Weak measurements, non-classicality and negative probability

Abstract

This paper establishes a direct, robust, and intimate connection among (i) non-classicality tests for various quantum features, e.g. non-Boolean logic, quantum coherence, nonlocality, quantum entanglement, quantum discord; (ii) negative probability, and, (iii) anomalous weak values. It has been shown (Adhikary et al. Eur Phys J D 74(68):68, 2020; Asthana et al. Quantum Inform Process 20(1):1–33, 2021) that the nonexistence of a classical joint probability scheme gives rise to sufficiency conditions for nonlocality, a nonclassical feature not restricted to quantum mechanics. The conditions for nonclassical features of quantum mechanics are obtained by employing pseudo-probabilities, which are expectation values of the parent pseudo-projections. The crux of the paper is that the pseudo-probabilities, which can take negative values, can be directly measured as anomalous weak values. We expect that this opens up new avenues for testing nonclassicality via weak measurements and also gives deeper insight into negative pseudo-probabilities, which become measurable. A quantum game, based on violation of a classical probability rule, is also proposed that can be played by employing weak measurements.

Appendices

Unit pseudo-projections have at least one negative eigenvalue.

Consider a unit PP, ${\varvec{\Pi }}$, constructed as the symmetrised product of N mutually non-commuting projections (of any ranks) in a Hilbert space of dimension D, as given by,

$$\begin{aligned} {\varvec{\Pi }} = \dfrac{1}{2}(\pi _1 \pi _2\ldots \pi _N + \mathrm {h. c. }). \end{aligned}$$

(58)

Let $\vert a \rangle ~ (\vert b \rangle )$ be a vector in the null space of $\pi _1 ~(\pi _N)$ but not in that of $\pi _N ~(\pi _1)$. Construct the orthogonal basis: $\{\vert e_1\rangle =\vert a \rangle$, $\vert e_2\rangle = \vert b\rangle -\langle a \vert b \rangle \vert a\rangle \}$ in the two-dimensional subspace spanned by the vectors. The determinant of the minor of ${\varvec{\Pi }}$ in this two-dimensional basis is evidently negative. Thus, ${\varvec{\Pi }}$ possesses at least one negative eigenvalue. In order to illustrate it, consider a unit PP, ${\varvec{\Pi }}$, constructed as symmetrised product of N mutually non-commuting projections $\pi _j = \sum _{i_j}|a_{i_j}\rangle \langle a_{i_j}|;~j \in \{1, \ldots , N\}$. Thus,

$$\begin{aligned} {\varvec{\Pi }} = \frac{1}{2}\sum _{i_1 \ldots i_N} \Big (|a^1_{i_1}\rangle \langle a^1_{i_1}|a^2_{i_2}\rangle \langle a^2_{i_2}|\cdots \langle a^{N-1}_{i_{N-1}}|a^N_{i_N}\rangle \langle a^N_{i_N}| +\mathrm{h.c.} \Big ). \end{aligned}$$

Let $i_1 \in \{1, \ldots , d\}$, where $d < D$. Since $\pi _1$ and $\pi _N$ are non-commuting, one can always find a state $|a^1_{d+1}\rangle$, lying in the null space of $\pi _1$, such that $\pi _N|a^1_{d+1}\rangle \ne 0$. We project ${\varvec{\Pi }}$ in the subspace spanned by $\{|a^1_d\rangle , |a^1_{d+1}\rangle \}$,

$$\begin{aligned}&\Big (|a^1_d\rangle \langle a^1_d| +|a^1_{d+1}\rangle \langle a^1_{d+1}|\Big ){\varvec{\Pi }}\Big (|a^1_d\rangle \langle a^1_d| +|a^1_{d+1}\rangle \langle a^1_{d+1}|\Big )\nonumber \\ =&\frac{1}{2}\sum _{i_1\ldots i_N}\Big \{|a^1_d\rangle \langle a^1_{i_1}|a^2_{i_2}\rangle \langle a^2_{i_2}|\ldots \langle a^{N-1}_{i_{N-1}}|a^N_{i_N}\rangle \Big (\langle |a^N_{i_N}|a^1_d\rangle \langle a^1_d| +\langle a^N_{i_N}|a^1_{d+1}\rangle \langle a^1_{d+1}|\Big ) + \mathrm{h.c.}\Big \}\nonumber \\ =&\frac{1}{2}|a^1_d\rangle \Big \{\sum _{i_1\ldots i_N}\langle a^1_{i_1}|a^2_{i_2}\rangle \langle a^2_{i_2}|\ldots \langle a^{N-1}_{i_{N-1}}|a^N_{i_N}\rangle \langle a^N_{i_N}|a^1_d\rangle + \mathrm{c.c.}\Big \}\langle a^1_d|\nonumber \\ =&\frac{1}{2}|a^1_d\rangle \Big \{\sum _{i_1\ldots i_N}\langle a^1_{i_1}|a^2_{i_2}\rangle \cdots \langle a^N_{i_{N-1}}|a^N_{i_N}\rangle \langle a^N_{i_N}|a^1_{d+1}\rangle \Big \}\langle a^1_{d+1}| +\mathrm{h.c.}, \end{aligned}$$

(59)

which has a negative determinant, thereby proving that ${\varvec{\Pi }}$ has a negative eigenvalue.

Derivation of coherence witness

In a fixed basis, all the diagonal states are considered to be incoherent and states having nonzero off-diagonal elements are designated as coherent [52]. PPs can act as witnesses for this coherence. To see this, let the state be diagonal in the eigenbasis of $\sigma _z$. We are interested in the PP, representing a joint outcome of two incompatible observables ${\varvec{\sigma }}\cdot \hat{a}_1$ and ${\varvec{\sigma }}\cdot \hat{a}_2$, which acquires negative values only for states $\rho = \frac{1}{2} ({1} + {\varvec{\sigma }}.{\varvec{p}})$ having nonzero off-diagonal term. Then, the PP, ${\varvec{\Pi }}_{a_1a_2} = \frac{1}{4} \Big ({1} + \hat{a}_1\cdot \hat{a}_2 + {\varvec{\sigma }}.(\hat{a}_1 + \hat{a}_2)\Big )$, has positive overlap with all diagonal states whenever $(\hat{a}_1 +\hat{a}_2)$ lies in the $x-y$ plane. Suppose $\hat{a}_1\cdot \hat{a}_2 = \cos \theta$ and we choose $(\hat{a}_1 + \hat{a}_2) \parallel (\cos \lambda \hat{x} + \sin \lambda \hat{y})$. Then,

$$\begin{aligned} {\varvec{\Pi }}_{a_1a_2} = \frac{1}{2}\cos \frac{\theta }{2}\Big \{ \cos \frac{\theta }{2} + (\cos \lambda \sigma _x + \sin \lambda \sigma _y)\Big \}. \end{aligned}$$

(60)

The expectation value of this operator with the state $\rho$ is,

$$\begin{aligned} \mathrm{Tr}({\varvec{\Pi }}_{a_1a_2} \rho ) = \frac{1}{2}\cos \frac{\theta }{2} \Big \{\cos \frac{\theta }{2} + (\cos \lambda p_x + \sin \lambda p_y)\Big \}. \end{aligned}$$

(61)

The operator ${\varvec{\Pi }}_{a_1a_2}$, for a proper choice of $\theta$ and $\lambda$, has negative overlap for all states with nonzero $p_x$ and $p_y$.

Derivation of CHSH inequality

The sum of pseudoprobabilities for CHSH nonlocality is as follows:

$$\begin{aligned}&{\mathcal {P}}(A_1=B_1=B_2)+{\mathcal {P}}(A_2=B_1={\bar{B}}_2). \end{aligned}$$

(62)

Employing dichotomic nature of the observables, we express each pseudoprobability in terms of expectation values of observables. Thus, the following expression results,

$$\begin{aligned} {\mathcal {P}}_\mathrm{NL} \equiv&{\mathcal {P}}(A_1=B_1=B_2)+{\mathcal {P}}(A_2=B_1={\bar{B}}_2)\nonumber \\ =&{\mathcal {P}}(A_1=+1;B_1=+1B_2=+1)+ {\mathcal {P}}(A_1=-1;B_1=-1B_2=-1)\nonumber \\ +&{\mathcal {P}}(A_2=+1;B_1=+1{B}_2=-1)+{\mathcal {P}}(A_2=-1;B_1=-1{B}_2=+1)\nonumber \\ =&\dfrac{1}{8}\Big \langle \big (1+A_1\big )\big (1+\{B_1, B_2\}+B_1+B_2\big )+\big (1-A_1\big )\big (1+\{B_1, B_2\}-B_1-B_2\big )\nonumber \\ +&\big (1+A_2\big )\big (1-\{B_1, B_2\}+B_1-B_2\big )+\big (1-A_2\big )\big (1-\{B_1, B_2\}-B_1+B_2\big )\Big \rangle \nonumber \\ =&\dfrac{1}{4}\Big \langle 2+A_1(B_1+B_2)+A_2(B_1-B_2)\Big \rangle . \end{aligned}$$

(63)

Imposing the nonclassicality criteria, ${\mathcal {P}}_\mathrm{NL}<0$, the CHSH inequality results,

$$\begin{aligned} \langle A_1(B_1+B_2)+A_2(B_1-B_2)\rangle <-2. \end{aligned}$$

(64)

For the following choice of the observables,

$$\begin{aligned}&A_1\equiv {\varvec{\sigma }}_1\cdot \hat{a}_1=\sigma _{1x}; A_2\equiv {\varvec{\sigma }}_1\cdot \hat{a}_2=\sigma _{1y};\nonumber \\&B_1\equiv \frac{\sigma _{2x}+\sigma _{2y}}{\sqrt{2}}; B_2\equiv \frac{\sigma _{2x}-\sigma _{2y}}{\sqrt{2}}, \end{aligned}$$

(65)

the values of the pseudo-probabilities, for two-qubit Werner states, $\rho = \frac{1}{4}(1-\eta {\varvec{\sigma }}_1\cdot {\varvec{\sigma }}_2)$, are given as below,

$$\begin{aligned}&{\mathcal {P}}(A_1;B_1B_2) ={\mathcal {P}}({\bar{A}}_1;{\bar{B}}_1{\bar{B}}_2)= {\mathcal {P}}(A_2;B_1{\bar{B}}_2) ={\mathcal {P}}({\bar{A}}_1;{\bar{B}}_1B_2) =\dfrac{1}{8}(1-\eta \sqrt{2}). \end{aligned}$$

Thus, all the four pseudoprobabilities turn negative for all the nonlocal Werner states ($\frac{1}{\sqrt{2}}<\eta \le 1$), thereby proving that all the weak values also turn negative, i.e. anomalous.

Derivation of linear entanglement inequalities

1.1 Expression of pseudoprojection operator for joint outcomes of two observables

First, we explicitly calculate the expression of PP representing a joint event, in which ${\varvec{\sigma }}\cdot \hat{m}_1$ and ${\varvec{\sigma }}\cdot \hat{m}_2$ both take value $+1$. The PP is given by the symmetrised rule,

$$\begin{aligned} {\varvec{\Pi }}&=\dfrac{1}{2}\Big \{\dfrac{1}{2}(1+{\varvec{\sigma }}\cdot \hat{m}_1), \dfrac{1}{2}(1+{\varvec{\sigma }}\cdot \hat{m}_2)\Big \} \nonumber \\&=\dfrac{1}{4}\Big (1+\frac{1}{2}\{{\varvec{\sigma }}\cdot \hat{m}_1, {\varvec{\sigma }}\cdot \hat{m}_2\}+{\varvec{\sigma }}\cdot (\hat{m}_1+\hat{m}_2)\Big )\nonumber \\&=\dfrac{1}{4}\Big (1+\hat{m}_1\cdot \hat{m}_2+{\varvec{\sigma }}\cdot \hat{m}_1+{\varvec{\sigma }}\cdot \hat{m}_2\Big ). \end{aligned}$$

(66)

If the included angle between $\hat{m}_1$ and $\hat{m}_2$ is given by $\alpha$, then,

$$\begin{aligned} \hat{m}_1\cdot \hat{m}_2 = \cos \alpha ~ \mathrm{and}~ \hat{m}+\hat{m}_2 = 2\cos \frac{\alpha }{2}\hat{m}. \end{aligned}$$

Here, $\hat{m}$ is a unit vector parallel to $(\hat{m}_1+\hat{m}_2)$. Plugging in these values in equation (66), we obtain,

$$\begin{aligned} {\varvec{\Pi }}&= \dfrac{1}{4}\Big (2\cos ^2\frac{\alpha }{2}+2\cos \frac{\alpha }{2}{\varvec{\sigma }}\cdot \hat{m}\Big ) \nonumber \\&= \dfrac{1}{2}\cos \frac{\alpha }{2}\Big (\cos \frac{\alpha }{2}+{\varvec{\sigma }}\cdot \hat{m}\Big ). \end{aligned}$$

(67)

We shall thoroughly use this expression to obtain various entanglement inequalities, conditions for discord and coherence witnesses. We start with the derivation of the first linear entanglement inequality.

1.2 Derivation of linear entanglement inequality: I

The sum of pseudoprobabilities for entanglement inequality is as follows:

$$\begin{aligned} {\mathcal {P}}_{E_1} = \sum _{i=1}^2{\mathcal {P}}(a_i=b^{(i)}_1=b^{(i)}_2). \end{aligned}$$

Writing each pseudo-probability in terms of expectation of PP, the following expression results,

$$\begin{aligned} {\mathcal {P}}(a_i = b_1^{(i)}&=b_2^{(i)})={\mathcal {P}}(a_i = +1;b_1^{(i)}=+1b_2^{(i)}=+1)\nonumber \\&\quad +{\mathcal {P}}(a_i =-1; b_1^{(i)}=-1b_2^{(i)}=-1)\nonumber \\&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)+(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\&=\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle . \end{aligned}$$

(68)

Substituting the values of pseudoprobabilities in terms of expectation values of observables, we obtain the following expression,

$$\begin{aligned} {\mathcal {P}}_{E_1} = \dfrac{1}{2}{\cos \frac{\alpha }{2}}\sum _{i=1}^2\big \langle 2\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle . \end{aligned}$$

(69)

Imposing the nonclassicality constraint, ${\mathcal {P}}_{E_1} <0$, the following inequality emerges,

$$\begin{aligned} \sum _{i=1}^2\big \langle 2\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle <0. \end{aligned}$$

(70)

The range of $\alpha$ is fixed since we demand all the separable states to violate the inequality. The maximum value of $\sum _{i=1}^2\langle {\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i \rangle$ for a separable state is 1. To show this, consider the two-qubit pure separable state $\rho = \dfrac{1}{4}(1+\sigma _{1z})(1+\sigma _{2z})$. Then,

$$\begin{aligned}&\sum _{i=1}^2\big \langle 2\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle _{\rho }\nonumber \\ =&2\cos \frac{\alpha }{2}+a_{1z}b_{1z}+a_{2z}b_{2z} \ge -1+2\cos \frac{\alpha }{2}. \end{aligned}$$

(71)

Note that, $\hat{a}_1\perp \hat{a}_2$ and $\hat{b}_1\perp \hat{b}_2$. It implies that $(a_{1z}b_{1z}+a_{2z}b_{2z})$ is upper bounded by 1 in magnitude. Thus, in order that the inequality (70) gets violated by all the separable states, $1 \le 2\cos \frac{\alpha }{2}< 2$, which, in turn, implies that $0\le \alpha <\frac{2\pi }{3}$.

In terms of weak values, the sum of pseudoprobabilities, ${\mathcal {P}}_{E_1}$, can be written as,

(72)

For the special choice of observables,

$$\begin{aligned} {\varvec{\sigma }}_1\cdot \hat{a}_1=\sigma _{1x}; {\varvec{\sigma }}_1\cdot \hat{a}_2=\sigma _{1y}; {\varvec{\sigma }}_2\cdot \hat{b}_1=\sigma _{2x}; {\varvec{\sigma }}_2\cdot \hat{b}_2=\sigma _{2y}, \end{aligned}$$

(73)

and for the $2\otimes 2$ Werner states, $\rho = \dfrac{1}{4}(1-\eta {\varvec{\sigma }}_1\cdot {\varvec{\sigma }}_2)$, the pseudoprobabilities and the inequality assumes the following form,

$$\begin{aligned}&{\mathcal {P}}(a_1 = +1;b_1^{(1)}=+1b_2^{(1)}=+1)= {\mathcal {P}}(a_1 =-1; b_1^{(1)}=-1b_2^{(1)}=-1) \nonumber \\ =&{\mathcal {P}}(a_2 = +1;b_1^{(2)}=+1b_2^{(2)}=+1)= {\mathcal {P}}(a_2 =-1; b_1^{(2)}=-1b_2^{(2)}=-1) \nonumber \\ =&\dfrac{1}{4}\cos \dfrac{\alpha }{2} \Big (\cos \dfrac{\alpha }{2}-\eta \Big ) \nonumber \\&2\cos \dfrac{\alpha }{2}+\langle \sigma _{1x}\sigma _{2x}+\sigma _{1y}\sigma _{2y}\rangle <0. \end{aligned}$$

(74)

We consider a particular value of $\alpha = \dfrac{2\pi }{3}$. For this value, all the four pseudoprobabilities are equal to the following value,

$$\begin{aligned}&{\mathcal {P}}(a_1 = +1;b_1^{(1)}=+1b_2^{(1)}=+1)= {\mathcal {P}}(a_1 =-1; b_1^{(1)}=-1b_2^{(1)}=-1)\nonumber \\ =&{\mathcal {P}}(a_2 = +1;b_1^{(2)}=+1b_2^{(2)}=+1)= {\mathcal {P}}(a_2 =-1; b_1^{(2)}=-1b_2^{(2)}=-1)\nonumber \\ =&\dfrac{1}{8}\Big (\dfrac{1}{2}-\eta \Big ). \end{aligned}$$

(75)

The Werner states are detected to be entangled in the range $\eta \in \Big (\dfrac{1}{2}, 1\Big ]$ by the inequality (41), which implies that all the four pseudoprobabilities will be negative. This, in turn, implies that all the four weak values are negative, i.e. anomalous.

1.3 Derivation of linear entanglement inequality: II

The sum of pseudoprobabilities for entanglement inequality is as follows:

$$\begin{aligned} {\mathcal {P}}_{E_2} = \sum _{i=1}^3{\mathcal {P}}(a_i=b^{(i)}_1=b^{(i)}_2). \end{aligned}$$

The pseudoprobabilities, as before, can be written as expectation values of Pauli operators as follows,

$$\begin{aligned} {\mathcal {P}}(a_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}\cos \frac{\alpha }{2}\Big \langle (1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\nonumber \\&\quad +(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\&=\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle . \end{aligned}$$

(76)

Substituting the values of pseudoprobabilities in terms of expectation values of observables, we obtain the following expression,

$$\begin{aligned} {\mathcal {P}}_{E_2} =\dfrac{1}{2} {\cos \frac{\alpha }{2}}\sum _{i=1}^3\big \langle 2\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle . \end{aligned}$$

(77)

The nonclassicality condition, ${\mathcal {P}}_{E_2}< 0$, yields the following inequality,

$$\begin{aligned} \sum _{i=1}^3\big \langle 2\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle <0. \end{aligned}$$

(78)

The range of $\alpha$ is fixed since we demand all the separable states to violate the inequality (78). The maximum value of $\sum _{i=1}^3\langle {\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i \rangle$ for a separable state is 1. To show this, consider the two-qubit pure separable state $\rho = \dfrac{1}{4}(1+\sigma _{1z})(1+\sigma _{2z})$. Then,

$$\begin{aligned}&\sum _{i=1}^3\big \langle 3\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\big \rangle _{\rho }\nonumber \\ =&3\cos \frac{\alpha }{2}+a_{1z}b_{1z}+a_{2z}b_{2z}+a_{3z}b_{3z} \ge -1+3\cos \frac{\alpha }{2}. \end{aligned}$$

(79)

Thus, in order that the inequality gets violated by all the separable states, $1 \le 3\cos \frac{\alpha }{2}< 3$, which, in turn, implies that $0\le \alpha <\mathrm{arccos}\big (-\frac{7}{9}\big )$.

In terms of weak values, the sum of pseudoprobabilities ${\mathcal {P}}_{E_2}$ can be written as,

(80)

For the special choices of observables,

$$\begin{aligned} {\varvec{\sigma }}_1\cdot \hat{a}_1&=\sigma _{1x}; {\varvec{\sigma }}_1\cdot \hat{a}_2=\sigma _{1y}; {\varvec{\sigma }}_1\cdot \hat{a}_3=\sigma _{1z}; {\varvec{\sigma }}_2\cdot \hat{b}_1=\sigma _{2x}; {\varvec{\sigma }}_2\cdot \hat{b}_2\nonumber \\&=\sigma _{2y}, {\varvec{\sigma }}_2\cdot \hat{b}_3=\sigma _{2z}, \end{aligned}$$

(81)

and for the $2\otimes 2$ werner states $\rho = \dfrac{1}{4}(1-\eta {\varvec{\sigma }}_1\cdot {\varvec{\sigma }}_2)$, the pseudoprobabilities and the inequality assume the following form,

$$\begin{aligned}&{\mathcal {P}}(a_1 = +1;b_1^{(1)}=+1b_2^{(1)}=+1)= {\mathcal {P}}(a_1 =-1; b_1^{(1)}=-1b_2^{(1)}=-1)\nonumber \\ =&{\mathcal {P}}(a_2 = +1;b_1^{(2)}=+1b_2^{(2)}=+1)= {\mathcal {P}}(a_2 =-1; b_1^{(2)}=-1b_2^{(2)}=-1)\nonumber \\ =&{\mathcal {P}}(a_3 = +1;b_1^{(3)}=+1b_2^{(3)}=+1)={\mathcal {P}}(a_3 =-1; b_1^{(3)}=-1b_2^{(3)}=-1)\nonumber \\ =&\dfrac{1}{4}\cos \dfrac{\alpha }{2} \Big (\cos \dfrac{\alpha }{2}-\eta \Big ) \nonumber \\&3\cos \dfrac{\alpha }{2}+\langle \sigma _{1x}\sigma _{2x}+\sigma _{1y}\sigma _{2y}+\sigma _{1z}\sigma _{2z}\rangle <0. \end{aligned}$$

(82)

We consider the case $\alpha =\mathrm{arccos} \Big (-\dfrac{7}{9}\Big )$, then all the six pseudoprobabilities are equal to the following value,

$$\begin{aligned}&{\mathcal {P}}(a_1 = +1;b_1^{(1)}=+1b_2^{(1)}=+1)= {\mathcal {P}}(a_1 =-1; b_1^{(1)}=-1b_2^{(1)}=-1)\nonumber \\ =&{\mathcal {P}}(a_2 = +1;b_1^{(2)}=+1b_2^{(2)}=+1)= {\mathcal {P}}(a_2 =-1; b_1^{(2)}=-1b_2^{(2)}=-1)\nonumber \\ =&{\mathcal {P}}(a_3 = +1;b_1^{(3)}=+1b_2^{(3)}=+1)= {\mathcal {P}}(a_3 =-1; b_1^{(3)}=-1b_2^{(3)}=-1)\nonumber \\ =&\dfrac{1}{8}\Big (\dfrac{1}{2}-\eta \Big ). \end{aligned}$$

(83)

The Werner states are detected to be entangled in the range $\eta \in \Big (\dfrac{1}{3}, 1\Big ]$ by the inequality (44), which implies that all the six pseudoprobabilities will be negative. This, in turn, implies that all the six weak values are negative, i.e. anomalous.

Derivation of nonlinear entanglement inequalities

1.1 Derivation of nonlinear entanglement inequality: I

The sum of pseudoprobabilities for entanglement inequality is as follows:

$$\begin{aligned} {\mathcal {S}}_1 = \sum _{i=1}^2\mathcal {P}(a_i=b^{(i)}_1=b^{(i)}_2)\mathcal {P}(\overline{a}_i =b^{(i)}_1=b^{(i)}_2). \end{aligned}$$

(84)

As before,

$$\begin{aligned} {\mathcal {P}}(a_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\nonumber \\&\quad +(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\ =&\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle , \end{aligned}$$

(85)

and,

$$\begin{aligned} {\mathcal {P}}({\bar{a}}_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big )\nonumber \\&\quad +\dfrac{1}{4}\Big ((1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\ =&\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle . \end{aligned}$$

(86)

Plugging in the expressions from equations (85) and (86) in Eq. (84),

$$\begin{aligned} {\mathcal {S}}_1 =\Big (\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big )^2\sum _{i=1}^2\Big (2\cos ^2\dfrac{\alpha }{2}-\langle {\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ^2\Big ). \end{aligned}$$

(87)

Imposing the non-classicality condition, ${\mathcal {S}}_1 < 0$, we arrive at the inequality,

$$\begin{aligned} 2\cos ^2\dfrac{\alpha }{2} -\sum _{i=1}^2 \langle {\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ^2 <0. \end{aligned}$$

(88)

The range of $\alpha$ can be fixed by demanding all the separable states to violate this inequality. We can consider the separable state, without any loss of generality, to be $\rho _s=\dfrac{1}{4}(1+\sigma _{1z})(1+\sigma _{2z})$. For this state, the LHS of the inequality (88) assumes the following form,

$$\begin{aligned} 2\cos ^2\dfrac{\alpha }{2} - \langle a_{1z}b_{1z}\rangle ^2-\langle a_{2z}b_{2z}\rangle ^2\ge 2\cos ^2\dfrac{\alpha }{2} - 1. \end{aligned}$$

(89)

The second step follows as, $\hat{a}_1\perp \hat{a}_2$ and $\hat{b}_1\perp \hat{b}_2$. In order that all the separable states violate this inequality, $1\le 2\cos ^2\dfrac{\alpha }{2} < 2$, which, in turn, implies that, $0< \alpha \le \frac{\pi }{2}$. In terms of weak values,

(90)

1.2 Derivation of nonlinear entanglement inequality: II

The sum of pseudoprobabilities for entanglement inequality is as follows:

$$\begin{aligned} {\mathcal {S}}_2 = \sum _{i=1}^3\mathcal {P}(a_i=b^{(i)}_1=b^{(i)}_2)\mathcal {P}(\overline{a}_i =b^{(i)}_1=b^{(i)}_2). \end{aligned}$$

(91)

As before,

$$\begin{aligned} {\mathcal {P}}(a_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\nonumber \\&\quad +(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\ =&\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle , \end{aligned}$$

(92)

and,

$$\begin{aligned} {\mathcal {P}}({\bar{a}}_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\nonumber \\&\quad +(1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\ =&\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle . \end{aligned}$$

(93)

Imposing the non-classicality condition, ${\mathcal {S}}_2 < 0$, we arrive at the inequality,

$$\begin{aligned} 3\cos ^2\dfrac{\alpha }{2} -\sum _{i=1}^3 \langle {\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ^2 <0. \end{aligned}$$

(94)

The range of $\alpha$ can be fixed by demanding all the separable states to violate this inequality. We can consider the separable state, without any loss of generality, to be $\rho _s=\dfrac{1}{4}(1+\sigma _{1z})(1+\sigma _{2z})$. For this state, the LHS of the inequality (88) assumes the following form,

$$\begin{aligned} 3\cos ^2\dfrac{\alpha }{2} - \langle a_{1z}b_{1z}\rangle ^2-\langle a_{2z}b_{2z}\rangle ^2-\langle a_{3z}b_{3z}\rangle ^2\ge 3\cos ^2\dfrac{\alpha }{2} - 1. \end{aligned}$$

(95)

The second step follows as $\hat{a}_1\perp \hat{a}_2 \perp \hat{a}_3$ and $\hat{b}_1\perp \hat{b}_2 \perp \hat{b}_3$. In order that all the separable states violate this inequality $1\le 3\cos ^2\dfrac{\alpha }{2} < 3$, which, in turn, implies $0< \alpha \le \frac{\pi }{2}$.

1.3 Proof of nonlinear entanglement inequality: III

The sum of pseudoprobabilities for entanglement inequality is as follows:

$$\begin{aligned} {\mathcal {S}}_3= & {} \sum _{i=1}^3 \Big [\mathcal {P}(a_i = b^{(i)}_1 = b^{(i)}_2) +\dfrac{1}{2}\Big \{ P(a_i)\mathcal {P}(\overline{a}^{(i)}_1, \overline{a}^{(i)}_2) \nonumber \\+ & {} P(\overline{a}_i)\mathcal {P}(a^{(i)}_1, a^{(i)}_2) + P(b_i)\mathcal {P}(\overline{b}^{(i)}_1, \overline{b}^{(i)}_2) \nonumber \\+ & {} P(\overline{b}_i)\mathcal {P}(b^{(i)}_1, b^{(i)}_2) + P(a_i)\mathcal {P}(\overline{b}^{(i)}_1, \overline{b}^{(i)}_2) \nonumber \\+ & {} P(\overline{a}_i)\mathcal {P}(b^{(i)}_1, b^{(i)}_2) + \mathcal {P}(\overline{a}^{(i)}_1, \overline{a}^{(i)}_2) P(b_i) \nonumber \\+ & {} \mathcal {P}(a^{(i)}_1, a^{(i)}_2) P(\overline{b}_i)\Big \}\Big ]. \end{aligned}$$

(96)

We next substitute the values of all the pseudoprobabilities as expectation values of corresponding PP operators as follows:

$$\begin{aligned} {\mathcal {S}}_3= & {} \sum _{i=1}^3 \dfrac{1}{2}\cos \frac{\alpha }{2}\Big (\cos \frac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big ) +\dfrac{1}{2}\Big \{\dfrac{1}{4}\cos \frac{\alpha }{2}(1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_1\cdot \hat{a}_i) \nonumber \\+ & {} \dfrac{1}{4}\cos \frac{\alpha }{2}(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i) + \dfrac{1}{4}\cos \frac{\alpha }{2}(1+{\varvec{\sigma }}_2\cdot \hat{b}_i)(\cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i) \nonumber \\+ & {} \dfrac{1}{4}\cos \frac{\alpha }{2}(1-{\varvec{\sigma }}_2\cdot \hat{b}_i)(\cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i) + \dfrac{1}{4}\cos \frac{\alpha }{2}(1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i) \nonumber \\+ & {} \dfrac{1}{4}\cos \frac{\alpha }{2}(1-{\varvec{\sigma }}_2\cdot \hat{a}_i)(\cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i) + \dfrac{1}{4}\cos \frac{\alpha }{2}(\cos \dfrac{\alpha }{2}-{\varvec{\sigma }}_1\cdot \hat{a}_i)(1+{\varvec{\sigma }}_2\cdot \hat{b}_i) \nonumber \\+ & {} \dfrac{1}{4}\cos \frac{\alpha }{2}(\cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i)(1-{\varvec{\sigma }}_2\cdot \hat{b}_i) \Big \}\nonumber \\= & {} \dfrac{1}{2}\cos \frac{\alpha }{2}\Big ( 9\cos \dfrac{\alpha }{2}+\sum _{i=1}^3{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i-\dfrac{1}{2}\langle {\varvec{\sigma }}_1\cdot \hat{a}_i+{\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ^2\Big ). \end{aligned}$$

(97)

Imposing the nonclassicality condition—${\mathcal {S}}_3 <0$, the following entanglement inequality emerges,

$$\begin{aligned} 9\cos \dfrac{\alpha }{2}+\sum _{i=1}^3\Big ({\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i-\dfrac{1}{2}\langle {\varvec{\sigma }}_1\cdot \hat{a}_i+{\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ^2\Big )<0. \end{aligned}$$

(98)

We demand that all the separable states should violate this inequality. It fixes the range of $\alpha$ to be $0 < \alpha \le \mathrm{arccos}\Big (-\frac{79}{81}\Big )$.

Proof of condition for quantum discord

The pseudoprobabilities are as follows:

$$\begin{aligned} {\mathcal {P}}_D ^{i}= & {} {\mathcal {P}}(a_i= b^{(i)}_1=b^{(i)}_2) + P(a_i) {\mathcal {P}}(\overline{b}^{(i)}_1\overline{b}^{(i)}_2) + P(\overline{a}_i) {\mathcal {P}}(b^{(i)}_1 b^{(i)}_2) ; i=1, 2.\nonumber \\ \end{aligned}$$

(99)

Note that the eigenbasis of $a_i \equiv {\varvec{\sigma }}_1\cdot \hat{a}_i$ is the same as that of the reduced density matrix of the first subsystem. We rewrite the pseudo-probabilities in terms of expectation values of Pauli observables as below,

$$\begin{aligned} {\mathcal {P}}(a_i = b_1^{(i)}=b_2^{(i)})&=\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \langle (1+{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}+{\varvec{\sigma }}_2\cdot \hat{b}_i)\nonumber \\&\quad +(1-{\varvec{\sigma }}_1\cdot \hat{a}_i)(\cos \frac{\alpha }{2}-{\varvec{\sigma }}_2\cdot \hat{b}_i)\Big \rangle \nonumber \\&=\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big \langle \cos \dfrac{\alpha }{2}+{\varvec{\sigma }}_1\cdot \hat{a}_i{\varvec{\sigma }}_2\cdot \hat{b}_i\Big \rangle , \end{aligned}$$

(100)

and,

$$\begin{aligned}&P(a_i) {\mathcal {P}}(\overline{b}_1^{(i)}\overline{b}_2^{(i)})+P(\overline{a}_i) {\mathcal {P}}(b^{(i)}_1 b^{(i)}_2)\nonumber \\&\quad =\dfrac{1}{4}{\cos \frac{\alpha }{2}}\Big \{(1+\langle {\varvec{\sigma }}_1\cdot \hat{a}_i\rangle )(\cos \frac{\alpha }{2}-\langle {\varvec{\sigma }}_2\cdot \hat{b}_i\rangle ) +{\cos \frac{\alpha }{2}}(1-\langle {\varvec{\sigma }}_1\cdot \hat{a}_i\rangle )(\cos \frac{\alpha }{2}+\langle {\varvec{\sigma }}_2\cdot \hat{b}_i\rangle )\Big \}\nonumber \\&\quad =\dfrac{1}{2}{\cos \frac{\alpha }{2}}\Big (\cos \dfrac{\alpha }{2}-\langle {\varvec{\sigma }}_1\cdot \hat{a}_i\rangle \langle {\varvec{\sigma }}_2\cdot \hat{b}_i\rangle \Big ). \end{aligned}$$

(101)

Thus,

$$\begin{aligned} {\mathcal {P}}_D ^{i}= & {} {\mathcal {P}}(a_i= b^{(i)}_1=b^{(i)}_2) + P(a_i) {\mathcal {P}}(\overline{b}^{(i)}_1\overline{b}^{(i)}_2) + P(\overline{a}_i) {\mathcal {P}}(b^{(i)}_1 b^{(i)}_2) \nonumber \\\equiv & {} \lambda \Big \{(16\lambda + 2\langle {\varvec{\sigma }}_{1}\cdot \hat{a}_i{\varvec{\sigma }}_{2}\cdot \hat{b}_i\rangle - 2 \langle {\varvec{\sigma }}_{1}\cdot \hat{a}_i\rangle \langle {\varvec{\sigma }}_{2}\cdot \hat{b}_i\rangle \Big \}, \end{aligned}$$

(102)

where $\lambda = \dfrac{1}{4}\cos \dfrac{\alpha }{2}.$ The two-qubit states having zero discord from ${\mathcal {D}}^{1\rightarrow 2}$ are of the following form:

$$\begin{aligned} \rho = \sum _{i}p_i|\phi _i\rangle \langle \phi _i|\otimes \rho _{2i}. \end{aligned}$$

(103)

Without any loss of generality, we may assume $|\phi _1\rangle = |0\rangle$ and $|\phi _2\rangle = |1\rangle$, where $|0\rangle (|1\rangle )$ are eigenvalues of $\sigma _{1z}$ with eigenvalues $+1(-1)$. Let $\rho _{2i} = \dfrac{1}{2}(1+{\varvec{\sigma }}_2\cdot \hat{c}_i)$. Thus, it is quite evident that all the correlation terms in equation (103) have the form $\sigma _{1z}{\varvec{\sigma }}_2\cdot \hat{c}_i$ and the local terms have the form $\sigma _{1z}$ and ${\varvec{\sigma }}_2\cdot \hat{c}_i$. Thus, in Eq. (102), if we choose $\hat{a}_1 =\hat{z}$ and $\hat{a}_2 = \hat{x}$ (say), only ${\mathcal {P}}^1_D$ turns negative and ${\mathcal {P}}^2_D$ is always non-negative for any value of $\alpha$ and any choice of $\hat{b}_1$ and $\hat{b}_2$ for a state of the form (103).

In order to show that there exist discordant states for which both the pseudoprobabilities, ${\mathcal {P}}_D^1$ and ${\mathcal {P}}_D^2$, are negative, consider the two-qubit Werner states $\rho _W = \frac{1}{4}(1-\alpha {\varvec{\sigma }}_1\cdot {\varvec{\sigma }}_2)$. The non-negativity of eigenvalues of $\rho _W$ demands that $\alpha \in \Big [-\dfrac{1}{3}, 1\Big ]$. For any nonzero value of $\alpha$, $\rho _W$ has nonzero discord [14]. If we choose, $\hat{a}_1 = \hat{x}, \hat{a}_2 = \hat{y}$ and $\hat{b}_1 = \hat{x}, \hat{b}_2 = \hat{y}$, both the pseudoprobabilities, ${\mathcal {P}}^1_D$ and ${\mathcal {P}}_D^2$, acquire negative values for some value of $\alpha$, which proves our claim.