Generalized Einstein-Podolsky-Rosen Steering Paradox

In order to investigate the existence of steering trait for a quantum state, we propose the generalized EPR steering paradox; the contradiction between the LHS model and EPR steering is used to determine whether one side, Alice, can steer the other side, Bob. The main result is summarized into a theorem:

Theorem 1.—In the two-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 $\hat{n}_{2}$, corresponding to the results $a$, and $a^{\prime}$; then Bob obtains the conditional states $\tilde{\rho}_{a}^{\hat{n}_{1}}$ and $\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}$, respectively. Considering the scenario where both sets of Bob’s conditional states $\{\tilde{\rho}_{a}^{\hat{n}_{1}}\}$ and $\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\}$ are all pure, there will be a contradiction of “$2_{Q}=\left(1+\delta\right)_{C}$” $\left(\text{with }0\leq\delta<1\right)$ if $\rho_{AB}$ satisfies “ the measurement requirement”, viz.,

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  Bob obtains two sets of results $\{\tilde{\rho}_{a}^{\hat{n}_{1}}\}$ and $\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\}$ that are not exactly the same, i.e., $\{\tilde{\rho}_{a}^{\hat{n}_{1}}\}\neq\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\}$.

The complete proof of the theorem is given in Appendix A. Notice that the EPR steering paradox “$2_{Q}=\left(1+\delta\right)_{C}$” mentioned in the theorem can be categorized into $\delta=0$ and $\delta\neq 0$, based on which some analysis and examples of the theorem are given below.

Case. 1—The sets of Bob’s conditional states $\{\tilde{\rho}_{a}^{\hat{n}_{1}}\}$ and $\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\}$ are all different.

Example 1.—We consider the example of an arbitrary two-qubit pure state presented by Chen et al. chenjl2016 . Alice prepares an arbitrary two-qubit pure state $\rho_{AB}=\left|\Psi\left(\theta\right)\right\rangle\left\langle\Psi\left(\theta\right)\right|$, where

with $\theta\in\left(0,\pi/2\right)$; she keeps one of the particle in Eq. (1), and sent the other to Bob. In the two-setting steering protocol $\left\{\hat{z},\hat{x}\right\}$, Alice performs four possible projective measurements

where $\left|\pm\right\rangle=1/\sqrt{2}\left(\left|0\right\rangle\pm\left|1\right\rangle\right)$. After Alice’s measurements, the four unnormalized conditional states corresponding to Bob are

where $\left|\lambda_{\pm}\right\rangle=\cos\theta\left|0\right\rangle\pm{\sin\theta}\left|1\right\rangle$. It is worth noting that Bob’s conditional states are all pure states and none of them are identical, which relate to the measurement requirement mentioned in Theorem 1.

The mathematical definition of the LHS model is given by Wiseman et al. Wiseman2007 . If the LHS model can describe the state assemblage of Bob’s unnormalized conditional states, then Bob’s state can be described by

and

Here $\wp_{\xi}$ and $\wp(a|\hat{n},\xi)$ are probabilities satisfying ${\sum_{\xi}\wp_{\xi}}=1$, and ${\sum_{a}}\wp(a|\hat{n},\xi)=1$, for a fixed $\xi$, and $\rho_{B}=\mathrm{tr}_{A}\left(\rho_{AB}\right)$ is Bob’s reduced density matrix. Because Bob’s unnormalized conditional states in Eq. (3) are four different pure states, the state ensemble can be taken as $\{\wp_{\xi}\rho_{\xi}\}=\{\wp_{1}\rho_{1},\wp_{2}\rho_{2},\wp_{3}\rho_{3},\wp_{4}\rho_{4}\}$. And it is a well-established tenet that a pure state cannot be obtained as a convex combination of other distinct states, which suggests

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 $\wp\left(a|\hat{n},\xi\right)=0$. Summing and taking the trace of Eq. (6) gives

on the left, and

on the right. Then the contradiction “$2_{Q}=1_{C}$” is obtained. In this case EPR steering paradox is formulated as “$2_{Q}=1_{C}$” with $\delta=0$.

Case. 2—Bob has the same conditional states in the same set of measurements.

Example 2.—We consider the example of a four-qubit mixed entangled state presented by Liu et al. Liu2021 . Alice prepares the state $\rho_{AB}=\cos^{2}\theta\left|LC_{4}\right\rangle\left\langle LC_{4}\right|+\sin^{2}\theta\left|LC_{4}^{\prime}\right\rangle\left\langle LC_{4}^{\prime}\right|$, where

are the linear cluster states clus2005 . She keeps the particles 1, 2 and sends the particles 3, 4 to Bob. In the two-setting steering protocol $\left\{\hat{n}_{1},\hat{n}_{2}\right\}$ $\left(\hat{n}_{1}\neq\hat{n}_{2}\right)$,

After Alice’s measurements, Bob’s unnormalized conditional states are

which are not all the same, and satisfy the requirement in Theorem 1.

If Bob’s states have a LHS description, they must satisfy Eqs. (4) and (5). Because the eight states in the set of Bob’s conditional states (11) are pure states, and there are only four different states therein; it is sufficient to take $\xi$ from $1$ to $4$. Similarly, a pure state cannot be obtained by the convex combination of other pure states, and thus one has

Since in the set Eq. (11), $\tilde{\rho}_{00}^{\hat{n}_{1}}=\tilde{\rho}_{01}^{\hat{n}_{1}}$, $\tilde{\rho}_{10}^{\hat{n}_{1}}=\tilde{\rho}_{11}^{\hat{n}_{1}}$, $\tilde{\rho}_{00}^{\hat{n}_{2}}=\tilde{\rho}_{11}^{\hat{n}_{2}}$, and $\tilde{\rho}_{01}^{\hat{n}_{2}}=\tilde{\rho}_{10}^{\hat{n}_{2}}$, in the LHS description we describe the identical terms by the same hidden state. And ${\sum_{a}}\wp\left(a|\hat{n},\xi\right)=1$, so we have $\wp\left(00|\hat{n}_{1},1\right)+\wp\left(01|\hat{n}_{1},1\right)=1$, $\wp\left(10|\hat{n}_{1},2\right)+\wp\left(11|\hat{n}_{1},2\right)=1$, $\wp\left(00|\hat{n}_{2},3\right)+\wp\left(11|\hat{n}_{2},3\right)=1$, $\wp\left(01|\hat{n}_{2},4\right)+\wp\left(10|\hat{n}_{2},4\right)=1$, and other $\wp\left(a|\hat{n},\xi\right)=0$. After taking the sum of Eq. (12) and taking the trace subsequently, we get “$2_{Q}=1_{C}$”. In the same way, we get the EPR steering paradox “$2_{Q}=1_{C}$” and $\delta=0$.

Case. 3—Bob has the same conditional states in different sets of measurements.

Example 3.—We consider the example of a 3-qubit pure entangled state $\rho_{AB}=\left|\Psi^{\prime}\right\rangle\left\langle\Psi^{\prime}\right|$, where

Alice prepares the state $\rho_{AB}$ as in Eq. (13). She keeps the particles 1, 2 and sends the particle 3 to Bob. In the two-setting steering protocol $\left\{\hat{n}_{1},\hat{n}_{2}\right\}$ $\left(\hat{n}_{1}\neq\hat{n}_{2}\right)$, with

after Alice’s measurements, Bob’s unnormalized conditional states are

Notice that all of Bob’s states are pure, while there are identical results in $\{\tilde{\rho}_{a}^{\hat{n}_{1}}\}$ and $\{\tilde{\rho}_{a^{\prime}}^{\hat{n}_{2}}\}$, which implies that Bob’s two sets of results are not identical, satisfying the requirement of Theorem 1.

Suppose Bob’s states have a LHS description, and then they must satisfy Eqs. (4) and (5). Because the eight states of Eq. (15) are pure states, and there are only six different states in the measurement result Eq. (15), it is sufficient to take $\xi$ from $1$ to $6$. Analogously, one has

Here we describe the identical terms by the same hidden states. $\wp\left(00|\hat{n}_{1},1\right)=\wp\left(01|\hat{n}_{1},2\right)=\wp\left(10|\hat{n}_{1},3\right)=\wp\left(11|\hat{n}_{1},4\right)=\wp\left(00|\hat{n}_{2},2\right)=\wp\left(01|\hat{n}_{2},3\right)=\wp\left(10|\hat{n}_{2},5\right)=\wp\left(11|\hat{n}_{2},6\right)=1$, and other $\wp\left(a|\hat{n},\xi\right)=0$. Summing Eq. (16), the quantum result is $2\rho_{B}$, while the classical result is $\rho_{B}+\wp_{2}\rho_{2}+\wp_{3}\rho_{3}$. Then taking the trace, we obtain “$2_{Q}=\left(1+\wp_{2}+\wp_{3}\right)_{C}$”. And $\sum\nolimits_{\xi=1}^{6}\wp_{\xi}=1$, $\wp_{1}$, $\wp_{2}$, $\wp_{3}$, $\wp_{4}$, $\wp_{5}$, together with $\wp_{6}$ are non-zero, thus $0<$ $\wp_{2}+\wp_{3}<1$. In this case, we get the EPR steering paradox “$2_{Q}=\left(1+\delta\right)_{C}$” and $\delta=\wp_{2}+\wp_{3}$, with $0<\delta<1$.

Case. 4—Bob has the same conditional states in the sets corresponding to both the same and different directions of Alice’s measurements.

Example 4.—We consider the W state W2000 shared by Alice and Bob. Suppose Alice prepares the W state $\rho_{AB}=\left|W\right\rangle\left\langle W\right|$, where

She keeps the particles 1, 2, and sends the other to Bob. In the two-setting steering protocol $\left\{\hat{n}_{1},\hat{n}_{2}\right\}$, where $\hat{n}_{1}$, $\hat{n}_{2}$ are shown in Eq. (14), after Alice’s measurements, Bob’s unnormalized conditional states are

namely, Bob obtains seven pure states. This satisfies the requirement in Theorem 1. Summing all the equations in Eq. (17) together, and then taking the trace, we get $\mathrm{tr}\left(2\rho_{B}\right)=2_{Q}$.

If Bob’s states have a LHS description, they satisfy Eqs. (4) and (5). In Eq. (17), there are four pure states, which means that it is sufficient to take $\xi$ from $1$ to $4$. Because the conditional states obtained by Bob are all pure states, and a pure state can only be expanded by itself, and then one selects

Perceive that in Eq. (17), $\tilde{\rho}_{00}^{zz}=\tilde{\rho}_{01}^{xx}=\tilde{\rho}_{10}^{xx}$, so we take the same hide state $\wp_{1}\rho_{1}$ to describe $\tilde{\rho}_{00}^{zz}$, $\tilde{\rho}_{01}^{xx}$, and $\tilde{\rho}_{10}^{xx}$. Similarly, $\tilde{\rho}_{01}^{zz}=\tilde{\rho}_{10}^{zz}$, and we take $\wp_{2}\rho_{2}$ to describe them. Because $\sum\nolimits_{a}\wp\left(a|\hat{n},\xi\right)=1$, we have $\wp\left(00|zz,1\right)=\wp\left(00|xx,3\right)=\wp\left(11|xx,4\right)=1$, $\wp\left(01|\hat{z}\hat{z},2\right)+\wp\left(10|\hat{z}\hat{z},2\right)=1$, $\wp\left(01|\hat{x}\hat{x},1\right)+\wp\left(10|\hat{x}\hat{x},1\right)=1$, and other $\wp\left(a|\hat{n},\xi\right)=0$. We sum up Eq. (18) and take the trace, which leads the right-hand side to $\left(1+\wp_{1}\right)_{C}$. Since $\tilde{\rho}_{00}^{zz}$, $\tilde{\rho}_{01}^{zz}$, $\tilde{\rho}_{10}^{zz}$, $\tilde{\rho}_{00}^{xx}$, $\tilde{\rho}_{01}^{xx}$, $\tilde{\rho}_{10}^{xx}$, and $\tilde{\rho}_{11}^{xx}$ are nonzero in Eq. (17), $\wp_{1}$, $\wp_{2}$, $\wp_{3}$, and $\wp_{4}$ are all nonzero, as well as $\sum\nolimits_{\xi}\wp_{\xi}=1$, which imply that $0<\wp_{1}<1$. However, the result of the left-hand side is $2$. Consequently, we obtain the paradox “$2_{Q}=\left(1+\wp_{1}\right)_{C}$”. In the case of W state, we get the contradiction “$2_{Q}=\left(1+\delta\right)_{C}$”, where $0<\delta<1$.

In Cases. 1 and 2, Bob’s two sets of results are completely different, with the contradiction “$2_{Q}=1_{C}$”, where $\delta=0$, consistent with the results in EPR paradox2024 . Notwithstanding, in Cases. 3 and 4, Bob’s two sets of results are not exactly the same, with the contradiction “$2_{Q}=\left(1+\delta\right)_{C}$”, where $0<\delta<1$. Generalized proof of each case and detailed analysis of the examples are provided in the Supplementary Information (SI) (See Supplementary Information for detail analyses of the theorem.).