# A relation among tangle, 3-tangle, and von Neumann entropy of entanglement
for three qubits

Dafa Li1∗, Maggie Cheng2, Xiangrong Li 3, Shuwang Li2 1Department of
Mathematical Sciences, Tsinghua University, Beijing, 100084, China.
Corresponding author<EMAIL_ADDRESS>
2 Department of Applied Mathematics, Illinois Institute of Technology,
Chicago, IL 60616, USA
3 Department of Mathematics, University of California-Irvine, Irvine, CA
92697, USA

###### Abstract

In this paper, we derive a general formula of the tangle for pure states of
three qubits, and present three explicit local unitary (LU) polynomial
invariants. Our result goes beyond the classical work of tangle, 3-tangle and
von Neumann entropy of entanglement for Acín et al.’ Schmidt decomposition
(ASD) of three qubits by connecting the tangle, 3-tangle, and von Neumann
entropy for ASD with Acín et al.’s LU invariants. In particular, our result
reveals a general relation among tangle, 3-tangle, and von Neumann entropy,
together with a relation among their averages. The relations can help us find
the entangled states satisfying distinct requirements for tangle, 3-tangle,
and von Neumann entropy. Moreover, we obtain all the states of three qubits of
which tangles, concurrence, 3-tangle and von Neumann entropy don’t vanish and
these states are endurable when one of three qubits is traced out. We indicate
that for the three-qubit W state, its average von Neumann entropy is maximal
only within the W SLOCC class, and that under ASD the three-qubit GHZ state is
the unique state of which the reduced density operator obtained by tracing any
two qubits has the maximal von Neumann entropy.

##  I Introduction

Quantum entanglement is considered as a unique quantum mechanical resource
Nielsen . Entanglement takes an important role in quantum information and
computation. Examples include quantum teleportation, quantum cryptography,
quantum metrology, and quantum key distribution. Considerable efforts have
been made to explore the entanglement classification via local unitary
operators (LU), local operations and classical communication (LOCC), and
Stochastic LOCC (SLOCC) Bennett -Kraus-PRA . It is known that any two states
of the same LU class have the same amount of entanglement Dur ; Verstraete ;
Kraus-prl . Under SLOCC, pure states of three qubits were partitioned into six
equivalence classes: GHZ, W, A-BC, B-AC, C-AB, and A-B-C Dur . It has been
established that two bipartite states are LU equivalent if and only if their
Schmidt coefficients coincide Nielsen ; Kraus-PRA . Acín et al. proposed the
Schmidt decomposition for three qubits Acin00 ; Acin01 . Kraus introduced a
standard form for multipartite systems and showed that two states are LU
equivalent if and only if their standard forms coincide Kraus-prl ; Kraus-PRA
.

Coffman et al. defined the tangle for the reduced density operator $\rho_{AB}$
(i.e., $tr_{C}\rho_{ABC}$) of a three-qubit state below Coffman . Let

$\overline{\rho_{AB}}=\sigma_{y}\otimes\sigma_{y}\rho_{AB}^{\ast}\sigma_{y}\otimes\sigma_{y},$
(1)

where $\rho_{AB}^{\ast}$ is the complex conjugate of $\rho_{AB}$ and
$\sigma_{y}$ is the Pauli matrix. Note that $\rho_{AB}\overline{\rho_{AB}}$
has only real and non-negative eigenvalues $\eta_{1}^{2}$, $\eta_{2}^{2}$,
$\eta_{3}^{2}$, and $\eta_{4}^{2}$, where
$\eta_{1}\geq\eta_{2}\geq\eta_{3}\geq\eta_{4}$. Then, the entanglement tangle
of $\rho_{AB}$ is defined as

$\tau_{AB}=[\max\\{\eta_{1}-\eta_{2}-\eta_{3}-\eta_{4},0\\}]^{2}.$ (2)

The tangle $\tau_{AC}$ of $\rho_{AC}$ ($=tr_{C}\rho_{ABC}$) and the tangle
$\tau_{BC}$ of $\rho_{BC}$ ($=tr_{A}\rho_{ABC}$) can be similarly defined.

The idea of taking average measure of entanglements comes from an extremely
useful technique in quantum information theory, the “average subsystem
approach” proposed by Page Page . For a pure state $|\psi\rangle$ of three
qubits, Dür et al. defined its average residual entanglement as
$\bar{\epsilon}(\psi)=\frac{1}{3}(\epsilon(\rho_{AB})+\epsilon(\rho_{AC})+\epsilon(\rho_{BC}))$,
where $\epsilon(\rho_{xy})$ is some entanglement measure, where notation $xy$
means $AB,AC,$ or $BC$ Dur . If $\epsilon(\rho_{xy})$ is the tangle (the von
Neumann entropy), then the average residual entanglement $\
\bar{\epsilon}(\psi)$ is the average tangle (von Neumann entropy). A state of
four qubits is defined to be maximally entangled if its average bipartite
entanglement (for example, the average tangle or the average entropy) with
respect to all possible bi-partite cuts is maximal Gour . Significant efforts
have contributed to the average tangle Dur ; Gour ; Chen ; Ana  and von
Neumann entropy, specially, the average entropy in bipartite, tripartite, and
multi-partite scenarios Soo ; Kallin ; Kumar ; You ; Koscik ; Ana ; Gour , and
reference PRL . Dür et al. showed that the W state of three qubits has the
maximal average tangles and indicated that for the GHZ state of three qubits,
the tangle vanishes Dur .

Considerable efforts have also been undertaken on the study of polynomial
invariants for $n$ qubits Linden ; Sudbery ; Leifer ; Fei ; Acin00  Poly .
SLOCC (LU) polynomial invariants can be used for SLOCC (LU) entanglement
classification of pure states of n qubits and can also be used as entanglement
measure.

In this paper, we derive a general formula of the tangle for three qubits, and
obtain three LU polynomial invariants of degree 4. We calculate tangle,
average tangle, 3-tangle, von Neumann entropy , and average von Neumann
entropy in the framework of ASD. We derive an equation which tangle, 3-tangle,
and von Neumann entropy satisfy. Via the equation, we present relations among
tangles and von Neumann entropy, among tangle, 3-tangle, and von Neumann
entropy , among the average tangles, the average von Neumann entropy , and
3-tangle, and among von Neumann entropy and Acín et al.’s LU invariants. The
relations can help us find the entangled states satisfying distinct
requirements for tangle, 3-tangle, and von Neumann entropy.

We obtain all three-qubit states whose tangle, concurrence, 3-tangle and von
Neumann entropy do not vanish. Via von Neumann entropy for ASD, we indicate
that the GHZ state is the unique three-qubit state under ASD that has the
maximal von Neumann entropy of $\ln 2$, and the average von Neumann entropy of
the W state is maximal only within the W SLOCC class.

##  II Tangles for ASD

In this section, we derive the formulas for tangles for pure states of three
qubits and for the ASD, and present a relation between tangles and Acín et
al.’s LU polynomial invariants.

###  II.1 Homogeneous polynomial of degree 4 for tangle

Let $|\psi\rangle=\sum_{i=0}^{7}c_{i}|i\rangle$. By solving Eq. (2), we obtain
$\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$ as follows,

$\displaystyle\tau_{AB}$ $\displaystyle=$
$\displaystyle\Delta-\frac{\tau_{ABC}}{2},$ (3) $\displaystyle\tau_{AC}$
$\displaystyle=$ $\displaystyle\Phi-\frac{\tau_{ABC}}{2},$ (4)
$\displaystyle\tau_{BC}$ $\displaystyle=$
$\displaystyle\Psi-\frac{\tau_{ABC}}{2},$ (5)

where $\Delta$, $\Phi$, $\Psi$ are defined in Eqs. (A3, LABEL:tang-ac, A12) in
Appendix A. Following Dli-12 , the 3-tangle $\tau_{ABC}$ can be written as

$\displaystyle\tau_{ABC}$ $\displaystyle=$ $\displaystyle
4|(c_{0}c_{7}-c_{2}c_{5}-c_{1}c_{6}+c_{3}c_{4})^{2}$ (6)
$\displaystyle-4(c_{0}c_{3}-c_{1}c_{2})(c_{4}c_{7}-c_{5}c_{6})|\text{.}$

Note that $\Delta$, $\Phi$, and $\Psi$ are LU homogeneous polynomial
invariants of degree 4 in the state coefficients and their complex conjugates.
Therefore, one needs only ”$+,-,\times$” operations of the state coefficients
and their complex conjugates to compute the tangles.

Williamson et al. obtained
$I(\tilde{{}_{a}}\tilde{{}_{b}})=tr[\rho_{ab}\tilde{\rho_{ab}}]=\tau_{ab}+\frac{1}{2}\tau_{abc}$,
where
$\tilde{\rho_{ab}}=\sigma_{y}\otimes\sigma_{y}\rho_{ab}^{T}\sigma_{y}\otimes\sigma_{y}$
Williamson . Note that Coffman et al. used $\rho_{ab}^{\ast}$ (which is the
complex conjugate of $\rho_{ab}$) in Eq. (1) rather than $\rho_{ab}^{T}$. It
is known that it is complicated to compute the reduced density operator
$\rho_{ab}$ and $tr[\rho_{ab}\tilde{\rho_{ab}}]$. Comparatively, it is easier
and simpler to compute $\Delta$ than $tr[\rho_{ab}\tilde{\rho_{ab}}]$.

###  II.2 Tangles for ASD

Acín et al. proposed the following Schmidt decomposition for pure states of
three qubits,

$\displaystyle|\psi\rangle$ $\displaystyle=$
$\displaystyle\lambda_{0}|000\rangle+\lambda_{1}e^{i\phi}|100\rangle+\lambda_{2}|101\rangle$
(7) $\displaystyle+\lambda_{3}|110\rangle+\lambda_{4}|111\rangle,$

where $\lambda_{i}\geq 0$, $i=0,1,\cdots,4$, $0\leq\phi<2\pi$, and
$\sum\lambda_{i}^{2}=1$ Acin00 ; Acin01 . Equation (7) is referred to the ASD
of $|\psi\rangle$. It is known that any state of three qubits is LU equivalent
to its ASD. For ASD, 3-tangle $\tau_{ABC}$ in Eq. (6) can be reduced to

$\tau_{ABC}=4\lambda_{0}^{2}\lambda_{4}^{2}.$ (8)

$\tau_{AB}$ in Eq. (3) and $\tau_{AC}$ in Eq. (4) can be reduced similarly,

$\displaystyle\tau_{AB}$ $\displaystyle=4\lambda_{0}^{2}\lambda_{3}^{2},$ (9)
$\displaystyle\tau_{AC}$ $\displaystyle=4\lambda_{0}^{2}\lambda_{2}^{2}.$ (10)

To reduce $\tau_{BC}$ in Eq. (5) for ASD, we write $\Psi$ in Eq. (A12) as
$\Psi=2\Pi$, where
$\Pi=[\lambda_{0}^{2}\lambda_{4}^{2}+2(\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{3})^{2}+8\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}\sin^{2}\frac{\phi}{2}].$
Then, we have

$\displaystyle\tau_{BC}$ $\displaystyle=$ $\displaystyle
2\left(\Pi-\lambda_{0}^{2}\lambda_{4}^{2}\right)$ (11) $\displaystyle=$
$\displaystyle
4[(\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{3})^{2}+4\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}\sin^{2}\frac{\phi}{2}]$
$\displaystyle=$ $\displaystyle
4|\lambda_{1}\lambda_{4}e^{i\phi}-\lambda_{2}\lambda_{3}|^{2}.$

In Table 1, we summarize the tangles $\tau_{AB}$, $\tau_{AC}$, $\tau_{BC}$,
and 3-tangle $\tau_{ABC}$ for ASD.

Table 1: $\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$ for ASD $\tau_{AB}=$ | $4\lambda_{0}^{2}\lambda_{3}^{2}=$ | $4J_{3}$
---|---|---
$\tau_{AC}=$ | $4\lambda_{0}^{2}\lambda_{2}^{2}=$ | $4J_{2}$
$\tau_{BC}=$ | $4|\lambda_{1}\lambda_{4}e^{i\phi}-\lambda_{2}\lambda_{3}|^{2}=$ | $4J_{1}$
$\tau_{ABC}=$ | $4\lambda_{0}^{2}\lambda_{4}^{2}=$ | $4J_{4}$

From Table 1, it is straightforward to calculate the tangles and 3-tangle for
any ASD. Because of the LU equivalence between a state and its ASD, a state
and its ASD have the same tangles, 3-tangle. For example, the ASD of the W
state is $(1/\sqrt{3})(|000\rangle+|101\rangle+|110\rangle)$. From Table 1, we
conclude that the tangles of the W state
$\displaystyle\tau_{AB}=\tau_{AC}=\tau_{BC}=\frac{4}{9}$, and the 3-tangle
$\tau_{ABC}$ vanishes.

For ASD, Acín et al. proposed five LU invariants $J_{i},i=1,2,3,4,5$, where
$J_{1}=|\lambda_{1}\lambda_{4}e^{i\phi}-\lambda_{2}\lambda_{3}|^{2}$, and
$J_{i}=(\lambda_{0}\lambda_{i})^{2}$ for $i=2,3,4$ Acin00 ; Acin01 . Thus, for
ASD, the tangles and 3-tangle in Eqs. (9, 10, 11, 8) can be rewritten in terms
of LU invariants $J_{i}$ (see Table 1),

$\displaystyle\tau_{AB}$ $\displaystyle=$ $\displaystyle 4J_{3},$ (12)
$\displaystyle\tau_{AC}$ $\displaystyle=$ $\displaystyle 4J_{2},$ (13)
$\displaystyle\tau_{BC}$ $\displaystyle=$ $\displaystyle 4J_{1},$ (14)
$\displaystyle\tau_{ABC}$ $\displaystyle=$ $\displaystyle 4J_{4}.$ (15)

##  III von Neumann entropy for ASD and a relation between tangles and von
Neumann entropy

###  III.1 von Neumann entropy for ASD

von Neumann entropy is defined as

$S(\rho)=-\sum\eta_{i}\ln\eta_{i},$ (16)

where $\eta_{i}\geq 0$ are the eigenvalues of $\rho$, and
$\sum_{i}\eta_{i}=1$. Note that $0\ln 0=0$. Via ASD, it is easy to verify that

$\displaystyle S(\rho_{A})$ $\displaystyle=$ $\displaystyle S(\rho_{BC}),$
(17) $\displaystyle S(\rho_{B})$ $\displaystyle=$ $\displaystyle
S(\rho_{AC}),$ (18) $\displaystyle S(\rho_{C})$ $\displaystyle=$
$\displaystyle S(\rho_{AB}).$ (19)

It is well known that the eigenvalues of $\rho$ is just the roots of the
characteristic polynomial of $\rho$ . A tedious calculation derives the
characteristic polynomials of the reduced density operators $\rho_{\mu}$,
$\mu=A$, $B$, $C$, which are $X^{2}-X+\alpha_{\mu}$, where $\alpha_{\mu}$ is
just the sum of tangles and 3-tangle, ref. Table II. Thus, we establish a
relation among the entanglement measures: the von Neumann entropy, tangle and
3-tangle.

From results in Table 1 and via ASD, a calculation yields Table 2, in which
abbreviation CP stands for the characteristic polynomial of the reduced
density operator, $\rho_{\mu}$.

Table 2: $S(\rho_{\mu})$ for ASD $\mu$ | RDO | CP of $\rho_{\mu}$ | $\alpha_{\mu}$ |
---|---|---|---|---
$A$ | $\rho_{A}$ | $X^{2}-X+\alpha_{A}$ | $\alpha_{A}$ | $=J_{2}+J_{3}+J_{4}$
|  |  |  | $=\frac{\tau_{AB}+\tau_{AC}+\tau_{ABC}}{4}$
|  |  |  | $=\frac{\tau_{A(BC)}}{4}$
$B$ | $\rho_{B}$ | $X^{2}-X+\alpha_{B}$ | $\alpha_{B}$ | $=J_{1}+J_{3}+J_{4}$
|  |  |  | $=\frac{\tau_{AB}+\tau_{BC}+\tau_{ABC}}{4}$
|  |  |  | $=\frac{\tau_{B(AC)}}{4}$
$C$ | $\rho_{C}$ | $X^{2}-X+\alpha_{C}$ | $\alpha_{C}$ | $=J_{1}+J_{2}+J_{4}$
|  |  |  | $=\frac{\tau_{AC}+\tau_{BC}+\tau_{ABC}}{4}$
|  |  |  | $=\frac{\tau_{C(AB)}}{4}$

In Table 2, the notation $\tau_{A(BC)}$ stands for the tangle between a qubit
A and a qubit pair BC, where the qubit pair BC is considered a single object
Coffman .

Let $\eta_{\mu}^{(1)}$ and $\eta_{\mu}^{(2)}$ be the two eigenvalues of
$\rho_{\mu}$. We have

$\displaystyle\eta_{\mu}^{(1)}$
$\displaystyle=\frac{1+\sqrt{1-4\alpha_{\mu}}}{2},$ (20)
$\displaystyle\eta_{\mu}^{(2)}$
$\displaystyle=\frac{1-\sqrt{1-4\alpha_{\mu}}}{2}.$ (21)

Then, from Eq. (16), one can see that

$\displaystyle S(\rho_{\mu})$ $\displaystyle=$
$\displaystyle-(\eta_{\mu}^{(1)}\ln\eta_{\mu}^{(1)}+\eta_{\mu}^{(2)}\ln\eta_{\mu}^{(2)}),$
(22)

where $\mu\in\\{A,B,C\\}$ and $0\leq\alpha_{\mu}\leq 1/4$.

Equation (22) and Table 2 reveal a relation among tangles, 3-tangle and von
Neumann entropy . In other words, tangles, 3-tangle, and von Neumann entropy
for any pure three-qubit state must satisfy Eq. (22). See Table 3 for the GHZ
state, the W state, as well as other states. We will explore the relation for
details below.

For $\mu\in\\{A,B,C\\}$, the derivative of the von Neumann entropy
$S(\rho_{\mu})$ with respect to $\alpha_{\mu}$ is given by

$(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}=-\frac{1}{\sqrt{1-4\alpha_{\mu}}}\ln\frac{1-\sqrt{1-4\alpha_{\mu}}}{1+\sqrt{1-4\alpha_{\mu}}}$
(23)

$S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}>0$ when $0<\alpha_{\mu}<1/4$. Hence,
$S(\rho_{\mu})$ increases monotonically as $\alpha_{\mu}$ increases. Recall
that $0\leq\alpha_{\mu}\leq 1/4$. Thus, $0\leq S(\rho_{\mu})\leq\ln 2$, and
$S(\rho_{\mu})=\ln 2$ iff $\alpha_{\mu}=1/4$, and $S(\rho_{\mu})=0\ $iff
$\alpha_{\mu}=0$. This means that the maximal von Neumann entropy for one
particle $\rho_{\mu}$ is $\ln 2$.

For the GHZ state, $\alpha_{\mu}=\frac{1}{4}$ and $S(\rho_{\mu})=\ln 2$; and
for the W state, $\alpha_{\mu}=\frac{2}{9}$ and $S(\rho_{\mu})=\frac{3\ln
3-2\ln 2}{3}$. Thus, for three qubits, the GHZ state has the maximal von
Neumann entropy for any kind of the reduced density operator $\rho_{\mu}$.

###  III.2  Relation between tangles and von Neumann entropy

Proposition 1. For any pure state of three qubits, two different types of
tangles are equal iff the corresponding the von Neumann entropy are equal.
That is,

$\displaystyle(1)\quad S(\rho_{AC})$ $\displaystyle=$ $\displaystyle
S(\rho_{BC})\quad\mathit{iff}\quad\tau_{AC}=\tau_{BC},$ $\displaystyle(2)\quad
S(\rho_{AB})$ $\displaystyle=$ $\displaystyle
S(\rho_{BC})\quad\mathit{iff}\quad\tau_{AB}=\tau_{BC},$ $\displaystyle(3)\quad
S(\rho_{AB})$ $\displaystyle=$ $\displaystyle
S(\rho_{AC})\quad\mathit{iff}\quad\tau_{AB}=\tau_{AC}.$

Clearly, $S(\rho_{AB})=S(\rho_{AC})=S(\rho_{BC})$ iff
$\tau_{AB}=\tau_{AC}=\tau_{BC}$ (see Table III).

Let $uvw$ and $xyz$ be two different three-character strings from the set
$\\{ABC,BAC,CAB\\}$. Then, also $\tau_{u(vw)}=\tau_{x(yz)}$ iff
$S(\rho_{u})=S(\rho_{x})$.

Proof of (1): If $\tau_{AC}=\tau_{BC}$, then from Table 2,
$\alpha_{A}=\alpha_{B}$, and then $S(\rho_{A})=S(\rho_{B})$, which leads to
$S(\rho_{BC})=S(\rho_{AC})$ from Eqs. (17 – 18). Conversely, if
$S(\rho_{AC})=S(\rho_{BC})$, then $S(\rho_{B})=S(\rho_{A})$, then
$\alpha_{B}=\alpha_{A}$ because $S(\rho)$ is strictly increasing, and then
$\tau_{AC}=\tau_{BC}$ from Table 2. Similarly, claims (2) and (3) also hold.

Proposition 2. For any pure state of three qubits, two different types of
tangles satisfy $\tau_{uv}>\tau_{xy}$ iff the corresponding the von Neumann
entropy satisfy $S(\rho_{uv})<S(\rho_{xy})$. That is,

$\displaystyle(1)\quad\tau_{AC}$ $\displaystyle>$
$\displaystyle\tau_{BC}\quad\mathit{iff}\quad S(\rho_{AC})<S(\rho_{BC}),$
$\displaystyle(2)\quad\tau_{AB}$ $\displaystyle>$
$\displaystyle\tau_{BC}\quad\mathit{iff}\quad S(\rho_{AB})<S(\rho_{BC}),$
$\displaystyle(3)\quad\tau_{AB}$ $\displaystyle>$
$\displaystyle\tau_{AC}\quad\mathit{iff}\quad S(\rho_{AB})<S(\rho_{AC}).$

Let $uvw$ and $xyz$ be two different three-character strings from the set
$\\{ABC,BAC,CAB\\}$. Then, also $\tau_{u(vw)}>\tau_{x(yz)}$ iff
$S(\rho_{u})>S(\rho_{x})$.

Proof of (1): If $\tau_{AC}>\tau_{BC}$, then $\alpha_{A}>\alpha_{B}$ from
Table II, and then $S(\rho_{A})>S(\rho_{B})$, i.e $S(\rho_{BC})>S(\rho_{AC})$,
because $S(\rho)$ is strictly increasing. Conversely, if
$S(\rho_{AC})<S(\rho_{BC})$, i.e. $S(\rho_{B})<S(\rho_{A})$, then
$\alpha_{B}<\alpha_{A}$ because $S(\rho)$ is strictly increasing, and then
$\tau_{AC}>\tau_{BC}$. Similarly, claims (2) and (3) also hold.

Next, we study the relation between the difference of two tangles and the
difference of the corresponding von Neumann entropy s. Using the differential
Mean Value theorem, we have

$\displaystyle S(\rho_{A})-S(\rho_{B})$ $\displaystyle=$
$\displaystyle(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{1})(\alpha_{A}-\alpha_{B})$
(24) $\displaystyle S(\rho_{A})-S(\rho_{C})$ $\displaystyle=$
$\displaystyle(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{2})(\alpha_{A}-\alpha_{C})$
(25) $\displaystyle S(\rho_{B})-S(\rho_{C})$ $\displaystyle=$
$\displaystyle(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{3})(\alpha_{B}-\alpha_{C})$
(26)

From Eq. (23), we have $(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{i})>0$,
$i=1,2,3$. Clearly, we can also use Eqs. (24, 25, 26) to prove Propositions 1
and 2. We next calculate $S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}$ using the
second order Taylor expansion of $\ln(1\pm x)$,

$\displaystyle(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1-4\alpha_{\mu}}}[\ln(1-\sqrt{1-4\alpha_{\mu}})-\ln(1+\sqrt{1-4\alpha_{\mu}})]$
$\displaystyle\approx$ $\displaystyle-\frac{1}{\sqrt{1-4\alpha_{\mu}}}\times$
$\displaystyle[(-\sqrt{1-4\alpha_{\mu}}-\frac{1-4\alpha_{\mu}}{2})-(\sqrt{1-4\alpha_{\mu}}-\frac{1-4\alpha_{\mu}}{2})]$
$\displaystyle=$ $\displaystyle 2.$

Therefore, $(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{i})\approx 2$. Via
Eqs. (24, 25, 26) we arrive at the following Proposition.

Proposition 3. For any pure state of three qubits, the difference of two
tangles is approximately twice as large as the difference of the corresponding
von Neumann entropy . We further explain this in details below.

Let different $uvw$ and $xyz$ belong to $\\{ABC,BAC,CAB\\}$. Then,

$\tau_{u(vw)}-\tau_{x(yz)}=\tau_{yz}-\tau_{vw}\approx
2[S(\rho_{vw})-S(\rho_{yz})].$

That is,

$\displaystyle\tau_{A(BC)}-\tau_{B(AC)}$ $\displaystyle=$
$\displaystyle\tau_{AC}-\tau_{BC}\approx 2[S(\rho_{BC})-S(\rho_{AC})],$
$\displaystyle\tau_{A(BC)}-\tau_{C(AB)}$ $\displaystyle=$
$\displaystyle\tau_{AB}-\tau_{BC}\approx 2[S(\rho_{BC})-S(\rho_{AB})],$
$\displaystyle\tau_{B(AC)}-\tau_{C(AB)}$ $\displaystyle=$
$\displaystyle\tau_{AB}-\tau_{AC}\approx 2[S(\rho_{AC})-S(\rho_{AB})].$

###  III.3 Relation between the von Neumann entropy and Acín et al.’s LU
invariants

It is well known that LU invariants are considered as entanglement measure
Sudbery ; Acin00 ; Acin01 . So far, no one discusses the relations between LU
invariants (as measures) and other entanglement measures. Here, we establish
the relation between von Neumann entropy and LU invariants (as entanglement
measure) $J_{i}$, $i=1,2,3,4$. From Eq. (22) and Table 2, we can write the von
Neumann entropy with Acín et al.’s LU invariants $J_{i}$, $i=1,2,3,4$. Thus,
we have the following immediate results.

(1). For any pure state of three qubits, two different types of Acín et al.’s
LU invariants are equal iff the corresponding the von Neumann entropy are
equal. That is, $S(\rho_{A})=S(\rho_{B})$ iff $J_{2}=J_{1}$;
$S(\rho_{A})=S(\rho_{C})$ iff $J_{3}=J_{1}$; $S(\rho_{B})=S(\rho_{C})$ iff
$J_{3}=J_{2}$.

(2). $S(\rho_{A})>S(\rho_{B})$ iff $J_{2}>J_{1}$; $S(\rho_{A})>S(\rho_{C})$
iff $J_{3}>J_{1}$; $S(\rho_{B})>S(\rho_{C})$ iff $J_{3}>J_{2}$.

Via Eqs. (24, 25, 26) and
$(S(\rho_{\mu}))_{\alpha_{\mu}}^{\prime}(\xi_{i})\approx 2$, then we obtain
the following.

(3). For any pure state of three qubits, the difference of two von Neumann
entropy is approximately twice as large as the difference of the corresponding
Acín et al.’s LU invariants.

$\displaystyle S(\rho_{A})-S(\rho_{B})$ $\displaystyle\approx$ $\displaystyle
2(J_{2}-J_{1})$ (27) $\displaystyle S(\rho_{A})-S(\rho_{C})$
$\displaystyle\approx$ $\displaystyle 2(J_{3}-J_{1})$ (28) $\displaystyle
S(\rho_{B})-S(\rho_{C})$ $\displaystyle\approx$ $\displaystyle 2(J_{3}-J_{2})$
(29)

##  IV Relation among tangle, 3-tangle, and von Neumann entropy

Equation (22) can be reduced to

$\displaystyle S(\rho_{\mu})$ $\displaystyle=$
$\displaystyle-[\eta_{\mu}^{(1)}\ln(1+\sqrt{1-4\alpha_{\mu}})$ (30)
$\displaystyle+\eta_{\mu}^{(2)}\ln(1-\sqrt{1-4\alpha_{\mu}})-\ln 2].$

Then by the second order Taylor expansion of $\ln(1\pm x)$, we approximate
$S(\rho_{\mu})$ as follows,

$\displaystyle S(\rho_{\mu})$ $\displaystyle\approx$
$\displaystyle-[\eta_{\mu}^{(1)}(\sqrt{1-4\alpha_{\mu}}-\frac{1-4\alpha_{\mu}}{2})$
(31)
$\displaystyle+\eta_{\mu}^{(2)}(-\sqrt{1-4\alpha_{\mu}}-\frac{1-4\alpha_{\mu}}{2})-\ln
2]$ $\displaystyle=$ $\displaystyle\ln 2-\frac{1}{2}+2\alpha_{\mu},$

where $\mu\in\\{A,B,C\\}$. From Eq. ( 31), we obtain the following relation
among tangle, 3-tangle, and the von Neumann entropy :

$\displaystyle 2S(\rho_{A})$ $\displaystyle\approx$ $\displaystyle 2\ln
2-1+\tau_{A(BC)}$ $\displaystyle=$ $\displaystyle 2\ln
2-1+\tau_{AB}+\tau_{AC}+\tau_{ABC},$ $\displaystyle 2S(\rho_{B})$
$\displaystyle\approx$ $\displaystyle 2\ln 2-1+\tau_{B(AC)}$ $\displaystyle=$
$\displaystyle 2\ln 2-1+\tau_{AB}+\tau_{BC}+\tau_{ABC},$ $\displaystyle
2S(\rho_{C})$ $\displaystyle\approx$ $\displaystyle 2\ln 2-1+\tau_{C(AB)}$
$\displaystyle=$ $\displaystyle 2\ln 2-1+\tau_{AC}+\tau_{BC}+\tau_{ABC}.$

The states in Table III satisfy the above relations. The above relations can
help us find the states which satisfy special requirements for tangle,
3-tangle, and von Neumann entropy, or the states of which tangle, 3-tangle,
and von Neumann entropy are as big as possible. For example, for the state
$|\kappa\rangle$ in Table III, $S(\rho_{A})=0.687$, $\tau_{AB}=\tau_{AC}=4/9$,
and $\tau_{ABC}=8/81$.

Next, we demonstrate how to use the above relation to explore properties of
tangle, 3-tangle, and von Neumann entropy. Let $S(\rho_{A})=\ln 2$ (maximum).
Then, tangles $\tau_{AB}$ and $\tau_{AC}$, and 3-tangle $\tau_{ABC}$ must
satisfy the following.

$\tau_{AB}+\tau_{AC}+\tau_{ABC}\approx 1.$ (32)

For example, if $\tau_{ABC}=1$, then $\tau_{AB}$ and $\tau_{AC}$ vanish; if
$\tau_{ABC}=1/2$, then $\tau_{AB}+\tau_{AC}=1/2$.

##  V Relation among the average tangle, the average von Neumann entropy, and
3-tangle

Lots of efforts have contributed to investigate the the average tangle Dur ;
Gour  and the average von Neumann entropy PRL . So far, no one discusses the
relation among them. In this subsection, we establish the relation.

###  V.1 Definition for the average tangle

Let $A(\psi)$ be the average tangles for the state $|\psi\rangle$. Then,

$A(\psi)=\frac{\tau_{AB}+\tau_{AC}+\tau_{BC}}{3}.$ (33)

We explain how to calculate the average tangle $A$. First, take partial traces
over qubit A (resp. B and C) to get the reduced density operators $\rho_{BC}$
(resp. $\rho_{AC}$, and $\rho_{AB}$), then by the definition of the tangle in
Eq. (2) calculate the tangle $\tau_{AB}$ (resp. $\tau_{AC}$, and $\tau_{BC}$)
for the reduced density operator $\rho_{AB}$ (resp. $\rho_{AC}$, and
$\rho_{BC}$). Finally make the average of the tangles $\tau_{AB}$,
$\tau_{AC}$, and $\tau_{BC}$ to get the average tangle$\ A$.

One can see that $A=0$ for the GHZ state and $A=4/9$ for the W state. It is
known that the W state has the maximal average tangles $A=4/9$ Dur . Thus,
$0\leq A\leq 4/9$.

###  V.2 Definition for the average von Neumann entropy

Let $m(\psi)$ be the average of the von Neumann entropy of all the reduced
density operators for the state $|\psi\rangle$. Then,

$m(\psi)=\frac{S(\rho_{A})+S(\rho_{B})+S(\rho_{C})}{3}.$ (34)

We explain how to calculate the average von Neumann entropy $m$. First, take
partial traces over qubits A and B (resp. A and C, and B and C) to get the
reduced density operators $\rho_{C}$ (resp. $\rho_{B}$, and $\rho_{A}$), then
calculate the von Neumann entropy $S(\rho_{A})$ (resp. $S(\rho_{B})$, and
$S(\rho_{C})$) of $\rho_{A}$(resp. $\rho_{B}$, and $\rho_{C}$) by the
definition in Eq. (16). Finally, make the average of $S(\rho_{A})$,
$S(\rho_{B})$, and $S(\rho_{C}))$ to get the average von Neumann entropy $m$.

It is easy to see that the GHZ state has the maximal average von Neumann
entropy of $\ln 2$, while the W state has the average von Neumann entropy of
$\displaystyle\frac{3\ln 3-2\ln 2}{3}$.Thus, $0\leq m\leq\ln 2$.

###  V.3 Relation among the average tangle, the average von Neumann entropy,
and 3-tangle

So far, no one explains why the GHZ state has the maximal 3-tangle but
vanishing tangle and conversely, and why the W state has the maximal average
tangle but vanishing 3-tangle. We will answer why the states GHZ and W have
the opposite properties.

Using Eq. (31), the average von Neumann entropy and the average tangle satisfy
the following equation for any pure state of three qubits.

$m-A-\frac{\tau_{ABC}}{2}\approx\ln 2-\frac{1}{2},$ (35)

where $0\leq m\leq\ln 2$ and $0\leq A\leq 4/9$.

Eq. (35) reveals a relation among the average von Neumann entropy $m$, the
average tangle $A$, and the 3-tangle $\tau_{ABC}$. Clearly, $m=\ln 2$,
$A=4/9$, $\tau_{ABC}$ $=1$ don’t satisfy Eq. (35). It means that for any
state, the average von Neumann entropy, the average tangle, and 3-tangle
cannot reach the maximum simultaneously. Below, we will investigate when $m$
(resp. $A$ and $\tau_{ABC}$) reaches the maximum, what happen to other two
measures.

Equation (35) implies that the value of $(m-A)$, i.e. the difference between
the average von Neumann entropy and the average tangle, increases linearly
with the 3-tangle $\tau_{ABC}$. For the GHZ SLOCC class, $0<\tau_{ABC}\leq 1$
and almost $\ln 2-1/2<m-A\leq\ln 2$. While for other SLOCC classes,
$\tau_{ABC}=0$ and $(m-A)\approx\ln 2-1/2$. Thus, we obtain almost $\ln
2-1/2\leq(m-A)\leq\ln 2$ for any state of three qubits.

###  V.4 The relation can help find the states which satisfy different
requirements for the average von Neumann entropy and the average tangle and
3-tangle.

Clearly, the requirements for the average tangle, the average von Neumann
entropy, and 3-tangle must satisfy Eq. (35). For example, $|G\rangle$ in Table
III has $A=1/4$, $m=0.56$, and $\tau_{ABC}=1/4$.

It is known that the GHZ state has vanishing tangle. This means that for the
GHZ state, if one of three qubits is traced out, the corresponding reduced
density operator becomes separable. In other words, the entanglement
properties of the GHZ state are fragile under particle losses Dur . One can
use Eq. (35) to find the states of which the average von Neumann entropy, the
average tangle, and 3-tangle are big enough. Clearly, the states are genuine
entangled ones even losing one particle.

###  V.5 The relation helps understand properties of the average tangle,
3-tangle, and the average von Neumann entropy

We next explore when a state has the maximal average von Neumann entropy, then
what Eq. (35) can tell us about the average tangle and the 3-tangle for this
state.

Let $m=\ln 2$ (for example, for the GHZ state). Then, Eq. (35) becomes

$A+\frac{\tau_{ABC}}{2}\approx\frac{1}{2}.$ (36)

Clearly, if $A=0$ then $\tau_{ABC}\approx 1$, and vice versa. Eq. (36) tells
us if a state has the maximal average von Neumann entropy, then the average
tangle and the 3-tangle must satisfy Eq. (36).

For example, Eq. (36) tells us there are the states with $m=\ln 2$,
$\tau_{ABC}\approx 1/2$, and $A\approx 1/4$. One can see that these states
belong to GHZ SLOCC class, but different from the states GHZ state. The three
entanglement measures: tangle, 3-tangle, and von Neumann entropy tell us that
the states are genuine entangled state even though one of three qubits is
traced over and have the maximal average von Neumann entropy.

We next explore when a state has the maximal 3-tangle, then what Eq. (35) can
tell us about the average von Neumann entropy and the average tangle for this
state.

Let $\tau_{ABC}=1$ (for example, for the GHZ state). For the case, Eq. (35)
becomes

$m-\ln 2\approx A.$ (37)

It is easy to see that $m=\ln 2$ and $A=0$ is the unique solution of Eq. (37)
because if $m<\ln 2$ then $A<0$.

This explains when a state has the maximal 3-tangle, then the state must have
the maximal average von Neumann entropy and vanishing the average tangle, i.e.
$\tau_{\mu\nu}=0$, where $\mu\nu=\\{AB,AC,BC\\}$. This is why the GHZ state
has the maximal 3-tangle and the maximal average von Neumann entropy but
vanishing tangle $\tau_{\mu\nu}$, where $\mu\nu=\\{AB,AC,BC\\}$.

We next explore when a state has the maximal average tangles of $4/9$, then
what Eq. (35) can tell us about the average von Neumann entropy and the
3-tangle for this state.

Let $A=4/9$ (for example, for the W state). Via Eq. (35), one can know that
$\max\tau_{ABC}\approx 1/9$ when $m=\ln 2$ and $\min m\approx\ln
2-1/18=\allowbreak 0.637\,59$ when $\tau_{ABC}=0$. It means the average von
Neumann entropy almost is maximal.

This explains when a state has the maximal average tangle of $4/9$, then the
state has almost vanishing 3-tangle and the almost maximal average von Neumann
entropy. This is why the W state has the maximal average tangles but vanishing
3-tangle.

##  VI Tangles and von Neumann entropy for GHZ SLOCC class

It is well known that an ASD state belongs to the GHZ SLOCC class iff
$\lambda_{0}\lambda_{4}\neq 0$ Dli-qip-18 ; Dli-jpa-20 . So, in this section
we assume $\lambda_{0}\lambda_{4}\neq 0$. We first discuss the properties of
tangle. It is known that if $\tau_{\mu\nu}$, vanishes, then $\rho_{\mu\nu}$ is
separable, where $\mu\nu\in\\{AB,AC,BC\\}$. From Table 1, one can see that the
tangles for the GHZ SLOCC class have the following properties.

Property (1). $0\leq\tau_{AB}$, $\tau_{AC}$, $\tau_{BC}<1$.

Property (2).

(2.1). $\tau_{AB}=0$ iff $\lambda_{3}=0$.

(2.2). $\tau_{AC}=0$ iff $\lambda_{2}=0$.

(2.3). $\tau_{BC}=0$ iff $\lambda_{1}\lambda_{4}=\lambda_{2}\lambda_{3}\neq 0$
and $\phi=0$ or $\lambda_{1}=0$ and $\lambda_{2}\lambda_{3}=0$.

Property (3). When only one of $\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$
vanishes,

(3.1). $\tau_{AB}=0$ and $\tau_{AC}\tau_{BC}\neq 0$ iff $\lambda_{3}=0$ and
$\lambda_{1}\lambda_{2}\neq 0$.

(3.2). $\tau_{AC}=0$ and $\tau_{AB}\tau_{BC}\neq 0$ iff $\lambda_{2}=0$ and
$\lambda_{1}\lambda_{3}\neq 0$.

(3.3). $\tau_{BC}=0$ and $\tau_{AB}\tau_{AC}\neq 0$ iff
$\lambda_{1}\lambda_{4}=\lambda_{2}\lambda_{3}\neq 0$ and $\phi=0$.

Proof of (3.3): Clearly, $\tau_{AB}\tau_{AC}\neq 0$ iff
$\lambda_{2}\lambda_{3}\neq 0$. If $\tau_{BC}=0$, then from Table 1,$\
\lambda_{1}\lambda_{4}e^{i\phi}-\lambda_{2}\lambda_{3}=0$, i.e.
$\lambda_{1}\lambda_{4}=\lambda_{2}\lambda_{3}\neq 0$ and $\phi=0$.
Conversely, if $\lambda_{1}\lambda_{4}=\lambda_{2}\lambda_{3}\neq 0$ and
$\phi=0$, then $\tau_{AB}\tau_{AC}\neq 0$ and $\tau_{BC}=0$.

Property (4). When only two of $\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$
vanish,

(4.1). $\tau_{AB}=\tau_{AC}=0$ and $\tau_{BC}\neq 0$ iff
$\lambda_{2}=\lambda_{3}=0$ and $\lambda_{1}\neq 0$.

(4.2). $\tau_{AB}=\tau_{BC}=0$ and $\tau_{AC}\neq 0$ iff
$\lambda_{1}=\lambda_{3}=0$ and $\lambda_{2}\neq 0$.

(4.3). $\tau_{AC}=\tau_{BC}=0$ and $\tau_{AB}\neq 0$ iff
$\lambda_{1}=\lambda_{2}=0$ and $\lambda_{3}\neq 0$.

Property (5). When all $\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$ vanish,

$\tau_{AB}=\tau_{AC}=\tau_{BC}=0$ iff $\lambda_{1}=\lambda_{2}=\lambda_{3}=0$,
i.e. the state is of the form $\lambda_{0}|000\rangle+\lambda_{4}|111\rangle$.
For instance, the GHZ state.

Property (6). When none of the tangles vanishes,
$\tau_{AB}\tau_{AC}\tau_{BC}>0$ iff (i). $\lambda_{2}\lambda_{3}\neq 0$ and
$\lambda_{1}=0$, or (ii).
$\lambda_{1}\lambda_{2}\lambda_{3}(\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{3})\neq
0$ and $\phi=0$, or (iii). $\lambda_{1}\lambda_{2}\lambda_{3}\neq 0$ and
$\phi\neq 0$.

###  VI.1 The GHZ state is the unique one under ASD which has the maximal von
Neumann entropy .

In Appendix B, we show that for any ASD state of three qubits, if
$S(\rho_{\mu})=\ln 2$, then the state must be GHZ. It means that the GHZ state
is the unique state of three qubits under ASD such that $S(\rho_{\mu})$
achieves the maximal value. Thus, for any state of GHZ LU class,
$S(\rho_{\mu})=\ln 2$, where $\mu\in\\{A,B,C\\}$.

##  VII Tangle and von Neumann entropy for W SLOCC class

Each state of the W SLOCC class is of the following form Dli-qip-18 ; Dli-
jpa-20 ,

$|\psi\rangle=\lambda_{0}|000\rangle+\lambda_{1}e^{i\phi}|100\rangle+\lambda_{2}|101\rangle+\lambda_{3}|110\rangle,$
(38)

where $\lambda_{0}\lambda_{2}\lambda_{3}\neq 0$ and $\lambda_{4}=0$.

###  VII.1 Tangle for the W SLOCC class

From Table 1, we obtain tangles for the W SLOCC class,

$\displaystyle\tau_{AB}$ $\displaystyle=$ $\displaystyle
4\lambda_{0}^{2}\lambda_{3}^{2},$ $\displaystyle\tau_{AC}$ $\displaystyle=$
$\displaystyle 4\lambda_{0}^{2}\lambda_{2}^{2},$ $\displaystyle\tau_{BC}$
$\displaystyle=$ $\displaystyle 4\lambda_{2}^{2}\lambda_{3}^{2},$
$\displaystyle\tau_{ABC}$ $\displaystyle=$ $\displaystyle 0.$

Clearly, $0<\tau_{AB},\tau_{AC},\tau_{BC}<1$.

###  VII.2 The W state is the unique state of the W SLOCC class under ASD
whose average von Neumann entropy is the maximal within the W SLOCC class.

Next, we show that the average von Neumann entropy of the W state is maximal
only within the W SLOCC class. From Table 2, for W SLOCC class, one can see
that

$\displaystyle\alpha_{A}$ $\displaystyle=$
$\displaystyle\lambda_{0}^{2}(\lambda_{2}^{2}+\lambda_{3}^{2}),$ (39)
$\displaystyle\alpha_{B}$ $\displaystyle=$
$\displaystyle\lambda_{3}^{2}(\lambda_{0}^{2}+\lambda_{2}^{2}),$ (40)
$\displaystyle\alpha_{C}$ $\displaystyle=$
$\displaystyle\lambda_{2}^{2}(\lambda_{0}^{2}+\lambda_{3}^{2}).$ (41)

Let $\mu\nu\upsilon$ be a string from the set $\\{ABC,BAC,CAB\\}$. When
$\alpha_{\mu}=1/4$, $S(\rho_{\mu})=\ln 2$, $S(\rho_{\nu})<\ln 2$ and
$S(\rho_{\upsilon})<\ln 2$. It means that the von Neumann entropy of the W
state is not maximal even within the W SLOCC class. We next show that the
average von Neumann entropy of the W state is maximal within the W SLOCC
class.

In Appendix C, we derive the extrema of $m$ with the constraint
$\sum_{i=0}^{3}\lambda_{i}^{2}=1$. A straightforward and tedious calculation
yields that $m$ has the extrema$\frac{3\ln 3-2\ln 2}{3}$ at $\lambda_{1}=0$
and $\lambda_{0}=\lambda_{2}=\lambda_{3}=1/\sqrt{3}$, which is just the ASD of
the W state, and the extrema is the maximum. Therefore, the W state is the
unique state under ASD of which the average von Neumann entropy of all kinds
of the reduced density operators is maximal within the W SLOCC class, although
the W state does not have the maximal average of the von Neumann entropy for
all states of three qubits.

##  VIII The states for which the tangle, concurrence, 3-tangle and von
Neumann entropy don’t vanish

It is known that $\tau_{ABC}$ vanishes for the SLOCC classes W, A-BC, B-AC,
C-AB, and A-B-C and the tangles $\tau_{AB}\geq 0,\tau_{AC}\geq 0,$ and
$\tau_{BC}\geq 0$ for the GHZ SLOCC class. Next, we present all the states for
which the tangles, concurrence, 3-tangle and von Neumann entropy do not
vanish. We know these states can only come from the GHZ SLOCC class. For these
states, when one of three qubits is traced out, their reduced density
operators are entangled while the reduced density operator of the GHZ state is
separable. It seems that the states are more entangled than the GHZ state
under tangle and the W state under 3-tangle, respectively. Note that the
tangle is the square of the concurrence. Therefore, if a tangle does not
vanish, the concurrence does not either, and hence, we do not need to discuss
the concurrence separately.

Therefore, if a state satisfies $\tau_{AB}\tau_{AC}\tau_{BC}\neq 0$,
$S(\rho_{A})S(\rho_{B})S(\rho_{C})\neq 0$, and $\tau_{ABC}\neq 0$, then the
state belongs to the GHZ SLOCC class and it falls in one of the three cases:
(i). $\lambda_{2}\lambda_{3}\neq 0$ and $\lambda_{1}=0$, i.e., the state is

$|\varpi_{1}\rangle=\lambda_{0}|000\rangle+\lambda_{2}|101\rangle+\lambda_{3}|110\rangle+\lambda_{4}|111\rangle,$
(42)

or (ii).
$\lambda_{1}\lambda_{2}\lambda_{3}(\lambda_{1}\lambda_{4}-\lambda_{2}\lambda_{3})\neq
0$ and $\phi=0$, i.e., the state is

$|\varpi_{2}\rangle=\lambda_{0}|000\rangle+\lambda_{1}|100\rangle+\lambda_{2}|101\rangle+\lambda_{3}|110\rangle+\lambda_{4}|111\rangle,$
(43)

where $\lambda_{1}\lambda_{4}\neq\lambda_{2}\lambda_{3}$, or (iii).
$\lambda_{1}\lambda_{2}\lambda_{3}\neq 0$ and $\phi\neq 0$, i.e., the state is

$\displaystyle|\varpi_{3}\rangle$ $\displaystyle=$
$\displaystyle\lambda_{0}|000\rangle+\lambda_{1}e^{i\phi}|100\rangle+\lambda_{2}|101\rangle$
(44) $\displaystyle+\lambda_{3}|110\rangle+\lambda_{4}|111\rangle.$

Thus, all states with non-vanishing tangles, 3-tangle, and von Neumann entropy
can be written in the form of $|\varpi_{1}\rangle$, $|\varpi_{2}\rangle$ or
$|\varpi_{3}\rangle$ with suitable $\lambda$’s. For example, the following
states in the form of $|\varpi_{1}\rangle$.

$\displaystyle|\kappa\rangle$ $\displaystyle=$
$\displaystyle\frac{2}{3}|000\rangle+\frac{1}{2}|101\rangle+\frac{1}{2}|110\rangle+\frac{\sqrt{2}}{6}|111\rangle,$
(45) $\displaystyle|G\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}(|000\rangle+|101\rangle+|110\rangle+|111\rangle),$
(46) $\displaystyle|\vartheta\rangle$ $\displaystyle=$
$\displaystyle\frac{\sqrt{5}}{10}(3|000\rangle+|101\rangle+|110\rangle+3|111\rangle).$
(47)

See Table 3 for details. Furthermore, there exist other interesting states
approximating the tangle and 3-tangle of the W state and the GHZ state.

(a). Let $\lambda_{2}=\lambda_{3}=\lambda_{0}$ in $|\varpi_{1}\rangle$, we
obtain a new state

$|\omega\rangle=\lambda_{0}|000\rangle+\lambda_{0}|101\rangle+\lambda_{0}|110\rangle+\lambda_{4}|111\rangle.$

The tangles are

$\tau_{AB}=\tau_{AC}=\tau_{BC}=A=4\lambda_{0}^{4}=\frac{4}{9}(1-\lambda_{4}^{2})^{2}.$
(48)

From Eq. (48), we know that when $\lambda_{4}$ is small enough, then the
tangles, the average tangle, 3-tangle, and von Neumann entropy of
$|\omega\rangle$ are almost equal to those of the W state.

(b). Let $\lambda_{4}=\lambda_{0}$, and let $\lambda_{3}=\lambda_{2}$, we
obtain a new state

$|\varkappa\rangle=\lambda_{0}|000\rangle+\lambda_{2}|101\rangle+\lambda_{2}|110\rangle+\lambda_{0}|111\rangle.$

The tangles and 3-tangle are given by

$\tau_{AB}=\tau_{AC}=4\lambda_{0}^{2}\lambda_{2}^{2},\tau_{BC}=4\lambda_{2}^{4},\tau_{ABC}=4\lambda_{0}^{4}.$
(49)

For $|\varkappa\rangle$, we have $\lim_{\lambda_{2}\rightarrow
0}S(\rho_{\mu})=\ln 2$, where $\mu\in\\{A,B,C\\}$,
$\lim_{\lambda_{2}\rightarrow 0}\tau_{ABC}=1$, and
$\lim_{\lambda_{2}\rightarrow 0}\tau_{\mu\upsilon}=0$, where
$\mu\upsilon\in\\{AB,AC,BC\\}$. Thus, when$\ \lambda_{2}$ is small enough,
then the tangles, the average tangle, 3-tangle, and von Neumann entropy of
$|\varkappa\rangle$ are almost equal to those of the GHZ state. See
$|\vartheta\rangle$ in Table 3.

Next, we demonstrate that tangles $\tau_{AB}$, $\tau_{AC}$, and $\tau_{BC}$
with $\tau_{AB}\tau_{AC}\tau_{BC}\neq 0$ determine a unique state of the form
of $|\varpi_{1}\rangle$. In other words, the LU invariants
$\\{J_{i}|i=1,2,3\\}$ with $J_{1}J_{2}J_{3}\neq 0$ can determine a unique
state of the form of $|\varpi_{1}\rangle$.

Suppose the tangles $\tau_{AB}=4\lambda_{0}^{2}\lambda_{3}^{2}=p^{4}$,
$\tau_{AC}=4\lambda_{0}^{2}\lambda_{2}^{2}=q^{4}$, and
$\tau_{BC}=4\lambda_{2}^{2}\lambda_{3}^{2}=r^{4}$, where $pqr\neq 0$. It is
known that $p^{4}+q^{4}+r^{4}<4/3$ from the CKW inequality
$\tau_{AB}+\tau_{AC}+\tau_{BC}<4/3$ (see Appendix B and also Dur ). The unique
state corresponding to the tangles is

$\frac{1}{\sqrt{2}}\frac{pq}{r}|000\rangle+\frac{1}{\sqrt{2}}\frac{qr}{p}|101\rangle+\frac{1}{\sqrt{2}}\frac{pr}{q}|110\rangle+\lambda_{4}|111\rangle,$

where
$\lambda_{4}^{2}=1-\frac{1}{2}((\frac{pq}{r})^{2}+(\frac{qr}{p})^{2}+(\frac{pr}{q})^{2})$.

On the other hand, the five LU invariants $\\{J_{i}|i=1,\ldots 5\\}$ cannot
uniquely determine a state of three qubits. For instance, for
$|\varpi_{1}\rangle$, from that
$\lambda_{0}^{2}+\sum_{i=2}^{4}\lambda_{i}^{2}=1$ we can derive
$J_{4}=\sqrt{J_{2}J_{3}/J_{1}}-(J_{2}J_{3}/J_{1}+J_{2}+J_{3})$. That is,
$J_{4}$ is a function of $J_{1}$, $J_{2}$, and $J_{3}$, which explains why
$\\{J_{i}|i=1,\cdots,5\\}$ cannot determine a unique state of three qubits
Acin01 .

Table 3: Tangle, 3-tangle and von Neumann entropy for some states ($\delta=\frac{3\ln 3-2\ln 2}{3}=\allowbreak 0.63651$) State | $\tau_{AB}$ | $\tau_{AC}$ | $\tau_{BC}$ | $S(\rho_{A})$ | $S(\rho_{B})$ | $S(\rho_{C})$ | $\tau_{ABC}$
---|---|---|---|---|---|---|---
GHZ | $0$ | $0$ | $0$ | $\ln 2$ | $\ln 2$ | $\ln 2$ | $1$
W | $\frac{4}{9}$ | $\frac{4}{9}$ | $\frac{4}{9}$ | $\delta$ | $\delta$ | $\delta$ | $0$
$|G\rangle$ | $\frac{1}{4}$ | $\frac{1}{4}$ | $\frac{1}{4}$ | $0.56$ | $0.56$ | $0.56$ | $\frac{1}{4}$
$|\kappa\rangle$ | $\frac{4}{9}$ | $\frac{4}{9}$ | $\frac{1}{4}$ | $0.687$ | $0.587$ | $0.587$ | $\frac{8}{81}$
$|\vartheta\rangle$ | $\frac{9}{100}$ | $\frac{9}{100}$ | $\frac{1}{100}$ | $0.688$ | $\allowbreak 0.647$ | $0.647$ | $\frac{81}{100}$

##  IX CKW inequalities for GHZ SLOCC class

For GHZ SLOCC class, $\lambda_{0}\lambda_{4}\neq 0$, we have the following CKW
inequalities. The arguments for CKW inequalities are summarized in Appendix D.

  1. 1.

If $\lambda_{1}=\lambda_{2}=\lambda_{3}=0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}=0$. This is the generated GHZ state
$p|000\rangle+q|111\rangle$.

  2. 2.

If $\lambda_{1}=\lambda_{2}=0$, and $\lambda_{3}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}<1$.

  3. 3.

If $\ \lambda_{1}=\lambda_{3}=0$, and $\lambda_{2}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}<1$.

  4. 4.

If $\lambda_{2}=\lambda_{3}=0$, and $\lambda_{1}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}<1$.

  5. 5.

If $\lambda_{1}=0$, and $\lambda_{2}\lambda_{3}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}<\frac{4}{3}$.

  6. 6.

If $\lambda_{2}=0$, and $\lambda_{1}\lambda_{3}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}\leq\frac{1}{2}$.

  7. 7.

If $\lambda_{3}=0$, and $\lambda_{1}\lambda_{2}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}\leq\frac{1}{2}$.

  8. 8.

If $\lambda_{1}\lambda_{2}\lambda_{3}\neq 0$, then
$\tau_{AB}+\tau_{AC}+\tau_{BC}\leq 1$.

##  X Summary

In this paper, we have given a general formula of the tangles for pure states
of three qubits, which are LU polynomial invariants of degree 4. We derived
tangles and von Neumann entropy for ASD, and presented a relation among
tangles, 3-tangle, and von Neumann entropy , as well as a relation among the
average tangle, the average von Neumann entropy , and 3-tangle.

Via the von Neumann entropy for ASD, we indicated that the GHZ state is the
unique state of three qubits under ASD that has the maximal von Neumann
entropy for all kinds of the reduced density operators, while the average von
Neumann entropy of the W state is maximal only within the W SLOCC class.

It is known that the tangles of the GHZ state and the 3-tangle of the W state
would vanish. We obtained all states of three qubits with non-vanishing
tangle, concurrence, 3-tangle and von Neumann entropy . For example,
$|\vartheta\rangle$ is such a state. It means that when one of the three
qubits is traced out, the remaining state from $|\vartheta\rangle$ is still
entangled, while the remaining state of the GHZ state becomes separable. From
Table 3, it is shown that $S(\rho_{\mu})$ of $|\vartheta\rangle$ is bigger
than that of the W state, where $\mu\in\\{A,B,C\\}$. Therefore, state
$|\vartheta\rangle$ seems to be more entangled than the GHZ state under tangle
and more tangled than the W state under von Neumann entropy and 3-tangle.

##  XI Appendix A. Calculation of tangles.

(A) Calculating $\tau_{AB}$

From the definition of $\tau_{AB}$ in Eq. (2), we obtain the characteristic
polynomial (CP) of $\rho_{AB}\overline{\rho_{AB}}$ as follows,

$(X^{2}-\Delta X+\Theta^{2})X^{2},$ (A1)

where

$\displaystyle\Theta$ $\displaystyle=$
$\displaystyle|(c_{0}c_{7}-c_{2}c_{5})^{2}+(c_{1}c_{6}-c_{3}c_{4})^{2}$ (A2)
$\displaystyle-2(c_{0}c_{7}+c_{2}c_{5})(c_{1}c_{6}+c_{3}c_{4})$
$\displaystyle+4c_{0}c_{3}c_{5}c_{6}+4c_{1}c_{2}c_{4}c_{7}|,$

$\displaystyle\Delta$ $\displaystyle=$ $\displaystyle
2\left(\left|c_{0}\right|^{2}+\left|c_{1}\right|^{2}\right)\left(\left|c_{6}\right|^{2}+\left|c_{7}\right|^{2}\right)$
(A3)
$\displaystyle+2\left(\left|c_{2}\right|^{2}+\left|c_{3}\right|^{2}\right)\left(\left|c_{4}\right|^{2}+\left|c_{5}\right|^{2}\right)$
$\displaystyle+2\left|c_{0}c_{6}^{\ast}+c_{1}c_{7}^{\ast}\right|^{2}+2\left|c_{2}c_{4}^{\ast}+c_{3}c_{5}^{\ast}\right|^{2}$
$\displaystyle-4\ast
re(\left(c_{0}c_{2}^{\ast}+c_{1}c_{3}^{\ast}\right)\left(c_{6}c_{4}^{\ast}+c_{7}c_{5}^{\ast}\right))$
$\displaystyle-4\ast
re(\left(c_{0}c_{4}^{\ast}+c_{1}c_{5}^{\ast}\right)\left(c_{6}c_{2}^{\ast}+c_{7}c_{3}^{\ast}\right)),$

and $re(c)$ indicates the real part of a complex number $c$. Moreover, one can
find

$4\Theta=\tau_{ABC}\text{.}$ (A4)

Hence, the eigenvalues of $\rho_{AB}\overline{\rho_{AB}}$ are $0,0,$ and
$\frac{\Delta\pm\sqrt{\Delta^{2}-4\Theta^{2}}}{2}$. It is known that
$\rho_{AB}\overline{\rho_{AB}}$ has only real and non-negative eigenvalues
Coffman . Then, by the definition of $\tau_{AB}$ in Eq. (2), we obtain

$\displaystyle\tau_{AB}$ $\displaystyle=$
$\displaystyle\left(\sqrt{\frac{\Delta+\sqrt{\Delta^{2}-4\Theta^{2}}}{2}}-\sqrt{\frac{\Delta-\sqrt{\Delta^{2}-4\Theta^{2}}}{2}}\right)^{2}$
(A5) $\displaystyle=$ $\displaystyle\Delta-2\Theta$ $\displaystyle=$
$\displaystyle\Delta-\frac{\tau_{ABC}}{2}.$

(B) Calculating $\tau_{AC}$

Similarly, we can obtain the CP of $\rho_{AC}\overline{\rho_{AC}}$ as follows,

$X^{2}(X^{2}-\Phi X+\Upsilon^{2}),$ (A6)

where

$\Upsilon=\Theta,$ (A7)

$\displaystyle\Phi$ $\displaystyle=$ $\displaystyle
2\left(\left|c_{0}\right|^{2}+\left|c_{2}\right|^{2}\right)\left(\left|c_{5}\right|^{2}+\left|c_{7}\right|^{2}\right)$
$\displaystyle+2\left(\left|c_{1}\right|^{2}+\left|c_{3}\right|^{2}\right)\left(\left|c_{4}\right|^{2}+\left|c_{6}\right|^{2}\right)$
$\displaystyle+2\left|c_{0}c_{5}^{\ast}+c_{2}c_{7}^{\ast}\right|^{2}+2\left|c_{1}c_{4}^{\ast}+c_{3}c_{6}^{\ast}\right|^{2}$
$\displaystyle-4\ast
re(\left(c_{0}c_{1}^{\ast}+c_{2}c_{3}^{\ast}\right)\left(c_{5}c_{4}^{\ast}+c_{7}c_{6}^{\ast}\right))$
$\displaystyle-4\ast
re(\left(c_{0}c_{4}^{\ast}+c_{2}c_{6}^{\ast}\right)\left(c_{5}c_{1}^{\ast}+c_{7}c_{3}^{\ast}\right)).$

Then, we obtain

$\tau_{AC}=\Phi-2\Upsilon=\Phi-\frac{\tau_{ABC}}{2}.$ (A9)

(C) Calculating $\tau_{BC}$

Similarly, the characteristic polynomial of $\rho_{BC}\overline{\rho_{BC}}$ is
given by

$X^{2}(X^{2}-\Psi X+\digamma^{2}),$ (A10)

where

$\digamma=\Theta,$ (A11) $\displaystyle\Psi$ $\displaystyle=$ $\displaystyle
2\left(\left|c_{0}\right|^{2}+\left|c_{4}\right|^{2}\right)\left(\left|c_{3}\right|^{2}+\left|c_{7}\right|^{2}\right)$
(A12)
$\displaystyle+2\left(\left|c_{1}\right|^{2}+\left|c_{5}\right|^{2}\right)\left(\left|c_{2}\right|^{2}+\left|c_{6}\right|^{2}\right)$
$\displaystyle+2\left|c_{0}c_{3}^{\ast}+c_{4}c_{7}^{\ast}\right|^{2}+2\left|c_{1}c_{2}^{\ast}+c_{5}c_{6}^{\ast}\right|^{2}$
$\displaystyle-4\ast
re(\left(c_{0}c_{1}^{\ast}+c_{4}c_{5}^{\ast}\right)\left(c_{3}c_{2}^{\ast}+c_{7}c_{6}^{\ast}\right))$
$\displaystyle-4\ast
re(\left(c_{0}c_{2}^{\ast}+c_{4}c_{6}^{\ast}\right)\left(c_{3}c_{1}^{\ast}+c_{7}c_{5}^{\ast}\right)).$

Then, we obtain

$\tau_{BC}=\Psi-2\digamma=\Psi-\frac{\tau_{ABC}}{2}.$ (A13)

(D) By solving CKW equations Coffman

$\displaystyle\tau_{AB}+\tau_{AC}+\tau_{ABC}$ $\displaystyle=$
$\displaystyle\tau_{A(BC)},$ (A14)
$\displaystyle\tau_{AB}+\tau_{BC}+\tau_{ABC}$ $\displaystyle=$
$\displaystyle\tau_{B(AC)},$ (A15)
$\displaystyle\tau_{AC}+\tau_{BC}+\tau_{ABC}$ $\displaystyle=$
$\displaystyle\tau_{C(AB)},$ (A16)

where $\tau_{A(BC)}=4\det\rho_{A}$, $\tau_{B(AC)}=4\det\rho_{B}$, and
$\tau_{C(AB)}=4\det\rho_{C}$, we obtain

$\displaystyle\tau_{AB}=$
$\displaystyle\frac{\tau_{A(BC)}+\tau_{B(AC)}-\tau_{C(AB)}-\tau_{ABC}}{2},$
(A17) $\displaystyle\tau_{AC}=$
$\displaystyle\frac{\tau_{A(BC)}-\tau_{B(AC)}+\tau_{C(AB)}-\tau_{ABC}}{2},$
(A18) $\displaystyle\tau_{BC}=$
$\displaystyle\frac{-\tau_{A(BC)}+\tau_{B(AC)}+\tau_{C(AB)}-\tau_{ABC}}{2}.$
(A19)

When the 3-tangle $\tau_{ABC}$ is zero, we obtain the following,

$\displaystyle\Delta$ $\displaystyle=$
$\displaystyle\frac{\tau_{A(BC)}+\tau_{B(AC)}-\tau_{C(AB)}}{2},$ (A20)
$\displaystyle\Phi$ $\displaystyle=$
$\displaystyle\frac{\tau_{A(BC)}-\tau_{B(AC)}+\tau_{C(AB)}}{2},$ (A21)
$\displaystyle\Psi$ $\displaystyle=$
$\displaystyle\frac{-\tau_{A(BC)}+\tau_{B(AC)}+\tau_{C(AB)}}{2}.$ (A22)

Obviously, $\Delta$, $\Phi$, $\Psi$ are simple polynomial of degree 4 although
it is hard to compute $\det\rho_{A}$, $\det\rho_{B}$, and $\det\rho_{C}$.

## Appendix B The GHZ state is unique state of three qubits which has
maximally von Neumann entropy

Claim: If a state of three qubits possesses the maximal von Neumann entropy ,
$S(\rho_{\mu})=\ln 2$, where $\mu\in\\{A,B,C\\}$, then the state must be GHZ.

Proof: Clearly, $S(\rho_{\mu})$ increases strictly monotonically as
$\alpha_{\mu}$ increases. Therefore, $S(\rho_{\mu})=\ln 2$ iff
$\alpha_{\mu}=1/4$. Thus, we have the following equations

$\displaystyle\alpha_{A}$ $\displaystyle=$ $\displaystyle
J_{2}+J_{3}+J_{4}=1/4,$ (B1) $\displaystyle\alpha_{B}$ $\displaystyle=$
$\displaystyle J_{1}+J_{3}+J_{4}=1/4,$ (B2) $\displaystyle\alpha_{C}$
$\displaystyle=$ $\displaystyle J_{1}+J_{2}+J_{4}=1/4,$ (B3)

and we obtain

$J_{1}=J_{2}=J_{3}.$ (B4)

Using Tables 1 and 2, equation (B1) leads to

$\lambda_{0}^{4}-\lambda_{0}^{2}(1-\lambda_{1}^{2})+1/4=0.$ (B5)

From (B5), we have a solution

$\displaystyle\lambda_{1}$ $\displaystyle=$ $\displaystyle 0,$ (B6)
$\displaystyle\lambda_{0}$ $\displaystyle=$ $\displaystyle 1/\sqrt{2},$ (B7)
$\displaystyle\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}$
$\displaystyle=$ $\displaystyle 1/2.$ (B8)

Using $J_{2}=J_{3}$ in Eq. (B4), we have

$\lambda_{2}=\lambda_{3}.$ (B9)

From that $J_{1}=J_{2}$ in Eq. (B4), we obtain

$\lambda_{2}\lambda_{3}=\lambda_{0}\lambda_{2}.$ (B10)

There are two scenarios for $\lambda_{2}$: $\lambda_{2}\neq 0$ and
$\lambda_{2}=0$. The first scenario is impossible since if $\lambda_{2}\neq
0$, then from Eqs. (B7, B9, B10), we obtain

$\lambda_{0}=\lambda_{2}=\lambda_{3}=1/\sqrt{2}.$ (B11)

From Eq. (B8), we know clearly that Eq. (B11) cannot hold. Therefore,
$\lambda_{2}$ must be zero.

With $\lambda_{2}=0$, from Eqs. (B8, B9) we obtain

$\displaystyle\lambda_{2}$ $\displaystyle=$ $\displaystyle\lambda_{3}=0,$
(B12) $\displaystyle\lambda_{4}$ $\displaystyle=$ $\displaystyle 1/\sqrt{2}.$
(B13)

The state satisfying Eqs. (B6, B7, B12, B13) is just GHZ.

## Appendix C The extrema for W SLOCC class

We next find an extrema of $m$ with the constraint of
$\sum_{i=0}^{3}\lambda_{i}^{2}=1$ for the W SLOCC class. For states of W SLOCC
class, $\alpha_{A}=\lambda_{0}^{2}(\lambda_{2}^{2}+\lambda_{3}^{2})$,
$\alpha_{B}=\lambda_{3}^{2}(\lambda_{0}^{2}+\lambda_{2}^{2})$, and
$\alpha_{C}=\lambda_{2}^{2}(\lambda_{0}^{2}+\lambda_{3}^{2})$

We define

$F=\frac{1}{3}(S(\rho_{A})+S(\rho_{B})+S(\rho_{C}))+\ell(\sum_{i=0}^{3}\lambda_{i}^{2}-1),$
(C1)

where $\ell$ is the Lagrange multiplier. In light of the constrained extreme
theorem, we need to solve the equations $\frac{\partial
F}{\partial\lambda_{i}}=0$, for $i=0,1,2,3$, and $\frac{\partial
F}{\partial\ell}=0$ to find the extrema. From $\frac{\partial
F}{\partial\lambda_{1}}=2\ell\lambda_{1}=0$, we obtain $\lambda_{1}=0$. Then,
$F$ is reduced to

$F=\frac{1}{3}(S(\rho_{A})+S(\rho_{B})+S(\rho_{C}))+\ell(\lambda_{0}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}-1).$
(C2)

From $\frac{\partial F}{\partial\ell}=0$ we obtain

$\lambda_{0}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}=1.$ (C3)

From Eq. (C2), we obtain

$\frac{\partial F}{\partial\lambda_{0}}=\frac{1}{3}\left[\frac{\partial
S(\rho_{A})}{\partial\lambda_{0}}+\frac{\partial
S(\rho_{B})}{\partial\lambda_{0}}+\frac{\partial
S(\rho_{C})}{\partial\lambda_{0}}\right]+2\ell\lambda_{0},$ (C4)

where

$\frac{\partial
S(\rho_{\mu})}{\partial\lambda_{0}}=\frac{dS(\rho_{\mu})}{d\alpha_{\mu}}\frac{\partial\alpha_{\mu}}{\partial\lambda_{0}},\mu\in\\{A,B,C\\}.$
(C5)

The derivative of $S(\rho_{A})$ is

$\frac{dS(\rho_{A})}{d\alpha_{A}}=-\left[\frac{d\eta_{A}^{(1)}}{d\alpha_{A}}(1+\ln\eta_{A}^{(1)})+\frac{d\eta_{A}^{(2)}}{d\alpha_{A}}(1+\ln\eta_{A}^{(2)})\right],$
(C6)

where

$\frac{d\eta_{A}^{(1)}}{d\alpha_{A}}=-\frac{1}{\sqrt{1-4\alpha_{A}}},\frac{d\eta_{A}^{(2)}}{d\alpha_{A}}=\frac{1}{\sqrt{1-4\alpha_{A}}}.$
(C7)

Thus,

$\displaystyle\frac{dS(\rho_{A})}{d\alpha_{A}}$ $\displaystyle=$
$\displaystyle-\left[-\frac{1+\ln\eta_{A}^{(1)}}{\sqrt{1-4\alpha_{A}}}+\frac{1+\ln\eta_{A}^{(2)}}{\sqrt{1-4\alpha_{A}}}\right]$
(C8) $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}.$

Similarly, we obtain

$\displaystyle\frac{dS(\rho_{B})}{d\alpha_{B}}$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}},$
(C9) $\displaystyle\frac{dS(\rho_{C})}{d\alpha_{C}}$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}.$
(C10)

From Eqs. (C5, C8, C9, C10), we obtain

$\displaystyle\frac{\partial S(\rho_{A})}{\partial\lambda_{0}}$
$\displaystyle=$
$\displaystyle-\frac{2\lambda_{0}(\lambda_{2}^{2}+\lambda_{3}^{2})}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}},$
(C11) $\displaystyle\frac{\partial S(\rho_{B})}{\partial\lambda_{0}}$
$\displaystyle=$
$\displaystyle-\frac{2\lambda_{0}\lambda_{3}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}},$
(C12) $\displaystyle\frac{\partial S(\rho_{C})}{\partial\lambda_{0}}$
$\displaystyle=$
$\displaystyle-\frac{2\lambda_{0}\lambda_{2}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}.$
(C13)

From Eqs. (C4, C11, C12, C13) and $\frac{\partial F}{\partial\lambda_{0}}=0$,
we obtain

$\displaystyle\ell$ $\displaystyle=$
$\displaystyle\frac{1}{3}\left(\frac{\lambda_{2}^{2}+\lambda_{3}^{2}}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}\right.$
(C14)
$\displaystyle+\frac{\lambda_{3}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}}$
$\displaystyle\left.+\frac{\lambda_{2}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}\right).$

Similarly, we consider

$\frac{\partial F}{\partial\lambda_{2}}=\frac{1}{3}\left(\frac{\partial
S(\rho_{A})}{\partial\lambda_{2}}+\frac{\partial
S(\rho_{B})}{\partial\lambda_{2}}+\frac{\partial
S(\rho_{C})}{\partial\lambda_{2}}\right)+2\ell\lambda_{2}.$ (C15)

Clearly,

$\frac{\partial
S(\rho_{\mu})}{\partial\lambda_{2}}=\frac{dS(\rho_{\mu})}{d\alpha_{\mu}}\frac{\partial\alpha_{\mu}}{\partial\lambda_{2}},\mu\in\\{A,B,C\\}.$
(C16)

From Eqs. (C8, C9, C10, C15, C16) and $\frac{\partial
F}{\partial\lambda_{2}}=0$, we obtain

$\displaystyle\ell$ $\displaystyle=$
$\displaystyle\frac{1}{3}\left(\frac{\lambda_{0}^{2}}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}\right.$
(C17)
$\displaystyle+\frac{\lambda_{3}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}}$
$\displaystyle\left.+\frac{\lambda_{0}^{2}+\lambda_{3}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}\right).$

Similarly, from $\frac{\partial F}{\partial\lambda_{3}}=0$ we obtain

$\displaystyle\ell$ $\displaystyle=$
$\displaystyle\frac{1}{3}\left(\frac{\lambda_{0}^{2}}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}\right.$
(C18)
$\displaystyle+\frac{\lambda_{0}^{2}+\lambda_{2}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}}$
$\displaystyle\left.+\frac{\lambda_{2}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}\right).$

From Eqs. (C3, C14, C17), we obtain

$\frac{1-2\lambda_{0}^{2}}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}=\frac{1-2\lambda_{2}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}.$
(C19)

From Eqs. (C3, C14, C18), we obtain

$\frac{1-2\lambda_{0}^{2}}{\sqrt{1-4\alpha_{A}}}\ln\frac{\eta_{A}^{(2)}}{\eta_{A}^{(1)}}=\frac{1-2\lambda_{3}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}}.$
(C20)

From Eqs. (C3, C17, C18), we obtain

$\frac{1-2\lambda_{3}^{2}}{\sqrt{1-4\alpha_{B}}}\ln\frac{\eta_{B}^{(2)}}{\eta_{B}^{(1)}}=\frac{1-2\lambda_{2}^{2}}{\sqrt{1-4\alpha_{C}}}\ln\frac{\eta_{C}^{(2)}}{\eta_{C}^{(1)}}.$
(C21)

When $\lambda_{0}=\lambda_{2}=\lambda_{3}$, Eqs. (C19, C20, C21) hold. Via Eq.
(C3), one can see that $\lambda_{0}=\lambda_{2}=\lambda_{3}=1/\sqrt{3}$ is an
extrema of $m$ with the constraint $\sum_{i=0}^{3}\lambda_{i}^{2}=1$.

##  XII Appendix D. CKW inequalities for GHZ SLOCC class.

Note that for the GHZ SLOCC class, $\lambda_{0}\lambda_{4}\neq 0$. There is
also an additional constraint $\sum_{i=0}^{4}\lambda_{i}^{2}=1$.

(A) When $\lambda_{1}=\lambda_{3}=0$ and $\lambda_{2}\neq 0$, we have

$A=\frac{4}{3}(\lambda_{0}^{2}\lambda_{2}^{2}).$

Clearly,

$A=\frac{4}{3}(\lambda_{0}^{2}\lambda_{2}^{2})\leq\frac{4}{3}\left(\frac{\lambda_{0}^{2}+\lambda_{2}^{2}}{2}\right)^{2}=\frac{1}{3}\left(1-\lambda_{4}^{2}\right)^{2}<\frac{1}{3}$

Similarly, we can obtain $A<\frac{1}{3}$ for the case with
$\lambda_{1}=\lambda_{2}=0,\lambda_{3}\neq 0$ and the case with
$\lambda_{2}=\lambda_{3}=0,\lambda_{1}\neq 0$.

(B) When $\lambda_{1}\lambda_{3}\neq 0$ and $\lambda_{2}=0$, $A$ is reduced to

$A=\frac{4}{3}(\lambda_{0}^{2}\lambda_{3}^{2}+\lambda_{1}^{2}\lambda_{4}^{2}).$

In light of constrained extreme theorem, we consider the following function

$U=\frac{4}{3}(\lambda_{0}^{2}\lambda_{3}^{2}+\lambda_{1}^{2}\lambda_{4}^{2})+q(\lambda_{0}^{2}+\lambda_{1}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}-1).$

From $\frac{\partial U}{\partial\lambda_{i}}=0$, obtain only one extreme
$\lambda_{0}=\lambda_{1}=\lambda_{3}=\lambda_{4}=\frac{1}{2}$, i.e.
$\frac{1}{2}(|000\rangle+|100\rangle+|110\rangle+|111\rangle)$, at which $\max
A=\frac{1}{6}$.

(C) When $\lambda_{1}\lambda_{2}\neq 0$ and $\lambda_{3}=0$, $A$ is reduced to

$A=\frac{4}{3}(\lambda_{0}^{2}\lambda_{2}^{2}+\lambda_{1}^{2}\lambda_{4}^{2}).$

The discussion is similar to (B), there is only one extreme
$\lambda_{0}=\lambda_{1}=\lambda_{2}=\lambda_{4}=\frac{1}{2}$, i.e.
$\frac{1}{2}(|000\rangle+|100\rangle+|101\rangle+|111\rangle)$, $\max
A=\frac{1}{6}$.

(D) When $\lambda_{1}=0$ and $\lambda_{2}\lambda_{3}\neq 0$, then

$A=\frac{4}{3}(\lambda_{0}^{2}\lambda_{3}^{2}+\lambda_{0}^{2}\lambda_{2}^{2}+\lambda_{2}^{2}\lambda_{3}^{2}).$

The constraint is

$\lambda_{0}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}=1,$

where $\lambda_{4}$ is considered as a parameter. In light of constrained
extreme theorem, for a fixed $\lambda_{4}$, when
$\lambda_{0}=\lambda_{2}=\lambda_{3}$, i.e.,
$\lambda_{0}|000\rangle+\lambda_{0}|101\rangle+\lambda_{0}|110\rangle+\lambda_{4}|111\rangle$,
then $\ A$ has the maximum
$A=4\lambda_{0}^{4}=\frac{4}{9}(1-\lambda_{4}^{2})^{2}<\frac{4}{9}$.

(E) When $\lambda_{1}\lambda_{2}\lambda_{3}\neq 0$, let

$f=\frac{4}{3}(\lambda_{0}^{2}\lambda_{3}^{2}+\lambda_{0}^{2}\lambda_{2}^{2}+(\lambda_{1}\lambda_{4}+\lambda_{2}\lambda_{3})^{2}).$

Clearly, $A\leq f$. We next calculate the maximum value of $f$. The constraint
reads
$\lambda_{0}^{2}+\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}=1.$
In light of the constrained extreme theorem, we consider the function

$F=f+g(\lambda_{0}^{2}+\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}-1).$

From $\frac{\partial F}{\partial\lambda_{i}}=0$, $i=0,1,2,3,4,$ we obtain
$\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}$ and $\lambda_{0}=0$. Thus,
$f$ has the maximum value $\frac{1}{3}$. However, we require that
$\lambda_{0}$ do not vanish. From $\sum_{i=0}^{4}\lambda_{i}^{2}=1$, we get
$\lambda_{0}^{2}+4\lambda_{4}^{2}=1$. Thus, when
$\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}$,
$A=\frac{4}{3}((2\lambda_{4}^{2}))(1-2\lambda_{4}^{2})<1/3.$ Clearly,
$\lim_{\lambda_{0}\rightarrow 0}A=\frac{1}{3}$. For example, when
$\lambda_{0}^{2}=\frac{1}{100}$, then
$A=\allowbreak\frac{3333}{10\,000}\approx\frac{1}{3}$.

Data Availability Statement: The datasets generated during and/or analyzed
during the current study are available from the corresponding author on
reasonable request.

Conflicts of interest/Competing interests: The authors have no conflicts of
interest to declare that are relevant to the content of this article.

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# Lower Bound on the Size-Ramsey Number of Tight Paths

Christian Winter University of Hamburg, Hamburg, Germany, E-mail:
<EMAIL_ADDRESS>
###### Abstract

The size-Ramsey number $\hat{R}^{(k)}(\mathcal{H})$ of a $k$-uniform
hypergraph $\mathcal{H}$ is the minimum number of edges in a $k$-uniform
hypergraph $\mathcal{G}$ with the property that each $2$-edge coloring of
$\mathcal{G}$ contains a monochromatic copy of $\mathcal{H}$. For $k\geq 2$
and $n\in\mathbb{N}$, a $k$-uniform tight path on $n$ vertices
$\mathcal{P}^{(k)}_{n,k-1}$ is defined as a $k$-uniform hypergraph on $n$
vertices for which there is an ordering of its vertices such that the edge set
consists of all $k$-element intervals of consecutive vertices in this
ordering.

We show a lower bound on the size-Ramsey number of $k$-uniform tight paths,
which is, considered assymptotically in both the uniformity $k$ and the number
of vertices $n$,
$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,k-1})=\Omega_{n,k}\big{(}\log(k)n\big{)}$.

##  1 Introduction

For a $k$-graph $\mathcal{G}=(V,E)$, i.e. a $k$-uniform hypergraph on a vertex
set $V$ and an edge set $E\subseteq\binom{V}{k}$, a $2$-edge coloring of
$\mathcal{G}$ is a function $c\colon
E(\mathcal{G})\to\\{\text{red},\text{blue}\\}$ that maps each edge to one of
the given colors red or blue. In the following refer to such a function simply
as a coloring of $\mathcal{G}$. We say that a $k$-graph $\mathcal{G}$ has the
Ramsey property $\mathcal{G}\rightarrow\mathcal{H}$ for some $k$-graph
$\mathcal{H}$ if each coloring of $\mathcal{G}$ contains a monochromatic copy
of $\mathcal{H}$. The size-Ramsey number of a $k$-graph $\mathcal{H}$ is
defined as

$\hat{R}^{(k)}(\mathcal{H})=\min\big{\\{}m\in\mathbb{N}\colon\exists\
k\text{-graph }\mathcal{G}\text{ with }|E(\mathcal{G})|=m\text{ and
}\mathcal{G}\rightarrow\mathcal{H}\big{\\}}.$

Size-Ramsey problems have been firstly introduced by Erdős, Faudree, Rousseau
and Schelp [7] for graphs. One of the focus points of studies on the graph
case is estimating the size-Ramsey number of paths. Beck [2] disproved a
conjecture of Erdős [6] by showing that $\hat{R}^{(2)}(P_{n})=O(n)$. Since
then, estimates on this number have been gradually improved, with the current
best known bounds being $\big{(}3.75-o(1)\big{)}n\leq\hat{R}^{(2)}(P_{n})\leq
74n$ given by Bal, DeBiasio [1] and Dudek, Pralat [5], respectively.

Let $n,k,\ell\in\mathbb{N}$ with $k\geq 2$, $1\leq\ell\leq k-1$ and
$\frac{n-\ell}{k-\ell}\in\mathbb{N}$. A $k$-uniform $\ell$-path on $n$
vertices $\mathcal{P}^{(k)}_{n,\ell}$ is a $k$-graph on $n$ vertices for which
there exists an ordering of its vertices such that every edge is a $k$-element
set of consecutive vertices (i.e. an interval of length $k$), two consecutive
edges have precisely $\ell$ vertices in common, and there are no isolated
vertices. A $k$-uniform $\ell$-path with $\ell=k-1$ is referred to as a
$k$-uniform tight path. If the uniformity is given from the context we omit
the term ‘$k$-uniform’.

Research on the size-Ramsey number of hypergraphs has been substantially
driven forward by Dudek, La Fleur, Mubayi and Rödl [4]. Among other results,
they conjectured that the size-Ramsey number of tight paths is linear in terms
of $n$. This conjecture was lately verified by Letzter, Pokrovskiy and
Yepremyan [8].

######  Theorem 1 ([8]).

Let $k\geq 2$ be fixed. Then

$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,k-1})=O(n).$

Regarding a lower bound on this number, it is a simple observation that the
following trivial bound holds.

######  Observation.

Let $n,k,\ell\in\mathbb{N}$, $k\geq 2$, $1\leq\ell\leq k-1$ with
$\tfrac{n-\ell}{k-\ell}\in\mathbb{N}$. Then

$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,\ell})\geq\frac{2(n-\ell)}{k-\ell}-1.$

In particular, $\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,k-1})\geq 2n-2k+1$ for any
$n,k\in\mathbb{N}$, $k\geq 2$.

In this paper we show an improved lower bound on the size-Ramsey number of
tight paths.

######  Theorem 2.

Let $n\geq 7$. Then

$\hat{R}^{(3)}(\mathcal{P}^{(3)}_{n,2})\geq\tfrac{8}{3}n-\tfrac{28}{3}.$

######  Theorem 3.

Let $k\geq 4$ and $n>\frac{k^{2}+k-2}{2}$. Then

$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,k-1})\geq\big{\lceil}\log_{2}(k+1)\big{\rceil}\cdot
n-2k^{2}.$

Section 2 discusses useful properties for the main proofs. In Section 3 the
proofs of Theorem 3 and Theorem 2 are presented. Lastly, Section 4 of this
paper briefly describes how the mentioned proofs can be applied to obtain a
lower bound for general $k$-uniform $\ell$-paths.

##  2 Preliminaries

Let $\mathcal{G}$ be a $k$-graph and $Z\subseteq E(\mathcal{G})$ be an edge
set. Let $\mathbin{\scalebox{1.2}{$\cup$}}Z=\\{v\in e\colon e\in Z\\}$ be the
set of vertices that are covered by $Z$. We say that the $k$-graph
$(\mathbin{\scalebox{1.2}{$\cup$}}Z,Z)$ is formed by $Z$. Given a vertex set
$W\subseteq V(\mathcal{G})$ the subhypergraph induced by $W$ is
$\mathcal{G}[W]=(W,\\{e\in E(\mathcal{G})\colon e\subseteq W\\})$. For
$q\in\mathbb{R}$, $0\leq q<k$, the $q$-neighborhood of $Z$ is the edge set

$N_{>q}(Z)=\big{\\{}e\in E(\mathcal{G})\colon\exists e^{\prime}\in Z\text{
with }|e\cap e^{\prime}|>q\big{\\}}.$

Note that we allow $e=e^{\prime}$, thus $Z\subseteq N_{>q}(Z)$ for all $0\leq
q<k$.

For each $k$-uniform $\ell$-path $\mathcal{P}$ on $n$ vertices we fix an
ordering of the vertices such that each edge is a set of consecutive vertices.
We say that such an enumeration $V(\mathcal{P})=\\{v_{1},\dots,v_{n}\\}$ is
according to $\mathcal{P}$. For a $k$-graph $\mathcal{G}$, we define
$e(\mathcal{G})=|E(\mathcal{G})|$. Observe that
$e(\mathcal{P}^{(k)}_{n,\ell})=\frac{n-\ell}{k-\ell}$ and in the tight case
$e(\mathcal{P}^{(k)}_{n,k-1})=n-k+1$. Furthermore, let $[n]=\\{1,\dots,n\\}$
for $n\in\mathbb{N}$. For any other notation, see Diestel [3].

######  Proposition 4.

Let $n,k\in\mathbb{N}$, $n\geq k\geq 2$ and let $\mathcal{P}$ be a $k$-uniform
tight path on $n$ vertices. Furthermore, let $\alpha\in\mathbb{R}$,
$1\leq\alpha\leq k$ be a constant and $W\subseteq V(\mathcal{P})$ be a vertex
set such that for each edge $e\in E(\mathcal{P})$ we have $|e\cap
W|\geq\alpha$. Then

$|W|\geq\frac{\alpha(n-k+1)}{k}.$

In particular, if for each $e\in E(\mathcal{P})$, $|e\cap W|>\frac{k+1}{2}$,
then for $n>\frac{k^{2}+k-2}{2}$,

$|W|>\frac{n}{2}.$

###### Proof.

We estimate the size of $W$ by double-counting ordered pairs $(v,e)$
consisting of a vertex $v\in W$ and an edge $e\in E(\mathcal{P})$ with $v\in
e$. Let $\rho_{(v,e)}$ be the number of such pairs.

Considering the edges of $\mathcal{P}$ it is immediate that

$\rho_{(v,e)}\geq\alpha\cdot e(\mathcal{P})=\alpha(n-k+1).$

Now consider the vertices in $W\subseteq V(\mathcal{P})$. The maximum degree
of the tight path $\mathcal{P}$ is at most $k$, so

$\rho_{(v,e)}\leq k\cdot|W|.$

Combining both inequations, we obtain

$|W|\geq\frac{\alpha(n-k+1)}{k}.$

Now consider the case that for each edge $e\in E(\mathcal{P})$ we have $|e\cap
W|>\frac{k+1}{2}$, then also $|e\cap W|\geq\frac{k+2}{2}$. Therefore we obtain
for sufficiently large $n$,

$|W|\geq\frac{k+2}{2}\cdot\frac{n-k+1}{k}>\frac{n}{2}.$

∎

##  3 Proofs of the main results

###### Proof of Theorem 3.

Let $\mathcal{G}$ be a $k$-uniform hypergraph with
$\mathcal{G}\rightarrow\mathcal{P}^{(k)}_{n,k-1}$, i.e. such that each
coloring contains a monochromatic $k$-uniform tight path on $n$ vertices. We
show that there are at least $\left\lceil\log_{2}(k+1)\right\rceil\cdot
n-2k^{2}$ many edges in $\mathcal{G}$ by iteratively constructing many edge-
disjoint tight paths. Let $\lambda=\left\lceil\log_{2}(k+1)\right\rceil-1$,
this number indicates how many iteration steps are executed. Additionally, we
define the function $q\colon\\{0,\dots,\lambda\\}\rightarrow\mathbb{R}$,

$q(i)=\left(1-\frac{1}{2^{i}}\right)(k+1),$

which will be the parameter of the $q$-neighborhoods considered in each
iteration step. Clearly, $q$ is an increasing function and $q(i)\geq 0$ for
$i\in\\{0,\dots,\lambda\\}$. For $i\leq\lambda$ (or equivalently
$i<\log_{2}(k+1)$) it can be seen that $q(i)<k$, which implies that the
$q(i)$-neighborhood is well-defined for all $i\in\\{0,\dots,\lambda\\}$.

As an initial step of the iteration, the Ramsey property
$\mathcal{G}\rightarrow\mathcal{P}^{(k)}_{n,k-1}$ provides that there is some
tight path on $n$ vertices in $\mathcal{G}$, which we denote by
$\mathcal{P}_{0}$.

From now on we proceed iteratively, so let $i=1,\dots,\lambda$ and suppose
that the iteration has been performed for all smaller values of $i$. In each
step of the iteration we construct the following:

  * •

Edge sets $Z^{1}_{i},Z^{2}_{i}\subseteq E(\mathcal{P}_{i-1})$ such that
$\mathbin{\scalebox{1.2}{$\cup$}}Z^{1}_{i}\cap\mathbin{\scalebox{1.2}{$\cup$}}Z^{2}_{i}=\varnothing$
and each of the sets forms a tight path in $\mathcal{G}$ on precisely
$\left\lfloor\frac{n}{2}\right\rfloor$ vertices.

  * •

A tight path $\mathcal{P}_{i}$ on $n$ vertices with $E(\mathcal{P}_{i})\cap
N_{>q(i)}(Z^{a}_{b})=\varnothing$ for all $a\in[2],b\in[i]$.

As a first step, we construct $Z^{1}_{i}$ and $Z^{2}_{i}$ by dividing the
tight path $\mathcal{P}_{i-1}$ into two parts of equal length and considering
the edge sets of the two created shorter tight paths. For this purpose,
consider an ordering of the vertices
$V(\mathcal{P}_{i-1})=\\{v_{1},\dots,v_{n}\\}$ according to
$\mathcal{P}_{i-1}$. Let

$V^{1}_{i}=\big{\\{}v_{1},\dots,v_{\left\lfloor\frac{n}{2}\right\rfloor}\big{\\}}\quad\text{
and }\quad
V^{2}_{i}=\big{\\{}v_{\left\lceil\frac{n}{2}\right\rceil+1},\dots,v_{n}\big{\\}}.$

Then $|V^{1}_{i}|=\left\lfloor\frac{n}{2}\right\rfloor=|V^{2}_{i}|$. Now let
$Z^{1}_{i}=E(\mathcal{P}_{i-1}[V^{1}_{i}])$ and
$Z^{2}_{i}=E(\mathcal{P}_{i-1}[V^{2}_{i}])$. Clearly, these two sets form
vertex-disjoint tight paths on $\left\lfloor\frac{n}{2}\right\rfloor$ vertices
in $\mathcal{G}$. The size of $Z^{1}_{i}$ and $Z^{2}_{i}$ is

$|Z^{1}_{i}|=|Z^{2}_{i}|=e(\mathcal{P}^{(k)}_{\left\lfloor\tfrac{n}{2}\right\rfloor,k-1})=\left\lfloor\frac{n}{2}\right\rfloor-k+1\geq\frac{n-2k+1}{2}.$

In a next step we show a key property of the edge sets $Z^{a}_{b}$ for
$a\in[2]$, $b\in[i]$.

Claim. Let $a_{1},a_{2}\in[2]$, $b_{1},b_{2}\in[i]$ such that
$(a_{1},b_{1})\neq(a_{2},b_{2})$. Then for any two edges $e_{1}\in
N_{>q(i)}(Z^{a_{1}}_{b_{1}})$ and $e_{2}\in N_{>q(i)}(Z^{a_{2}}_{b_{2}})$ we
have

$|e_{1}\cap e_{2}|<k-1.$

Proof of the claim. Assume that there are edges $e_{1}\in
N_{>q(i)}(Z^{a_{1}}_{b_{1}})$, $e_{2}\in N_{>q(i)}(Z^{a_{2}}_{b_{2}})$ with
$|e_{1}\cap e_{2}|\geq k-1$. By definition, there is an edge $z_{1}\in
Z^{a_{1}}_{b_{1}}$ such that $|e_{1}\cap z_{1}|>q(i)$ and an edge $z_{2}\in
Z^{a_{2}}_{b_{2}}$ with $|e_{2}\cap z_{2}|>q(i)$.

Figure 1: Possible constellation of the edges in iteration step $i=1$ where
$k=6$

We estimate the size of $z_{1}\cap z_{2}$ in order to find a contradiction to
our assumption. Since $|e_{1}\cap e_{2}|\geq k-1$, we have $|e_{1}\backslash
e_{2}|\leq 1$ and so $|e_{2}\cap z_{1}|>q(i)-1$. Applying this, we obtain:

$|z_{1}\cap z_{2}|\geq|e_{2}\cap z_{1}\cap z_{2}|\geq|e_{2}|-|e_{2}\backslash
z_{1}|-|e_{2}\backslash z_{2}|=-|e_{2}|+|e_{2}\cap z_{1}|+|e_{2}\cap z_{2}|\\\
>-k+q(i)-1+q(i)=\left(1-\frac{1}{2^{i-1}}\right)(k+1)=q(i-1).$

If $b_{1}=b_{2}$, we have
$\mathbin{\scalebox{1.2}{$\cup$}}Z^{a_{1}}_{b_{1}}\cap\mathbin{\scalebox{1.2}{$\cup$}}Z^{a_{2}}_{b_{2}}=\varnothing$
by construction. Then $|z_{1}\cap z_{2}|=0\leq q(i-1)$, which is a
contradiction.

We suppose that $b_{1}\neq b_{2}$, then without loss of generality
$b_{1}>b_{2}$ (and by this $b_{1}-1\geq 1$). By construction we know $z_{1}\in
Z^{a_{1}}_{b_{1}}\subseteq E(\mathcal{P}_{b_{1}-1})$. In the iteration step
$b_{1}-1$ the tight path $\mathcal{P}_{b_{1}-1}$ was chosen to be edge-
disjoint from $\bigcup_{a\in[2],b<{b_{1}}}N_{>q(b_{1}-1)}(Z^{a}_{b})$. This
yields that $z_{1}\notin N_{>q(b_{1}-1)}(Z^{a_{2}}_{b_{2}})$ and so

$|z_{1}\cap z_{2}|\leq q(b_{1}-1)\leq q(i-1),$

since $q$ is an increasing function, and we again reach a contradiction. This
concludes the proof of the claim.

Now we find the next tight path $\mathcal{P}_{i}$ in $\mathcal{G}$ by
considering the following coloring of $\mathcal{G}$. For all $a\in[2]$ and
$b\in[i]$, assign the color red to each edge in $N_{>q(i)}(Z^{a}_{b})$. The
remaining edges are colored blue. We will prove that there is a monochromatic
blue $\mathcal{P}^{(k)}_{n,k-1}$ in this coloring. We shall let
$\mathcal{P}_{i}$ be that path. With this in mind, assume for a contradiction
that there is a monochromatic red tight path $\mathcal{R}$ on $n$ vertices in
$\mathcal{G}$.

Clearly, each edge in $E(\mathcal{R})$ is in some neighborhood
$N_{>q(i)}(Z^{a}_{b})$, $a\in[2],b\in[i]$. Now the above claim provides that
any two edges which are consecutive in $\mathcal{R}$, so intersect in
precisely $k-1$ vertices, belong to the same neighborhood
$N_{>q(i)}(Z^{a}_{b})$ for some $a\in[2],b\in[i]$. By repeating this argument,
we obtain that $E(\mathcal{R})\subseteq N_{>q(i)}(Z^{a}_{b})$. This implies
that for all $e\in E(\mathcal{R})$,

$|e\cap\mathbin{\scalebox{1.2}{$\cup$}}Z^{a}_{b}|>q(i)\geq
q(1)=\tfrac{k+1}{2}.$

Then applying Proposition 4 for the tight path $\mathcal{R}$ and the vertex
set $\mathbin{\scalebox{1.2}{$\cup$}}Z^{a}_{b}$ yields
$|\mathbin{\scalebox{1.2}{$\cup$}}Z^{a}_{b}|>\frac{n}{2}$. But by construction
$Z^{a}_{b}$ forms a $k$-graph on precisely
$\left\lfloor\frac{n}{2}\right\rfloor$ vertices, a contradiction.

Consequently, there is no red tight path on $n$ vertices in the coloring, so
the Ramsey property $\mathcal{G}\rightarrow\mathcal{P}^{(k)}_{n,k-1}$ implies
the existence of a monochromatic blue $\mathcal{P}^{(k)}_{n,k-1}$, which we
denote $\mathcal{P}_{i}$. Observe that for each $e\in E(\mathcal{P}_{i})$,
$e\notin N_{>q(i)}(Z^{a}_{b})$ for all $a\in[2],b\in[i]$, since all edges in
these neighborhoods are colored in red.

By iterating the described procedure for $i=1,\dots,\lambda$, we obtain edge
sets $Z^{a}_{b}$ for $a\in[2]$, $b\in[\lambda]$ which are pairwise disjoint
and additionally a tight path $\mathcal{P}_{\lambda}$ on $n$ vertices such
that each edge in $E(\mathcal{P}_{\lambda})$ is not contained in any set
$Z^{a}_{b}$. This allows for the following estimate on the number of edges in
$\mathcal{G}$

$e(\mathcal{G})\geq\sum_{b\in[\lambda]}\big{(}|Z^{1}_{b}|+|Z^{2}_{b}|\big{)}+e(\mathcal{P}_{\lambda})\geq\lambda(n-2k-1)+(n-k+1)\geq\left\lceil\log_{2}(k+1)\right\rceil\cdot
n-2k^{2}.$

∎

We point out that the above proof also applies to $3$-uniform tight paths, but
does not yield an improvement of the trivial bound. In order to obtain a
refined bound in this case, we instead use a non-iterative adaption of the
above proof.

###### Proof of Theorem 2.

Let $\mathcal{G}$ be an arbitrary $3$-uniform hypergraph which has the Ramsey
property $\mathcal{G}\rightarrow\mathcal{P}^{(3)}_{n,2}$. As before, we show
that $\mathcal{G}$ is a $3$-graph on at least $\tfrac{8}{3}n-\tfrac{28}{3}$
many edges. Using the Ramsey property
$\mathcal{G}\rightarrow\mathcal{P}^{(3)}_{n,2}$, there exists some tight path
on $n$ vertices in $\mathcal{G}$. In particular, we find a shorter tight path
$\mathcal{P}_{0}$ on only
$n_{0}\coloneqq\left\lceil\tfrac{2}{3}n-\tfrac{4}{3}\right\rceil-1$ many
vertices. Observe that
$e(\mathcal{P}_{0})=n_{0}-2\geq\tfrac{2}{3}n-\tfrac{13}{3}$.

In order to find a tight path $\mathcal{P}_{1}$ which is edge-disjoint from
$\mathcal{P}_{0}$, we consider the following coloring. Color all edges in the
$1$-neighborhood $N_{>1}\big{(}E(\mathcal{P}_{0})\big{)}$ in red and the
remaining edges in blue. Assume for a contradiction that in this coloring
there is a monochromatic red tight path on $n$ vertices, say $\mathcal{R}$.
Then Proposition 4 applied to the tight path $\mathcal{R}$ and the vertex set
$V(\mathcal{P}_{0})$ provides a contradiction. Since
$\mathcal{G}\rightarrow\mathcal{P}^{(3)}_{n,2}$, there is a monochromatic blue
tight path on $n$ vertices in $\mathcal{G}$. This implies that there is also a
blue tight path on $n_{1}\coloneqq n-1$ vertices and by this $n-3$ edges. We
fix such a tight path $\mathcal{P}_{1}$ with $e(\mathcal{P}_{1})=n-3$. Note
that $N_{>1}\big{(}E(\mathcal{P}_{0})\big{)}$ and $E(\mathcal{P}_{1})$ are
disjoint edge sets.

In the following, in order to find a third edge-disjoint tight path, we
consider another coloring of $\mathcal{G}$. From now on, let each edge in
$E(\mathcal{P}_{0})\cup E(\mathcal{P}_{1})$ be colored red and all other edges
blue. Assume for a contradiction that there is a red tight path $\mathcal{R}$
on $n$ vertices in this coloring. Then neither $E(\mathcal{R})\subseteq
E(\mathcal{P}_{0})$ nor $E(\mathcal{R})\subseteq E(\mathcal{P}_{1})$ since the
two edge sets have a size strictly less than $e(\mathcal{P}^{(3)}_{n,2})$.
Therefore, $\mathcal{R}$ consists of edges of both $E(\mathcal{P}_{0})$ and
$E(\mathcal{P}_{1})$. Both of these edge sets are disjoint, so there exist two
edges $e_{1}\in E(\mathcal{P}_{0})\cap E(\mathcal{R}),e_{2}\in
E(\mathcal{P}_{1})\cap E(\mathcal{R})$ which are consecutive in $\mathcal{R}$,
i.e. $|e_{1}\cap e_{2}|=2$. But that is a contradiction to the fact that
$N_{>1}\big{(}E(\mathcal{P}_{0})\big{)}$ and $E(\mathcal{P}_{1})$ are
disjoint. Consequently, there is no red $\mathcal{P}^{(3)}_{n,2}$ in this
coloring. By the same argument as before, there is a blue tight path
$\mathcal{P}_{2}$ on $n_{2}\coloneqq n$ vertices in $\mathcal{G}$.

Then the three edge sets
$E(\mathcal{P}_{0}),E(\mathcal{P}_{1}),E(\mathcal{P}_{2})$ are pairwise
disjoint. Thus,

$e(\mathcal{G})\geq
e(\mathcal{P}_{0})+e(\mathcal{P}_{1})+e(\mathcal{P}_{2})\geq\tfrac{8}{3}n-\tfrac{28}{3}.$

∎

##  4 Generalization to $k$-uniform $\ell$-paths

In order to obtain a lower bound for all $k$-uniform $\ell$-paths, both proofs
presented in the previous section can be generalized.

######  Theorem 5.

Let $k\geq 2$ and $1\leq\ell\leq\frac{2}{3}k$. Let $n\in\mathbb{N}$ be
sufficiently large such that $\tfrac{n-\ell}{k-\ell}\in\mathbb{N}$. Then

$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,\ell})\geq\left(2+\frac{\ell}{k+c}\right)\cdot\frac{n-\ell}{k-\ell}-4,$

where $c$ is the smallest non-negative integer with
$\frac{k+c}{k-\ell}\in\mathbb{N}$.

######  Theorem 6.

Let $k\geq 4$ and $\frac{2}{3}k<\ell\leq k-1$. Let $n\in\mathbb{N}$ be
sufficiently large with $\tfrac{n-\ell}{k-\ell}\in\mathbb{N}$. Then

$\hat{R}^{(k)}(\mathcal{P}^{(k)}_{n,\ell})\geq\left\lceil\log_{2}\left(\frac{2k-\ell}{k-\ell}\right)\right\rceil\cdot\frac{n-\ell}{k-\ell}-4k^{2}.$

Detailed proofs of these results are given in [9], in the following we outline
the proof ideas of both theorems. Similarly to Proposition 4, the following
generalization holds.

######  Proposition 7.

Let $k\geq 2$, $1\leq\ell\leq k-1$ and $n\in\mathbb{N}$ with
$\tfrac{n-\ell}{k-\ell}\in\mathbb{N}$. Let $\mathcal{P}$ be a $k$-uniform
$\ell$-path on $n$ vertices and let $c\geq 0$ be the smallest non-negative
integer with $\frac{k+c}{k-\ell}\in\mathbb{N}$. Furthermore, let
$\alpha\in\mathbb{R}$, $1\leq\alpha\leq k$ be a constant and $W\subseteq
V(\mathcal{P})$ be a vertex set such that for each edge $e\in E(\mathcal{P})$
we have $|e\cap W|\geq\alpha$. Then

$|W|\geq\frac{\alpha(n-\ell)}{k+c}.$

∎

A lower bound for $\ell$-paths with $\ell>\frac{2}{3}k$, Theorem 6, can be
proven using an iteration similarly to Theorem 3 with some minor
modifications. Since the above proposition is weaker than the bound obtained
in Proposition 4, it is required in each iterative step to divide the previous
chosen $\ell$-path $\mathcal{P}_{i-1}$ not into two parts
$Z^{1}_{i},Z^{2}_{i}$ on roughly $\frac{n}{2}$ vertices, but into three parts
$Z^{1}_{i},Z^{2}_{i},Z^{3}_{i}$ each of which has a size of about
$\frac{n}{3}$. The remaining arguments are executed analogously to the proof
stated above.

For Theorem 5, when $\ell\leq\frac{2}{3}k$, the proof of Theorem 2 can be
applied. Like in the mentioned proof we construct three edge-disjoint
$\ell$-paths of different lengths. As sizes for the $\ell$-paths we choose
$n_{0}=\left\lceil\frac{\ell(n-\ell)}{k+c}\right\rceil-1$, $n_{1}=n-1$,
$n_{2}=n$. Additionally, as a coloring for finding $\mathcal{P}_{1}$ we assign
the color red to the neighborhood $N_{>\ell-1}(E(\mathcal{P}_{0}))$ (instead
of $N_{>1}(E(\mathcal{P}_{0}))$ in the above proof). By the same arguments as
before, we obtain the desired bound.

## Acknowledgments

The author would like to thank Mathias Schacht for his helpful comments and
support with the thesis on which this paper is based as well as Maria
Axenovich for comments on the manuscript.

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