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

1. 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?”,¹ 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.^2,3 Quantum entanglement distinguishes quantum theory from classical theory. 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.^4,5 So far, quantum steering has been widely applied in various fields.^6,7,8,9

Steering is a quantum correlation between entanglement¹⁰ and Bell nonlocality.^11,12,13 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. Quantum steering is an asymmetric quantum nonlocality. That is, in some cases Alice can steer Bob, but in turn Bob cannot steer Alice.^{14,15,16,17,18} Based on some properties of quantum steering, Chen et al. proposed the EPR steering paradox “$2=1$”,¹⁹ 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.²⁰ 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. This paper is organized as follows. In Sec. 2, we propose a theorem for N-qubit quantum states that contains two requirements: “the pure state requirement” and “the measurement requirement”. In Sec. 3, we obtain the EPR steering paradox “$2=1$” for the N-qubit states under Bell-like basis measurement. In Sec. 4, 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. 5. Some detailed proofs are given in Appendix A, B, and C.

2. EPR Steering Paradox “$2=1$” for N-Qubit States

2.1. “$2=1$” 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.¹⁹ They analyzed a 2-qubit pure entangled state given by

where with $\theta \in (0,\pi ∕2)$. They chose the 2-setting protocol as $\{ẑ,\widehat{x}\}$. 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. where $|\pm 〉=(1∕\sqrt{2})(|0〉\pm |1〉)$, and to inform him of the measurement results of “a” (where $a=0,1$). Then Bob’s four unnormalized conditional states are

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

and Here ${\wp }_{\xi }$ and $\wp (a|\widehat{n},\xi )$ are probabilities satisfying and for a fixed $\xi$, and ${\rho }_B={\mathrm{\text{tr}}}_A({\rho }_{AB})$ is Bob’s reduced density matrix.^4,5

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

Because the four states of Eq. (4) are pure states, it is sufficient to take $\xi$ from 1 to 4. Equation (9) can be written as Owing to the fact that a pure state cannot be obtained by convex combination of other pure states, one has Here and other $\wp (a|\widehat{n},\xi )=0$. By summing four terms up in Eq. (11) and taking trace, the left side gives $\mathrm{\text{tr}}({\tilde{\rho }}^{ẑ}+{\tilde{\rho }}^{\widehat{x}})=2\mathrm{\text{tr}}({\rho }_B)=2$. But the right side gives $\mathrm{\text{tr}}({\wp }_1{\rho }_1+{\wp }_2{\rho }_2+{\wp }_3{\rho }_3+{\wp }_4{\rho }_4)=\mathrm{\text{tr}}({\rho }_B)=1$, 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

where are the linear cluster states.²⁰ Alice prepares the state $\rho (\theta )$ as in Eq. (12). She keeps 1,2 particles and sends 3,4 particles to Bob. In the 2-setting steering protocol $\{{\widehat{n}}_1,{\widehat{n}}_2\}$$({\widehat{n}}_1\ne {\widehat{n}}_2)$, with In the protocol, Bob asks Alice to carry out either one of two possible projective measurements on her qubits, i.e. where $|\pm 〉=(1∕\sqrt{2})(|0〉\pm |1〉)$, $|\updownarrow 〉=(1∕\sqrt{2})(|0〉\pm \mathrm{\text{i}}|1〉)$. After Alice’s measurement, Bob’s unnormalized conditional states are

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

By summing eight terms up in Eq. (17) and taking trace, the left side gives $\mathrm{\text{tr}}({\tilde{\rho }}^{{\widehat{n}}_1}+{\tilde{\rho }}^{{\widehat{n}}_2})=2\mathrm{\text{tr}}({\rho }_B)=2$. But the right side gives $\mathrm{\text{tr}}({\sum }_{\xi =1}^8{\wp }_{\xi }{\rho }_{\xi })=\mathrm{\text{tr}}({\rho }_B)=1$, then one can obtained an EPR steering paradox “$2=1$” for the specific 4-qubit mixed entangled state.

2.2. “$2=1$” 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 $\{{\widehat{n}}_1,{\widehat{n}}_2\}$, Alice and Bob share an N-qubit state ${\rho }_{AB}$. Assume that Alice measures along ${\widehat{n}}_1$ and ${\widehat{n}}_2$, and then Bob obtains ${\tilde{\rho }}_a^{{\widehat{n}}_1}$ and ${\tilde{\rho }}_{a^{\prime }}^{{\widehat{n}}_2}$, 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)	The pure state requirement: Bob’s unnormalized conditional states $\{{\tilde{\rho }}_a^{{\widehat{n}}_1}\}$ and $\{{\tilde{\rho }}_{a^{\prime }}^{{\widehat{n}}_2}\}$ are all pure states.
(2)	The measurement requirement: any one of $\{{\tilde{\rho }}_a^{{\widehat{n}}_1}\}$ is different from any one of $\{{\tilde{\rho }}_{a^{\prime }}^{{\widehat{n}}_2}\}$.

Let Alice and Bob share an N-qubit entangled state

Alice has $M(M<N)$ particles and Bob has $(N-M)$ particles. In the 2-setting steering protocol $\{{\widehat{n}}_1,{\widehat{n}}_2\}$ (with ${\widehat{n}}_1\ne {\widehat{n}}_2$), Alice performs $2^{M+1}$ projective measurements, each of them measuring M particles of Alice. For each projective measurement ${\widehat{P}}_a^{{\widehat{n}}_k}$, Bob obtains the corresponding unnormalized state ${\tilde{\rho }}_a^{{\widehat{n}}_k}={\mathrm{\text{tr}}}_A[(P_a^{{\widehat{n}}_k}\otimes 𝟙){\rho }_{AB}]$, with ${\widehat{n}}_k$ the measurement direction, a the Alice’s measurement result, $𝟙$ the $2^{N-M}\times 2^{N-M}$ identity matrix. The wave-function $|{\psi }_{AB}^{(\alpha )}〉$ may be written as or where $i,j=1,2,\dots ,2^{M-1}$. Equations (19) and (20) are two representations of $|{\psi }_{AB}^{\alpha }〉$ in different bases. Here $|\pm {\varphi }_i〉$ and $|\pm {\phi }_j〉$ are the eigenstates of the operator ${\widehat{P}}_a^{{\widehat{n}}_k}$$(\mathrm{\text{with}}k=1,2)$, respectively. And $\{|\pm {\varphi }_i〉\}$ and $\{|\pm {\phi }_j〉\}$ are two sets of complete basis of $2^M$-dimensional Hilbert space, respectively. At the same time, $|\pm {\varphi }_i〉$ and $|\pm {\phi }_j〉$ also represent Alice’s particles. $|{\eta }_{i\pm }^{(\alpha )}〉$ and $|{\epsilon }_{j\pm }^{(\alpha )}〉$ are Bob’s collapsed states (unnormalized), where and where $𝟙$ is a $2^{(N-M)}\times 2^{(N-M)}$ identity matrix. $s_{i\pm }^{(\alpha )}$ and $t_{j\pm }^{(\alpha )}$ are complex numbers satisfying and

The two requirements can be rewritten as

(1)	The pure state requirement:$|{\eta }_{i\pm }^{(\alpha )}〉$and$|{\epsilon }_{j\pm }^{(\alpha )}〉$are independent of$\alpha$.
(2)	The measurement requirement: for the result obtained by Bob, any one of$\{|{\eta }_{i\pm }〉\}$is different from any one of$\{|{\epsilon }_{j\pm }〉\}.$

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

and Since $p_{\alpha }$, $|s_{i\pm }^{(\alpha )}{|}^2$ and $|t_{j\pm }^{(\alpha )}{|}^2$ are all non-negative, any terms cannot cancel. Bob’s un-normalized conditional states are pure if and only if where $c_{{\alpha }^{\prime }(i\pm )}^{\alpha }$ and $d_{{\alpha }^{\prime }(j\pm )}^{\alpha }$ are arbitrary complex numbers related to $i\pm$ and $j\pm$, respectively. For convenience, we analyze in the representation with $\{|\pm {\varphi }_i〉\otimes |{\eta }_{i\pm }〉\}$ as basis. The projection of $|{\psi }_{AB}^{(\alpha )}〉$ in Eq. (19) onto $|\pm {\phi }_j〉$ is

$$\begin{array}{c}〈\pm {\phi }_{j^{\prime }}|{\psi }_{AB}^{(\alpha )}〉=\sum _i{s}_{i\pm }^{(\alpha )}〈\pm {\phi }_{j^{\prime }}|\pm {\varphi }_i〉|{\eta }_{i\pm }^{(\alpha )}〉\hfill \\ \equiv \sum _i{V}_{(j^{\prime }\pm )(i\pm )}{s}_{i\pm }^{(\alpha )}|{\eta }_{i\pm }^{(\alpha )}〉,\hfill \end{array}$$(28)

where $V_{(j^{\prime }\pm )(i\pm )}\equiv 〈\pm {\phi }_{j^{\prime }}|\pm {\varphi }_i〉$, $\{V_{(j^{\prime }\pm )(i\pm )}\}$ can be written as Then we have The pure state requirement equation (27) can be expressed as where $|{\chi }_{j\pm }^{(\alpha )}〉\equiv {\sum }_i{V}_{(j\pm )(i\pm )}{s}_{i\pm }^{(\alpha )}|{\eta }_{i\pm }^{(\alpha )}〉$.

The measurement requirement suggests that if Alice chooses different measurements ${\widehat{P}}_a^{{\widehat{n}}_1}$ or ${\widehat{P}}_{a^{\prime }}^{{\widehat{n}}_2}$, 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. “The different measurements” refer to the measurements in different directions (${\widehat{n}}_1$ and ${\widehat{n}}_2$). After Alice’s measurements, Bob obtains $s_{i\pm }^{(\alpha )}|{\eta }_{i\pm }^{(\alpha )}〉$ and $|{\chi }_{j\pm }^{(\alpha )}〉={\sum }_i{V}_{(j\pm )(i\pm )}{s}_{i\pm }^{(\alpha )}|{\eta }_{i\pm }^{(\alpha )}〉$. It can be seen that $|{\chi }_{j\pm }^{(\alpha )}〉$ is obtained by superposition of $|{\eta }_{i\pm }^{(\alpha )}〉$. If Bob’s two results are required to be different, $|{\chi }_{j\pm }^{(\alpha )}〉$ contains at least two summation terms. This also requires: (1) At least two terms in the summation in $|{\psi }_{AB}^{(\alpha )}〉$ are nonzero. That is, at least two $s_{i\pm }^{(\alpha )}$ in $|{\psi }_{AB}^{(\alpha )}〉$ 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 $P_a^{{\widehat{n}}_1}$ and $P_{a^{\prime }}^{{\widehat{n}}_2}$ are different. (3) $|{\psi }_{AB}^{(\alpha )}〉\ne \left[\right.{\sum }_i(s_{i+}^{(\alpha )}|+{\varphi }_i〉+s_{i-}^{(\alpha )}|-{\varphi }_i〉)]\otimes |{\eta }_{\ell }^{(\alpha )}〉$, where $|{\eta }_{\ell }^{(\alpha )}〉$ is one of $\left\{\right.|{\eta }_{i\pm }^{(\alpha )}〉\left\}\right.$ That is, each $|{\psi }_{AB}^{(\alpha )}〉$ 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 ${\widehat{P}}_a^{{\widehat{n}}_1}$ and ${\widehat{P}}_{a^{\prime }}^{{\widehat{n}}_2}$, 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

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

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 $\wp (a|{\widehat{n}}_k,\xi )$, 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 ${\tilde{\rho }}_a^{{\widehat{n}}_k}$ in Eq. (32) contains only one term. So for the LHS models results, we have

Finally, we sum up terms in Eq. (33) and take the trace. The left side gives $\mathrm{\text{tr}}({\tilde{\rho }}^{{\widehat{n}}_1}+{\tilde{\rho }}^{{\widehat{n}}_2})=2\mathrm{\text{tr}}({\rho }_B)=2$, the result of quantum. While the right side gives $\mathrm{\text{tr}}({\wp }_1{\rho }_1+\cdots +{\wp }_{2^{M+1}}{\rho }_{2^{M+1}})=\mathrm{\text{tr}}({\rho }_B)=1$, 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 we only need to verify whether it satisfies the measurement requirement. It can be seen that in the 2-setting protocol $\{ẑ,\widehat{x}\}$, Bob’s results are $\{|0〉〈0|,|1〉〈1|\}$ and $\{|+〉〈+|,|-〉〈-|\}$, respectively.¹⁹ Obviously, this satisfies the measurement requirement. According to our analysis, such state equation (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 $\{{\widehat{n}}_1,{\widehat{n}}_2\}=\{ẑẑ,ŷ\widehat{x}\}$, Bob’s eight conditional states are all pure states as shown in Eq. (16). For two different measurements $P_a^{{\widehat{n}}_1}$ and $P_{a^{\prime }}^{{\widehat{n}}_2}$, 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 equation (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. Our results are completely consistent with the previous conclusions. This confirms the correctness of our conclusion.

3. 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 ${\rho }_{AB}={\sum }_{\alpha }{p}_{\alpha }|{\psi }_{AB}^{(\alpha )}〉〈{\psi }_{AB}^{(\alpha )}|$, in which $|{\psi }_{AB}^{(\alpha )}〉$ may be written as

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 $\{{\widehat{n}}_1,{\widehat{n}}_2\}$ (with ${\widehat{n}}_1\ne {\widehat{n}}_2$), 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

in which ${\beta }_k\in [0,2\pi )$. Alice performs the measurement along ${\widehat{n}}_k$ directions (with $k=1,2$). And $|\pm {\varphi }_i〉$ ($i=1,2,\dots ,2^{M-1}$) is a set of complete basis of $2^M$-dimensional Hilbert space. We prove that in Appendix C. After Alice’s measurement, Bob obtains with

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

It is obvious that there are only two nonzero matrix elements $V_{jq+}$ as well as $V_{jq-}$ (with $q=1,2,\dots ,2^{M-1}$) in each row of matrix $\{V_{(j\pm )(i\pm )}\}$, and $|{\chi }_{j\pm }^{(\alpha )}〉$ contains only two terms. Similarly, only $|{\eta }_{q+}^{(\alpha )}〉$ and $|{\eta }_{q-}^{(\alpha )}〉$ contribute to $|{\psi }_{AB}^{(\alpha )}〉$, and we can consider $|{\psi }_{AB}^{(\alpha )}〉$ as The measurement requirement also requires so $|{\eta }_{q+}^{(\alpha )}〉\ne |{\eta }_{q-}^{(\alpha )}〉$. In this way, the measurement requirement is satisfied. Second, the pure state requirement requires that $|{\eta }_{q\pm }^{(\alpha )}〉$ is independent of $\alpha$, i.e.

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

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

$$V_{jp+}{s}_{p+}^{(\alpha )}|{\eta }_{p+}〉+V_{jp-}{s}_{ip-}^{(\alpha )}|{\eta }_{p-}〉=d_{{\alpha }^{\prime }j}^{\alpha }(V_{jp+}{s}_{p+}^{({\alpha }^{\prime })}|{\eta }_{p+}〉+V_{jp-}{s}_{p-}^{({\alpha }^{\prime })}|{\eta }_{p-}〉).$$(43)

The measurement requirement requires $|{\eta }_{p+}^{(\alpha )}〉\ne |{\eta }_{p-}^{(\alpha )}〉$. According to Eqs. (42) and (43), we have So that means $|{\psi }_{AB}^{(\alpha )}〉$ and $|{\psi }_{AB}^{({\alpha }^{\prime })}〉$ are the same state. In summary, $|{\psi }_{AB}^{(\alpha )}〉$ and $|{\psi }_{AB}^{({\alpha }^{\prime })}〉$ cannot contain the same state.

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

with $q\ne q^{\prime }$ and $q,q^{\prime }=1,2,\dots ,2^{M-1}$. After Alice’s measurement, Bob’s results are Similarly, suppose every Bob’s state has a LHS description. Bob can check the following set of $2^{M+1}$ equations :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 By summing them up and taking the trace, the left side gives $\mathrm{\text{tr}}({\tilde{\rho }}^{{\widehat{n}}_1}+{\tilde{\rho }}^{{\widehat{n}}_2})=2\mathrm{\text{tr}}({\rho }_B)=2$. But the right side gives $\mathrm{\text{tr}}({\wp }_1{\rho }_1+\cdots +{\wp }_{2^{M+1}}{\rho }_{2^{M+1}})=\mathrm{\text{tr}}({\rho }_B)=1$. This leads to the contradiction “$2=1$”

In summary, we discuss the steering paradox with the Bell-like basis measurement. 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, $|{\psi }_{AB}^{(\alpha )}〉$ and $|{\psi }_{AB}^{({\alpha }^{\prime })}〉$ 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.

4. 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 ${\rho }_{AB}={\sum }_{\alpha }{p}_{\alpha }|{\psi }_{AB}^{(\alpha )}〉〈{\psi }_{AB}^{(\alpha )}|$, in which

Here we take $|{\eta }_1^{(\alpha )}〉=|{\eta }_1〉$ and $|{\eta }_2^{(\alpha )}〉=|{\eta }_2〉$ to satisfy the pure state requirement. The measurement requirement requests $s_1^{(\alpha )},s_2^{(\alpha )}\ne 0$, and $|{\eta }_1〉\ne |{\eta }_2〉$. Alice takes one particle and the other belongs to Bob. Similarly, in the 2-setting steering protocol $\{{\widehat{n}}_1,{\widehat{n}}_2\}$, suppose that Alice’s projective measurements are $\{|{\varphi }_i〉〈{\varphi }_i|\}$ and $\{|{\phi }_j〉〈{\phi }_j|\}$, where $|〈{\varphi }_i|{\phi }_j〉|<1$ and $i,j=1,2$. Then Bob’s unnormalized conditional states are where $|{\chi }_j^{(\alpha )}〉={\sum }_i{V}_{ji}{s}_i^{(\alpha )}|{\eta }_i^{(\alpha )}〉$, $V_{ji}=〈{\phi }_j|{\varphi }_i〉$. In order to get the contradiction, for any $\alpha$ and ${\alpha }^{\prime }$, the conditional state of Bob needs to satisfy 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 $|{\psi }_{AB}^{(\alpha )}〉$ and $|{\psi }_{AB}^{({\alpha }^{\prime })}〉$ are the same state, similarly. Therefore, there is no contradiction “$2=1$” in 2-qubit mixed states. □

5. 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. “$k=1$”.

Acknowledgments

J.L.C. is supported by the National Natural Science Foundation of China (Grant Nos. 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.

ORCID

Zhi-Jie Liu  https://orcid.org/0009-0008-0373-1796

Jie Zhou  https://orcid.org/0000-0002-7518-3716

Hui-Xian Meng  https://orcid.org/0000-0002-3807-7201

Xing-Yan Fan  https://orcid.org/0000-0002-0702-4094

Mi Xie  https://orcid.org/0000-0003-2482-121X

Fu-Lin Zhang  https://orcid.org/0000-0001-7077-1001

Jing-Ling Chen  https://orcid.org/0000-0002-3411-6015