# Einstein-Podolsky-Rosen steering paradox “2=1” for $N$ qubits

Zhi-Jie Liu  Theoretical Physics Division, Chern Institute of Mathematics and
LPMC, Nankai University, Tianjin 300071, People’s Republic of China  Jie Zhou
College of Physics and Materials Science, Tianjin Normal University, Tianjin
300382, People’s Republic of China  Hui-Xian Meng  School of Mathematics and
Physics, North China Electric Power University, Beijing 102206, People’s
Republic of China  Xing-Yan Fan  Theoretical Physics Division, Chern Institute
of Mathematics and LPMC, Nankai University, Tianjin 300071, People’s Republic
of China  Mi Xie<EMAIL_ADDRESS>Department of Physics, School of Science,
Tianjin University, Tianjin 300072, People’s Republic of China  Fu-lin Zhang
<EMAIL_ADDRESS>Department of Physics, School of Science, Tianjin
University, Tianjin 300072, People’s Republic of China  Jing-Ling Chen
<EMAIL_ADDRESS>Theoretical Physics Division, Chern Institute of
Mathematics and LPMC, Nankai University, Tianjin 300071, People’s Republic of
China

###### Abstract

Einstein-Podolsky-Rosen (EPR) paradox highlights the absence of a local
realistic explanation for quantum mechanics, and shows the incompatibility of
the local-hidden-state models with quantum theory. For $N$-qubit states, or
more importantly, the $N$-qubit mixed states, we present the EPR steering
paradox in the form of the contradictory equality “2=1”. We show that the
contradiction holds for any $N$-qubit state as long as both “the pure state
requirement” and “the measurement requirement” are satisfied. This also
indicates that the EPR steering paradox exists in more general cases. Finally,
we give specific examples to demonstrate and analyze our arguments.

Keywords: Einstein-Podolsky-Rosen steering; quantum paradox; $N$ qubits
states.

##  I Introduction

The quantum paradox serves as a powerful tool in elucidating the fundamental
distinction between the quantum theory and the classical theory. Quantum
correlations play a central role in the study of quantum information and
quantum mechanics. Among the quantum correlations, quantum entanglement and
Bell’s nonlocality are the first to be proposed and studied. In 1935,
Einstein, Podolsky, and Rosen (EPR) published their famous article “ _Can
Quantum-Mechanical Description of Physical Reality Be Considered Complete?_ ”
EPR1935  , which questioned the completeness of quantum mechanics under the
assumptions of locality and reality. This is nowadays well-known as the EPR
paradox. Soon after the publication of the EPR paper, Schrödinger introduced
two important concepts, namely, quantum entanglement and quantum steering
Schr1935 ; Schr1936 . Quantum entanglement distinguishes quantum theory from
classical theory. And quantum steering is closely related to “the spooky
action at a distance”. However, the idea of steering did not receive
considerable attention or advancement until the year 2007, at which point
Wiseman _et al._ presented a meticulous definition by utilizing quantum
information concepts Wiseman2007 ; jones2007 . So far, quantum steering has
been widely applied in various fields preskill1998 ; Galindo2002 ; Ekert1996 ;
Childs2010 .

Steering is a quantum correlation between entanglement Quantum
entanglement2009  and Bell nonlocality Bell nonlocality2014 ; Quantum machine
learning2017 ; bell1964 . Steering is used to describe the situation in a
bipartite system. When people use different observables to detect one of the
particles, it will cause the corresponding other particle to collapse to a
certain state. In practice, Alice prepares a bipartite quantum state, and she
sends one of the particles to Bob. They each measure the particles in their
hands and communicate over a classical channel. It is Alice’s task to convince
Bob that Alice has prepared a pair of entangled state. In the process, Bob
needs to assess the correctness of the assumptions of quantum mechanics and
acknowledge the measurements as described by quantum mechanics. Specifically,
Bob can disbelieve Alice’s equipment and measurements. However, in this case,
Bob needs to rule out the influence of hidden variables on the measurement
results by the measurements he has in hand. Bob can fully trust his own
equipment and results. If Bob cannot explain the measurement results on his
side with the local-hidden states (LHS), he must recognize that Alice has
prepared an entangled bipartite state. Only EPR steering states can accomplish
this task. And quantum steering is an asymmetric quantum nonlocality. That is,
in some cases Alice can steer Bob, but in turn Bob cannot steer Alice
Asymmetric2010 ; Bowles2014 ; Quintino2015 ; Wollmann2016 ; HV2012 . Based on
some properties of quantum steering, Chen _et al._ proposed the EPR steering
paradox “2=1” chenjl2016 , where “2” is the quantum result and “1” is the
corresponding result of LHS models. They verified the EPR steering state by
the contradiction between quantum mechanics and classical theory. In the
2-setting EPR steering protocol, they found that any 2-qubit entangled pure
state possesses the contradiction. Thereafter, Liu _et al._ found that such a
contradiction was also valid for a specific 4-qubit entangled mixed state
Liu2021 . In other words, the discussion of the EPR steering paradox “2=1” has
been limited to arbitrary 2-qubit pure state and a special 4-qubit mixed
state.

The purpose of this paper is to study the EPR steering paradox “2=1” for $N$
qubits. Based on the 2-setting steering protocol, we have obtained such EPR
steering paradox “2=1” for $N$-qubit entangled states. In this work, we
demonstrate that any $N$-qubit state can lead to the contradiction, provided
that both “the pure state requirement” and “the measurement requirement” are
fulfilled simultaneously. The paper is organized as follows. In Sec. II, we
propose a theorem for $N$-qubit quantum states that contains two requirements
: “the pure state requirement” and “the measurement requirement”. In Sec. III,
We obtain the EPR steering paradox “2=1” for the $N$-qubit states under Bell-
like basis measurement. In Sec. IV, we apply the results to the 2-qubit mixed
states and obtain a corollary that there is no EPR steering paradox “2=1” for
2-qubit mixed states. Finally, we conclude with a summary in Sec. V. Some
detailed proofs are given in the Appendix A and B.

##  II EPR steering paradox “2=1” for $N$-qubit states

###  II.1 “2=1” for the 2-qubit pure state and the 4-qubit mixed state

To make the paper be self-contained, in this subsection let us make a brief
review. In 2016, Chen _et al._ first simplified the EPR steering paradox as a
contradiction “2=1” for any 2-qubit pure entangled state chenjl2016 . They
analyzed a 2-qubit pure entangled state given by

$\rho_{AB}=\left|\Psi\left(\theta\right)\right\rangle\left\langle\Psi\left(\theta\right)\right|,$
(1)

where

$\left|\Psi\left(\theta\right)\right\rangle=\cos\theta\left|0\right\rangle\left|0\right\rangle+\sin\theta\left|1\right\rangle\left|1\right\rangle,$
(2)

with $\theta\in\left(0,\pi/2\right)$. They chose the 2-setting protocol as
$\left\\{\hat{z},\hat{x}\right\\}$. In the protocol, Bob asks Alice to carry
out either one of two possible projective measurements on her qubit along the
$z$-direction and the $x$-direction, i.e.,

$\begin{array}[c]{l}{{P}_{0}^{\hat{z}}=}\left|0\right\rangle\left\langle
0\right|,\\\ {{P}_{1}^{\hat{z}}=}\left|1\right\rangle\left\langle 1\right|,\\\
{{P}_{0}^{\hat{x}}=}\left|+\right\rangle\left\langle+\right|,\\\
{{P}_{1}^{\hat{x}}=}\left|-\right\rangle\left\langle-\right|,\end{array}$ (3)

where
$\left|\pm\right\rangle=(1/\sqrt{2})(\left|0\right\rangle\pm\left|1\right\rangle)$,
and to inform him of the measurement results of “$a$” (where $a=0,1$). Then
Bob’s four unnormalized conditional states are

$\begin{array}[c]{l}\tilde{\rho}{{}_{0}^{\hat{z}}=\cos}^{2}\theta\left|0\right\rangle\left\langle
0\right|,\quad\tilde{\rho}{{}_{1}^{\hat{z}}=\sin}^{2}{\theta}\left|1\right\rangle\left\langle
1\right|,\\\
\tilde{\rho}{{}_{0}^{\hat{x}}=}\dfrac{1}{2}\left(\cos\theta\left|0\right\rangle+{\sin\theta}\left|1\right\rangle\right)\left(\cos\theta\left\langle
0\right|+{\sin\theta}\left\langle 1\right|\right),\\\\[9.0pt]
\tilde{\rho}{{}_{1}^{\hat{x}}=}\dfrac{1}{2}\left(\cos\theta\left|0\right\rangle-{\sin\theta}\left|1\right\rangle\right)\left(\cos\theta\left\langle
0\right|-{\sin\theta}\left\langle 1\right|\right).\end{array}$ (4)

If Bob’s states have a LHS description, then there exists an ensemble
$\\{\wp_{\xi}\rho_{\xi}\\}$ and a stochastic map $\wp(a|\hat{n},\xi)$
satisfying

$\tilde{\rho}_{a}^{\hat{n}}={\displaystyle\sum\limits_{\xi}}\wp\left(a|\hat{n},\xi\right)\wp_{\xi}\rho_{\xi},$
(5)

and

${\displaystyle\sum\limits_{\xi}}\wp_{\xi}\rho_{\xi}=\rho_{B}.$ (6)

Here $\wp_{\xi}$ and $\wp\left(a|\hat{n},\xi\right)$ are probabilities
satisfying

${\sum\limits_{\xi}}\wp_{\xi}=1,$ (7)

and

${\sum\limits_{a}}\wp\left(a|\hat{n},\xi\right)=1,$ (8)

for a fixed $\xi$, and $\rho_{B}=\mathrm{tr}_{A}\left(\rho_{AB}\right)$ is
Bob’s reduced density matrix Wiseman2007 ; jones2007 .

Then Bob’s four unnormalized conditional states satisfy Eq. (5), and one has

$\begin{array}[c]{c}\widetilde{\rho}_{0}^{\hat{z}}={\displaystyle\sum\limits_{\xi}}\wp\left(0|\hat{z},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\widetilde{\rho}_{1}^{\hat{z}}={\displaystyle\sum\limits_{\xi}}\wp\left(1|\hat{z},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\widetilde{\rho}_{0}^{\hat{x}}={\displaystyle\sum\limits_{\xi}}\wp\left(0|\hat{x},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\widetilde{\rho}_{1}^{\hat{x}}={\displaystyle\sum\limits_{\xi}}\wp\left(1|\hat{x},\xi\right)\wp_{\xi}\rho_{\xi}.\end{array}$
(9)

Because the four states of Eq. (4) are pure states, it is sufficient to take
$\xi$ from $1$ to $4$. Eq. (9) can be write as

$\begin{array}[c]{c}\widetilde{\rho}_{0}^{\hat{z}}=\wp\left(0|\hat{z},1\right)\wp_{1}\rho_{1}+\wp\left(0|\hat{z},2\right)\wp_{2}\rho_{2}+\wp\left(0|\hat{z},3\right)\wp_{3}\rho_{3}+\wp\left(0|\hat{z},4\right)\wp_{4}\rho_{4},\\\
\widetilde{\rho}_{1}^{\hat{z}}=\wp\left(1|\hat{z},1\right)\wp_{1}\rho_{1}+\wp\left(1|\hat{z},2\right)\wp_{2}\rho_{2}+\wp\left(1|\hat{z},3\right)\wp_{3}\rho_{3}+\wp\left(1|\hat{z},4\right)\wp_{4}\rho_{4},\\\
\widetilde{\rho}_{0}^{\hat{x}}=\wp\left(0|\hat{x},1\right)\wp_{1}\rho_{1}+\wp\left(0|\hat{x},2\right)\wp_{2}\rho_{2}+\wp\left(0|\hat{x},3\right)\wp_{3}\rho_{3}+\wp\left(0|\hat{x},4\right)\wp_{4}\rho_{4},\\\
\widetilde{\rho}_{1}^{\hat{x}}=\wp\left(1|\hat{x},1\right)\wp_{1}\rho_{1}+\wp\left(1|\hat{x},2\right)\wp_{2}\rho_{2}+\wp\left(1|\hat{x},3\right)\wp_{3}\rho_{3}+\wp\left(1|\hat{x},4\right)\wp_{4}\rho_{4}.\end{array}$
(10)

And owing to the fact that a pure state cannot be obtained by convex
combination of other pure states, one has

$\begin{array}[c]{l}\tilde{\rho}{{}_{0}^{\hat{z}}=}\wp_{1}\rho_{1},\\\
\tilde{\rho}{{}_{1}^{\hat{z}}=}\wp_{2}\rho_{2},\\\
\tilde{\rho}{{}_{0}^{\hat{x}}=}\wp_{3}\rho_{3},\\\
\tilde{\rho}{{}_{1}^{\hat{x}}=}\wp_{4}\rho_{4}.\end{array}$ (11)

Here

$\wp\left(0|\hat{z},1\right)=\wp\left(1|\hat{z},2\right)=\wp\left(0|\hat{x},3\right)=\wp\left(1|\hat{x},4\right)=1,$

and other $\wp\left(a|\hat{n},\xi\right)=0$. By summing four terms up in Eq.
(11) and taking trace, the left side gives
$\mathrm{tr}(\tilde{\rho}{{}^{\hat{z}}}+\tilde{\rho}{{}^{\hat{x}}})=2\mathrm{tr}(\rho_{B})=2$.
But the right side gives
$\mathrm{tr}(\wp_{1}\rho_{1}+\wp_{2}\rho_{2}+\wp_{3}\rho_{3}+\wp_{4}\rho_{4})=\mathrm{tr}(\rho_{B})=1$,
then one can obtain the contradiction “2=1”, i.e., the EPR steering paradox,
for any 2-qubit pure entangled state.

After that, in 2021, Liu _et al._ found a 4-qubit mixed entangled state

$\rho\left(\theta\right)=\cos^{2}\theta\left|LC_{4}\right\rangle\left\langle
LC_{4}\right|+\sin^{2}\theta\left|LC_{4}^{\prime}\right\rangle\left\langle
LC_{4}^{\prime}\right|,$ (12)

where

$\begin{array}[c]{c}\left|LC_{4}\right\rangle=\dfrac{1}{2}\left(\left|0000\right\rangle+\left|1100\right\rangle+\left|0011\right\rangle-\left|1111\right\rangle\right),\\\\[9.0pt]
\left|LC_{4}^{\prime}\right\rangle=\dfrac{1}{2}\left(\left|0100\right\rangle+\left|1000\right\rangle+\left|0111\right\rangle-\left|1011\right\rangle\right),\end{array}$
(13)

are the linear cluster states Liu2021 . Alice prepares the state
$\rho\left(\theta\right)$ as in Eq. (12). She keeps 1,2 particles and sends
3,4 particles to Bob. In the 2-setting steering protocol
$\left\\{\hat{n}_{1},\hat{n}_{2}\right\\}$
$\left(\hat{n}_{1}\neq\hat{n}_{2}\right)$, with

$\begin{array}[c]{c}\hat{n}_{1}=\sigma_{z}\sigma_{z}\equiv zz,\\\
\hat{n}_{2}=\sigma_{y}\sigma_{x}\equiv yx.\end{array}$ (14)

In the protocol, Bob asks Alice to carry out either one of two possible
projective measurements on her qubits, i.e.,

$\begin{array}[c]{l}P_{00}^{\hat{n}_{1}}=\left|00\right\rangle\left\langle
00\right|,\\\ P_{01}^{\hat{n}_{1}}=\left|01\right\rangle\left\langle
01\right|,\\\ P_{10}^{\hat{n}_{1}}=\left|10\right\rangle\left\langle
10\right|,\\\ P_{11}^{\hat{n}_{1}}=\left|11\right\rangle\left\langle
11\right|,\\\
P_{00}^{\hat{n}_{2}}=\left|\uparrow+\right\rangle\left\langle\uparrow+\right|,\\\
P_{01}^{\hat{n}_{2}}=\left|\uparrow-\right\rangle\left\langle\uparrow-\right|,\\\
P_{10}^{\hat{n}_{2}}=\left|\downarrow+\right\rangle\left\langle\downarrow+\right|,\\\
P_{11}^{\hat{n}_{2}}=\left|\downarrow-\right\rangle\left\langle\downarrow-\right|,\end{array}$
(15)

where $|\pm\rangle=(1/\sqrt{2})(\left|0\right\rangle\pm\left|1\right\rangle)$,
$|\updownarrow\rangle=(1/\sqrt{2})(\left|0\right\rangle\pm{\rm
i}\left|1\right\rangle)$. After Alice’s measurement, Bob’s unnormalized
conditional states are

$\begin{array}[c]{l}\tilde{\rho}_{00}^{\hat{n}_{1}}=\dfrac{1}{4}\cos^{2}\theta\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{01}^{\hat{n}_{1}}=\dfrac{1}{4}\sin^{2}\theta\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{10}^{\hat{n}_{1}}=\dfrac{1}{4}\sin^{2}\theta\left(\left|00\right\rangle-\left|11\right\rangle\right)\left(\left\langle
00\right|-\left\langle 11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{11}^{\hat{n}_{1}}=\dfrac{1}{4}\cos^{2}\theta\left(\left|00\right\rangle-\left|11\right\rangle\right)\left(\left\langle
00\right|-\left\langle 11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{00}^{\hat{n}_{2}}=\dfrac{1}{8}\left(\left|00\right\rangle+{\rm
i}\left|11\right\rangle\right)\left(\left\langle 00\right|-{\rm i}\left\langle
11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{01}^{\hat{n}_{2}}=\dfrac{1}{8}\left(\left|00\right\rangle-{\rm
i}\left|11\right\rangle\right)\left(\left\langle 00\right|+{\rm i}\left\langle
11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{10}^{\hat{n}_{2}}=\dfrac{1}{8}\left(\left|00\right\rangle-{\rm
i}\left|11\right\rangle\right)\left(\left\langle 00\right|+{\rm i}\left\langle
11\right|\right),\\\\[9.0pt]
\tilde{\rho}_{11}^{\hat{n}_{2}}=\dfrac{1}{8}\left(\left|00\right\rangle+{\rm
i}\left|11\right\rangle\right)\left(\left\langle 00\right|-{\rm i}\left\langle
11\right|\right).\end{array}$ (16)

If Bob’s states have a LHS description, they must satisfy Eq. (5) and Eq. (6).
Because the eight states of Eq. (16) are pure states, it is sufficient to take
$\xi$ from $1$ to $8$. And a pure state cannot be obtained by convex
combination of other pure states, one has

$\begin{array}[c]{c}\tilde{\rho}_{00}^{\hat{n}_{1}}=\wp_{1}\rho_{1},\\\
\tilde{\rho}_{01}^{\hat{n}_{1}}=\wp_{2}\rho_{2},\\\
\tilde{\rho}_{10}^{\hat{n}_{1}}=\wp_{3}\rho_{3},\\\
\tilde{\rho}_{11}^{\hat{n}_{1}}=\wp_{4}\rho_{4},\\\
\tilde{\rho}_{00}^{\hat{n}_{2}}=\wp_{5}\rho_{5},\\\
\tilde{\rho}_{01}^{\hat{n}_{2}}=\wp_{6}\rho_{6},\\\
\tilde{\rho}_{10}^{\hat{n}_{2}}=\wp_{7}\rho_{7},\\\
\tilde{\rho}_{11}^{\hat{n}_{2}}=\wp_{8}\rho_{8}.\end{array}$ (17)

By summing eight terms up in Eq. (17) and taking trace, the left side gives
$\mathrm{tr}(\tilde{\rho}^{\hat{n}_{1}}+\tilde{\rho}^{\hat{n}_{2}})=2\mathrm{tr}\left(\rho_{B}\right)=2$.
But the right side gives
$\mathrm{tr}\big{(}{\sum_{\xi=1}^{8}}\wp_{\xi}\rho_{\xi}\big{)}=\mathrm{tr}(\rho_{B})=1$,
then one can obtained an EPR steering paradox “2=1” for the specific 4-qubit
mixed entangled state.

###  II.2 “2=1” for $N$-qubit states

In this subsection, we study the EPR steering paradox “2=1” for $N$-qubit
states. Our result is the following theorem.

######  Theorem 1.

In the 2-setting steering protocol $\left\\{\hat{n}_{1},\hat{n}_{2}\right\\}$,
Alice and Bob share an $N$-qubit state $\rho_{AB}$. Assume that Alice measures
along $\hat{n}_{1}$ and $\hat{n}_{2}$, and then Bob obtains
$\tilde{\rho}_{a}^{\hat{n}_{1}}$ and
$\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}$, respectively, where $a,a^{\prime}$
is the measurement result of Alice. There will be a contradiction of “2=1” if
$\rho_{AB}$ satisfies simultaneously “the pure state requirement” and “the
measurement requirement”. The two requirements are:

  1. 1.

The pure state requirement: Bob’s unnormalized conditional states
$\\{\tilde{\rho}_{a}^{\hat{n}_{1}}\\}$ and
$\\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\\}$ are all pure states.

  2. 2.

The measurement requirement: any one of $\\{\tilde{\rho}_{a}^{\hat{n}_{1}}\\}$
is different from any one of $\\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\\}$.

Let Alice and Bob share an $N$-qubit entangled state

$\rho_{AB}=\sum_{\alpha}p_{\alpha}\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle\left\langle\psi_{AB}^{\left(\alpha\right)}\right|.$
(18)

Alice has $M\left(M<N\right)$ particles and Bob has $(N-M)$ particles. In the
2-setting steering protocol $\left\\{\hat{n}_{1},\hat{n}_{2}\right\\}$
$\left(\text{with }\hat{n}_{1}\neq\hat{n}_{2}\right)$, Alice performs
$2^{M+1}$ projective measurements, each of them measuring $M$ particles of
Alice. For each projective measurement $\hat{P}_{a}^{\hat{n}_{k}}$, Bob
obtains the corresponding unnormalized state
$\tilde{\rho}_{a}^{\hat{n}_{k}}=\mathrm{tr}_{A}\left[\left({{P}_{a}^{\hat{n}_{k}}\otimes\mathds{1}}\right)\rho_{AB}\right]$,
with $\hat{n}_{k}$ the measurement direction, $a$ the Alice’s measurement
result, ${\mathds{1}}$ the $2^{N-M}\times 2^{N-M}$ identity matrix. The wave-
function $|\psi_{AB}^{\left(\alpha\right)}\rangle$ may be written as

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=\sum_{i}\left(s_{i+}^{\left(\alpha\right)}\left|+\phi_{i}\right\rangle\left|\eta_{i+}^{\left(\alpha\right)}\right\rangle+s_{i-}^{\left(\alpha\right)}\left|-\phi_{i}\right\rangle\left|\eta_{i-}^{\left(\alpha\right)}\right\rangle\right),$
(19)

or

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=\sum_{j}\left(t_{j+}^{\left(\alpha\right)}\left|+\varphi_{j}\right\rangle\left|\varepsilon_{j+}^{\left(\alpha\right)}\right\rangle+t_{j-}^{\left(\alpha\right)}\left|-\varphi_{j}\right\rangle\left|\varepsilon_{j-}^{\left(\alpha\right)}\right\rangle\right),$
(20)

where $i,j=1,2,\cdots,2^{M-1}$. Eq. (19) and Eq. (20) are two representations
of $\left|\psi_{AB}^{\alpha}\right\rangle$ in different bases. Here
$\left|\pm\phi_{i}\right\rangle$ and $\left|\pm\varphi_{j}\right\rangle$ are
the eigenstates of the operator $\hat{P}_{a}^{\hat{n}_{k}}$ $\left(\text{with
}k=1,2\right)$, respectively. And
$\left\\{\left|\pm\phi_{i}\right\rangle\right\\}$ and
$\left\\{\left|\pm\varphi_{j}\right\rangle\right\\}$ are two sets of complete
basis of $2^{M}$-dimensional Hilbert space, respectively. At the same time,
$\left|\pm\phi_{i}\right\rangle$ and $\left|\pm\varphi_{j}\right\rangle$ also
represent Alice’s particles. $|\eta_{i\pm}^{(\alpha)}\rangle$ and
$|\varepsilon_{j\pm}^{(\alpha)}\rangle$ are Bob’s collapsed states
(unnormalized), where

$\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle\left\langle\eta_{i\pm}^{\left(\alpha\right)}\right|=\text{tr}_{A}\bigg{[}\left(\left|\pm\phi_{i}\right\rangle\left\langle\pm\phi_{i}\right|\otimes{\mathds{1}}\right)\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle\left\langle\psi_{AB}^{\left(\alpha\right)}\right|\bigg{]},$
(21)

and

$\left|\varepsilon_{j\pm}^{\left(\alpha\right)}\right\rangle\left\langle\varepsilon_{j\pm}^{\left(\alpha\right)}\right|=\text{tr}_{A}\bigg{[}\left(\left|\pm\varphi_{j}\right\rangle\left\langle\pm\varphi_{j}\right|\otimes{\mathds{1}}\right)\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle\left\langle\psi_{AB}^{\left(\alpha\right)}\right|\bigg{]},$
(22)

where ${\mathds{1}}$ is a $2^{(N-M)}\times 2^{(N-M)}$ identity matrix.
$s_{i\pm}^{(\alpha)}$ and $t_{j\pm}^{(\alpha)}$ are complex numbers satisfying

$\sum_{i}\left(\left|s_{i+}^{\left(\alpha\right)}\right|^{2}+\left|s_{i-}^{\left(\alpha\right)}\right|^{2}\right)=1,$
(23)

and

$\sum_{j}\left(\left|t_{j+}^{\left(\alpha\right)}\right|^{2}+\left|t_{j-}^{\left(\alpha\right)}\right|^{2}\right)=1.$
(24)

_The two requirements can be rewritten as:_

  1. 1.

_The pure state requirement:_ $|\eta_{i\pm}^{(\alpha)}\rangle$_and_
$|\varepsilon_{j\pm}^{(\alpha)}\rangle$_are independent of_ $\alpha$_._

  2. 2.

_The measurement requirement: for the result obtained by Bob, any one of_
$\\{|\eta_{i\pm}\rangle\\}$_is different from any one of_
$\\{|\varepsilon_{j\pm}\rangle\\}$

The pure state requirement guarantees that Bob’s unnormalized conditional
states are all pure states. After Alice’s measurement, Bob obtains the states

$\tilde{\rho}_{a_{i\pm}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{i\pm}^{\left(\alpha\right)}\right|^{2}\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle\left\langle\eta_{i\pm}^{\left(\alpha\right)}\right|,$
(25)

and

$\tilde{\rho}_{a_{j\pm}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{j\pm}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{j\pm}^{\left(\alpha\right)}\right\rangle\left\langle\varepsilon_{j\pm}^{\left(\alpha\right)}\right|.$
(26)

Since $p_{\alpha}$, $|s_{i\pm}^{(\alpha)}|^{2}$ and
$|t_{j\pm}^{(\alpha)}|^{2}$ are all non-negative, any terms cannot cancell.
Bob’s un-normalized conditional states are pure if and only if

$\left\\{\begin{array}[c]{l}s_{i\pm}^{\left(\alpha\right)}\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle=c_{\alpha^{\prime}\left(i\pm\right)}^{\alpha}s_{i\pm}^{(\alpha^{\prime})}\left|\eta_{i\pm}^{(\alpha^{\prime})}\right\rangle,\\\\[9.0pt]
t_{j\pm}^{\left(\alpha\right)}\left|\varepsilon_{j\pm}^{\left(\alpha\right)}\right\rangle=d_{\alpha^{\prime}\left(j\pm\right)}^{\alpha}t_{j\pm}^{(\alpha^{\prime})}\left|\varepsilon_{j\pm}^{(\alpha^{\prime})}\right\rangle,\end{array}\right.$
(27)

where $c_{\alpha^{\prime}\left(i\pm\right)}^{\alpha}$ and
$d_{\alpha^{\prime}\left(j\pm\right)}^{\alpha}$ are arbitrary complex numbers
related to $i\pm$ and $j\pm$, respectively. For convenience, we analyze in the
representation with
$\left\\{\left|\pm\phi_{i}\right\rangle\otimes\left|\eta_{i\pm}\right\rangle\right\\}$
as basis. The projection of $|\psi_{AB}^{(\alpha)}\rangle$ in Eq. (19) onto
$|\pm\varphi_{j}\rangle$ is

$\displaystyle\left\langle\pm\varphi_{j^{\prime}}|\psi_{AB}^{\left(\alpha\right)}\right\rangle$
$\displaystyle=\sum_{i}s_{i\pm}^{\left(\alpha\right)}\left\langle\pm\varphi_{j^{\prime}}\left|\pm\phi_{i}\right.\right\rangle\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle,$
$\displaystyle\equiv\sum_{i}V_{\left(j^{\prime}\pm\right)\left(i\pm\right)}s_{i\pm}^{\left(\alpha\right)}\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle,$
(28)

where
$V_{(j^{\prime}\pm)(i\pm)}\equiv\langle\pm\varphi_{j^{\prime}}|\pm\phi_{i}\rangle$,
$\\{V_{(j^{\prime}\pm)(i\pm)}\\}$ can be written as

$V=\left(\begin{array}[c]{cccccc}V_{\left(1+\right)\left(1+\right)}&\cdots&V_{\left(1+\right)\left(2^{M-1}+\right)}&V_{\left(1+\right)\left(1-\right)}&\cdots&V_{\left(1+\right)\left(2^{M-1}-\right)}\\\
\vdots&&\vdots&\vdots&&\vdots\\\
V_{\left(2^{M-1}+\right)\left(1+\right)}&\cdots&V_{\left(2^{M-1}+\right)\left(2^{M-1}+\right)}&V_{\left(2^{M-1}+\right)\left(1-\right)}&\cdots&V_{\left(2^{M-1}+\right)\left(2^{M-1}-\right)}\\\
V_{\left(1-\right)\left(1+\right)}&\cdots&V_{\left(1-\right)\left(2^{M-1}+\right)}&V_{\left(1-\right)\left(1-\right)}&\cdots&V_{\left(1-\right)\left(2^{M-1}-\right)}\\\
\vdots&&\vdots&\vdots&&\vdots\\\
V_{\left(2^{M-1}-\right)\left(1+\right)}&\cdots&V_{\left(2^{M-1}-\right)\left(2^{M-1}+\right)}&V_{\left(2^{M-1}-\right)\left(1-\right)}&\cdots&V_{\left(2^{M-1}-\right)\left(2^{M-1}-\right)}\end{array}\right).$

Then we have

$t_{j^{\prime}\pm}^{\left(\alpha\right)}\left|\varepsilon_{j^{\prime}\pm}^{\left(\alpha\right)}\right\rangle=\sum_{i}V_{\left(j^{\prime}\pm\right)\left(i\pm\right)}s_{i\pm}^{\left(\alpha\right)}\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle.$
(29)

The pure state requirement Eq. (27) can be expressed as

$\left\\{\begin{array}[c]{l}s_{i\pm}^{\left(\alpha\right)}\left|\eta_{i\pm}^{\left(\alpha\right)}\right\rangle=c_{\alpha^{\prime}\left(i\pm\right)}^{\alpha}s_{i\pm}^{(\alpha^{\prime})}\left|\eta_{i\pm}^{(\alpha^{\prime})}\right\rangle,\\\
\left|\chi_{j\pm}^{\left(\alpha\right)}\right\rangle=d_{\alpha^{\prime}\left(j\pm\right)}^{\alpha}\left|\chi_{j\pm}^{\left(\alpha^{\prime}\right)}\right\rangle.\end{array}\right.$
(30)

Where
$|\chi_{j\pm}^{(\alpha)}\rangle\equiv\sum_{i}V_{(j\pm)(i\pm)}s_{i\pm}^{(\alpha)}|\eta_{i\pm}^{(\alpha)}\rangle$.

The measurement requirement suggests that if Alice chooses different
measurements $\hat{P}_{a}^{\hat{n}_{1}}$ or
$\hat{P}_{a^{\prime}}^{\hat{n}_{2}}$, Bob cannot get the same result. We prove
that in Appendix A and Appendix B the results obtained by Bob cannot be the
same in different measurements. And “the different measurements” refers to the
measurements in different directions ($\hat{n}_{1}$ and $\hat{n}_{2}$). After
Alice’s measurements, Bob obtains
$s_{i\pm}^{(\alpha)}|\eta_{i\pm}^{(\alpha)}\rangle$ and
$|\chi_{j\pm}^{(\alpha)}\rangle=\sum_{i}V_{(j\pm)(i\pm)}s_{i\pm}^{(\alpha)}|\eta_{i\pm}^{(\alpha)}\rangle$.
It can be seen that $|\chi_{j\pm}^{(\alpha)}\rangle$ is obtained by
superposition of $|\eta_{i\pm}^{(\alpha)}\rangle$. If Bob’s two results are
required to be different, $|\chi_{j\pm}^{(\alpha)}\rangle$ contains at least
two summation terms. This also requires: (1) At least two terms in the
summation in $|\psi_{AB}^{(\alpha)}\rangle$ are nonzero. That is, at least two
$s_{i\pm}^{(\alpha)}$ in $|\psi_{AB}^{(\alpha)}\rangle$ are nonzero. (2) The
matrix $\\{V_{(j\pm)(i\pm)}\\}$ has at least two nonzero matrix elements in
each row. That is, the two measurements ${{P}_{a}^{\hat{n}_{1}}}$ and
${{P}_{a^{\prime}}^{\hat{n}_{2}}}$ are different. (3)
$|\psi_{AB}^{(\alpha)}\rangle\neq\big{[}\sum_{i}(s_{i+}^{(\alpha)}|+\phi_{i}\rangle+s_{i-}^{(\alpha)}|-\phi_{i}\rangle)\big{]}\otimes|\eta_{\ell}^{(\alpha)}\rangle$,
where $|\eta_{\ell}^{(\alpha)}\rangle$ is one of
$\big{\\{}|\eta_{i\pm}^{(\alpha)}\rangle\big{\\}}$. That is, each
$|\psi_{AB}^{(\alpha)}\rangle$ is an entangled state.

###### Proof.

Here, we prove that for $N$-qubit state $\rho_{AB}$, the difference between
quantum theory and classical theory can be expressed as “2=1”, as long as the
pure state requirement and the measurement requirement are satisfied. It is
well known that if $\rho_{AB}$ satisfies two requirements at the same time,
Bob’s unnormalized conditional states are all pure states. And for different
measurements $\hat{P}_{a}^{\hat{n}_{1}}$ and
$\hat{P}_{a^{\prime}}^{\hat{n}_{2}}$, Bob ’s results are different. Without
loss of generality, we assume that Bob’s $2^{M+1}$ unnormalized conditional
states are different. Then for the quantum results we have

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{1+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{1+}\right\rangle\left\langle\eta_{1+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{2^{M-1}+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{2^{M-1}+}\right\rangle\left\langle\eta_{2^{M-1}+}\right|,\\\
\tilde{\rho}_{a_{2^{M-1}-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{2^{M-1}-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{2^{M-1}-}\right\rangle\left\langle\eta_{2^{M-1}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{1-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{1-}\right\rangle\left\langle\eta_{1-}\right|,\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{1+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{1+}\right\rangle\left\langle\varepsilon_{1+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{2^{M-1}+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{2^{M-1}+}\right\rangle\left\langle\varepsilon_{2^{M-1}+}\right|,\\\
\tilde{\rho}_{a_{2^{M-1}-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{2^{M-1}-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{2^{M-1}-}\right\rangle\left\langle\varepsilon_{2^{M-1}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{1-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{1-}\right\rangle\left\langle\varepsilon_{1-}\right|.\end{array}\right.$
(31)

Suppose Bob’s states have a LHS description, they must satisfy Eq. (5) and Eq.
(6). Then, Bob will check the following set of $2^{M+1}$ equations:

$\tilde{\rho}_{a}^{\hat{n}_{k}}={\displaystyle\sum\limits_{\xi}}\wp\left(a|\hat{n}_{k},\xi\right)\wp_{\xi}\rho_{\xi}.$
(32)

If these $2^{M+1}$ equations have a contradiction, that is they cannot have a
common solution for the sets $\\{\wp_{\xi}\rho_{\xi}\\}$ and
$\wp(a|\hat{n}_{k},\xi)$, then Bob is convinced that the LHS models are non-
existent and that Alice can steer the state of his qubits.

In the quantum result, there are $2^{M+1}$ pure states in Eq. (31). Then in
the LHS description, it is sufficient to take $\xi$ from 1 to $2^{M+1}$. It is
a fact that a density matrix of pure state can only be expanded by itself.
Therefore, any $\tilde{\rho}_{a}^{\hat{n}_{k}}$ in the equation (32) contains
only one term. So for the LHS models results we have

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\wp_{1}\rho_{1},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}}^{\hat{n}_{1}}=\wp_{2^{M-1}}\rho_{2^{M-1}},\\\
\tilde{\rho}_{a_{2^{M-1}-}}^{\hat{n}_{1}}=\wp_{2^{M-1}+1}\rho_{2^{M-1}+1},\\\
\cdots\\\ \tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\wp_{2^{M}}\rho_{2^{M}},\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}+1}\rho_{2^{M}+1},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}+2^{M-1}}\rho_{2^{M}+2^{M-1}},\\\
\tilde{\rho}_{a_{2^{M-1}-}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}+2^{M-1}+1}\rho_{2^{M}+2^{M-1}+1},\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M+1}}\rho_{2^{M+1}}.\end{array}\right.$
(33)

Finally, we sum up terms in Eq. (33) and take the trace. The left side gives
$\mathrm{tr}(\tilde{\rho}^{\hat{n}_{1}}+\tilde{\rho}^{\hat{n}_{2}})=2\mathrm{tr}(\rho_{B})=2$,
the result of quantum. While the right side gives
$\mathrm{tr}(\wp_{1}\rho_{1}+\cdots+\wp_{2^{M+1}}\rho_{2^{M+1}})=\mathrm{tr}(\rho_{B})=1$,
the result of the classical LHS models. This leads to the contradiction “2=1”,
which represents the EPR paradox in the 2-setting steering protocol. ∎

######  Remark 1.

For 2-qubit pure state Eq. (2), Bob’s unnormalized conditional states are
always pure. So only need to verify whether it satisfies the measurement
requirement. It can be seen that in the 2-setting protocol
$\\{\hat{z},\hat{x}\\}$, Bob ’s results are $\\{|0\rangle\langle
0|,|1\rangle\langle 1|\\}$ and $\\{|+\rangle\langle+|,|-\rangle\langle-|\\}$,
respectively chenjl2016 . Obviously, this satisfies the measurement
requirement. According to our analysis, such state Eq. (2) can get the
contradiction “2=1”.

######  Remark 2.

For the 4-qubit mixed state Eq. (12), it is necessary to analyze whether it
satisfies both the pure state requirement and the measurement requirement. In
the 2-setting protocol
$\\{\hat{n}_{1},\hat{n}_{2}\\}=\\{\hat{z}\hat{z},\hat{y}\hat{x}\\}$, Bob’s
eight conditional states are all pure states as shown in Eq. (16). And for two
different measurements $P_{a}^{\hat{n}_{1}}$ and
$P_{a^{\prime}}^{\hat{n}_{2}}$, Bob’s results are different. It is obvious
that the pure state requirement and the measurement requirement are satisfied
at the same time, so the specific 4-qubit mixed state Eq. (12) can also get
the contradiction “2=1”.

In the EPR steering paradox, we propose a theorem for $N$-qubit quantum states
which contains two requirements: the pure state requirement and the
measurement requirement. If Alice and Bob share an $N$-qubit mixed state,
there will be a contradiction of “2=1” only when the pure state requirement
and the measurement requirement are satisfied at the same time. If they share
an $N$-qubit pure state, the pure state requirement is automatically
satisfied. In this situation, after Alice’s measurement, Bob must get a pure
state, which only needs to meet the measurement requirement. And our results
are completely consistent with the previous conclusions. This confirms the
correctness of our conclusion.

##  III Bell-like basis measurement

Here we show a more specific example of the Bell-like basis measurement for
the $N$-qubit mixed states. Let us consider Alice and Bob share an $N$-qubit
entangled state
$\rho_{AB}=\sum_{\alpha}p_{\alpha}|\psi_{AB}^{(\alpha)}\rangle\langle\psi_{AB}^{(\alpha)}|$,
in which$|\psi_{AB}^{(\alpha)}\rangle$ may be written as

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=\sum_{i}\left(s_{i+}^{\left(\alpha\right)}\left|+\phi_{i}\right\rangle\left|\eta_{i+}^{\left(\alpha\right)}\right\rangle+s_{i-}^{\left(\alpha\right)}\left|-\phi_{i}\right\rangle\left|\eta_{i-}^{\left(\alpha\right)}\right\rangle\right).$
(34)

and $\sum_{i}(|s_{i+}^{(\alpha)}|^{2}+|s_{i-}^{(\alpha)}|^{2})=1$. Alice has
$M(M<N)$ particles and Bob has $(N-M)$ particles. In particular, in the
2-setting steering protocol $\left\\{\hat{n}_{1},\hat{n}_{2}\right\\}$
$\left(\text{with }\hat{n}_{1}\neq\hat{n}_{2}\right)$, Alice performs the
Bell-like basis measurement on her qubits. Then according to the theorem, we
analyze whether this example can obtain the contradiction “2=1”, and if yes,
what conditions $\rho_{AB}$ need to meet with the Bell-like basis measurement.

The Bell-like basis measurement can be written as

$\left\\{\begin{array}[c]{l}P_{a_{1+}}^{\hat{n}_{k}}=\left(\cos\beta_{k}\left|+\phi_{1}\right\rangle+\sin\beta_{k}\left|-\phi_{1}\right\rangle\right)\left(\cos\beta_{k}\left\langle+\phi_{1}\right|+\sin\beta_{k}\left\langle-\phi_{1}\right|\right),\\\
\vdots\\\
P_{a_{2^{M-1}+}}^{\hat{n}_{k}}=\left(\cos\beta_{k}\left|+\phi_{2^{M-1}}\right\rangle+\sin\beta_{k}\left|-\phi_{2^{M-1}}\right\rangle\right)\left(\cos\beta_{k}\left\langle+\phi_{2^{M-1}}\right|+\sin\beta_{k}\left\langle-\phi_{2^{M-1}}\right|\right),\\\
P_{a_{2^{M-1}-}}^{\hat{n}_{k}}=\left(\sin\beta_{k}\left|+\phi_{2^{M-1}}\right\rangle-\cos\beta_{k}\left|-\phi_{2^{M-1}}\right\rangle\right)\left(\sin\beta_{k}\left\langle+\phi_{2^{M-1}}\right|-\cos\beta_{k}\left\langle-\phi_{2^{M-1}}\right|\right),\\\
\vdots\\\
P_{a_{1-}}^{\hat{n}_{k}}=\left(\sin\beta_{k}\left|+\phi_{1}\right\rangle-\cos\beta_{k}\left|-\phi_{1}\right\rangle\right)\left(\sin\beta_{k}\left\langle+\phi_{1}\right|-\cos\beta_{k}\left\langle-\phi_{1}\right|\right),\end{array}\right.$
(35)

in which $\beta_{k}\in\left[0,2\pi\right)$. Alice performs the measurement
along $\hat{n}_{k}$ directions (with $k=1,2$). And $|\pm\phi_{i}\rangle$ (
$i=1,2,\cdots,2^{M-1}$) is a set of complete basis of $2^{M}$-dimensional
Hilbert space. We prove that in Appendix C. After Alice’s measurement, Bob
obtains

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{i+}}^{\hat{n}_{k}}=\sum_{\alpha}p_{\alpha}\left|\omega_{i+}^{k\left(\alpha\right)}\right\rangle\left\langle\omega_{i+}^{k\left(\alpha\right)}\right|,\\\\[6.0pt]
\tilde{\rho}_{a_{i-}}^{\hat{n}_{k}}=\sum_{\alpha}p_{\alpha}\left|\omega_{i-}^{k\left(\alpha\right)}\right\rangle\left\langle\omega_{i-}^{k\left(\alpha\right)}\right|,\end{array}\right.$
(36)

with

$\begin{array}[c]{c}\left|\omega_{i+}^{k\left(\alpha\right)}\right\rangle=s_{i+}^{\left(\alpha\right)}\cos\beta_{k}\left|\eta_{i+}^{\left(\alpha\right)}\right\rangle+s_{i-}^{\left(\alpha\right)}\sin\beta_{k}\left|\eta_{i-}^{\left(\alpha\right)}\right\rangle,\\\\[6.0pt]
\left|\omega_{i-}^{k\left(\alpha\right)}\right\rangle=s_{i+}^{\left(\alpha\right)}\sin\beta_{k}\left|\eta_{i+}^{\left(\alpha\right)}\right\rangle-
s_{i-}^{\left(\alpha\right)}\cos\beta_{k}\left|\eta_{i-}^{\left(\alpha\right)}\right\rangle.\end{array}$
(37)

Firstly, in Bell-like basis measurement, the transformation matrix
$\\{V_{(j\pm)(i\pm)}\\}$ is

$V=\left(\begin{array}[c]{cccccc}\cos\left(\beta_{1}-\beta_{2}\right)&0&\cdots&\sin\left(\beta_{1}-\beta_{2}\right)&0&\cdots\\\
0&\cos\left(\beta_{1}-\beta_{2}\right)&\cdots&0&\sin\left(\beta_{1}-\beta_{2}\right)&\cdots\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
-\sin\left(\beta_{1}-\beta_{2}\right)&0&\cdots&\cos\left(\beta_{1}-\beta_{2}\right)&0&\cdots\\\
0&-\sin\left(\beta_{1}-\beta_{2}\right)&\cdots&0&\cos\left(\beta_{1}-\beta_{2}\right)&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\end{array}\right).$ (38)

It is obvious that there are only two nonzero matrix elements $V_{jq+}$ as
well as $V_{jq-}$ (with $q=1,2,\cdots,2^{M-1}$) in each row of matrix
$\\{V_{(j\pm)(i\pm)}\\}$, and $|\chi_{j\pm}^{(\alpha)}\rangle$ contains only
two terms. Similarly, only $|\eta_{q+}^{(\alpha)}\rangle$ and
$|\eta_{q-}^{(\alpha)}\rangle$ contribute to $|\psi_{AB}^{(\alpha)}\rangle$,
and we can consider $|\psi_{AB}^{(\alpha)}\rangle$ as

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=s_{q+}^{\left(\alpha\right)}\left|+\phi_{q}\right\rangle\left|\eta_{q+}^{\left(\alpha\right)}\right\rangle+s_{q-}^{\left(\alpha\right)}\left|-\phi_{q}\right\rangle\left|\eta_{q-}^{\left(\alpha\right)}\right\rangle.$
(39)

The measurement requirement also requires
$|\psi_{AB}^{(\alpha)}\rangle\neq\big{[}\sum_{i}(s_{i+}^{(\alpha)}|+\phi_{i}\rangle+s_{i-}^{(\alpha)}|-\phi_{i}\rangle)\big{]}\otimes|\eta_{\ell}^{(\alpha)}\rangle$,
so $|\eta_{q+}^{(\alpha)}\rangle\neq|\eta_{q-}^{(\alpha)}\rangle$. In this
way, the measurement requirement is satisfied. Secondly, the pure state
requirement requires that $|\eta_{q\pm}^{(\alpha)}\rangle$ is independent of
$\alpha$, i.e.

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=s_{q+}^{\left(\alpha\right)}\left|+\phi_{q}\right\rangle\left|\eta_{q+}\right\rangle+s_{q-}^{\left(\alpha\right)}\left|-\phi_{q}\right\rangle\left|\eta_{q-}\right\rangle.$
(40)

It is apparent that after a series of analysis, the form of
$|\psi_{AB}^{(\alpha)}\rangle$ is simple and only contains two terms. There is
an interesting question worthy of our further analysis, that is, whether
$|\psi_{AB}^{(\alpha)}\rangle$ and $|\psi_{AB}^{(\alpha^{\prime})}\rangle$ can
contain the same states? Suppose that $|\psi_{AB}^{(\alpha)}\rangle$ and
$|\psi_{AB}^{(\alpha^{\prime})}\rangle$ are

$\left\\{\begin{array}[c]{l}\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=s_{p+}^{\left(\alpha\right)}\left|+\phi_{p}\right\rangle\left|\eta_{p+}\right\rangle+s_{p-}^{\left(\alpha\right)}\left|-\phi_{p}\right\rangle\left|\eta_{p-}\right\rangle,\\\\[6.0pt]
\left|\psi_{AB}^{(\alpha^{\prime})}\right\rangle=s_{p+}^{(\alpha^{\prime})}\left|+\phi_{p}\right\rangle\left|\eta_{p+}\right\rangle+s_{p-}^{(\alpha^{\prime})}\left|-\phi_{p}\right\rangle\left|\eta_{p-}\right\rangle,\end{array}\right.$
(41)

with $p\in\\{1,2,\cdots,2^{m}\\}$. The pure state requirement requires

$\left\\{\begin{array}[c]{c}s_{p+}^{(\alpha^{\prime})}=c_{\alpha^{\prime}\left(p+\right)}^{\alpha}s_{p+}^{\left(\alpha\right)},\\\
s_{p-}^{(\alpha^{\prime})}=c_{\alpha^{\prime}\left(p-\right)}^{\alpha}s_{p-}^{\left(\alpha\right)},\end{array}\right.$
(42)

and

$\begin{array}[c]{l}V_{jp+}s_{p+}^{\left(\alpha\right)}\left|\eta_{p+}\right\rangle+V_{jp-}s_{ip-}^{\left(\alpha\right)}\left|\eta_{p-}\right\rangle\\\
=d_{\alpha^{\prime}j}^{\alpha}\left(V_{jp+}s_{p+}^{(\alpha^{\prime})}\left|\eta_{p+}\right\rangle+V_{jp-}s_{p-}^{(\alpha^{\prime})}\left|\eta_{p-}\right\rangle\right).\end{array}$
(43)

The measurement requirement requires
$|\eta_{p+}^{(\alpha)}\rangle\neq|\eta_{p-}^{(\alpha)}\rangle$. According to
Eq. (42) and Eq. (43), we have

$c_{\alpha^{\prime}\left(p+\right)}^{\alpha}=c_{\alpha^{\prime}\left(p-\right)}^{\alpha}=\frac{1}{d_{\alpha^{\prime}j}^{\alpha}}\equiv
c_{\alpha^{\prime}}^{\alpha}.$ (44)

So

$\left|\psi_{AB}^{(\alpha^{\prime})}\right\rangle=c_{\alpha^{\prime}}^{\alpha}\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle,$
(45)

that means $|\psi_{AB}^{(\alpha)}\rangle$ and
$|\psi_{AB}^{(\alpha^{\prime})}\rangle$ are the same state. In summary,
$|\psi_{AB}^{(\alpha)}\rangle$ and $|\psi_{AB}^{(\alpha^{\prime})}\rangle$
cannot contain the same state.

Therefore, for arbitrary $\alpha$ and $\alpha^{\prime}$, we have

$\left\\{\begin{array}[c]{l}\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=s_{q+}^{\left(\alpha\right)}\left|+\phi_{q}\right\rangle\left|\eta_{q+}\right\rangle+s_{q-}^{\left(\alpha\right)}\left|-\phi_{q}\right\rangle\left|\eta_{q-}\right\rangle,\\\
\left|\psi_{AB}^{(\alpha^{\prime})}\right\rangle=s_{q^{\prime}+}^{(\alpha^{\prime})}\left|+\phi_{q^{\prime}}\right\rangle\left|\eta_{q^{\prime}+}\right\rangle+s_{q^{\prime}-}^{(\alpha^{\prime})}\left|-\phi_{q^{\prime}}\right\rangle\left|\eta_{q^{\prime}-}\right\rangle,\end{array}\right.$
(46)

with $q\neq q^{\prime}$ and $q,q^{\prime}=1,2,\cdots,2^{M-1}$. And after
Alice’s measurement, Bob’s results are

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{q+}}^{\hat{n}_{k}}=p_{\alpha}\left|\omega_{q+}^{k\left(\alpha\right)}\right\rangle\left\langle\omega_{q+}^{k\left(\alpha\right)}\right|,\\\\[6.0pt]
\tilde{\rho}_{a_{q-}}^{\hat{n}_{k}}=p_{\alpha}\left|\omega_{q-}^{k\left(\alpha\right)}\right\rangle\left\langle\omega_{q-}^{k\left(\alpha\right)}\right|.\end{array}\right.$
(47)

Similarly, suppose every Bob’s state has a LHS description. Bob can check the
following set of $2^{M+1}$ equations:

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{k}}={\displaystyle\sum\limits_{\xi}}\wp\left(a_{1+}|\hat{n}_{k},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}}^{\hat{n}_{k}}={\displaystyle\sum\limits_{\xi}}\wp\left(a_{2^{M-1}+}|\hat{n}_{k},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{k}}={\displaystyle\sum\limits_{\xi}}\wp\left(a_{1-}|\hat{n}_{k},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}-}}^{\hat{n}_{k}}={\displaystyle\sum\limits_{\xi}}\wp\left(a_{2^{M-1}-}|\hat{n}_{k},\xi\right)\wp_{\xi}\rho_{\xi}.\end{array}\right.$
(48)

According to Eq. (47) , Bob’s unnormalized conditional states are pure. A
density matrix of pure state can only be expanded by itself, therefore, from
Eq. (47) we have

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{k}}=\wp_{1}\rho_{1},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}}^{\hat{n}_{k}}=\wp_{2^{M-1}}\rho_{2^{M-1}},\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{k}}=\wp_{2^{M-1}+1}\rho_{2^{M-1}+1},\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}-}}^{\hat{n}_{k}}=\wp_{2^{M}}\rho_{2^{M}}.\end{array}\right.k=1,2$
(49)

By summing them up and taking the trace, the left side gives
$\mathrm{tr}(\tilde{\rho}^{\hat{n}_{1}}+\tilde{\rho}^{\hat{n}_{2}})=2\mathrm{tr}(\rho_{B})=2$.
But the right side gives
$\mathrm{tr}(\wp_{1}\rho_{1}+\cdots+\wp_{2^{M+1}}\rho_{2^{M+1}})=\mathrm{tr}(\rho_{B})=1$.
This leads to the contradiction “2=1”

In summary, we discuss the steering paradox with the Bell-like basis
measurement. And it shows that for Bell-like basis measurement, when
$\rho_{AB}$ satisfied both the pure state requirement and the measurement
requirement, we can obtain the contradiction “2=1”. In this case,
$|\psi_{AB}^{(\alpha)}\rangle$ and $|\psi_{AB}^{(\alpha^{\prime})}\rangle$
cannot contain the same state, and only contain two items. It is evident that
the maximum value range of $\alpha$ is from $1$ to $2^{M-1}$ ($M$ is the
number of particles of Alice), that is, the maximum rank of $\rho_{AB}$ is
$2^{M-1}$ for the Bell-like basis measurement.

##  IV Nonexistence of contradiction “2 = 1” for 2-qubit mixed states

In this section, we apply the results to the 2-qubit mixed states. We get the
following corollary.

######  Corollary 1.

If Alice and Bob share a 2-qubit mixed state, there is no contradiction “2 =
1”.

From the above analysis, we know that for the Bell-like basis measurement, the
maximum rank of $\rho_{AB}$ is $2^{M-1}$. Suppose Alice and Bob share a
2-qubit entangled state $\rho_{AB}$ and each of them has a qubit, then $M=1$.
If Alice performs the Bell-like basis measurement, the maximum rank of
$\rho_{AB}$ is $1$. That is, for 2-qubit entangled states with the Bell-like
basis measurement, only the $2$-qubit prue entangled states have the
contradiction “2=1”. So is there a contradiction “2=1” for 2-qubit mixed
states? Next, we prove that there is no contradiction “2 = 1” for 2-qubit
mixed states

###### Proof.

Suppose Alice and Bob share a 2-qubit mixed state
$\rho_{AB}=\sum_{\alpha}p_{\alpha}|\psi_{AB}^{(\alpha)}\rangle\langle\psi_{AB}^{(\alpha)}|$,
in which

$\left|\psi_{AB}^{\left(\alpha\right)}\right\rangle=s_{1}^{\left(\alpha\right)}\left|\phi_{1}\right\rangle\left|\eta_{1}\right\rangle+s_{2}^{\left(\alpha\right)}\left|\phi_{2}\right\rangle\left|\eta_{2}\right\rangle.$
(50)

Here we take $|\eta_{1}^{(\alpha)}\rangle$ $=$ $|\eta_{1}\rangle$ and
$|\eta_{2}^{(\alpha)}\rangle$ $=$ $|\eta_{2}\rangle$ to satisfy the pure state
requirement. And the measurement requirement requests
$s_{1}^{(\alpha)},s_{2}^{(\alpha)}\neq 0$, and
$|\eta_{1}\rangle\neq|\eta_{2}\rangle$. Alice takes one particle and the other
belongs to Bob. Similaily, in the 2-setting steering protocol
$\\{\hat{n}_{1},\hat{n}_{2}\\}$, suppose that Alice’ s projective measurements
are $\\{|\phi_{i}\rangle\langle\phi_{i}|\\}$ and
$\\{|\varphi_{j}\rangle\langle\varphi_{j}|\\}$, where
$|\langle\phi_{i}|\varphi_{j}\rangle|<1$ and $i,j=1,2$. Then Bob’s
unnormalized conditional states are

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{i}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{i}^{\left(\alpha\right)}\right|^{2}\left|\eta_{i}\right\rangle\left\langle\eta_{i}\right|,\\\\[6.0pt]
\tilde{\rho}_{a_{j}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|\chi_{j}^{\left(\alpha\right)}\right\rangle\left\langle\chi_{j}^{\left(\alpha\right)}\right|,\end{array}\right.$
(51)

in which
$|\chi_{j}^{\left(\alpha\right)}\rangle=\sum_{i}V_{ji}s_{i}^{\left(\alpha\right)}|\eta_{i}^{\left(\alpha\right)}\rangle$,
$V_{ji}=\left\langle\varphi_{j}|\phi_{i}\right\rangle$. In order to get the
contradiction, for any $\alpha$ and $\alpha^{\prime}$, the conditional state
of Bob needs to satisfy

$\left\\{\begin{array}[c]{l}\left|\chi_{j}^{\left(\alpha\right)}\right\rangle=d_{\alpha^{\prime}j}^{\alpha}\left|\chi_{j}^{(\alpha^{\prime})}\right\rangle,\\\\[6.0pt]
s_{i}^{\left(\alpha\right)}=c_{\alpha^{\prime}i}^{\alpha}s_{i}^{\left(\alpha^{\prime}\right)}.\end{array}\right.$
(52)

According to Eq. (52) , we have
$c_{\alpha^{\prime}(i)}^{\alpha}=c_{\alpha^{\prime}(i)}^{\alpha}=1/{d_{\alpha^{\prime}j}^{\alpha}}$.
Then we can obtain $|\psi_{AB}^{(\alpha)}\rangle$ and
$|\psi_{AB}^{(\alpha^{\prime})}\rangle$ are the same state, similarly.
Therefore, there is no contradiction “2=1” in 2-qubit mixed states. ∎

##  V Conclusion

We have presented a simple EPR steering paradox that shows the incompatibility
of the local-hidden-state models with quantum theory for any $N$-qubit
entangled state based on a 2-setting steering protocol. The argument is valid
for any $N$-qubit entangled state, not only $N$-qubit pure entangled states,
but more importantly, $N$-qubit mixed entangled states. We propose a simple
theorem and prove that for any $N$-qubit state satisfying simultaneously “the
pure state requirement” and “the measurement requirement”, then the
contradiction “2=1” can be obtained. In the example of Bell-like basis
measurement, we obtain that the maximum rank of the $N$-qubit mixed state is
$2^{M-1}$ ($M$ is the number of particles of Alice), and prove that there is
no contradiction “2=1” in the 2-qubit mixed state. In general, we obtain the
contradiction “2=1” in a more general case. Furthermore, if one considers the
EPR steering scenario in $k$-setting for arbitrary $N$-qudit entangled mixed
state, then following the similar approach one can arrive at a full
contradiction, i.e., “$k=1$”.

###### Acknowledgements.

J.L.C. is supported by the National Natural Science Foundation of China
(Grants No. 12275136 and 12075001) and the 111 Project of B23045. H.X.M. is
supported by the National Natural Science Foundation of China (Grant No.
11901317). Z.J.L. is supported by the Nankai Zhide Foundation.

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##  Appendix A Can Bob have the same state in the same measurement?

In order to obtain the contradiction “2=1”, we analyze whether Bob can have
the same pure state in the same measurement, and whether Bob can have the same
pure state in different measurements.

After Alice’s measurement, Bob obtains

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{1+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{1+}\right\rangle\left\langle\eta_{1+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{m+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{m+}\right\rangle\left\langle\eta_{m+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{n+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{n+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{n+}\right\rangle\left\langle\eta_{n+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{2^{M-1}+}^{\left(\alpha\right)}\right|^{2}\left|\eta_{2^{M-1}+}\right\rangle\left\langle\eta_{2^{M-1}+}\right|,\\\
\tilde{\rho}_{a_{2^{M-1}-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{2^{M-1}-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{2^{M-1}-}\right\rangle\left\langle\eta_{2^{M-1}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{n-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{n-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{n-}\right\rangle\left\langle\eta_{n-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{m-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{m-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{m-}\right\rangle\left\langle\eta_{m-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\sum_{\alpha}p_{\alpha}\left|s_{1-}^{\left(\alpha\right)}\right|^{2}\left|\eta_{1-}\right\rangle\left\langle\eta_{1-}\right|,\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{1+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{1+}\right\rangle\left\langle\varepsilon_{1+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{m^{\prime}+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|s_{m^{\prime}+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{m^{\prime}+}\right\rangle\left\langle\varepsilon_{m^{\prime}+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{n^{\prime}+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|s_{n^{\prime}+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{n^{\prime}+}\right\rangle\left\langle\varepsilon_{n^{\prime}+}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{2^{M-1}+}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{2^{M-1}+}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{2^{M-1}+}\right\rangle\left\langle\varepsilon_{2^{M-1}+}\right|,\\\
\tilde{\rho}_{a_{2^{M-1}-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{2^{M-1}-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{2^{M-1}-}\right\rangle\left\langle\varepsilon_{2^{M-1}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{n^{\prime}-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|s_{n^{\prime}-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{n^{\prime}-}\right\rangle\left\langle\varepsilon_{n^{\prime}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{m^{\prime}-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|s_{m^{\prime}-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{m^{\prime}-}\right\rangle\left\langle\varepsilon_{m^{\prime}-}\right|,\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\sum_{\alpha}p_{\alpha}\left|t_{1-}^{\left(\alpha\right)}\right|^{2}\left|\varepsilon_{1-}\right\rangle\left\langle\varepsilon_{1-}\right|.\end{array}\right.$
(53)

In Eq. (53), assuming that
$\left|\eta_{m+}\right\rangle=\left|\eta_{n+}\right\rangle$ , i.e., only
$2^{M+1}-1$ different pure states appear in the quantum result of Eq. (53), so
that it is sufficient to take $\xi$ from $1$ to $2^{M+1}-1$, namely, one can
take the ensemble as

$\left\\{\wp_{\xi}\rho_{\xi}\right\\}=\left\\{\wp_{1}\rho_{1},\wp_{2}\rho_{2},\wp_{3}\rho_{3},\cdots,\wp_{2^{M+1}-1}\rho_{2^{M+1}-1}\right\\}.$
(54)

Then Eq. (53) can be written as

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{1+}|\hat{n}_{1},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{m+}|\hat{n}_{1},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{n+}}^{\hat{n}_{1}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{n+}|\hat{n}_{1},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{1-}|\hat{n}_{1},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{1+}^{\prime}|\hat{n}_{2},\xi\right)\wp_{\xi}\rho_{\xi},\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}={\displaystyle\sum\limits_{\xi=1}^{2^{M+1}-1}}\wp\left(a_{1-}^{\prime}|\hat{n}_{2},\xi\right)\wp_{\xi}\rho_{\xi}.\end{array}\right.$
(55)

Since the $2^{M+1}$ states on the left-hand side of Eq. (53) are all pure
states, a pure state cannot be obtained by a convex sum of other different
states. Therefore,

$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\wp\left(a_{1+}|\hat{n}_{1},1\right)\wp_{1}\rho_{1},\\\
\wp\left(a_{1+}|\hat{n}_{1},2\right)=\wp\left(a_{1+}|\hat{n}_{1},3\right)=\cdots=\wp\left(a_{1+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(56)

Similarly, one has

$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{2+}}^{\hat{n}_{1}}=\wp\left(a_{2+}|\hat{n}_{1},2\right)\wp_{2}\rho_{2},\\\
\wp\left(a_{2+}|\hat{n}_{1},1\right)=\wp\left(a_{2+}|\hat{n}_{1},3\right)=\cdots=\wp\left(a_{2+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(57) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}=\wp\left(a_{m+}|\hat{n}_{1},m\right)\wp_{m}\rho_{m},\\\
\wp\left(a_{m+}|\hat{n}_{1},1\right)=\wp\left(a_{m+}|\hat{n}_{1},2\right)=\cdots=\wp\left(a_{m+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(58) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{n+}}^{\hat{n}_{1}}=\wp\left(a_{n+}|\hat{n}_{1},m\right)\wp_{m}\rho_{m},\\\
\wp\left(a_{n+}|\hat{n}_{1},1\right)=\wp\left(a_{n+}|\hat{n}_{1},2\right)=\cdots=\wp\left(a_{n+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(59) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\wp\left(a_{1-}|\hat{n}_{1},2^{M}-1\right)\wp_{2^{M}-1}\rho_{2^{M}-1},\\\
\wp\left(a_{2^{M}}|\hat{n}_{1},1\right)=\wp\left(a_{2^{M}}|\hat{n}_{1},2\right)=\cdots=\wp\left(a_{2^{M}}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(60)
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},1\right)\wp_{2^{M}}\rho_{2^{M}},\\\
\wp\left(a_{1+}^{\prime}|\hat{n}_{2},2\right)=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},3\right)=\cdots=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},2^{M+1}-1\right)=0.\end{array}\right.$
(61) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2^{M+1}-1\right)\wp_{2^{M+1}-1}\rho_{2^{M+1}-1},\\\
\wp\left(a_{1-}^{\prime}|\hat{n}_{2},1\right)=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2\right)=\cdots=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2^{M+1}-2\right)=0.\end{array}\right.$
(62)

Because ${\sum\limits_{a}}\wp(a|\hat{n},\xi)=1$, one has

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\wp_{1}\rho_{1},\\\
\cdots\\\
\tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}=\wp\left(a_{m+}|\hat{n}_{1},m\right)\wp_{m}\rho_{m},\\\
\cdots\\\
\tilde{\rho}_{a_{n+}}^{\hat{n}_{1}}=\wp\left(a_{n+}|\hat{n}_{1},n\right)\wp_{n}\rho_{n},\\\
\cdots\\\ \tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\wp_{2^{M}-1}\rho_{2^{M}-1},\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}}\rho_{2^{M}},\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M+1}-1}\rho_{2^{M+1}-1}.\end{array}\right.$
(63)

where
$\wp\left(a_{m+}|\hat{n}_{1},m\right)+\wp\left(a_{n+}|\hat{n}_{1},n\right)=1$,
The sum on the left-hand side of Eq. (63) is $2\rho_{B}$, and the sum on the
right-hand side is $\rho_{B}$. By summing terms in Eq. (63) and taking trace,
we arrive at the contradiction “2=1”. So in the same measurement, if Bob gets
the same pure state, we can get the contradiction “2=1”.

##  Appendix B Can Bob have the same state in different measurements?

In Eq. (53), assuming that
$\left|\eta_{m+}\right\rangle=\left|\varepsilon_{m^{\prime}+}\right\rangle$ ,
i.e., only $2^{M+1}-1$ different pure states appear in the quantum result of
Eq. (53). Similarly, it is sufficient to take $\xi$ from $1$ to $2^{M+1}-1$,
one can take the ensemble as Eq. (54). Then one has Eq. (55). The $2^{M+1}$
states on the left-hand side of Eq. (53) are all pure states. The same
reasoning can be used to obtain

$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1+}}^{\hat{n}_{1}}=\wp\left(a_{1+}|\hat{n}_{1},1\right)\wp_{1}\rho_{1},\\\
\wp\left(a_{1+}|\hat{n}_{1},2\right)=\wp\left(a_{1+}|\hat{n}_{1},3\right)=\cdots=\wp\left(a_{1+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(64)

Similarly, one has

$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{2+}}^{\hat{n}_{1}}=\wp\left(a_{2+}|\hat{n}_{1},2\right)\wp_{2}\rho_{2},\\\
\wp\left(a_{2+}|\hat{n}_{1},1\right)=\wp\left(a_{2+}|\hat{n}_{1},3\right)=\cdots=\wp\left(a_{2+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(65) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}=\wp\left(a_{m+}|\hat{n}_{1},m\right)\wp_{m}\rho_{m},\\\
\wp\left(a_{m+}|\hat{n}_{1},1\right)=\wp\left(a_{m+}|\hat{n}_{1},2\right)=\cdots=\wp\left(a_{m+}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(66) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\wp\left(a_{1-}|\hat{n}_{1},2^{M}\right)\wp_{2^{M}}\rho_{2^{M}},\\\
\wp\left(a_{2^{M}}|\hat{n}_{1},1\right)=\wp\left(a_{2^{M}}|\hat{n}_{1},2\right)=\cdots=\wp\left(a_{2^{M}}|\hat{n}_{1},2^{M+1}-1\right)=0.\end{array}\right.$
(67)
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},1\right)\wp_{2^{M}+1}\rho_{2^{M}+1},\\\
\wp\left(a_{1+}^{\prime}|\hat{n}_{2},2\right)=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},3\right)=\cdots=\wp\left(a_{1+}^{\prime}|\hat{n}_{2},2^{M+1}-1\right)=0.\end{array}\right.$
(68) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{{}_{m^{\prime}+}}^{\prime}}^{\hat{n}_{2}}=\wp\left(a_{{}_{m^{\prime}+}}^{\prime}|\hat{n}_{2},m^{\prime}\right)\wp_{2^{M}+m^{\prime}}\rho_{2^{M}+m^{\prime}},\\\
\wp\left(a_{1}^{\prime}|\hat{n}_{2},1\right)=\wp\left(a_{1}^{\prime}|\hat{n}_{2},2\right)=\cdots=\wp\left(a_{1}^{\prime}|\hat{n}_{2},2^{M+1}-1\right)=0.\end{array}\right.$
(69) $\cdots$
$\left.\begin{array}[c]{l}\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2^{M+1}-1\right)\wp_{2^{M+1}-1}\rho_{2^{M+1}-1},\\\
\wp\left(a_{1-}^{\prime}|\hat{n}_{2},1\right)=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2\right)=\cdots=\wp\left(a_{1-}^{\prime}|\hat{n}_{2},2^{M+1}-2\right)=0.\end{array}\right.$
(70)

Because ${\sum\limits_{a}}\wp(a|\hat{n},\xi)=1$, one has

$\left\\{\begin{array}[c]{l}\tilde{\rho}_{a_{1}}^{\hat{n}_{1}}=\wp_{1}\rho_{1},\\\
\cdots\\\ \tilde{\rho}_{a_{m+}}^{\hat{n}_{1}}=\wp_{m}\rho_{m},\\\ \cdots\\\
\tilde{\rho}_{a_{1-}}^{\hat{n}_{1}}=\wp_{2^{M}}\rho_{2^{M}},\\\
\tilde{\rho}_{a_{1+}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}+1}\rho_{2^{M}+1},\\\
\cdots\\\
\tilde{\rho}_{a_{{}_{m^{\prime}+}}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M}+m^{\prime}}\rho_{2^{M}+m^{\prime}},\\\
\cdots\\\
\tilde{\rho}_{a_{1-}^{\prime}}^{\hat{n}_{2}}=\wp_{2^{M+1}-1}\rho_{2^{M+1}-1}.\end{array}\right.$
(71)

Here $\wp(a_{m+}|\hat{n}_{1},m)=1$,
$\wp(a_{{}_{m^{\prime}+}}^{\prime}|\hat{n}_{2},m^{\prime})=1$, no more
$\wp(a_{m+}|\hat{n}_{1},m)+\wp(a_{{}_{m^{\prime}+}}^{\prime}|\hat{n}_{2},m^{\prime})=1$.
The sum on the left-hand side of Eq. (71) is $2\rho_{B}$, and the sum on the
right-hand side is $\rho_{B}+\wp_{i}\rho_{i}$. By summing terms in Eq. (71)
and taking trace, we no longer get the contradiction “2=1”. So in the
different measurements, if Bob gets the same pure state, we can’t get the
contradiction “2=1”.

##  Appendix C $\left|\pm\phi_{i}\right\rangle$ is a set of complete basis of
Hilbert space

Here we show that $|\pm\phi_{i}\rangle$ ( $i=1,2,\cdots,2^{M-1}$) which in the
Bell-like basis measurement is a set of complete basis of $2^{M}$-dimensional
Hilbert space. For any $k$( $k=1,2$), we have
$\sum_{i=1}^{2^{M-1}}(P_{a_{i+}}^{\hat{n}_{k}}+P_{a_{i-}}^{\hat{n}_{k}})=1$.
That can be expanded as

$\displaystyle{\displaystyle\sum\limits_{i=1}^{2^{M-1}}}\left(P_{a_{i+}}^{\hat{n}_{k}}+P_{a_{i-}}^{\hat{n}_{k}}\right)$
$\displaystyle=$
$\displaystyle{\displaystyle\sum\limits_{i=1}^{2^{M-1}}}\left(\cos^{2}\beta_{k}\left|+\phi_{i}\right\rangle\left\langle+\phi_{i}\right|+\cos\beta_{k}\sin\beta_{k}\left|+\phi_{i}\right\rangle\left\langle-\phi_{i}\right|+\sin\beta_{k}\cos\beta_{k}\left|-\phi_{i}\right\rangle\left\langle+\phi_{i}\right|+\sin^{2}\beta_{k}\left|-\phi_{i}\right\rangle\left\langle-\phi_{i}\right|\right.$
$\displaystyle\left.+\sin^{2}\beta_{k}\left|+\phi_{i}\right\rangle\left\langle+\phi_{i}\right|-\sin\beta_{k}\cos\beta_{k}\left|+\phi_{i}\right\rangle\left\langle-\phi_{i}\right|-\cos\beta_{k}\sin\beta_{k}\left|-\phi_{i}\right\rangle\left\langle+\phi_{i}\right|+\cos^{2}\beta_{k}\left|-\phi_{i}\right\rangle\left\langle-\phi_{i}\right|\right)$
$\displaystyle=$
$\displaystyle{\displaystyle\sum\limits_{i=1}^{2^{M-1}}}\left(\left|+\phi_{i}\right\rangle\left\langle+\phi_{i}\right|+\left|-\phi_{i}\right\rangle\left\langle-\phi_{i}\right|\right)$
$\displaystyle=$ $\displaystyle{\mathds{1}}$ (72)

It is obvious that $|\pm\phi_{i}\rangle$ is a set of complete basis of
$2^{M}$-dimensional Hilbert space.