Nonlinear steering criteria for arbitrary two-qubit quantum systems

Abstract

By employing Pauli measurements, we present some nonlinear steering criteria applicable for arbitrary two-qubit quantum systems and optimized ones for symmetric quantum states. These criteria provide sufficient conditions to witness steering, which can recover the previous elegant results for some well-known states. Compared with the existing linear steering criterion and entropic criterion, ours can certify more steerable states without selecting measurement settings or correlation weights, which can also be used to verify entanglement as all steerable quantum states are entangled.

Appendix: Proof of the equation \(\sum _{k=1}^{9}\lambda _{k}=\sum _{i=1}^{3}\sum _{j=1}^{3}\delta ^{2}(\sigma _{i}\otimes \sigma _{j})\)

In order to prove the equation $\sum _{k=1}^{9}\lambda _{k}=\sum _{i=1}^{3} \sum _{j=1}^{3}\delta ^{2}(\sigma _{i}\otimes \sigma _{j})$, we extend principal components analysis to quantum correlation matrix $\gamma _{mm^{'}}(\rho _{AB})$ of local observables $\{O_{m}\}=\{\sigma _{i}\otimes \sigma _{j}\} (i,j=1,2,3, m=3(i-1)+j)$. As in classical correlation analysis, the principal components on a matrix space can be expressed as

$$\begin{aligned} P_{j}=a_{1j}O_{1}+a_{2j}O_{2}+\cdots +a_{9j}O_{9}, \end{aligned}$$

(14)

where $j=1,2,...,9$. $\sum _{i}a_{ij}^{*}a_{ij}=1$, and $\sum _{i}a_{ij}^{*}a_{ik}=0$ for $j\ne k$.

To achieve the first principal component, we use the Lagrange multiplier technique to find the maximum of a function. The Lagrangian function is defined as

$$\begin{aligned} L(a)\!=\! & {} tr[\rho (a_{11}O_{1}\!+\!a_{21}O_{2}\!+\!\cdots \!+\!a_{91}O_{9})^{2}] \!-\!\{tr[\rho (a_{11}O_{1}\!+\!a_{21}O_{2}\!+\!\cdots +a_{91}O_{9})]\}^{2}\nonumber \\&+\lambda _{1}(1-a_{11}^{2}-a_{21}^{2}-\cdots -a_{91}^{2}), \end{aligned}$$

(15)

where $\lambda _{1}$ are the Lagrange multipliers. The necessary conditions for the maximum are

$$\begin{aligned} \frac{\partial L}{\partial a_{11}}=0; \frac{\partial L}{\partial a_{21}}=0;...; \frac{\partial L}{\partial a_{91}}=0. \end{aligned}$$

(16)

By using the properties of the trace, we obtain

$$\begin{aligned} \frac{\partial L}{\partial a_{i1}}= & {} 2a_{i1}tr(\rho O_{i}^{2}) -2a_{i1}[tr(\rho O_{i})]^{2}+\mathop {\sum }\limits _{k=1,...,9,k \ne i}a_{k1}[tr(\rho O_{i}O_{k})+tr(\rho O_{k}O_{i})]\nonumber \\&-\mathop {\sum }\limits _{k=1,...,9,k\ne i}a_{k1}[tr(\rho O_{i})tr(\rho O_{k})+tr(\rho O_{k})tr(\rho O_{i})]-2\lambda _{1} a_{i1}=0. \end{aligned}$$

(17)

By rearranging the above expression, we get

$$\begin{aligned}&a_{i1}[tr(\rho O_{i}^{2})-(tr(\rho O_{i}))^{2}] +\left\{ \mathop {\sum }\limits _{k=1,...,9,k\ne i}a_{k1} [tr(\rho O_{i}O_{k})+tr(\rho O_{k}O_{i})]\right\} /2\nonumber \\&-\mathop {\sum }\limits _{k=1,...,9,k\ne i}a_{k1}tr(\rho O_{i})tr(\rho O_{k})=\lambda _{1} a_{i1}. \end{aligned}$$

(18)

For $i=1,...,9$, the following eigenvalue problem is obtained in compact form:

$$\begin{aligned} \gamma {{\varvec{a}}}_{1}=\lambda _{1} {{\varvec{a}}}_{1}, \end{aligned}$$

(19)

where ${{\varvec{a}}}_{1}=(a_{11},a_{21},...,a_{91})'$, $\gamma _{ij}=(\langle O_{i}O_{j}\rangle +\langle O_{j}O_{i}\rangle )/2-\langle O_{i}\rangle \langle O_{j}\rangle$, which is exactly the quantum covariance matrix as defined in Eq.(6). It shows that ${{\varvec{a}}}_{1}$ should be chosen to be an eigenvector of the covariance matrix $\gamma$, with eigenvalue $\lambda _{1}$. The variance of the first principal component is

$$\begin{aligned} V(P_{1})=tr({{\varvec{a}}}_{1}^{\dagger } \gamma {{\varvec{a}}}_{1})=\lambda _{1}. \end{aligned}$$

(20)

Therefore, in order to obtain the maximum of the variance, ${{\varvec{a}}}_{1}$ should be chosen as the eigenvector corresponding to the largest eigenvalue $\lambda _{1}$ of the covariance matrix. Similarly, for the second principal component, in order to obtain the second maximum of the variance, ${{\varvec{a}}}_{2}$ should be chosen as the eigenvector corresponding to the second largest eigenvalue $\lambda _{2}$ of the covariance matrix. This is fully consistent with the classical principal components analysis since the variances correspond to the eigenvalues of the covariance matrix.

For a arbitrary covariance matrix $\gamma _{ij}(\rho _{AB})$ of local observables $\{O_{m}\}=\{\sigma _{i}\otimes \sigma _{j}\} (i,j=1,2,3, m=3(i-1)+j)$, the variance of the observables $O_{m}$ can be analytically given as $\sum _{m=1}^{9}\delta ^{2}(O_{m})=\sum _{i=1}^{N}\delta ^{2}P_{i}$ due to the fact $\sum _{j}a_{ij}^{*}a_{ij}=1$. As $\sum _{i=1}^{N}\delta ^{2}P_{i}=\sum _{i=1}^{N}\lambda _{i}$, one achieves $\sum _{m=1}^{9}\delta ^{2}(O_{m})=\sum _{i=1}^{9}\lambda _{i}$.