#  Reducing the detection of genuine entanglement of n qubits to two qubits

Dafa Li1,2 1Department of Mathematical Sciences, Tsinghua University, Beijing,
100084, China
2Center for Quantum Information Science and Technology, Tsinghua National
Laboratory for Information Science and Technology (TNList), Beijing, 100084,
China

###### Abstract

We propose a criterion for the detection of genuine entanglement of pure
multiqubit states. To this aim, we define an operator called the losing one
qubit operator, which is different from the reduced density operator. The
states obtained from a multiqubit state by applying the losing one qubit
operator are referred to as its projected states. We show that all of the
projected states of a pure product n-qubit state are pure product states
provided that it cannot be written as a product of a single qubit state and a
genuinely entangled (n-1)-qubit state. We also show that a pure n-qubit state
is genuinely entangled provided that the state has at least two genuinely
entangled (n-1)-qubit projected states. By repeating the losing process, we
reduce the detection of entanglement of pure n-qubit states to the one of pure
two-qubit states. Also we write a LISP program for the reduction process.

##  I Introduction

Quantum entanglement is considered as a unique quantum mechanical resource
Nielsen . It is well known that entanglement takes a key rule in quantum
information processing tasks, for e.g., quantum teleportation, quantum
cryptography and quantum key distribution.

It is known that for GHZ of three qubits, tracing out of qubit i, the reduced
density operator becomes completely unentangled, while for W of three qubits,
tracing out of qubit i the reduced density operator remains entangled Dur . It
is indicated that many physical implementations of qubits, for example ion
traps, optical lattices and linear optics, suffer from loss of qubits Wallman
. The entanglement resistant to particles loss via tracing out the particles
was investigated Quinta .

Many efforts have been devoted to exploring criteria for detection of quantum
entanglement Peres -Osterloh .A necessary condition for separability of a
quantum system consisting of two subsystems is that a matrix, obtained by
partial transposition of $\rho$, has only non-negative eigenvalues Peres . The
criterion is referred to as PPT (Positive partial transpose). For $2\times\ 2$
and $2\times\ 3$ systems, the positivity of the partial transposition of a
state is necessary and sufficient for its separability Horodecki .

It was shown that for any separable state of a bipartite system, the sum of
the singular values of the realigned matrix constructed from the density
matrix is necessarily not greater than 1 Chen . The generalized reduction
criterion for the separability of bipartite system in arbitrary dimensions was
proposed Albeverio . The cross norm necessary criterion for the separability
of density matrices for bipartite systems was studied Rudolph .

Recently, Zwerger et al. showed that genuine entanglement of all multipartite
pure states can be detected in a device-independent way via bipartite Bell
inequalities Zwerger . The genuine $n$-qubit entanglement can be detected via
the proportionality of two coefficient vectors DLI-QIP-19 .

In this paper, we propose the losing one qubit operator. Applying the losing
one qubit operator to a state of $n$ qubits, we can obtain $n$ states of
$(n-1)$ qubits which are referred to as the projected states. We demonstrate
that the losing one qubit operator can reduce the detection of genuine
entanglement of n qubits to two qubits. In section 2, we give necessary and
sufficient conditions for two, three, and four qubits to be product states. In
section 3, we show that all of the projected states of an$\ n$-qubit product
state are product states provided that the $n$-qubit product state cannot be
written as a product of a single qubit state and a genuinely entangled
$(n-1)$-qubit state. Thus, the losing one qubit operator can reduce the
detection of the entanglement of n-qubit states to $(n-1)$-qubit ones and
finally to two-qubit ones.

##  II The necessary and sufficient conditions for product states of two,
three, and four qubits

For given two vectors $v_{1}$, and $v_{2}$, if $v_{1}=kv_{2}$, where $k$ is a
complex number, then we say that $v_{1}$ is proportional to $v_{2}$.
Specially, when $v_{1}=0$, then $k=0$. For a set of vectors $v_{1}$,
$v_{2}$,$\cdots$, and $v_{m}$, if there is a vector $v_{i}\neq 0$ of the set
of vectors such that any vector $v_{\ell}$ of the set of vectors is
proportional to $v_{i}$, then we say that the set of vectors are proportional.
Clearly, if at least two non-zero vectors of the set of the vectors are not
proportional, then clearly, the set of the vectors are not proportional.

###  II.1 Two qubits,

For two qubits, we can write any pure state of two qubits as
$|\psi\rangle_{12}=\sum_{i=0}^{3}c_{i}|i\rangle$.

Result 1. It is well known that $|\psi\rangle_{12}$ is a product state iff the
following two vectors are proportional DLI-QIP-19

$\left(\begin{array}[]{cc}c_{0}&c_{1}\end{array}\right)^{T},\left(\begin{array}[]{cc}c_{2}&c_{3}\end{array}\right)^{T}.$
(1)

Note that the above two vectors are proportional iff the following equality
holds.

$c_{0}c_{3}=c_{1}c_{2}.$ (2)

Note also that $|c_{0}c_{3}-c_{1}c_{2}|^{2}$ is just the concurrence of
$|\psi\rangle_{12}$.

###  II.2 Three qubits

Result 2. Let $|\psi\rangle_{123}=\sum_{i=0}^{7}c_{i}|i\rangle$ be any pure
state of three qubits. Then, $|\psi\rangle_{123}$ is a product state iff at
least one of the following three pairs of vectors is proportional. Otherwise,
it is genuinely entangled DLI-QIP-19 .

$\left(\begin{array}[]{cccc}c_{0}&c_{1}&c_{2}&c_{3}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{4}&c_{5}&c_{6}&c_{7}\end{array}\right)^{T}$
(3)

$\left(\begin{array}[]{cccc}c_{0}&c_{1}&c_{4}&c_{5}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{2}&c_{3}&c_{6}&c_{7}\end{array}\right)^{T}$
(4)

$\left(\begin{array}[]{cccc}c_{0}&c_{2}&c_{4}&c_{6}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{1}&c_{3}&c_{5}&c_{7}\end{array}\right)^{T}$
(5)

###  II.3 Four qubits

Result 3. Let $|\psi\rangle_{1234}=\sum_{i=0}^{15}c_{i}|i\rangle$. A pure
product state $|\psi\rangle$ of four qubits can be written as (1).
$|\varphi\rangle_{1}|\phi\rangle_{234}$, (2).
$|\varphi\rangle_{2}|\phi\rangle_{134}$, (3).
$|\varphi\rangle_{3}|\phi\rangle_{124}$, (4).
$|\varphi\rangle_{4}|\phi\rangle_{123}$, (5).
$|\varphi\rangle_{12}|\phi\rangle_{34}$, (6).
$|\varphi\rangle_{13}|\phi\rangle_{24}$, or (7).
$|\varphi\rangle_{14}|\phi\rangle_{23}$ iff the following corresponding set of
vectors are proportional.

$\displaystyle\text{(1).}\left(\begin{array}[]{cccccccc}c_{0}&c_{1}&c_{2}&c_{3}&c_{4}&c_{5}&c_{6}&c_{7}\end{array}\right)^{T},$
(7)
$\displaystyle\left(\begin{array}[]{cccccccc}c_{8}&c_{9}&c_{10}&c_{11}&c_{12}&c_{13}&c_{14}&c_{15}\end{array}\right)^{T};$
(9)

$\displaystyle\text{(2).
}\left(\begin{array}[]{cccccccc}c_{0}&c_{1}&c_{2}&c_{3}&c_{8}&c_{9}&c_{10}&c_{11}\end{array}\right)^{T},$
(11)
$\displaystyle\left(\begin{array}[]{cccccccc}c_{4}&c_{5}&c_{6}&c_{7}&c_{12}&c_{13}&c_{14}&c_{15}\end{array}\right)^{T};$
(13)

$\displaystyle\text{(3).
}\left(\begin{array}[]{cccccccc}c_{0}&c_{1}&c_{4}&c_{5}&c_{8}&c_{9}&c_{12}&c_{13}\end{array}\right)^{T},$
(15)
$\displaystyle\left(\begin{array}[]{cccccccc}c_{2}&c_{3}&c_{6}&c_{7}&c_{10}&c_{11}&c_{14}&c_{15}\end{array}\right)^{T};$
(17)

$\displaystyle\text{(4).
}\left(\begin{array}[]{cccccccc}c_{0}&c_{2}&c_{4}&c_{6}&c_{8}&c_{10}&c_{12}&c_{14}\end{array}\right)^{T},$
(19)
$\displaystyle\left(\begin{array}[]{cccccccc}c_{1}&c_{3}&c_{5}&c_{7}&c_{9}&c_{11}&c_{13}&c_{15}\end{array}\right)^{T};$
(21) $\displaystyle\text{(5).
}\left(\begin{array}[]{cccc}c_{0}&c_{1}&c_{2}&c_{3}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{4}&c_{5}&c_{6}&c_{7}\end{array}\right)^{T},$
(24)
$\displaystyle\left(\begin{array}[]{cccc}c_{8}&c_{9}&c_{10}&c_{11}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{12}&c_{13}&c_{14}&c_{15}\end{array}\right)^{T};$
(27)

$\displaystyle\text{(6).
}\left(\begin{array}[]{cccc}c_{0}&c_{1}&c_{4}&c_{5}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{2}&c_{3}&c_{6}&c_{7}\end{array}\right)^{T},$
(30)
$\displaystyle\left(\begin{array}[]{cccc}c_{8}&c_{9}&c_{12}&c_{13}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{10}&c_{11}&c_{14}&c_{15}\end{array}\right)^{T};$
(33)

$\displaystyle\text{(7).
}\left(\begin{array}[]{cccc}c_{0}&c_{1}&c_{8}&c_{9}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{2}&c_{3}&c_{10}&c_{11}\end{array}\right)^{T},$
(36)
$\displaystyle\left(\begin{array}[]{cccc}c_{4}&c_{5}&c_{12}&c_{13}\end{array}\right)^{T},\left(\begin{array}[]{cccc}c_{6}&c_{7}&c_{14}&c_{15}\end{array}\right)^{T}.$
(39)

Proof. We only prove Case (1): a pure product state
$|\psi\rangle=\sum_{i=0}^{15}c_{i}|i\rangle$ of four qubits can be written as
$|\varphi\rangle_{1}|\phi\rangle_{234}$ iff the the set of vectors in Eq. (9)
are proportional.

($\Longrightarrow$). Let $|\psi\rangle=|\varphi\rangle_{1}|\phi\rangle_{234}$,
where $|\varphi\rangle_{1}=(\alpha|0\rangle_{1}+\beta|1\rangle_{1})$ and
$|\phi\rangle_{234}=\sum_{i=0}^{7}a_{i}|i\rangle_{234}$. Clearly,
$|\psi\rangle=$
$\alpha\sum_{i=0}^{7}a_{i}|0\rangle_{1}|i\rangle_{234}+\beta\sum_{i=0}^{7}a_{i}|1\rangle_{1}|i\rangle_{234}$.
Let $v_{1}$ be the first vector and $v_{2}$ be the second vector in Eq. (9).
Then,

$\displaystyle v_{1}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccccc}\alpha a_{0}&\alpha a_{1}&\alpha
a_{2}&\alpha a_{3}&\alpha a_{4}&\alpha a_{5}&\alpha a_{6}&\alpha
a_{7}\end{array}\right)^{T},$ (41) $\displaystyle v_{2}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccccc}\beta a_{0}&\beta a_{1}&\beta
a_{2}&\beta a_{3}&\beta a_{4}&\beta a_{5}&\beta a_{6}&\beta
a_{7}\end{array}\right)^{T}.$ (43)

Thus, one can see that $v_{1}$ and $v_{2}$ are proportional.

($\Longleftarrow$). Conversely, if $v_{1}$ and $v_{2}$ are proportional, then
we can write $v_{2}=kv_{1}$ or $v_{1}=kv_{2}$. Assume that $v_{2}=kv_{1}$ and
$v_{1}=\left(\begin{array}[]{cccc}c_{0}&c_{1}&\cdots&c_{7}\end{array}\right)^{T}$.
Then, via $v_{1}$ and $v_{2}$ we can write

$\displaystyle|\psi\rangle$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{7}c_{i}|0\rangle_{1}|i\rangle_{234}+k\sum_{i=0}^{7}c_{i}|1\rangle_{1}|i\rangle_{234}$
(44) $\displaystyle=$ $\displaystyle|\varphi\rangle_{1}|\phi\rangle_{234},$
(45)

where $|\varphi\rangle_{1}=$ $(|0\rangle_{1}+k|1\rangle_{1})$ and
$|\phi\rangle_{234}=\sum_{i=0}^{7}c_{i}|i\rangle_{234}$.

Thus, to determine that a state of four qubits is genuinely entangled, we need
to determine that each of above seven sets of vectors are not proportional. We
investigated the separability of four qubits by using permutations of qubits
DLI-QIP-19 .

##  III Losing one qubit operator

###  III.1 Losing one qubit operator for three qubits

First we use the following example to demonstrate the losing one qubit
operator. We consider a product state of three qubits, for e.g.,
$|\psi\rangle_{123}=|\phi\rangle_{2}|\varphi\rangle_{13}$, where
$|\phi\rangle_{2}=(\alpha|0\rangle_{2}+\beta|1\rangle_{2})$ and
$|\varphi\rangle_{13}=(a|00\rangle_{13}+b|01\rangle_{13}+c|10\rangle_{13}+d|11\rangle_{13})$.

There are the following cases for losing qubits.

Losing qubit 1:

Let $|\psi^{\ast}\rangle_{123/1}=_{1}\langle 0|\psi\rangle_{123}+_{1}\langle
1|\psi\rangle_{123}$. Then, a calculation yields that

$|\psi^{\ast}\rangle_{123/1}=(\alpha|0\rangle_{2}+\beta|1\rangle_{2})((a+c)|0\rangle_{3}+(b+d)|1\rangle_{3})$.
Note that if the coefficients of $|\psi^{\ast}\rangle_{123/1}$ vanish, then we
also call $|\psi^{\ast}\rangle_{123/1}$ a product state. Thus, clearly
$|\psi^{\ast}\rangle_{123/1}$ is a product state.

Losing qubit 2:

Let $|\psi^{\ast}\rangle_{123/2}=_{2}\langle 0|\psi\rangle_{123}+_{2}\langle
1|\psi\rangle_{123}$. Then, a calculation yields that
$|\psi^{\ast}\rangle_{123/2}=(\alpha+\beta)|\varphi\rangle_{13}$. Then,
$|\psi^{\ast}\rangle_{123/2}$ is an entangled state iff $|\varphi\rangle_{13}$
is an entangled state whenever $\alpha+\beta\neq 0$.

Losing qubit 3:

Let $|\psi^{\ast}\rangle_{123/3}=$ ${}_{3}\langle
0|\psi\rangle_{123}+_{3}\langle 1|\psi\rangle_{123}$. Then, a calculation
yields that $|\psi^{\ast}\rangle_{123/3}=$
$(a+b)(\alpha|00\rangle_{12}+\beta|01\rangle_{12})+(c+d)(\alpha|10\rangle_{12}+\beta|11\rangle_{12})$.
It is easy to see that the concurrence vanishes for
$|\psi^{\ast}\rangle_{123/3}$. Via Eq. (2), therefore
$|\psi^{\ast}\rangle_{123/3}$ is a product state.

So, for a product state of three qubits, there is at most one qubit $i$,
$i=1,2,$ or $3$, such that $|\psi^{\ast}\rangle_{123/i}$ is entangled.

A calculation yields the Table 1.

Table 1.

3-qubits | $|\psi^{\ast}\rangle_{123/1}$ | $|\psi^{\ast}\rangle_{123/2}$ | $|\psi^{\ast}\rangle_{123/3}$
---|---|---|---
$|000\rangle$ | product | product | product
$|0\rangle_{1}|EPR\rangle_{23}$ | entangled | product | product
GHZ | entangled | entangled | entangled
W | entangled | entangled | entangled

###  III.2 Reducing the detection of the entanglement of n qubits to (n-1)
qubits

Definition: Let $|\psi\rangle_{1\cdots n}$ be a state of $n(\geq 3)$ qubits
and $|\psi^{\ast}\rangle_{1\cdots n/i}=_{i}\langle 0|\psi\rangle_{1\cdots
n}+_{i}\langle 1|\psi\rangle_{1\cdots n}$. We call the $(n-1)$-qubit state
$|\psi^{\ast}\rangle_{1\cdots n/i}$ a projected state of
$|\psi\rangle_{1\cdots n}$ obtained by losing qubit $i$.

We can conclude the following theorem.

Theorem 1. If $|\psi\rangle_{1\cdots n}$ is a product state of $n$ qubits,
then there exists at most one qubit $i$ such that
$|\psi^{\ast}\rangle_{1\cdots n/i}$ is genuinely entangled. That is, for
$j\neq i$, $|\psi^{\ast}\rangle_{1\cdots n/j}$ are product states or zero.

Proof. It is trivial for $n=2$. For $n\geq 3$ qubits, the proof is put in
Appendix A.

Corollary 1. Let $|\psi\rangle_{1\cdots n}$ be a state of $n$ qubits. If there
are at least two qubits $i$ and $j$ such that $(n-1)$-qubit projected states
$|\psi^{\ast}\rangle_{1\cdots n/i}$ and $|\psi^{\ast}\rangle_{1\cdots n/j}$
are genuinely entangled, then the $n$-qubit state $|\psi\rangle_{1\cdots n}$
is genuinely entangled.

We give the following example to show that Corollary 1 is not necessary.

Let $|\psi\rangle_{123}=|001\rangle+|010\rangle+|100\rangle+|111\rangle$
(belonging to W SLOCC class). Then,
$|\psi^{\ast}\rangle_{123/1}=|\psi^{\ast}\rangle_{123/2}=|\psi^{\ast}\rangle_{123/3}=(|0\rangle+|1\rangle)(|0\rangle+|1\rangle)$.
Though $|\psi\rangle_{123}$ is genuinely entangled,
$|\psi^{\ast}\rangle_{123/i}$, $i=1,2,3$, are product states.

Corollary 2. If $|\psi\rangle_{1\cdots n}$ is an $n$-qubit product state but
not as a product of a single qubit state and an ($n-1$)-qubit genuinely
entangled state, then the projected state $|\psi^{\ast}\rangle_{1\cdots n/i}$
is a product state for any qubit $i$.

Proof. Ref. the proof of Theorem 1.

From Corollary 2, it is easy to show that the following Corollary 3 is true.

Corollary 3. If $|\psi\rangle_{1\cdots n}$, which is not as a product of a
single qubit state and an $(n-1)$-qubit genuinely entangled state, has a
genuinely entangled $(n-1)$-qubit projected state, then $|\psi\rangle_{1\cdots
n}$ is genuinely entangled.

Via Corollary 1, for three qubits, we derive a simple sufficient condition for
genuinely entangled states in Appendix B.

###  III.3 Reducing the detection of the entanglement of n qubits to two
qubits

For any pure state $|\psi\rangle_{1\cdots
n}=\sum_{i=0}^{2^{n}-1}a_{i}|i\rangle$, in light of Corollary 1, if there are
at least two $(n-1)$-qubit projected states which are genuinely entangled,
then the $n$-qubit state $|\psi\rangle_{1\cdots n}$ is genuinely entangled.
For the $(n-1)$-qubit projected state $|\psi^{\ast}\rangle_{1\cdots n/i}$, in
light of Corollary 1, if it has at least two $(n-2)$-qubit projected states
which are genuinely entangled, then the $(n-1)$-qubit state
$|\psi^{\ast}\rangle_{1\cdots n/i}$ is genuinely entangled. The process
repeats until we get $4$ (3, or 2)-qubit projected states. Therefore,
detecting genuine entanglement of n qubits may reduce to four, three, or two
qubits.

For four, three and two qubits, we have given the simple necessary and
sufficient conditions for detecting genuine entanglement in Result 1, Result 2
and Result 3.

Example 1. For the $n$-qubit W state, all the ($n-1$) qubit projected states
are of the form $\sqrt{\frac{n-1}{n}}$W${}_{n-1}+\frac{1}{\sqrt{n}}|0\cdots
0\rangle$, where Wn-1 is the $(n-1)$-qubit W state. The $k$-qubit projected
states are of the form
$\sqrt{\frac{k}{n}}$W${}_{k}+\frac{n-k}{\sqrt{n}}|0\cdots 0\rangle$.
Specially, 2-qubit projected states have the form
$\sqrt{\frac{2}{n}}(|01\rangle+|10\rangle)+\frac{n-2}{\sqrt{n}}|00\rangle$.
The latter state has nonzero concurrence and therefore is entangled.
Consequently, via Corollary 1, the n-qubit W state is genuinely entangled.

Example 2. Let the $n$-qubit state $|\psi\rangle_{1\cdots n}=$
$\alpha|i_{1}i_{2}\cdots
i_{n}\rangle+\beta|\overline{i_{1}}\overline{i_{2}}\cdots\overline{i_{n}}\rangle$,
where $\alpha\beta\neq 0$, $i_{l}=0$ or 1 and $\overline{i_{l}}=1-i_{l}$.
Clearly, after losing one qubit $k$, we obtain the projected state
$|\psi^{\ast}\rangle_{1\cdots n/k}=\alpha|i_{1}\cdots i_{(k-1)}i_{(k+1)}\cdots
i_{n}\rangle+\beta|\overline{i_{1}}\cdots\overline{i_{(k-1)}}\overline{i_{(k+1)}}\cdots\overline{i_{n}}\rangle$.
We can continue applying the losing one qubit operator to
$|\psi^{\ast}\rangle_{1\cdots n/k}$. Finally, we obtain two-qubit projected
states
$\alpha|z_{1}z_{2}\rangle+\beta|\overline{z_{1}}\overline{z_{2}}\rangle$. It
is easy to know that
$\alpha|z_{1}z_{2}\rangle+\beta|\overline{z_{1}}\overline{z_{2}}\rangle$ is
entangled. Therefore $|\psi\rangle_{1\cdots n}$ is genuinely entangled via
Corollary 1. Specially, the $n$-qubit GHZ is genuinely entangled.

Example 3. Let us check that
$|\psi\rangle_{1234}=|0000\rangle+|0111\rangle-|1111\rangle$ is genuinely
entangled. A calculation produces that
$|\psi^{\ast}\rangle_{1234/1}=|000\rangle$, which is a product state, and
$|\psi^{\ast}\rangle_{1234/i}=|000\rangle+|011\rangle-|111\rangle$, $i=2,3,4$.
Next, we show that $|\psi^{\ast}\rangle_{1234/i}$, $i=2,3,4$, are genuinely
entangled. Let $|\omega\rangle_{123}=|000\rangle+|011\rangle-|111\rangle$. A
calculation yields that $|\omega^{\ast}\rangle_{123/1}=$ $|00\rangle$ and
$|\omega^{\ast}\rangle_{123/2}=|\omega^{\ast}\rangle_{123/3}=|00\rangle+|01\rangle-|11\rangle$,
which is an entangled state of two qubits. So, via Corollary 1, the three-
qubit states $|\psi^{\ast}\rangle_{1234/i}$, $i=2,3,4$, are genuinely
entangled, and then $|\psi\rangle_{1234}$ is genuinely entangled.

###  III.4 Maximally entangled states (MES)

MES can be defined via different ways such as entropy and LOCC. It is well
known that the GHZ state can be regarded as the maximally entangled state of
three qubits in several aspects.

Let $|\psi\rangle_{1\cdots n}$ be a state of $n$ qubits. If $k$ of all the
$(n-1)$-qubit projected states $|\psi^{\ast}\rangle_{1\cdots n/i}$ are
genuinely entangled, then we say that $|\psi\rangle_{1\cdots n}$ has the
entanglement measure of $k$. If all the $(n-1)$-qubit projected states are
genuinely entangled, then the state is called MES.

We next demonstrate the entanglement measure of some entangled states below.

(1). After losing one qubit, the projected states of the $n$-qubit GHZ state
are just the $(n-1)$-qubit GHZ state. It means that losing one qubit operator
preserves the GHZ-constructions. Clearly, GHZ is a MES.

(2). For the W state of $n$ qubits, all the $(n-1)$-qubit projected states are
$|P\rangle=$ $\sqrt{\frac{n-1}{n}}$W${}_{n-1}+\frac{1}{\sqrt{n}}$ $|0\cdots
0\rangle$, which is not the W-states. It means that losing one qubit operator
does not preserve the W-construction. Thus, the W-construction is fragile
under losing one qubit operator. From Example 1, $|P\rangle$ is genuinely
entangled. Therefore, W is a MES.

(3). $|\Phi_{4}\rangle=|0001\rangle+|0010\rangle+|1100\rangle+|1111\rangle$ is
genuinely entangled Osterloh .
$|\Phi_{4}^{\ast}\rangle_{1234/4}=|\Phi_{4}^{\ast}\rangle_{1234/3}=(|00\rangle+|11\rangle)(|0\rangle+|1\rangle)$
which are product states while
$|\Phi_{4}^{\ast}\rangle_{1234/2}=|\psi^{\ast}\rangle_{1234/1}=|001\rangle+|010\rangle+|100\rangle+|111\rangle$
which are genuinely entangled. Thus, $|\Phi_{4}\rangle$ has the entanglement
measure 2.

###  III.5 A program for losing one qubit operator

We write a LISP program for the formula for the projected states
$|\psi^{\ast}\rangle_{1\cdots n/i}$ and the procedure reducing detection of
entanglement of n qubits to two qubits in Appendix C. The program can detect
the genuinely entangled states with the measure $\geq 2$. For example, the
program can detect the following genuinely entangled states.

(1) The states GHZ and W of three qubits

(2). For four qubits, GHZ, W, Cluster state, Dicke state $|2,4\rangle$, and
Osterloh’s $|\Phi_{5}\rangle$ and $|\Phi_{4}\rangle$ states Osterloh .

(3). For five qubits, Osterloh’s $|\Psi_{2}\rangle,|\Psi_{5}\rangle,$ and
$|\Psi_{6}\rangle$ states Osterloh .

(4). For six qubits, Osterloh’s
$|\Xi_{2}\rangle,|\Xi_{5}\rangle,|\Xi_{6}\rangle,$ and $|\Xi_{7}\rangle$
states Osterloh .

###  III.6 Comparing reduced density operator with losing one qubit operator

(1). Reduced density operator and losing one qubit operator are different.

For the Bell state $|$Bell$\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$,
the reduced density is $tr_{2}(|$Bell$\rangle\langle$Bell$|)=\frac{1}{2}I$.
After losing one qubit,
$|\psi^{\ast}\rangle_{12/2}=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$.
Therefore, losing a qubit operator is different from the reducing density
operator.

(2). $|\psi^{\ast}\rangle_{1\cdots n/i}$ is a pure state while the reducing
density $tr_{i}(|\psi\rangle_{1\cdots n}\langle\psi|)$ may become a mixed
state. For example, $tr_{2}(|$Bell$\rangle\langle$Bell$|)=\frac{1}{2}I$, which
is a mixed state.

(3). After losing one qubit, the $(n-1)$-qubit projected states of the
$n$-qubit GHZ state are still the $(n-1)$-qubit GHZ state while for GHZ of
three qubits, tracing out of a qubit, the reduced density operator becomes
completely unentangled. It means that the entanglement property of the state
GHZ is fragile under tracing out a qubit. Ref. Table 2.

Table 2.

3 qubits | $\rho^{12},\rho^{23},\rho^{13}$ | $|\psi^{\ast}\rangle_{123/i},i=1,2,3$
---|---|---
GHZ | separable | entangled, GHZ
W | entangled | entangled, not W

##  IV Summary

In this paper, we show that all the projected states obtained via the losing
one qubit operator are product states for a product state of $n$ qubits but
not as a product of a single qubit state and a genuinely entangled
$(n-1)$-qubit state. Thus, if there are two $(n-1)$-qubit projected states
which are genuinely entangled, then the state of $n$ qubit is genuinely
entangled. We can repeat the process until we get $2$ (3,or 4)-qubit projected
states for which we have the necessary and sufficient conditions to detect
their separability. Thus, the losing one qubit operator can reduce the
detection of entanglement for $n$-qubits to two qubits. We have written a LISP
program for detection of entanglement for $n$-qubits.

##  V Appendix A The proof of Theorem 1

The proof of Theorem is follows.

For $n\geq 3$ qubits, there are three cases.

Case 1. $|\psi\rangle_{1\cdots n}=|\phi\rangle|\varphi\rangle|\omega\rangle$.
Clearly, it is easy to see that $|\psi^{\ast}\rangle_{12\cdots n/i}$ is a
product state for any $i$.

Case 2. $|\psi\rangle_{1\cdots n}=|\phi\rangle_{q_{1}\cdots
q_{i}}|\varphi\rangle_{q_{i+1}\cdots q_{n}}$, where $i\geq 2$, $n-i\geq 2$.

Clearly, $|\psi\rangle_{1\cdots n/q_{k}}^{\ast}=$ $|\phi\rangle_{q_{1}\cdots
q_{i}/q_{k}}^{\ast}|\varphi\rangle_{q_{i+1}\cdots q_{n}}$, where $1\leq k\leq
i$. One can see that $|\psi\rangle_{1\cdots n/q_{k}}^{\ast}$ is a product
state of $(n-1)$ qubits or zero.

Case 3. $|\psi\rangle_{1\cdots
n}=(\alpha|0\rangle_{q_{1}}+\beta|1\rangle_{q_{1}})|\varphi\rangle_{q_{2}\cdots
q_{n}}$, where $|\varphi\rangle_{q_{2}\cdots q_{n}}$ is genuinely entangled.
For example, $|\psi\rangle_{123}=|0\rangle_{1}|$EPR$\rangle_{23}$,
$|\psi^{\ast}\rangle_{123/1}=|$EPR$\rangle_{23}$.

Case 3.1. $|\psi^{\ast}\rangle_{12\cdots
n/q_{1}}=(\alpha+\beta)|\varphi\rangle_{q_{2}\cdots q_{n}}$, which is
genuinely entangled when $\alpha+\beta\neq 0$ or zero when $\alpha+\beta=0$.

Case 3.2. $|\varphi\rangle_{q_{2}\cdots
q_{n}}=|0\rangle_{q_{k}}|\varphi^{(0)}\rangle_{q_{2}\cdots
q_{n}/q_{k}}+|1\rangle_{q_{k}}|\varphi^{(1)}\rangle_{q_{2}\cdots
q_{n}/q_{k}}$, $k\neq 1$.

Thus, $|\psi^{\ast}\rangle_{12\cdots
n/q_{k}}=(\alpha|0\rangle_{q_{1}}+\beta|1\rangle_{q_{1}})|\varphi^{\ast}\rangle_{q_{2}\cdots
q_{n}/q_{k}}$ which is a product state.

From the above three cases, we can conclude If $|\psi\rangle_{1\cdots n}$ is a
product state of $n$ qubits, then there exists at most one qubit $q_{i}$ such
that $|\psi^{\ast}\rangle_{12\cdots n/q_{i}}$ is genuinely entangled. That is,
for other $q_{j}$, $|\psi^{\ast}\rangle_{12\cdots n/q_{j}}$ are product
states.

##  VI Appendix B. A simple sufficient condition for genuinely entangled
states of three qubits

Via Corollary 1, for three qubits, we derive a simple sufficient condition for
genuinely entangled states by losing one qubit operator. Let
$|\psi\rangle_{123}=\sum_{i=0}^{7}c_{i}|i\rangle$ be any pure state of three
qubits.

A calculation yields that

$\displaystyle|\psi^{\ast}\rangle_{123/1}$ $\displaystyle=$
$\displaystyle(c_{0}+c_{4})|00\rangle+(c_{1}+c_{5})|01\rangle$
$\displaystyle+(c_{2}+c_{6})|10\rangle+(c_{3}+c_{7})|11\rangle.$

Via Eq. (2), clearly, $|\psi^{\ast}\rangle_{123/1}$ is entangled iff

$(c_{0}+c_{4})(c_{3}+c_{7})\neq(c_{1}+c_{5})(c_{2}+c_{6}).$ (46)

A calculation yields that

$\displaystyle|\psi^{\ast}\rangle_{123/2}$ $\displaystyle=$
$\displaystyle(c_{0}+c_{2})|00\rangle+(c_{1}+c_{3})|01\rangle$
$\displaystyle+(c_{4}+c_{6})|10\rangle+(c_{5}+c_{7})|11\rangle.$

Via Eq. (2), $|\psi^{\ast}\rangle_{123/2}$ is entangled iff

$(c_{0}+c_{2})(c_{5}+c_{7})\neq(c_{1}+c_{3})(c_{4}+c_{6}).$ (47)

A calculation yields that

$\displaystyle|\psi^{\ast}\rangle_{123/3}$ $\displaystyle=$
$\displaystyle(c_{0}+c_{1})|00\rangle+(c_{2}+c_{3})|01\rangle$
$\displaystyle+(c_{4}+c_{5})|10\rangle+(c_{6}+c_{7})|11\rangle.$

Via Eq. (2), $|\psi^{\ast}\rangle_{123/3}$ is entangled iff

$(c_{0}+c_{1})(c_{6}+c_{7})\neq(c_{2}+c_{3})(c_{4}+c_{5}).$ (48)

Via Corollary 1, $|\psi\rangle_{123}=\sum_{i=0}^{7}c_{i}|i\rangle$ is
genuinely entangled if at least two of Eqs. (46, 47, 48) hold.

##  VII Appendix C A formula for the projected states

Let $|\psi\rangle_{1\cdots n}=\sum_{i=0}^{2^{n}-1}a_{i}|i\rangle$ be any pure
state of n qubits. Then, we can write $|\psi\rangle_{1\cdots
n}=|0\rangle_{k}|\psi^{(0)}\rangle_{1\cdots
n/k}+|1\rangle_{k}|\psi^{(1)}\rangle_{1\cdots n/k}$. Therefore,
$|\psi^{\ast}\rangle_{1,\cdots,n/k}=|\psi^{(0)}\rangle_{1\cdots
n/k}+|\psi^{(1)}\rangle_{1\cdots n/k}$. Denoting $\ell=2^{n-k}$, then
$|\psi^{\ast}\rangle_{1\cdots n/k}$

$=\sum_{i=0}^{\ell-1}(a_{0\times\ell+i}+a_{1\times\ell+i})|\underbrace{0\cdots
0}_{k-1}\rangle|i\rangle$

$+\sum_{i=0}^{\ell-1}(a_{2\times\ell+i}+a_{3\times\ell+i})|\underbrace{0\cdots
01}_{k-1}\rangle|i\rangle+\cdots$

$+\sum_{i=0}^{\ell-1}(a_{(2^{k}-4)\times\ell+i}+a_{(2^{k}-3)\times\ell+i})|\underbrace{1\cdots
10}_{k-1}\rangle|i\rangle$

$+\sum_{i=0}^{\ell-1}(a_{(2^{k}-2)\times\ell+i}+a_{(2^{k}-1)\times\ell+i})|\underbrace{1\cdots
10}_{k-1}\rangle|i\rangle$.

## References

  * (1) M.A. Nielsen, Chuang, I.L.: Quantum Computation and Quantum Information (Cambridge Univ. Press, Cambridge, 2000).
  * (2) W. Dür et al., Phys. Rev. A 62, 062314 (2000).
  * (3) J. J. Wallman et al., Phys. Rev. Lett. 115, 060501 (2015).
  * (4) G.M. Quinta et al., Phys. Rev. A 100, 062329 (2019).
  * (5) A. Peres, Phys. Lett. A 202, 16-17 (1995).
  * (6) R. Horodecki et al., Phys. Lett. A 223, 1 (1996).
  * (7) K. Chen and L.A. Wu, Quant. Inf. Comput. 3, 193 (2003).
  * (8) S. Albeverio, K. Chen, and S.-M. Fei, Phys. Rev. A 68, 062313 (2003).
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stochasticity inherent in model training, thus reporting only the total ACC
value without standard deviation. Moreover, owing to its direct utilization of
embeddings and labels for classification, $k$-NN does not furnish a reliable
AUC score. The overall best result across all training schemes, models, and
resolutions is highlighted with a background color; the best result per
resolution across all training schemes and models is highlighted with
underline; and the best result per training scheme and resolution is
highlighted in bold. TissueMNIST
---
Methods | Accuracy (ACC) |  | Area Under the ROC Curve (AUC)
$28\times 28$ | $64\times 64$ | $128\times 128$ | $224\times 224$ |  | $28\times 28$ | $64\times 64$ | $128\times 128$ | $224\times 224$
End-to-End |  |  |  |  |  |  |  |  |
VGG16 | 67.75$\pm$0.46 |  71.29$\pm$0.81 |  71.62$\pm$0.16 |  70.57$\pm$0.35 |  | 92.67$\pm$0.12 |  94.12$\pm$0.23 |  94.37$\pm$0.10 |  94.05$\pm$0.06
AlexNet |  59.80$\pm$0.43 |  64.12$\pm$0.25 |  67.06$\pm$0.06 |  69.25$\pm$0.33 |  |  88.79$\pm$0.11 |  91.16$\pm$0.04 |  92.56$\pm$0.07 |  93.50$\pm$0.08
ResNet-18 |  63.02$\pm$0.10 |  67.36$\pm$0.20 |  70.17$\pm$0.73 |  69.35$\pm$0.67 |  |  90.59$\pm$0.08 |  92.65$\pm$0.09 |  93.65$\pm$0.21 |  93.57$\pm$0.13
DenseNet-121 |  66.53$\pm$0.37 | 71.54$\pm$0.71 | 74.25$\pm$0.39 | 74.08$\pm$0.32 |  |  92.47$\pm$0.04 | 94.46$\pm$0.24 | 95.35$\pm$0.07 | 95.25$\pm$0.04
EfficientNet-B4 |  59.92$\pm$0.55 |  65.16$\pm$1.63 |  71.35$\pm$0.24 |  69.31$\pm$1.50 |  |  88.97$\pm$0.50 |  91.59$\pm$0.71 |  94.15$\pm$0.06 |  93.35$\pm$0.53
ViT-B/16 |  60.55$\pm$0.56 |  66.72$\pm$0.71 |  71.29$\pm$0.40 |  72.89$\pm$0.70 |  |  88.97$\pm$0.18 |  92.62$\pm$0.31 |  94.32$\pm$0.17 |  94.84$\pm$0.16
CLIP ViT-B/16 |  56.63$\pm$0.53 |  62.97$\pm$0.67 |  66.47$\pm$0.29 |  66.25$\pm$0.28 |  |  86.44$\pm$0.27 |  90.55$\pm$0.35 |  92.32$\pm$0.25 |  92.20$\pm$0.11
EVA-02 ViT-B/16 |  57.60$\pm$1.00 |  64.42$\pm$0.67 |  70.42$\pm$0.87 |  70.34$\pm$0.93 |  |  87.31$\pm$0.58 |  91.21$\pm$0.29 |  93.92$\pm$0.33 |  93.97$\pm$0.35
DINO ViT-B/16 |  59.44$\pm$0.31 |  65.80$\pm$1.01 |  69.35$\pm$1.72 |  70.38$\pm$0.99 |  |  88.55$\pm$0.27 |  91.98$\pm$0.44 |  93.48$\pm$0.77 |  94.02$\pm$0.37
SAM ViT-B/16 |  58.86$\pm$0.20 |  66.19$\pm$0.70 |  69.42$\pm$0.60 |  71.47$\pm$0.41 |  |  88.07$\pm$0.32 |  92.25$\pm$0.33 |  93.70$\pm$0.23 |  94.40$\pm$0.13
Linear Probing |  |  |  |  |  |  |  |  |
VGG16 |  53.19$\pm$0.00 |  53.53$\pm$0.02 |  55.50$\pm$0.08 |  58.12$\pm$0.13 |  |  83.95$\pm$0.00 |  84.03$\pm$0.00 |  86.36$\pm$0.03 |  87.89$\pm$0.04
AlexNet |  49.10$\pm$0.10 |  53.80$\pm$0.06 |  55.75$\pm$0.06 |  59.57$\pm$0.05 |  |  80.35$\pm$0.13 |  84.74$\pm$0.06 |  86.47$\pm$0.03 |  88.98$\pm$0.01
ResNet-18 |  51.06$\pm$0.00 |  53.43$\pm$0.02 |  54.60$\pm$0.02 |  56.46$\pm$0.01 |  |  82.44$\pm$0.00 |  84.76$\pm$0.00 |  85.65$\pm$0.00 |  86.97$\pm$0.00
DenseNet-121 |  55.93$\pm$0.01 |  59.46$\pm$0.02 |  60.86$\pm$0.02 |  61.09$\pm$0.02 |  |  86.45$\pm$0.00 |  88.64$\pm$0.00 |  89.38$\pm$0.01 |  89.64$\pm$0.01
EfficientNet-B4 |  54.24$\pm$0.01 |  56.91$\pm$0.00 |  57.88$\pm$0.02 |  58.47$\pm$0.01 |  |  85.29$\pm$0.00 |  87.17$\pm$0.00 |  87.63$\pm$0.00 |  88.33$\pm$0.00
ViT-B/16 |  53.91$\pm$0.06 |  60.69$\pm$0.09 |  62.79$\pm$0.02 |  63.80$\pm$0.08 |  |  84.95$\pm$0.03 |  89.27$\pm$0.03 |  90.39$\pm$0.01 |  90.88$\pm$0.06
CLIP ViT-B/16 |  55.28$\pm$0.03 |  59.39$\pm$0.04 |  61.03$\pm$0.08 |  61.50$\pm$0.06 |  |  86.02$\pm$0.01 |  88.60$\pm$0.02 |  89.62$\pm$0.01 |  89.76$\pm$0.08
EVA-02 ViT-B/16 |  54.23$\pm$0.01 |  58.35$\pm$0.02 |  59.71$\pm$0.00 |  60.51$\pm$0.02 |  |  85.14$\pm$0.00 |  87.93$\pm$0.00 |  88.68$\pm$0.00 |  89.24$\pm$0.00
DINO ViT-B/16 | 57.46$\pm$0.22 | 63.09$\pm$0.02 | 63.92$\pm$0.05 | 64.04$\pm$0.12 |  | 87.22$\pm$0.10 | 90.51$\pm$0.02 | 91.02$\pm$0.01 | 90.98$\pm$0.03
SAM ViT-B/16 |  37.70$\pm$0.00 |  41.70$\pm$0.00 |  46.05$\pm$0.01 |  46.66$\pm$0.01 |  |  68.59$\pm$0.01 |  73.73$\pm$0.01 |  76.61$\pm$0.00 |  76.89$\pm$0.00
$k$-NN ($k=11$) |  |  |  |  |  |  |  |  |
VGG16 | 47.30 | 47.20 | 48.55 | 51.30 |  | - | - | - | -
AlexNet | 45.97 | 49.96 | 50.79 | 54.11 |  | - | - | - | -
ResNet-18 | 48.25 | 48.89 | 49.05 | 51.54 |  | - | - | - | -
DenseNet-121 | 48.66 | 50.10 | 51.04 | 52.67 |  | - | - | - | -
EfficientNet-B4 | 49.79 | 51.14 | 50.35 | 51.26 |  | - | - | - | -
ViT-B/16 | 47.47 | 50.73 | 52.84 | 54.33 |  | - | - | - | -
CLIP ViT-B/16 | 48.04 | 50.48 | 53.29 | 53.46 |  | - | - | - | -
EVA-02 ViT-B/16 | 49.74 | 50.92 | 52.52 | 53.86 |  | - | - | - | -
DINO ViT-B/16 | 51.56 | 56.10 | 57.39 | 57.12 |  | - | - | - | -
SAM ViT-B/16 | 48.00 | 49.67 | 49.64 | 46.97 |  | - | - | - | -