Quantifying Contextuality of Empirical Models in Terms of Trace-Distance

Abstract

In order to quantify contextuality of empirical models, the quantity of contextuality (QoC) of empirical models is introduced in terms of the trace-distance. Let Q C(e) denote the QoC of an empirical model e. The following conclusions are proved. (i) An empirical model e is non-contextual if and only if Q C(e)=0, and then it is contextual if and only if Q C(e)>0; (ii) the QoC function QC is convex, contractive and continuous. Finally, the QoC of some famous models is computed, including PM-isotropic boxes P M ^α, M-isotropic boxes M ^α, C H _n -isotropic boxes $CH_{n}^{\alpha }$ as well as K box, where α∈[0,1]. Moreover, P M ^α is non-contextual if and only if $\alpha \in [\frac {1}{6},\frac {5}{6}]$; M ^α is non-contextual if and only if $\alpha \in [0,\frac {4}{5}]$; when n is even, $CH_{n}^{\alpha }$ is non-contextual if and only if $\alpha \in [\frac {1}{n},\frac {n-1}{n}]$, and when n is odd, $CH_{n}^{\alpha }$ is non-contextual if and only if $\alpha \in [0,\frac {n-1}{n}]$. The most important thing is that it is very easy to compare the QoC of any two isotropic boxes discussed in the above.

Appendix

In this section, we show the (3.3), (3.4), (3.6), (3.7), (3.9), (3.10) and (3.13). First, we prove that (3.3) and (3.4) hold, i.e. for any α∈[0,1],it holds that

$${QC}_{u}({PM^{\alpha}})=\frac{1}{6}\min \left\{{\sum}_{i=1}^{6}{T(PM^{\alpha}_{C_{i}},PM^{\beta}_{C_{i}})}:\ PM^{\beta}\in NCEM \right\}, $$

(A1)

$${QC}({PM^{\alpha}})={QC}_{u}({PM^{\alpha}}), $$

(A2)

where $u\in \mathcal {D}(\mathcal {M})$ satisfies that $u(C_{i})=\frac {1}{6}$ for any i=1,2,…,6.

Proof

By (2.7), we obtain that ${QC}_{u}({PM^{\alpha }})\leq \frac {1}{6}\min \left \{{\sum }_{i=1}^{6}{T(PM^{\alpha }_{C_{i}},PM^{\beta }_{C_{i}})}: \ PM^{\beta }\in NCEM \right \}.$ To show (A1), it is sufficient to show that there exists P M ^β∈N C E M such that

$${QC}_{u}({PM^{\alpha}})\geq \frac{1}{6}{\sum}_{i=1}^{6}{T(PM^{\alpha}_{C_{i}},PM^{\beta}_{C_{i}})}. $$

(A3)

By (2.7), we have that there exists $d^{\ast }\in \mathcal {D}(\mathcal {E}(X))$ such that ${QC}_{u}({PM^{\alpha }})=\frac {1}{6}{\sum }_{i=1}^{6}{T(PM^{\alpha }_{C_{i}},d^{\ast }|{C_{i}})}.$ From Tables 4 and 5, we see that when $h_{i}\in \mathcal {L}_{PM}$, $h_{i}^{-1}=h_{i}\in \mathcal {L}_{PM}$. By Theorem 2 in [25, Appendix], we see that $\mathcal {G}_{\mathcal {L}_{PM}}$ is a finite group and

$$\mathcal{I}_{\mathcal{L}_{PM}}=\{B\in CSB:h_{i}(B)=B,\forall i\}. $$

(A4)

Lemma 3 in [25, Appendix] said that

$$\mathcal{I}_{\mathcal{L}_{PM}}=\{PM^{\beta}:\beta\in[0,1]\}. $$

(A5)

Since $\mathcal {G}_{\mathcal {L}_{PM}}$ is a finite group generated by h _i (1≤i≤8), (A4) and (A5) imply that

$$\ell(PM^{\beta})=PM^{\beta},\ \forall \ell\in \mathcal{G}_{\mathcal{L}_{PM}}, \beta\in[0,1].$$

Since $\{d^{\ast }|C_{i}\}_{i=1}^{6}$ is non-contextual and each $\ell \in \mathcal {G}_{\mathcal {L}_{PM}}$ is non-contextuality preserving, it follows from [31, Theorem 2.2] that the convex combination

$$e:={\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{PM}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{PM}}|} \ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})$$

is also non-contextual. On the other hand, we see from (3.2) that $e\in \mathcal {I}_{\mathcal {L}_{PM}}$ and so (A5) implies that there exists β ₀∈[0,1] such that $PM^{\beta _{0}}=e$, which is non-contextual. Since every element of $\mathcal {L}_{PM}$ is a composition of the following two types of linear maps: (i) π _i -permutations of observables, and (ii) b _i -negations of outputs of observables, and $\mathcal {G}_{\mathcal {L}_{PM}}$ is a finite group generated by $\mathcal {L}_{PM}$, we obtain that

$${QC}_{u}({PM^{\alpha}})\,=\,\frac{1}{6}{\sum}_{i=1}^{6}{T(PM^{\alpha}_{C_{i}},d^{\ast}|C_{i})}\,=\,\frac{1}{6}{\sum}_{i=1}^{6} T\left( [\ell(PM^{\alpha})]_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})]_{C_{i}}\right), \ \forall \ell\!\in\! \mathcal{G}_{\mathcal{L}_{PM}}. $$

Combining the fact that ℓ(P M ^α) = P M ^α, we compute that

$$\begin{array}{@{}rcl@{}} {QC}_{u}({PM^{\alpha}}) &=&\frac{1}{6}{\sum}_{i=1}^{6} T\left( [\ell(PM^{\alpha})]_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})]_{C_{i}}\right)\\ &=&\frac{1}{6}{\sum}_{i=1}^{6} T\left( PM^{\alpha}_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})]_{C_{i}}\right)\\ &=&\frac{1}{6}{\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{PM}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{PM}}|}{\sum}_{i=1}^{6} T\left( PM^{\alpha}_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})]_{C_{i}}\right)\\ &=&\frac{1}{6}{\sum}_{i=1}^{6}{\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{PM}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{PM}}|} T\left( PM^{\alpha}_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})]_{C_{i}}\right)\\ &\geq&\frac{1}{6}{\sum}_{i=1}^{6} T\left( PM^{\alpha}_{C_{i}}, \left[{\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{PM}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{PM}}|} \ell(\{d^{\ast}|C_{i}\}_{i=1}^{6})\right]_{C_{i}}\right)\\ &=&\frac{1}{6}{\sum}_{i=1}^{6} T\left( PM^{\alpha}_{C_{i}},e_{C_{i}}\right)\\ &=&\frac{1}{6}{\sum}_{i=1}^{6}T\left( PM^{\alpha}_{C_{i}},PM^{\beta_{0}}_{C_{i}})\right), \end{array} $$

where the inequality holds because of the convexity of trace-distance, and so inequality (3) holds. Thus, (A1) holds. Since $T({PM}^{\alpha }_{C_{i}},{PM}^{\beta }_{C_{i}})=\frac { 1}{2}\left (\frac {4|\alpha -\beta |}{4}+\frac {4|1-\alpha -(1-\beta )|}{4}\right )=|\alpha -\beta |$ for any α,β∈[0,1] and $C_{i}\in \mathcal {M}$, we obtain by (A1) that

$${QC}_{u}({PM}^{\alpha})=\min \left\{|\alpha-\beta|:\ PM^{\beta}\in NCEM \right\}. $$

Hence, there exists $\widetilde {\beta }\in [0,1]$ such that $PM^{\widetilde {\beta }}\in NCEM$ and ${QC}_{u}({PM}^{\alpha })=\left |\alpha -\widetilde {\beta }\right |.$ By (2.7), we obtain that

$${QC}_{q}({PM}^{\alpha})\!\leq\! {\sum}_{i=1}^{6}q(C_{i})T(PM^{\alpha}_{C_{i}},PM^{\beta}_{C_{i}})\,=\, {\sum}_{i=1}^{6}q(C_{i})\left|\alpha-\widetilde{\beta}\right|=\left|\alpha-\widetilde{\beta}\right|={QC}_{u}({PM}^{\alpha}) $$

for any $q\in \mathcal {D}(\mathcal {M}).$ Now, (2.8) yields that (A2) holds. □

By similarly discussing, we can obtain that (3.6), (3.7), (3.9), (3.10) hold. Next, we shall prove that (3.13) holds, i.e.

$${QC}({K})={QC}_{u}({K})=\min_{d}\ T(K_{C_{1}},d|C_{1}),$$

where $u(C_{i})=\frac {1}{5}$ for any i=1,2,…,5 and the min was taken over all $d\in \mathcal {D}(\mathcal {E}(X))$ such that d is a convex combination of 4 distributions d ¹,d ²,d ³ and d ⁴ with

$$\begin{array}{@{}rcl@{}} d^{1}|C_{i}(00)&=&1,\ d^{2}|C_{i}(11)=1,\ d^{3}|C_{i}(00)=\frac{1}{5},\ d^{3}|C_{i}(01)=d^{3}|C_{i}(1 0)=\frac{2}{5}, \\d^{4}|C_{i}(11)&=&\frac{1}{5},\ d^{4}|C_{i}(01)=d^{4}|C_{i}(1 0)=\frac{2}{5} \end{array} $$

for all i.

Proof

By (2.7), we obtain that

$$\begin{array}{@{}rcl@{}} {QC}_{u}(K)&\leq& \frac{1}{5}\min \left\{{\sum}_{i=1}^{5}{T(K_{C_{i}},d|{C_{i}})} :d=xd^{1}+yd^{2}+zd^{3}\right.\\ &&\left.+\eta d^{4},x,y,z,\eta\in[0,1], x+y+z+\eta=1 \right\}. \end{array} $$

To show the second equality in (3.13), it suffices to show that there exist x ₀,y ₀,z ₀,η ₀∈[0,1] such that x ₀ + y ₀ + z ₀ + η ₀=1 and

$${QC}_{u}(K)\geq \frac{1}{5}{\sum}_{i=1}^{5}{T(K_{C_{i}},d_{0}|{C_{i}})}, $$

(A6)

with d ₀ = x ₀ d ¹ + y ₀ d ² + z ₀ d ³ + η ₀ d ⁴. By (2.7), we obtain that there exists $d^{\ast }\in \mathcal {D}(\mathcal {E}(X))$ such that ${QC}_{u}(K)=\frac {1}{5}{\sum }_{i=1}^{5}{T(K_{C_{i}},d^{\ast }|{C_{i}})}.$ Let $\mathcal {L}_{K}$ be the dihedral group D ₅ consisting of d=10 permutations [25, Appendix E, Section 4]. Then group $\mathcal {G}_{\mathcal {L}_{K}}=D_{5}$. Section 4 in [25, Appendix E] said that

$$\begin{array}{@{}rcl@{}} \mathcal{I}_{\mathcal{L}_{K}}\bigcap NCEM&=&\left\{\{d|C_{i}\}_{i=1}^{5}: d=xd^{1}+yd^{2}+zd^{3}\right.\\ &&\left.+\eta d^{4},x,y,z,\eta\in[0,1], x+y+z+\eta=1\right\}. \end{array} $$

(A7)

Since $\{d^{\ast }|C_{i}\}_{i=1}^{5}$ is non-contextual and each $\ell \in \mathcal {G}_{\mathcal {L}_{K}}$ is non-contextuality preserving, it follows from [31, Theorem 2.2] that the convex combination ${\sum }_{\ell \in \mathcal {G}_{\mathcal {L}_{K}}}\frac {1}{|\mathcal {G}_{\mathcal {L}_{K}}|} \ell (\{d^{\ast }|C_{i}\}_{i=1}^{5})$ is also non-contextual. By (3.2), we see that

$${\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{K}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{K}}|} \ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})\in \mathcal{I}_{\mathcal{L}_{K}}.$$

By (A7), there exist x ₀,y ₀,z ₀,η ₀∈[0,1] with x ₀ + y ₀ + z ₀ + η ₀=1 and d ₀ = x ₀ d ¹ + y ₀ d ² + z ₀ d ³ + η ₀ d ⁴ such that

$$\{d_{0}|C_{i}\}_{i=1}^{5}= {\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{K}}} \frac{1}{|\mathcal{G}_{\mathcal{L}_{K}}|}\ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})\in NCEM.$$

Since $\mathcal {L}_{K}$ is a set of permutations of observables, and $\mathcal {G}_{\mathcal {L}_{K}}$ is a finite group generated by $\mathcal {L}_{K}$, we obtain that

$${\sum}_{i=1}^{5}{T(K_{C_{i}},d^{\ast}|C_{i})}={\sum}_{i=1}^{5} T\left( [\ell(K)]_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})]_{C_{i}}\right),\ \forall \ell\in \mathcal{G}_{\mathcal{L}_{K}}.$$

Based on the fact that ℓ(K) = K for any $\ell \in \mathcal {G}_{\mathcal {L}_{K}},$ we compute that

$$\begin{array}{@{}rcl@{}} {QC}_{u}(K)&=&\frac{1}{5}{\sum}_{i=1}^{5}{T(K_{C_{i}},d^{\ast}|C_{i})}\\ &=&\frac{1}{5}{\sum}_{i=1}^{5} T\left( [\ell(K)]_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})]_{C_{i}}\right)\\ &=&\frac{1}{5}{\sum}_{i=1}^{5}{\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{K}}}\frac{1}{|\mathcal{G}_{\mathcal{L}_{K}}|} T\left( K_{C_{i}},[\ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})]_{C_{i}}\right)\\ &\geq&\frac{1}{5}{\sum}_{i=1}^{5} T\left( K_{C_{i}}, \left[{\sum}_{\ell\in \mathcal{G}_{\mathcal{L}_{K}}} \frac{1}{|\mathcal{G}_{\mathcal{L}_{K}}|}\ell(\{d^{\ast}|C_{i}\}_{i=1}^{5})\right]_{C_{i}}\right)\\ &=&\frac{1}{5}{\sum}_{i=1}^{5}T\left( K_{C_{i}},d_{0}|{C_{i}})\right) \end{array} $$

where the inequality holds because of the convexity of trace-distance, and so inequality (A6) holds.

For any d = x d ¹ + y d ² + z d ³ + η d ⁴ with x,y,z,η≥0 and x + y + z + η=1, we obtain that

$$\begin{array}{@{}rcl@{}} T(K_{C_{i}},d|C_{i}) &=&\frac{1}{2}\cdot\left( \left|1-\frac{2}{\sqrt{5}}-x-\frac{1}{5}z\right|+ 2\left|\frac{1}{\sqrt{5}}-\frac{2}{5}(z+\eta)\right|+y+\frac{1}{5}\eta\right). \end{array} $$

Hence,

$$\begin{array}{@{}rcl@{}} {QC}_{u}(K)&=&\frac{1}{2}\min \left\{\left|1-\frac{2}{\sqrt{5}}-x-\frac{1}{5}z\right|+ 2\left|\frac{1}{\sqrt{5}}-\frac{2}{5}(z+\eta)\right|\right.\\ &&\left.+y+\frac{1}{5}\eta:\ x,y,z,\eta\in[0,1], x+y+z+\eta=1 \right\}. \end{array} $$

Thus, there exist x ₀,y ₀,z ₀,η ₀∈[0,1] with x ₀ + y ₀ + z ₀ + η ₀=1 such that d ₀ = x ₀ d ¹ + y ₀ d ² + z ₀ d ³ + η ₀ d ⁴ and

$${QC}_{u}(K)=\frac{1}{2}\cdot\left( \left|1-\frac{2}{\sqrt{5}}-x_{0}-\frac{1}{5}z_{0}\right|+ 2\left|\frac{1}{\sqrt{5}}-\frac{2}{5}(z_{0}+\eta_{0})\right|+y_{0}+\frac{1}{5}\eta_{0}\right).$$

By (2.7), we obtain that

$$\begin{array}{@{}rcl@{}} {QC}_{q}(K) &\leq& {\sum}_{i=1}^{5}q(C_{i})T(K_{C_{i}},d_{0}|{C_{i}})\\ &=&{\sum}_{i=1}^{5}q(C_{i})\frac{1}{2}\cdot\left( \left|1-\frac{2}{\sqrt{5}}-x_{0}-\frac{1}{5}z_{0}\right|+ 2\left|\frac{1}{\sqrt{5}}-\frac{2}{5}(z_{0}+\eta_{0})\right|+y_{0}+\frac{1}{5}\eta_{0}\right)\\ &=&\frac{1}{2}\cdot\left( \left|1-\frac{2}{\sqrt{5}}-x_{0}-\frac{1}{5}z_{0}\right|+ 2\left|\frac{1}{\sqrt{5}}-\frac{2}{5}(z_{0}+\eta_{0})\right|+y_{0}+\frac{1}{5}\eta_{0}\right)\\ &=&{QC}_{u}(K) \end{array} $$

for any $q\in \mathcal {D}(\mathcal {M}).$ Now, (2.8) yields that (3.13) holds. □