|---|---|---|---
Stage 1 | -29.81 | -30.26 | -14.14 | 57.55 | 1.46 | 1.41 | 0.82 | 2.48 | 227.64 | 306.03 | 492.95 | 605.07
Stage 2 | -31.26 | -30.58 | -14.31 | 54.33 | 1.20 | 1.11 | 0.75 | 2.32 | 82.71 | 84.39 | 41.13 | 108.17
Stage 3 | -22.50 | -32.44 | -14.83 | 48.90 | 1.10 | 1.00 | 0.71 | 2.16 | 100.16 | 33.01 | 49.41 | 107.41
GB | -18.44 | -34.94 | -11.36 | 47.33 | 1.12 | 1.85 | 0.78 | 3.20 | 24.56 | 4.65 | 14.26 | 10.78

Upon analysing Table 3 it can be seen that every GB criterion that was applied
to XRAVE catalogue also applies a selection effect kinematically, which can be
observed from the decreasing space velocity dispersions in every step. The GB
members’ low space velocity dispersions and their low errors show that those
stars have similar kinematic characteristics. In order to demonstrate that we
plotted a Toomre diagram, which shows the kinematical similarities or the lack
thereof for star samples, of the entire XRAVE sample (Fig. 7). Every step
reduces the scatter of stars in the diagram noticeably and ultimately yields a
very narrow interval for the GB members (red circles in Fig. 7), which
verifies the dispersion decrease in Table 3. This points out that GB can be
detected kinemetically as well, however, the lack of number of stars in the
sample prevents the determination of a distinguishable feature. This can be
countered by working with larger samples and thus obtaining bigger samples of
GB members. After that is done kinematical constraints can be determined for
the GB.

Fig. 7 : Toomre diagram for the entire XRAVE sample. Grey circles are non GB
stars, blue circles denote Stage 1 candidates, green circles are for Stage 2,
whereas purple circles represent Stage 3 and big red circles are the GB
members. The x-axis is the $V_{LSR}$, the y-axis is
$\sqrt{U_{LSR}^{2}+W_{LSR}^{2}}$ both in km s-1.

Our analyses show that there are 10 GB members which appear in both RAVE DR 5
and 3XMM DR 8 catalogues. We tested our subsample photometrically and
kinematically. Our findings indicate that the 10 stars are part of a single
structure. The majority of the members are regular F or G spectral type main
sequence stars as expected from RAVE’s selection criteria. There are two BY
Dra type variables which are chromospherically active stars. A NGC2451 open
cluster member and two giants also belong in the member subsample. As giants
of F or G spectral type need to be at least billions of years old it is
possible that faulty atmospheric parameter detection threw off the
calculations. We also found that a kinematical GB membership test can be
derived with a large enough GB member sample. If such a test would exist it
would help in defining the GB better and more precisely. Comerón (2001) also
remarked that the GB’s inclination to the Galactic plane might show up as a
systematic pattern in the radial velocities of the HII regions, which would
help in deriving the aforementioned kinematical membership test. In order to
obtain such a sample larger spectroscopic surveys such as Mauna Kea
Spectroscopic Explorer (MSE) used in conjunction with future data releases of
GAIA are required.

## Acknowledgements

We would like to thank Dr. Sinan Aliş for his numerous contributions towards
improving the paper. Also, we thank Prof. A. T. Saygaç and Dr. G. Gün for
their input towards developing the study.

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# Entanglement of Quantum States which are Zero on the Symmetric Sector

D. D’Alessandro Department of Mathematics, Iowa State University, Ames, IA
50011<EMAIL_ADDRESS>
###### Abstract

We consider a quantum system of $n$ qudits and the Clebsch-Gordan
decomposition of the associated Hilbert space $(\mathbb{C}^{d})^{\otimes n}$.
In this decomposition, one of the subspaces is the so-called symmetric
subspace or symmetric sector, that is, the subspace of all states that are
invariant under the action of the symmetric group. We prove that any separable
state must have a nonzero component on the symmetric sector, or, equivalently,
any state which has zero component on the symmetric sector must be entangled.
For the cases of $n=2,3$ particles, and in arbitrary dimension $d$, this
result can be refined by providing sharp lower bounds on the size of the
component of separable states on the symmetric sector. This leads us to
identify a class of entanglement witnesses for these systems. We provide an
example showing that in the multipartite case, this class of witnesses detects
PPT entangled states.

Keywords: Geometry of Quantum States, Quantum Entanglement, Symmetric Sector,
Entanglement Witnesses, Partial Transposition.

##  1 Introduction

The study of the geometry of quantum states is at the heart of quantum
mechanics and quantum information theory. In this context, when considering
multipartite quantum systems, the states are naturally classified in separable
states, that is, pure states that can be written as products
$|\psi\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes\cdots\otimes|\psi_{n}\rangle$,
where $|\psi_{j}\rangle$ is a state of the $j$-th subsystem, and entangled
states, that is, states that cannot be written as products.111The definition
is a bit more elaborate for mixed states (see, e.g., [13]) and we shall
consider it later in the paper. Detecting entangled states and classifying
them according to the type and amount of entanglement (using a specific
entanglement measure) is one of the most studied and fundamental topics in
quantum information [8].

When considering a multipartite system of $n$, $d$-dimensional, subsystems
(qudits), the Hilbert space ${\cal H}:=(\mathbb{C}^{d})^{\otimes n}$, splits
according to the Clebsch-Gordan (CG) decomposition into irreducible
representations of $SU(d)$. This decomposition, in the case of $d=2$ (qubits)
is well studied in the theory of angular momentum addition (see, e.g., [17]),
and it has several important applications in the general case $d\geq 2$ (see,
e.g., [2] and references therein). Irreducible representations of $SU(d)$
appear with various multiplicities in ${\cal H}$. One representation which
always appears with multiplicity one is spanned by states which are invariant
under the permutation group $S_{n}$. This is referred to as the symmetric
subspace or symmetric sector.

Symmetric states have many applications in quantum information theory [5] and
therefore they have attracted great attention in recent years both at the
theoretical and at the experimental level (see, e.g., [1], [9], [11], [15]).
In particular, they may arise as ground states of Hamiltonians in nuclear and
trapped ion and atom physics such as the Lipkin-Meshkov-Glick Hamiltonian [10]
or as examples of violation of Bell’s inequalities in quantum nonlocality
[18]. They can also form a set of structured data in geometric quantum machine
learning protocols [12]. Tests of entanglement and separability have been
established for such states (see, e.g., [4], [14], [16], [20]).

In this paper, we want to establish yet another property of the symmetric
sector and therefore of the CG decomposition of ${\cal
H}:=(\mathbb{C}^{d})^{\otimes n}$. We look at the orthogonal complement of the
symmetric sector in ${\cal H}$ and prove that it does not contain any
separable state. It only contains entangled states. Equivalently, any
separable state in ${\cal H}$ must have a non zero component on the symmetric
sector. We will prove this as a consequence of a general property of products
of elements of a matrix and will establish it first for pure states and then
naturally extend it to mixed states by convex combination. For the cases $n=3$
particles and arbitrary dimension $d$, we shall obtain a more detailed result,
giving a (nonzero) sharp lower bound on the size of the component of any
separable state on the symmetric sector. A bound for the bipartite case was
already known (see, e.g., [3]) and we shall recall such a result also. This
will lead us to define a family of entanglement witnesses for these systems.
We will show with an example that this class of entanglement witnesses can
detect entangled states that cannot be detected using the standard partial
transposition test.

The paper is organized as follows. In section 2, we describe the states in the
symmetric sector and in particular the Dicke states which form an orthonormal
basis of this subspace of ${\cal H}:=(\mathbb{C}^{d})^{\otimes n}$. In section
3 we prove a technical linear algebra lemma which is the key step to prove the
above described property of the symmetric sector. This property is proved in
section 4. In section 5, we obtain sharp lower bounds for the component of any
separable state on the symmetric sector for the cases of $n=2$ and $n=3$.
Bounds for $n>3$ can be obtained numerically after setting up an appropriate
multivariable optimization problem and we give an example in the appendix for
the case $n=4$, $d=2$. The whole treatment is carried out for pure states, but
the extension to mixed states is natural by convex combination and it is
discussed in section 6. In section 7 we interpret the above bounds as
specifying entanglement witnesses. After setting up the appropriate
definitions, we use an example to show that such a class of entanglement
witnesses are not decomposable and detect states that cannot be detected using
the (multipartite version) of the standard Peres-Horodecki partial
transposition criterion (see, e.g., [7]), that is, Positive Partial Transposed
(PPT) entangled states.

##  2 The Symmetric Sector and Dicke States

When dealing with systems in quantum information theory, it is customary to
work with an orthonormal basis for $\mathbb{C}^{d}$, often referred to as the
computational basis, given by $\\{|0\rangle,|1\rangle,...,|d-1\rangle\\}$.
Accordingly, an orthonormal basis of ${\cal H}:=(\mathbb{C}^{d})^{\otimes n}$
is given by $|j_{1}\,j_{2}\,\cdots j_{n}\rangle$, with
$j_{1},j_{2},...,j_{n}\in\\{0,1,...,d-1\\}$. The symmetric sector is the
subspace of states invariant under any element of the permutation group
$S_{n}$.

Consider now the set $\Delta_{d,n}$ of $d$-tuples $[k_{0},k_{1},...,k_{d-1}]$
with natural numbers $k_{j}\geq 0$ and $k_{0}+k_{1}+\cdots+k_{d-1}=n$. We have
that $|\Delta_{d,n}|=\begin{pmatrix}n+d-1\cr d-1\end{pmatrix}$. For any
element $[k_{0},...,k_{d-1}]$ in $\Delta_{d,n}$ denote by
$G_{[k_{0},...,k_{d-1}]}$ the set of $n$-ples, $(j_{1},,...,j_{n})$ with
$k_{0}$ $0$’s, $k_{1}$, $1$’s,…,$k_{d-1}$, $d-1$’s. For example set $d=3$,
$n=4$, and consider $[1,0,3]\in\Delta_{3,4}$. Then every $4$-ple with one
‘$0$’ and three ‘$2$’s and zero $1$’s belongs to $G_{[1,0,3]}$. For example
$(0,2,2,2)$, but also $(2,0,2,2)$ etc. The set $G_{[k_{0},k_{1},...,k_{d-1}]}$
contains
$\left|G_{[k_{0},k_{1},...,k_{d-1}]}\right|=\frac{n!}{k_{0}!k_{1}!\cdots
k_{d-1}!}$ elements. To any $d$-tuple
$[k_{0},k_{1},...,k_{d-1}]\in\Delta_{d,n}$ we associate a Dicke state,
$|\phi_{[k_{0},k_{1},...,k_{d-1}]}\rangle$, defined as

$|\phi_{[k_{0},k_{1},...,k_{d-1}]}\rangle:=\frac{1}{\sqrt{|G_{[k_{0},k_{1},...,k_{d-1}]}|}}\sum_{(j_{1},...,j_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}|j_{1},...,j_{n}\rangle=$ (1)
$\sqrt{\frac{k_{0}!k_{1}!\cdots k_{d-1}!}{n!}}\sum_{(j_{1},...,j_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}|j_{1},...,j_{n}\rangle.$

For example, for $d=3$, $n=3$, the state $|\phi_{[1,0,2]}\rangle$ is
$|\phi_{[1,0,2]}\rangle:=\frac{1}{\sqrt{3}}\left(|0\,2\,2\rangle+|2\,0\,2\rangle+|2\,2\,0\rangle\right).$
Dicke states form an orthonormal basis of the symmetric sector, which has
therefore dimension $|\Delta_{d,n}|=\begin{pmatrix}n+d-1\cr d-1\end{pmatrix}$.
This is the basis we shall refer to in the following.

In terms of the Dicke states, we can write the orthogonal projection, $\Pi$,
onto the symmetric sector as,

$\Pi:=\sum_{[k_{0},k_{1},...,k_{d-1}]\in\Delta_{d,n}}|\phi_{[k_{0},k_{1},...,k_{d-1}]}\rangle\langle\phi_{[k_{0},k_{1},...,k_{d-1}]}|.$
(2)

The projection can also be written in terms of the permutations in the
symmetric group $S_{n}$, that is, as

$\Pi=\frac{1}{n!}\sum_{S\in S_{n}}S=\frac{1}{n!}\left(\sum_{S\in
S_{n}}\,\sum_{(j_{1},...,j_{n})\in\\{0,1,...,d-1\\}}|S(j_{1},...,j_{n})\rangle\langle
j_{1},...,j_{n}|\right).$ (3)

##  3 A Technical Lemma

Consider now a (real or complex) $n\times d$ matrix $F_{n\times d}$ with
entries $f_{l,m}$, $l=1,2,...,n$, $m=0,1,...,d-1$, and consider for
$F_{n\times d}$ the following property.

Property A For any element $[k_{0},k_{1},...,k_{d-1}]\in\Delta_{d,n}$ we have

$\sum_{(j_{1},j_{2},...,j_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}f_{1,j_{1}}f_{2,j_{2}}\cdots f_{n,j_{n}}=0.$ (4)

If a matrix $F_{n\times d}$ has a zero row, then, automatically it satisfies
Property A, since the products that appear in the sums in (4) always have one
zero element, the one corresponding to the zero row. The following lemma shows
that the converse fact is also true, that is, if a matrix satisfies Property A
then one of its rows must be zero. Notice also (something we will use) that
both properties are independent of the ordering of the rows, or, in other
terms, they are invariant if we replace $F_{n\times d}$ by $PF_{n\times d}$
where $P$ is a permutation matrix, since this would only change the order in
the products in (4).

######  Lemma 1.

Assume a matrix $F_{n\times d}$ has Property A. Then $F_{n\times d}$ has one
zero row.

###### Proof.

The proof is by induction on $d$. For $d=1$ the statement is true. In this
case $\Delta_{1,n}$ has only one element $[n]$ and $G_{[n]}$ has only the
element $(0,0,...,0)$ with $n$ zeros. Property A has only one condition

$f_{1,0}f_{2,0}\cdots f_{n,0}=0,$ (5)

which implies that one among $\\{f_{1,0},f_{2,0},...,f_{n,0}\\}$ is zero.
Since each one of these entries is the only entry in a row (there is only one
column), the base step is proven.

We now assume the Lemma to be valid for $d-1$. In order to prove it for
general $d$ we prove the following Claim: If $F_{n\times d}$ has no zero row
we must have $f_{1,0}=f_{2,0}=\cdots=f_{n,0}=0$, that is, the first column is
zero. To prove this, we prove by induction on $l$ that $f_{l,0}=0$, for
$l=1,...,n$. We start by using property (4) for $[n,0,...,0]$ which gives,
with the only element $(0,0,...,0)\in G_{[n,0,...,0]}$, formula (5). Therefore
one among $\\{f_{1,0},f_{2,0},...,f_{n,0}\\}$ is zero and we choose $f_{1,0}$
since we can always reorder the rows. This gives the base step. Assume now
that $f_{1,0}=f_{2,0}=\cdots=f_{l,0}=0$, for $l$, $n>l\geq 1$. We want to show
that $f_{l+1,0}=0$. Consider the conditions (4) for an element of
$\Delta_{d,n}$ of the form $[n-l,k_{1},k_{2},...,k_{d-1}]$ with
$k_{1}+k_{2}+\cdots+k_{d-1}=l$. Every multi-index $(j_{1},j_{2},...,j_{n})$ in
the sum has $n-l$ zeros. All the elements that have a $0$ in the first $l$
positions give a zero element in the sum, since by the inductive assumption,
we have $f_{1,0}=f_{2,0}=\cdots=f_{l,0}=0$. The only possibly nonzero terms
are the ones where the zeros appear all in the last $n-l$ position. Therefore
the sum can be written as

$\sum_{(j_{1},j_{2},...,j_{n})\in
G_{[n-l,k_{1},...,k_{d-1}]}}f_{1,j_{1}}f_{2,j_{2}}\cdots
f_{l,j_{l}}f_{l+1,j_{l+1}}\cdots f_{n,j_{n}}=$
$\sum_{(j_{1},j_{2},...,j_{l},0,...,0)\in
G_{[n-l,k_{1},...,k_{d-1}]}}f_{1,j_{1}}f_{2,j_{2}}\cdots
f_{l,j_{l}}f_{l+1,0}\cdots f_{n,0}=$ $f_{l+1,0}\cdots
f_{n,0}\left(\sum_{(j_{1},...,j_{l})\in
G_{[k_{1},...,k_{d-1}]}}f_{1,j_{1}}\cdots f_{l,j_{l}}\right)=0.$

Notice that, in the second to last element of the above string of equalities,
$[k_{1},...,k_{d-1}]$ is an element of $\Delta_{d-1,l}$. Since the above
equalities have to hold for any element of $\Delta_{d-1,l}$, either
$f_{l+1,0}\cdots f_{n,0}=0$ or $\sum_{(j_{1},...,j_{l})\in
G_{[k_{1},...,k_{d-1}]}}f_{1,j_{1}}\cdots f_{l,j_{l}}=0$ for any
$[k_{1},...,k_{d-1}]\in\Delta_{d-1,l}$. However, in the second case, this
would mean that the matrix obtained from $F_{n\times d}$ selecting the first
$l$ rows and neglecting the column $0$ has, by the inductive assumption
applied to $d-1$, a zero row. This would mean, since
$f_{1,0}=f_{2,0}=\cdots=f_{l,0}=0$, that the original matrix has a zero row,
which we have excluded (and would conclude the proof of the lemma). Therefore
we have

$f_{1,0}=f_{2,0}=\cdots=f_{n,0}=0.$ (6)

Now consider the equations (4) corresponding to elements
$[0,k_{1},...,k_{d-1}]$ in $\Delta_{d,n}$, that is, the elements
$(j_{1},...,j_{n})\in G_{[0,k_{1},...,k_{d-1}]}$ are $n$-ples which have no
$0$’s. We have, specializing (4),

$0=\sum_{(j_{1},...,j_{n})\in G_{[0,k_{1},...,k_{d-1}]}}f_{1,j_{1}}\cdots
f_{n,j_{n}}=\sum_{(j_{1},...,j_{n})\in
G_{[k_{1},...,k_{d-1}]}}f_{1,j_{1}}\cdots f_{n,j_{n}}.$

In the last sum, $[k_{1},...,k_{d-1}]$ is an arbitrary element of
$\Delta_{d-1,n}$. Therefore, by the inductive assumption, the matrix
$\\{f_{l,m}\\}$, $l=1,...,n$, $m=1,...,d-1$ has one zero row. This matrix is
given by the last $d-1$ columns of $F_{n\times d}$. Since the first column is
also zero from (6), $F_{n\times d}$ has one zero row, and the lemma is proved.

∎

##  4 Entanglement of States that are Zero on the Symmetric Sector

Assume now that we have a separable state $|\psi\rangle$ in ${\cal H}$, that
is,
$|\psi\rangle:=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes\cdots\otimes|\psi_{n}\rangle$.
Writing each $|\psi_{m}\rangle$, $m=1,...,n$, in the computational basis as
$|\psi_{m}\rangle:=\sum_{j_{m}=0}^{d-1}f_{m,j_{m}}|j_{m}\rangle$, we can write

$|\psi\rangle=\sum_{j_{1},j_{2},...,j_{n}=0}^{d-1}f_{1,j_{1}}f_{2,j_{2}}\cdots
f_{n,j_{n}}|j_{1}\rangle\otimes|j_{2}\rangle\otimes\cdots\otimes|j_{n}\rangle.$
(7)

Now consider the inner product of $|\psi\rangle$ with a Dicke state
$|\phi_{[k_{0},k_{1},...,k_{d-1}]}\rangle$ in (1). We have

$\sqrt{|G_{[k_{0},k_{1},...,k_{d-1}]}|}\langle\phi_{[k_{0},k_{1},...,k_{d-1}]}|\psi\rangle=$
(8) $\left(\sum_{(l_{1},...,l_{n})\in G_{[k_{0},k_{1},...,k_{d-1}]}}\langle
l_{1}|\otimes\langle l_{2}|\otimes\cdots\otimes\langle
l_{n}|\right)\left(\sum_{j_{1},...,j_{n}=0}^{d-1}f_{1,j_{1}}\cdots
f_{n,j_{n}}|j_{1}\rangle\otimes|j_{2}\rangle\otimes\cdots\otimes|j_{n}\rangle.\right)=$
$\sum_{(l_{1},...,l_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}\sum_{j_{1},...,j_{n}=0}^{d-1}f_{1,j_{1}}\cdots
f_{n,j_{n}}\delta_{l_{1},j_{1}}\delta_{l_{2},j_{2}}\cdots\delta_{l_{n},j_{n}}=$
$\sum_{(j_{1},...,j_{n})\in G_{[k_{0},k_{1},...,k_{d-1}]}}f_{1,j_{1}}\cdots
f_{n,j_{n}}.$

Using this, if the separable state $|\psi\rangle$ has zero component in the
symmetric sector, since all the inner products
$\langle\phi_{[k_{0},k_{1},...,k_{d-1}]}|\psi\rangle$ are zero, the
coefficients $f_{1,j_{1}},f_{2,j_{2}},...,f_{n,j_{n}}$,
$j_{1},j_{2},...,j_{n}=0,1,...,d-1$ must satisfy Property A with (4).
Therefore, we can apply Lemma 1 to conclude that for at least one $k=1,...,n$,
$f_{k,j_{k}}=0$, for all $j_{k}=0,1,...,d-1$. This means that one of the
factors, $|\psi_{k}\rangle$, $k=1,...,n$, in the factorization
$|\psi\rangle=|\psi_{1}\otimes|\psi_{2}\rangle\otimes\cdots\otimes|\psi_{n}\rangle$
is zero. However this is impossible for a meaningful quantum state
$|\psi\rangle$. Therefore, we have proven any one of the equivalent statements
in the following theorem.

######  Theorem 2.

  1. 1.

The orthogonal complement of the symmetric sector in ${\cal
H}:=(\mathbb{C}^{d})^{\otimes n}$ contains no separable state.

  2. 2.

Any separable state in ${\cal H}:=(\mathbb{C}^{d})^{\otimes n}$ must have a
nonzero orthogonal component on the symmetric sector.

  3. 3.

Any quantum state that has zero component on the symmetric sector must be
entangled.

This result may be suggested by and generalizes the situation in the two
qubits case where the symmetric sector is spanned by
$\\{|0,0\rangle,\frac{1}{\sqrt{2}}(|0,1\rangle+|1,0\rangle),|1,1\rangle\\}$
and its orthogonal complement is the antisymmetric subspace spanned by
$\frac{1}{\sqrt{2}}(|0,1\rangle-|1,0\rangle)$, an entangled state.

Some remarks that will be used in the following sections are in order. The
$n\times d$ matrix $F$ with entries $f_{m,j}$, $m=1,...,n$, $j=0,1,...,d-1$,
can be used to represent an $n$-ple of states $\\{|\psi_{m}\rangle\\}$,
$m=1,...,n$, which appear in the product $\otimes_{m=1}^{n}|\psi_{m}\rangle$
for a separable state. Every row represents the components of each of the
factor states in the computational basis. We shall refer to this matrix as the
$F$-matrix associated to the separable state
$\otimes_{m=1}^{n}|\psi_{m}\rangle$. Rows can be modified by multiplying all
the elements by a phase factor $e^{i\phi}$, without modifying the physical
state. The component of the product state $|\psi\rangle$ onto the Dicke state
$|\phi_{[k_{0},k_{1},...,k_{d-1}]}\rangle$ is given, according to formula (8),
by

$\langle\phi_{[k_{0},k_{1},...,k_{d-1}]}|\psi\rangle=\sqrt{\frac{k_{0}!k_{1}!\cdots
k_{d-1}!}{n!}}\sum_{(j_{1},...,j_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}f_{1,j_{1}}f_{2,j_{2}}\cdots f_{n,j_{n}}.$ (9)

Thus the square of the norm of the orthogonal component onto the symmetric
sector, $\||\psi\rangle\|_{SS}^{2}$, is

$\||\psi\rangle\|_{SS}^{2}=\sum_{[k_{0},k_{1},...,k_{d-1}]\in\Delta_{d,n}}\frac{k_{0}!k_{1}!\cdots
k_{d-1}!}{n!}\left|\sum_{(j_{1},j_{2},...,j_{n})\in
G_{[k_{0},k_{1},...,k_{d-1}]}}f_{1,j_{1}}f_{2,j_{2}}\cdots
f_{n,j_{n}}\right|^{2}.$ (10)

The quantity $\||\psi\rangle\|_{SS}^{2}$ can also be written in terms of the
expression (3) for the projection $\Pi$, i.e.,

$\||\psi\rangle\|_{SS}^{2}=\langle\psi|\Pi|\psi\rangle=\frac{1}{n!}\sum_{S\in
S_{n}}\langle\psi|S|\psi\rangle.$ (11)

To compute $\langle\psi|S|\psi\rangle$ in the case where
$|\psi\rangle:=|\psi_{1}\rangle\otimes|\psi_{2}\otimes\cdots\otimes|\psi_{n}\rangle$
we can use the fact that every permutation $S$ is the product of disjoint
cycles, $C_{1},C_{2},...,C_{m}$ so that,
$\langle\psi|S|\psi\rangle=\prod_{j=1}^{m}\langle\psi|C_{j}|\psi\rangle$,
where if $C$ is the cycle $C=(i_{1}\,i_{2}\,\cdots\,i_{k})$,
$\langle\psi|C|\psi\rangle=\langle\psi_{i_{1}}|\psi_{k}\rangle\langle\psi_{i_{2}}|\psi_{i_{1}}\rangle\langle\psi_{i_{3}}|\psi_{i_{2}}\rangle\cdots\langle\psi_{k}|\psi_{i_{k-1}}\rangle$.
In the following section we shall calculate the minimum value for
$\||\psi\rangle\|_{SS}^{2}$ for $n=2,3$ and we shall use mostly the method
based on (10). In the appendix we shall consider the case $n=4$, $d=2$ and we
shall use the expression (11). This makes a connection with the structure
theory of the symmetric group $S_{n}$. If $X\in U(d)$ and we define
$Y:=X\otimes X\otimes\cdots\otimes X$, the tensor product of $n$ equal
factors, we have $Y^{\dagger}\Pi Y=\Pi$.222To see this, just apply left hand
side and right hand side to an arbitrary vector in $(\mathbb{C}^{d})^{\otimes
n}$ written as the sum of its components in the symmetric sector and in the
orthogonal complement and use the fact that both these spaces are invariant
under $Y$ and $\Pi$ is the identity on the symmetric sector and zero on its
orthogonal complement. If $|\psi\rangle$ is separable, we can choose $X$ to
make the first factor equal to $|0\rangle$, and then the second factor of the
form $\alpha|0\rangle+\beta|1\rangle$ (with $\alpha$ and $\beta$ real), and
then the third factor of the form
$\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle$ and (possibly) so on, without
loss of generality.

##  5 Lower bounds for $\||\psi\rangle\|_{SS}^{2}$ for separable
$|\psi\rangle$

###  5.1 Case $n=2$

The bipartite case $n=2$ is treated as an example in [3] (Example 3.1). In
this case $\Pi$ in (3) is $\Pi=\frac{1}{2}\left({\bf 1}+(1\,2)\right)$, where
$(1\,2)$ is the SWAP operator permuting the first and second factor. If
$|\psi\rangle$ is separable,
$|\psi\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle$, then

$\||\psi\rangle\|_{SS}^{2}=\langle\psi|\Pi|\psi\rangle=\frac{1}{2}+\frac{1}{2}|\langle\psi_{1}|\psi_{2}\rangle|^{2}\geq\frac{1}{2},$

where equality is achieved when $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ are
orthogonal to each other.

###  5.2 Case $n=3$; $d=2$

To compute the minimum $\||\psi\rangle\|_{SS}^{2}$ for the case $n=3$, $d=2$,
we shall use the expression (10) for $\||\psi\rangle\|_{SS}^{2}$. According to
the discussion at the end of the previous section, we can apply a common
unitary on the qubits and collect the phase factors, so that the resulting
$F-$matrix has, without loss of generality, the form,

$F=\begin{pmatrix}1&0\cr\cos(\theta)&\sin(\theta)\cr\cos(\eta)&\sin(\eta)e^{i\alpha}\end{pmatrix}.$
(12)

Following (9) The components along the Dicke states in this case are

$\langle\phi_{[3,0]}|\psi\rangle=f_{1,0}f_{2,0}f_{3,0}=\cos(\theta)\cos(\eta),$
$\langle\phi_{[2,1]}|\psi\rangle=\frac{1}{\sqrt{3}}\left(f_{1,1}f_{2,0}f_{3,0}+f_{1,0}f_{2,1}f_{3,0}+f_{1,0}f_{2,0}f_{3,1}\right)=\frac{1}{\sqrt{3}}\left(\sin(\theta)\cos((\eta)+\cos(\theta)\sin(\eta)e^{i\alpha}\right),$
$\langle\phi_{[1,2]}|\psi\rangle=\frac{1}{\sqrt{3}}\left(f_{1,1}f_{2,1}f_{3,0}+f_{1,0}f_{2,1}f_{3,1}+f_{1,1}f_{2,0}f_{3,1}\right)=\frac{1}{\sqrt{3}}\sin(\theta)\\\
sin(\eta)e^{i\alpha},$
$\langle\phi_{[0,3]}|\psi\rangle=f_{1,1}f_{2,1}f_{3,1}=0.$

Therefore we have

$\||\psi\rangle\|_{SS}^{2}=\cos^{2}(\theta)\cos^{2}(\eta)+$ (13)
$\frac{1}{3}\left(\sin^{2}(\theta)\cos^{2}(\eta)+\cos^{2}(\theta)\sin^{2}(\eta)+2\sin(\theta)\cos(\eta)\cos(\theta)\sin(\eta)\cos(\alpha)\right)+\frac{1}{3}\sin^{2}(\theta)\sin^{2}(\eta)=$
$\frac{1}{3}+\frac{2}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{6}\sin(2\theta)\sin(2\eta)\cos(\alpha)=$
$\frac{1}{3}+\frac{2}{3}\left(\frac{1+\cos(2\theta)}{2}\right)\left(\frac{1+\cos(2\eta)}{2}\right)+\frac{1}{6}\sin(2\theta)\sin(2\eta)\cos(\alpha)=$
$\frac{1}{2}+\frac{1}{6}\left(\cos(2\eta)+(1+\cos(2\eta))\cos(2\theta)+\sin(2\theta)\sin(2\eta)\cos(\alpha)\right).$

Now fix $\eta$ and $\alpha$ and minimize this expression over $\theta$, using
the fact that the minimum of $a\cos(x)+b\sin(x)$ is $-\sqrt{a^{2}+b^{2}}$.
This gives

$\||\psi\rangle\|_{SS}^{2}\geq\frac{1}{2}+\frac{1}{6}\left(\cos(2\eta)-\sqrt{(1+\cos(2\eta))^{2}+\sin^{2}(2\eta)\cos^{2}(\alpha)}\right)\geq$
$\frac{1}{2}+\frac{1}{6}\left(\cos(2\eta)-\sqrt{(1+\cos(2\eta))^{2}+\sin^{2}(2\eta)}\right)=$
$\frac{1}{2}+\frac{1}{6}\left(\cos(2\eta)-\sqrt{2+2\cos(2\eta)}\right)=\frac{1}{2}+\frac{1}{6}\left(2\cos^{2}(\eta)-1-2|\cos(\eta)|\right)=\frac{1}{3}+\frac{1}{3}\left(\cos^{2}(\eta)-|\cos(\eta)\right)\geq\frac{1}{4},$

where we used the fact that the minimum of $x^{2}-x$ in $[0,1]$ is
$-\frac{1}{4}$. This bound of $\frac{1}{4}$ is achieved by setting in (13)
$\alpha=0$, $\theta=\frac{2\pi}{3}$, $\eta=\frac{\pi}{3}$.

###  5.3 Case $n=3$; $d\geq 3$

In the case $n=3$ and $d\geq 3$, we use again formulas (9) and (10). By
applying a common unitary following the discussion at the end of the previous
section, and collecting a phase factor, we can assume the $F$-matrix, without
loss of generality, of the form

$F=\begin{pmatrix}1&0&0&0&\cdots&0\cr\cos(\theta)&\sin(\theta)&0&0&\cdots&0\cr\cos(\eta)&\sin(\eta)\cos(\gamma)e^{i\rho}&\sin(\eta)\sin(\gamma)&0&\cdots&0\end{pmatrix}.$
(14)

Consider now a $d$-tuple $[k_{0},k_{1},k_{2},...,k_{d-1}]$ with one $k_{l}$
for $l\geq 3$ different from zero. Then every triple $(j_{1},j_{2},j_{3})$ in
the sum in (9) would have an element equal to $l$, which would make every term
in the sum equal to zero from the form of the matrix $F$ in (14). Similarly,
every $d$-tuple such that $k_{0}=0$ will be such that $(j_{1},j_{2},j_{3})\in
G_{[k_{0},k_{1},...,k_{d-1}]}$ implies $j_{1}\not=0$. Therefore, all the terms
in the sum (9) will be zero, from the $F$-matrix in (14). Therefore the only
components on Dicke states we have to consider are

$\langle\phi_{[3,0,0,0,...,0]}|\psi\rangle=f_{1,0}f_{2,0}f_{3,0}=\cos(\theta)\cos(\eta),$
$\langle\phi_{[2,1,0,0,...,0]}|\psi\rangle=\frac{1}{\sqrt{3}}\left(f_{1,0}f_{2,0}f_{3,1}+f_{1,0}f_{2,1}f_{3,0}+f_{1,1}f_{2,0}f_{3,0}\right)=\frac{1}{\sqrt{3}}\left(\cos(\theta)\sin(\eta)\cos(\gamma)e^{i\rho}+\sin(\theta)\cos(\eta)\right),$
$\langle\phi_{[2,0,1,0,...,0]}|\psi\rangle=\frac{1}{\sqrt{3}}\left(f_{1,0}f_{2,0}f_{3,2}+f_{1,0}f_{2,2}f_{3,0}+f_{1,2}f_{2,0}f_{3,0}\right)=\frac{1}{\sqrt{3}}\cos(\theta)\sin(\eta)\sin(\gamma),$
$\langle\phi_{[1,2,0,0,...,0]}|\psi\rangle=\frac{1}{\sqrt{3}}\left(f_{1,0}f_{2,1}f_{3,1}+f_{1,1}f_{2,1}f_{3,0}+f_{1,1}f_{2,0}f_{3,1}\right)=\frac{1}{\sqrt{3}}\sin(\theta)\sin(\eta)\cos(\gamma)e^{i\rho},$
$\langle\phi_{[1,1,1,0,...,0]}|\psi\rangle=\frac{1}{\sqrt{6}}\left(f_{1,0}f_{2,1}f_{3,2}+f_{1,0}f_{2,2}f_{3,1}+f_{1,1}f_{2,0}f_{3,2}+f_{1,1}f_{2,2}f_{3,0}+f_{1,2}f_{2,1}f_{3,0}+f_{1,2}f_{2,0}f_{3,1}\right)=$
$\frac{1}{\sqrt{6}}\sin(\theta)\sin(\eta)\sin(\gamma).$

We have

$\||\psi\rangle\|_{SS}^{2}=\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)\sin^{2}(\eta)\cos^{2}(\gamma)+\frac{1}{3}\sin^{2}(\theta)\cos^{2}(\eta)+\frac{1}{6}\sin(2\theta)\sin(2\eta)\cos(\gamma)\cos(\rho)+$
$\frac{1}{3}\cos^{2}(\theta)\sin^{2}(\eta)\sin^{2}(\gamma)+\frac{1}{3}\sin^{2}(\theta)\sin^{2}(\eta)\cos^{2}(\gamma)+\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\eta)\sin^{2}(\gamma)=$

$\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)\sin^{2}(\eta)+\frac{1}{3}\sin^{2}(\theta)\cos^{2}(\eta)+$
$\frac{1}{6}\sin(2\theta)\sin(2\eta)\cos(\gamma)\cos(\rho)+\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\eta)\cos^{2}(\gamma)+\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\eta)=$
$\frac{1}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)+\frac{1}{3}\cos^{2}(\eta)+$
$\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\eta)+\frac{1}{6}\left[\sin(2\theta)\sin(2\eta)\cos(\gamma)\cos(\rho)+\sin^{2}(\theta)\sin^{2}(\eta)\cos^{2}(\gamma)\right].$

For a fixed $\theta$, $\eta$ and $\rho$, let us minimize the quantity in
square brackets with respect to $\gamma$. In the simple case where
$\sin(\theta)=0$ and/or $\sin(\eta)=0$, the quantity in square brackets is
identically zero. Let us assume, without loss of generality that
$\sin(\theta)=0$ since $\theta$ and $\eta$ are intercheangable. The function
becomes equal to $\frac{1}{3}+\frac{2}{3}\cos^{2}(\eta)\geq\frac{1}{3}$ . Let
us now consider the case where $\sin(\theta)\not=0$ and $\sin(\eta)\not=0$. A
calculus exercise shows that the minimum of a function $f(x)=bx^{2}+ax$ with
$b>0$ is $-\frac{a^{2}}{4b}$ if $\frac{|a|}{2b}<1$ and $b-|a|$ if
$\frac{|a|}{2b}\geq 1$. Applying this, to the function above with
$b=\sin^{2}(\theta)\cos^{2}(\theta)$, $a=\sin(2\theta)\sin(2\eta)\cos(\rho)$,
we have to consider two situations:

  1. 1.

$\frac{|a|}{2b}=\frac{|sin(2\theta)\sin(2\eta)\cos(\rho)|}{2\sin^{2}(\theta)\sin^{2}(\eta)}=\frac{|2\cos(\theta)\cos(\eta)\cos(\rho)|}{|\sin(\theta)\sin(\eta)|}<1$

In this case, using the expression $-\frac{a^{2}}{4b}$ for the minimum of the
function in parenthesis, we get,

$\|\psi\|_{SS}^{2}\geq\frac{1}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)+\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\eta)-\frac{1}{6}\left(\frac{\sin^{2}(2\theta)\sin^{2}(2\eta)\cos^{2}(\rho)}{4\sin^{2}(\theta)\sin^{2}(\eta)}\right)\geq$
$-\frac{1}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)+\frac{1}{6}\sin^{2}(\theta)\sin^{2}(\theta)=$
$\frac{1}{6}+\frac{1}{6}\left(\cos^{2}(\eta)+\cos^{2}(\theta)-\cos^{2}(\theta)\cos^{2}(\eta)\right)\geq\frac{1}{6}.$

Here, in the last inequality, we used the fact that the minimum of the
function $f(x,y)=x+y-xy$ in the square $0\leq x\leq 1$, $0\leq y\leq 1$, is
zero. Such a minimum is achieved with $x=y=0$ and therefore
$\theta=\frac{\pi}{2}$, $\eta=\frac{\pi}{2}$. Notice that this is also
consistent with the assumption
$\frac{|a|}{2b}=\frac{|2\cos(\theta)\cos(\eta)\cos(\rho)|}{|\sin(\theta)\sin(\eta)|}<1$.

  2. 2.

$\frac{|a|}{2b}=\frac{|sin(2\theta)\sin(2\eta)\cos(\rho)|}{2\sin^{2}(\theta)\sin^{2}(\eta)}=\frac{|2\cos(\theta)\cos(\eta)\cos(\rho)|}{|\sin(\theta)\sin(\eta)|}\geq
1$ (15)

We have, using the value
$b-|a|=\sin^{2}(\theta)\sin^{2}(\eta)-|\sin(2\theta)\sin(2\eta)\cos(\rho)|$
for the minimum,

$\|\psi_{SS}\|^{2}\geq\frac{1}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)+\frac{1}{3}\sin^{2}(\theta)\sin^{2}(\eta)-\frac{1}{6}|\sin(2\theta)\sin(2\eta)\cos(\rho)|$
$\||\psi\rangle\|_{SS}^{2}\geq_{\rho=0}\frac{1}{3}\cos^{2}(\theta)\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\eta)+\frac{1}{3}\cos^{2}(\theta)+\frac{1}{3}\sin^{2}(\theta)\sin^{2}(\eta)-\frac{1}{6}|\sin(2\theta)\sin(2\eta)|.$

Using Young inequality
$|sin(2\theta)\sin(2\eta)|\leq\frac{1}{2}\left(\sin^{2}(2\theta)+\sin^{2}(2\eta)\right)$
and trigonometric identities we get

$\||\psi\rangle\|_{SS}^{2}\geq$
${\frac{1}{3}}\left[\cos^{2}(\theta)\cos^{2}(\eta)+\cos^{2}(\eta)+\cos^{2}(\theta)+\sin^{2}(\theta)\sin^{2}(\eta)-\sin^{2}(\theta)\cos^{2}(\theta)-\sin^{2}(\eta)\cos^{2}(\eta)\right]=$
$\frac{1}{3}\left[(\cos^{2}(\theta)+\cos^{2}(\eta))^{2}+1-(\cos^{2}(\eta)+\cos^{2}(\theta))\right].$

We then minimize this subject to (15) which for $\rho=0$ gives

$3\cos^{2}(\theta)\cos^{2}(\eta)\geq 1-\cos^{2}(\eta)-\cos^{2}(\theta),$ (16)

or, with notational changes, we consider the minimization of a function
$f(x,y)=$

$\frac{1}{3}\left((x+y)^{2}+1-x-y\right)$ subject to $0\leq x,y,\leq 1$ and
$3xy\geq 1-x-y$. This minimum is achieved for $x=y=\frac{1}{3}$ and it is
equal to $\frac{7}{27}$.333There is no stationary point for the function in
the allowed set since stationary points are where $x+y=\frac{1}{2}$ which is
incompatible with the given constraints. So the minimum must be on the
boundary of the set which is where $x\equiv 1$, $y\equiv 1$ or where
$y=\frac{1-x}{3x+1}$ with $0\leq x\leq 1$. With $x\equiv 1$ ($y\equiv 1$) the
minimum is for $y=0$ ($x=0$) and it is equal to $\frac{1}{3}$. For
$y=\frac{1-x}{3x+1}$ with $0\leq x\leq 1$, the minimum is obtained for
$x=\frac{1}{3}$ (which also gives $y=\frac{1}{3}$) and it is equal to
$\frac{7}{27}$ which is smaller than $\frac{1}{3}$.

In conclusion, comparing all cases, the minimum is $\frac{1}{6}$.

###  5.4 Summary

We summarize the lower bounds we found, which are all sharp, in the following
table

|  |
---|---|---
$\||\psi\rangle\|_{SS}^{2}\leq$ | $d=2$ | $d\geq 3$
|  |
|  |
$n=2$ | ${{\frac{1}{2}}}$ | $\frac{1}{2}$
|  |
|  |
$n=3$ | $\frac{1}{4}$ | $\frac{1}{6}$
|  |

.

##  6 Extension to Mixed States

The extension of the result of Theorem 2 to the case of mixed states and in a
density matrix description of quantum states comes naturally once we recall
the appropriate definitions. Recall that a general (possibly mixed) quantum
state can be described by a density matrix operator as a statistical mixture
of pure states, i.e., (see, e.g., [17])

$\rho:=\sum_{s}\lambda_{s}|\psi_{s}\rangle\langle\psi_{s}|,\quad\lambda_{s}>0,\quad\sum_{s}\lambda_{s}=1.$
(17)

In this context, $\rho$ is called separable if it is a statistical mixture of
product states, that can be assumed, without loss of generality, to be pure,
i.e.,

$\rho=\sum_{j}{\omega_{j}}|\psi_{1,j}\rangle\langle\psi_{1,j}|\otimes|\psi_{2,j}\rangle\langle\psi_{2,j}|\otimes\cdots\otimes|\psi_{n,j}\rangle\langle\psi_{n,j}|=$
(18)
$\sum_{j}{\omega_{j}}\left(|\psi_{1,j}\rangle\otimes|\psi_{2,j}\rangle\otimes\cdots\otimes|\psi_{n,j}\rangle\right)\left(\langle\psi_{1,j}|\otimes\langle\psi_{2,j}|\otimes\cdots\otimes\langle\psi_{n,j}|\right)$

with $\sum_{j}\omega_{j}=1,$ $\omega_{j}>0$. States $\rho$ that cannot be
written as in (18) are called entangled. With these definitions, the extension
of Theorem 2 to general mixed states reads as follows.

######  Theorem 3.

Assume for every state $|\phi\rangle$ in the symmetric sector
$\rho|\phi\rangle=0$, or, equivalently $Tr(\Pi\rho)=0$ with $\Pi$ in (2) (3).
Then $\rho$ is an entangled state.

###### Proof.

If for every $|\phi\rangle$ in the symmetric sector $\rho|\phi\rangle=0$ and
$\rho$ is separable, using (18), we get
$\sum_{j}\omega_{j}\left|\langle\phi|\psi_{1,j}\rangle\otimes|\psi_{2,j}\rangle\otimes\cdots\otimes|\psi_{n,j}\rangle\right|^{2}=0$
which implies, for every $j$,
$\left|\langle\phi|\psi_{1,j}\rangle\otimes|\psi_{2,j}\rangle\otimes\cdots\otimes|\psi_{n,j}\rangle\right|^{2}=0$,
which is impossible according to Theorem 2. ∎

Let us consider again $\Pi$, the orthogonal projection onto the symmetric
sector as defined in (2), (3). If we have a lower bound $B$ on the component
square of any separable state on the symmetric sector, we have that, for any
separable state $\rho$,
$\rho:=\sum_{j}\omega_{j}|\psi_{j}\rangle\langle\psi_{j}|$,

$Tr(\Pi\rho)=\sum_{j}\omega_{j}\langle\psi_{j}|\Pi|\psi_{j}\rangle\geq B.$

Therefore, for any separable state $\rho$ we have (${\bf 1}$ denotes the
identity)

$Tr\left[\left(\Pi-B{\bf 1}\right)\rho\right]=Tr(\Pi\rho)-B\geq 0.$

Therefore $W:=\Pi-B{\bf 1}$ is an entanglement witness [3] for multipartite
systems on $(\mathbb{C}^{d})^{\otimes n}$ (see, e.g., [13]) that is an
operator $W$ such that $Tr(W\rho)\geq 0$ for any separable state $\rho$. The
condition $Tr(W\rho)<0$ therefore indicates that $\rho$ is an entangled state.

##  7 Entanglement witnesses and PPT states

It is interesting to know whether the above class of entanglement witnesses is
able to detect entangled states which are not detected by the popular Peres-
Horodecki Positive Partial Transposition (PPT) criterion (see, e.g., [8]). In
this section we explore this issue after setting up the appropriate
definitions. We shall show with an example for a system of three qubits that
this is actually the case.

The PPT criterion of entanglement was originally proposed for bipartite
quantum systems (of possibly different dimensions) and it turned out to be not
only avery easy to use criterion but also a very powerful one. In the
bipartite case it detects all entangled states when the product of the
dimensions of the two subsystems is less than or equal to $6$ [7]. We discuss
it here in the context of a (natural) extension to the multipartite case,
assuming, as in the rest of the paper, that all the $n$ subsystems have the
same dimension $d$. Let ${\cal I}$ be a subset of ${\cal N}:=\\{1,...,n\\}$,
and denote by $\rho^{\Gamma_{\cal I}}$ the partial transpose of $\rho$ with
respect to the factors labeled by the elements in ${\cal I}$.444That is,
expand $\rho$ as a linear combination of tensor products and then take the
transpose of only the elements in the products with positions given by the
indexes in ${\cal I}$ The state $\rho$ is called ${\cal I}$-PPT if
$\rho^{\Gamma_{\cal I}}\geq 0$, that is, $\rho^{\Gamma_{\cal I}}$ is a state
as well (since the trace equal to 1 condition is automatically satisfied).
Notice that since $(\rho^{\Gamma_{\cal I}})^{T}=\rho^{\Gamma_{{\cal N}-{\cal
I}}}$, $\rho$ is ${\cal I}$-PPT if and only if it is (${\cal N}-{\cal
I}$)-PPT. Every separable state $\rho$ is ${\cal I}$-PPT, no matter what
${\cal I}$ is. Therefore if there exists a set ${\cal I}$ such that $\rho$ is
not ${\cal I}$-PPT, $\rho$ is entangled. In our multipartite setting, we shall
call PPT states, states that are ${\cal I}$-PPT for any ${\cal I}$. Separable
states are PPT. As stated in [3], ‘the hard problem in entanglement theory is
to detect entangled PPT states’. Notice that, to check that a state is PPT, we
need to check that it is ${\cal I}$-PPT for any ${\cal I}\subseteq{\cal N}$,
that is, for $2^{n}$ combinations. However ${\cal I}=\\{\emptyset\\}$ and
${\cal I}={\cal N}$ do not need to be checked (since $\rho\geq 0$ and
$\rho^{T}\geq 0$) and ${\cal I}$ and ${\cal N}-{\cal I}$ give the same result.
Thus, we need to check a total of $2^{n-1}-1$ subsets in ${\cal N}$, which
reduces to the known single test in the bipartite case.

We now present a class of PPT states which is detected by the entanglement
witnesses found in this paper. We consider three qubits ($n=3$, $d=2$) and the
class of generalized Werner states

$\rho=\frac{p}{4}\Pi+\frac{1-p}{4}\left({\bf 1}-\Pi\right),$ (19)

where $\Pi$ is the projection onto the symmetric sector and $p\in[0,1]$.
According to the results of section 5, $\Pi-\frac{1}{4}{\bf 1}$ is an
entanglement witness for the system $n=3$ $d=2$, which applied to the states
$\rho$ in ([19]) implies that $\rho$ is entangled for $p<\frac{1}{4}$. When
calculating $\rho^{\Gamma_{\cal I}}$ for $\rho$ in (19) and ${\cal I}$, to
check whether $\rho$ is PPT, we only need to check for $|{\cal I}|=1$ since
$\rho^{\Gamma_{{\cal N}-{\cal I}}}\geq 0\leftrightarrow\rho^{\Gamma_{{\cal
I}}}\geq 0$. Furthermore, let $P$ any permutation matrix. Since $\rho$
commutes with $P$ we have that for ${\cal I}_{1}$ and ${\cal I}_{2}$ of
cardinality $1$,

$\rho^{\Gamma_{{\cal I}_{2}}}=P\rho^{\Gamma_{{\cal I}_{1}}}P^{T}$ (20)

for some permutation $P$.555In general for ${\cal I}_{1}$ and ${\cal I}_{2}$
of the same cardinality we have for an appropriate permutation matrix $P$,
$\rho^{\Gamma_{{\cal I}_{2}}}=P\left(P^{T}\rho P\right)^{\Gamma_{{\cal
I}_{1}}}P^{T}$ which reduces to (20), if $\rho$ commutes with $P$ as in our
case. Therefore, we need to check the PPT property only for one set ${\cal I}$
with cardinality $1$ and we take, without loss of generality, ${\cal
I}=\\{1\\}$, denoting simply by $\Gamma$ the corresponding $\Gamma_{\cal I}$.
Applying (3) this to the case $n=3$, $d=2$ leads to

$\Pi=\frac{1}{6}\begin{pmatrix}6&0&0&0&0&0&0&0\cr 0&2&2&0&2&0&0&0\cr
0&2&2&0&2&0&0&0\cr 0&0&0&2&0&2&2&0\cr 0&2&2&0&2&0&0&0\cr 0&0&0&2&0&2&2&0\cr
0&0&0&2&0&2&2&0\cr 0&0&0&0&0&0&0&6\end{pmatrix}.$ (21)

Partial transposition with respect to the first factor corresponds to swapping
the off diagonal $4\times 4$ blocks of a partitioned matrix
$\begin{pmatrix}A&B\cr C&D\end{pmatrix}\rightarrow\begin{pmatrix}A&C\cr
B&D\end{pmatrix}$. Using this and (21) in (19), we obtain

$\rho^{\Gamma}=\frac{p}{24}\begin{pmatrix}6&0&0&0&0&2&2&0\cr
0&2&2&0&0&0&0&2\cr 0&2&2&0&0&0&0&2\cr 0&0&0&2&0&0&0&0\cr 0&0&0&0&2&0&0&0\cr
2&0&0&0&0&2&2&0\cr 2&0&0&0&0&2&2&0\cr
0&2&2&0&0&0&0&6\end{pmatrix}+\frac{1-p}{24}\begin{pmatrix}0&0&0&0&0&-2&-2&0\cr
0&4&-2&0&0&0&0&{-2}\cr 0&-2&4&0&0&0&0&-2\cr 0&0&0&4&0&0&0&0\cr
0&0&0&0&4&0&0&0\cr{-2}&0&0&0&0&4&{-2}&0\cr{-2}&0&0&0&0&{-2}&4&0\cr
0&{-2}&{-2}&0&0&0&0&0\end{pmatrix}.$ (22)

Application of the Sylvester criterion (see, e.g., [6]) shows that
$\rho^{\Gamma}$ is positive definite if and only if $p>0.2$. Therefore $\rho$
is PPT for every $p>0.2$. If $0.25>p>0.2$, the entanglement witnesses proposed
in this paper detect entangled PPT Werner states in (19).

######  Remark 4.

Werner states were introduced in [19] for the $n=2$ case and $d$ arbitrary and
considered in [3] (Example 3.2). It is known that they are separable if and
only if they are PPT. Our example shows that this property does not extend to
the $n=3$ case.

######  Remark 5.

In the bipartite case, an entanglement witness $W$ is called decomposable if
there exist positive semidefinite Hermitian matrices $X$ and $Y$ such that
$W=X^{\Gamma}+Y$. Such definition naturally extends to the multipartite case
by saying that $W$ is decomposable iff there exists a set ${\cal
I}\subseteq{\cal N}$ and positive semidefinite $X$ and $Y$ such that
$W=X^{\Gamma_{\cal{I}}}+Y$. Decomposable entanglement witness are not able to
detect entangled PPT states because if $\rho^{\Gamma_{\cal I}}\geq 0$,666Here
we use the property of the partial transposition $Tr(A^{\Gamma_{\cal
I}}B)=Tr(AB^{\Gamma_{\cal I}})$.

$Tr\left(W\rho\right)=Tr\left(((X^{\Gamma_{\cal
I}}+Y)\rho\right)=Tr\left(X^{\Gamma_{\cal
I}}\rho\right)+Tr\left(Y\rho\right)=Tr(X\rho^{\Gamma_{\cal I}})+Tr(Y\rho)\geq
0.$

Therefore, the entanglement witnesses proposed here are, in general, not
decomposable in the sense of the above definition.777One can show them to be
decomposable in the $n=2$, bipartite case [3].

## Acknowledgement

This research was supported by the ARO MURI grant W911NF-22-S-0007.

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##  Appendix A Bound for $\|\langle\psi\|^{2}_{SS}$ in the case $n=4$, $d=2$

We consider in this appendix a numerical calculation of the minimum for
$\||\psi\rangle\|_{SS}^{2}$ for the case of $n=4$ qubits ($d=2$). Besides the
interest in finding such a bound, which, as we have seen in the main text,
gives an entanglement witness for such systems, the treatment we shall present
also gives us an example of using the expression
$\||\psi\rangle\|_{SS}^{2}=\langle\psi|\Pi|\psi\rangle$ with $\Pi$ in (3) to
set up the function to minimize. This requires some knowledge of the structure
of the symmetric group $S_{4}$.

The group $S_{4}$ contains $24$ elements which are usually described as
(products of) elementary cycles: the identity ${\bf 1}$, six simple
transpositions (cycles with length $2$), three double transpositions,
$(1\,2)(3\,4)$, $(1\,3)(2\,4)$, $(1\,4)(2\,3)$, eight cycles of length three,
six cycles of length four. In the following subsections, we consider these
classes of permutations and their contributions to
$\||\psi\rangle\|_{SS}^{2}$. For every permutation $S$ in one class the
inverse is in the same class, and this guarantees that contributions from each
class are real. As done in the main text in section 5, we can apply a common
unitary transformation and collect a common phase factor which leads us to
assume, without loss of generality, that
$|\psi\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes|\psi_{3}\rangle\otimes|\psi_{4}\rangle$,
with

$|\psi_{1}\rangle=|0\rangle,\quad|\psi_{2}\rangle=\cos(\theta_{2})|0\rangle+\sin(\theta_{2})|1\rangle,\quad$
$|\psi_{3}\rangle=\cos(\theta_{3})|0\rangle+e^{i\alpha_{3}}\sin(\theta_{3})|1\rangle,\quad|\psi_{4}\rangle=\cos(\theta_{4})|0\rangle+e^{i\alpha_{4}}\sin(\theta_{4})|1\rangle$

In the following we list the contributions to the sum (11) given by the
various sets of transformations in $S_{4}$.

###  A.1 Identity

The identity ${\bf 1}$ gives a contribution $\langle\psi|{\bf
1}|\psi\rangle=1$.

###  A.2 Six transpositions

Consider the six transpositions $(1\,2)$, $(1\,3)$, $(1\,4)$, $(2\,3)$,
$(2\,4)$, $(3\,4)$. The transposition $(j\,k)$, gives a contribution
$|\langle\psi_{j}|\psi_{k}\rangle|^{2}$. We have therefore the following
transpositions with the associated contributions

$(1\,2)\leftrightarrow\cos^{2}(\theta_{2}),\quad(1\,3)\leftrightarrow\cos^{2}(\theta_{3}),\quad(1\,4)\leftrightarrow\cos^{2}(\theta_{4}),$
$(2\,3)\leftrightarrow\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})+\sin^{2}(\theta_{2})\sin^{2}(\theta_{3})+\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3}),$
$(2\,4)\leftrightarrow\cos^{2}(\theta_{2})\cos^{2}(\theta_{4})+\sin^{2}(\theta_{2})\sin^{2}(\theta_{4})+\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4}),$
$(3\,4)\leftrightarrow\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\sin^{2}(\theta_{3})\sin^{2}(\theta_{4})+\frac{1}{2}\sin(2\theta_{3})\sin(2\theta_{4})\cos(\alpha_{3}-\alpha_{4})$

###  A.3 Three double transpositions

For the three double transpositions the contributions to the sum (11) are

$(1\,2)(3\,4)\leftrightarrow\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\cos^{2}(\theta_{2})\sin^{2}(\theta_{3})\sin^{2}(\theta_{4})+\frac{\cos^{2}(\theta_{2})}{2}\sin(2\theta_{3})\sin(2\theta_{4})\cos(\alpha_{3}-\alpha_{4}),$
$(1\,3)(2\,4)\leftrightarrow\cos^{2}(\theta_{3})\cos^{2}(\theta_{2})\cos^{2}(\theta_{4})+\cos^{2}(\theta_{3})\sin^{2}(\theta_{2})\sin^{2}(\theta_{4})+\frac{\cos^{2}(\theta_{3})}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4})$
$(1\,4)(2\,3)\leftrightarrow\cos^{2}(\theta_{4})\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})+\cos^{2}(\theta_{4})\sin^{2}(\theta_{2})\sin^{2}(\theta_{3})+\frac{\cos^{2}(\theta_{4})}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3}),$

###  A.4 Eight 3-cycles

It is convenient to group the eight three cycles in pairs where one cycle is
the inverse of the other. The corresponding contributions to the sum in (3)
are one the complex conjugate of the other and thus are grouped together to
give a real value. The four pairs are $(1\,2\,3)+(2\,1\,3)$,
$(1\,2\,4)+(2\,1\,4)$, $(1\,3\,4)+(3\,1\,4)$, $(2\,3\,4)+(3\,2\,4)$. Let us
calculate the contribution of the first pair, $(1\,2\,3)+(2\,1\,3)$, which is,

$\langle\psi_{2}|\psi_{1}\rangle\langle\psi_{3}|\psi_{2}\rangle\langle\psi_{1}|\psi_{3}\rangle+\langle\psi_{1}|\psi_{2}\rangle\langle\psi_{3}|\psi_{1}\rangle\langle\psi_{2}|\psi_{3}\rangle=2\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})+\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3}).$

Analogously, we can obtain the contributions of the other pairs which we list
below

$(1\,2\,4)+(2\,1\,4)\leftrightarrow
2\cos^{2}(\theta_{4})\cos^{2}(\theta_{2})+\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{4})cos(\alpha_{4}),$
$(1\,3\,4)+(3\,1\,4)\leftrightarrow
2\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\frac{1}{2}\sin(2\theta_{3})\sin(2\theta_{4})cos(\alpha_{3}-\alpha_{4}),$
$(2\,3\,4)+(3\,2\,4)\leftrightarrow
2\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+2\sin^{2}(\theta_{2})\sin^{2}(\theta_{3})\sin^{2}(\theta_{4})+\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4})+$
$\frac{1}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3})+\frac{1}{2}\sin(2\theta_{3})\sin(2\theta_{4})\cos(\alpha_{3}-\alpha_{4})$

###  A.5 Six 4-cycles

It is convenient to group the three $4$-cycles in pairs where one cycle is the
inverse of the other so that the corresponding contributions are one the
complex conjugate of the other, giving a real value. The three four cycles
pairs are $(1\,2\,3\,4)+(4\,3\,2\,1)$, $(1\,2\,4\,3)+(3\,4\,2\,1)$,
$(1\,3\,2\,4)+(4\,2\,3\,1)$. Let us calculate the contribution of the pair
$(1\,2\,3\,4)+(4\,3\,2\,1)$, which is

$2Re\left(\langle\psi_{1}|\psi_{2}\rangle\langle\psi_{2}|\psi_{3}\rangle\langle\psi_{3}|\psi_{4}\rangle\langle\psi_{4}|\psi_{1}\rangle\right)=$
$2\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\frac{\cos^{2}(\theta_{2})}{2}\sin(2\theta_{4})\sin(2\theta_{3})\cos(\alpha_{4}-\alpha_{3})+$
$\frac{\cos^{2}(\theta_{4})}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3})+\frac{\sin^{2}(\theta_{3})}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4}).$

We also have the contributions

$(1\,2\,4\,3)+(3\,4\,2\,1)\leftrightarrow
2\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\frac{\cos^{2}(\theta_{2})}{2}\sin(2\theta_{4})\sin(2\theta_{3})\cos(\alpha_{3}-\alpha_{4})+$
$\frac{\cos^{2}(\theta_{3})}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4})+\frac{\sin^{2}(\theta_{4})}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3}),$

and

$(1\,3\,2\,4)+(4\,2\,3\,1)\leftrightarrow
2\cos^{2}(\theta_{2})\cos^{2}(\theta_{3})\cos^{2}(\theta_{4})+\frac{\cos^{2}(\theta_{3})}{2}\sin(2\theta_{2})\sin(2\theta_{4})\cos(\alpha_{4})+$
$\frac{\cos^{2}(\theta_{4})}{2}\sin(2\theta_{2})\sin(2\theta_{3})\cos(\alpha_{3})+\frac{\sin^{2}(\theta_{2})}{2}\sin(2\theta_{3})\sin(2\theta_{4})\cos(\alpha_{4}-\alpha_{3})$

###  A.6 Putting all together

Let us sum all the contributions listed in the previous subsections. We shall
write $4!\langle\psi|\Pi|\psi\rangle$ as
$4!\langle\psi|\Pi|\psi\rangle=R+A\left(\alpha_{3},\alpha_{4}\right)$ where
$R$ and $A$ are the terms that do not, respectively, do depend on the
$\alpha$’s, i.e., $\alpha_{3}$ and $\alpha_{4}$. Direct calculations and the
use of standard trigonometric identities (by expressing everything in terms of
the cosines only) lead to (use the shorthand notation
$c_{j}:=\cos(\theta_{j})$)

$R=4-c_{2}^{2}-c_{3}^{2}-c_{4}^{2}+3c_{2}^{2}c_{3}^{2}+3c_{2}^{2}c_{4}^{2}+3c_{3}^{2}c_{4}^{2}+13c_{2}^{2}c_{3}^{2}c_{4}^{2},$
(23)
$A=\cos(\alpha_{3})(2+c_{4}^{2})\sin(2\theta_{2})\sin(2\theta_{3})+\cos(\alpha_{4})(2+c_{3}^{2})\sin(2\theta_{2})\sin(2\theta_{4})+$
(24)
$\cos(\alpha_{3}-\alpha_{4})(2+c_{2}^{2})\sin(2\theta_{3})\sin(2\theta_{4})$

###  A.7 Minimizing with respect to $\alpha_{4}$’s

We first minimize the part $A$ with respect to $\alpha_{4}$. Set

$a:=\sin(2\theta_{2})\sin(2\theta_{3})(2+c_{4}^{2}),\quad
b:=\sin(2\theta_{2})\sin(2\theta_{4})(2+c_{3}^{2}),\quad
c=\sin(2\theta_{3})\sin(2\theta_{4})(2+c_{2}^{2}),$ (25)

so that $A=a\cos(\alpha_{3})+b\cos(\alpha_{4})+c\cos(\alpha_{3}-\alpha_{4})$.
Minimizing with respect to $\alpha_{4}$, using the fact that the minimum over
$\alpha$ of $k_{1}\cos(\alpha)+k_{2}\sin(\alpha)$ is
$-\sqrt{k_{1}^{2}+k_{2}^{2}}$ and standard trigonometric identities, we can
replace $A$ with $a\cos(\alpha_{3})-\sqrt{b^{2}+c^{2}+2bc\cos(\alpha_{3})}$.
In our case, from (25), $a$ and $bc$ have the same sign and since
$\cos(\alpha_{3})$ only multiplies these two terms we can assume without loss
of generality that $a\geq 0$, $bc\geq 0$, and also replace $b$ with $|b|$ and
$c$ with $|c|$. Notice also that the sign of $a$ does not influence the $R$
part of the function to minimize. In the following, it is convenient to write
$a$, $b$ and $c$ in (25) and the function to be minimized in terms of the
variables $x:=c_{2}^{2}$, $y:=c_{3}^{2}$, $z:=c_{4}^{2}$, with
$x,y,z\in[0,1]$, and the cases $\sin(2\theta_{2,3,4})=0$ corresponding to
$x,y,z=0,1$. With this definition, we write, $a=a(x,y,z)$, $b=b(x,y,z)$, and
$c=c(x,y,z)$ as

$a:=4\sqrt{(x-x^{2})(y-y^{2})}(2+z),\quad
b=4\sqrt{(x-x^{2})(z-z^{2})}(2+y),\quad c=4\sqrt{(y-y^{2})(z-z^{2})}(2+x).$
(26)

We also write $\alpha_{3}:=2w\pi$, with $w\in[0,1]$.

###  A.8 Numerical Minimization

With the above notations and using the expression $R$ in (23), we need to
minimize over $x,y,z,w$ in the box $[0,1]^{4}$, the function

$f(x,y,z,w)=4-x-y-z+3xy+3xz+3yz+13xyz+a\cos(2w\pi)-\sqrt{b^{2}+c^{2}+2bc\cos(2w\pi)}.$

We report below a MATLAB code that carries out such a minimization searching
on a grid in $[0,1]^{4}$. Our program is an exhaustive search over a grid with
stepsize $10^{-3}$. More sophisticated optimization methods can definitely be
used although their use is complicated by the fact that the function $f$ is
not convex. We initialize the minimum of the function (the variable ‘mini’ )
at the value achieved for $(x,y,z,w)=(0,0,0,0)$, which is $4$.

    mini=4
    for x=0:0.001:1,
      for y=0:0.001:1,
       for z=0:0.001:1,
        for w=0:0.001:1,
          a=4*(2+z)*sqrt((x-x^2)*(y-y^2));
          b=4*(2+y)*sqrt((x-x^2)*(z-z^2));
          c=4*(2+x)*sqrt((y-y^2)*(z-z^2));
          curr=4-x-y-z+3*x*y+3*x*z+3*y*z+13*x*y*z+a*cos(2*w*pi)-sqrt(b^2+c^2+2*b*c*cos(2*w*pi)
          if curr < mini,
             mini=curr
          end
       end
      end
     end
    end

The approximate minimum calculated with this routine is $\approx 1.3572$
therefore an approximate lower bound for $\||\psi\rangle\|_{SS}$ is
$\approx\frac{1.3572}{24}$.