Operational characterization of weight-based resource quantifiers via exclusion tasks in general probabilistic theories

Abstract

The formalism of general probabilistic theories (GPTs), which include classical probability theory and quantum mechanics, has provided us with physical theories beyond quantum mechanics. In this work, we focus on studying the connection between several classes of exclusion tasks and core problems of resource theories-weight-based resource quantifiers and resource manipulations in any GPT. First, we introduce two resourceful quantifiers called the weight of state (WOS) and the weight of measurement (WOM) in any GPT; then, we show that the WOS accurately quantifies the best advantage that a given resource state offers over resourceless states in all channel exclusion tasks. Meanwhile, a similar conclusion can be drawn for the WOM. Second, we introduce the weight-generating power of a channel (WGPC) in any GPT, based on which the resource content of a nonfree channel can be quantified by understanding the number of resources produced by it. It is proven that the WGPC can be considered as the best advantage provided by a given nonfree channel when considering a class of free-state exclusion tasks. In the context of quantum mechanics, we show that the best advantage that a given resource channel provides over resourceless channels in a class of entanglement-assisted state exclusion tasks can be accurately quantified by the weight of channel (WOC). In addition, we introduce the maximum WOC ensemble (MWCE) and find that the MWCE admits an operational interpretation as the best advantage that a given resource channel ensemble provides over free channel ensembles in a class of specific free-channel exclusion tasks. Finally, we show that several classes of channel and state exclusion tasks can constitute complete sets of monotones, completely describing the transformations of states and measurements in any GPT, respectively.

Appendices

Appendix A: The proof of Lemma 1

Equation (10) can be rewritten as the following form:

$$\begin{aligned} {\text {minimize}} \quad&w ,&\end{aligned}$$

(A1)

$$\begin{aligned} \text{ subject } \text{ to }\quad&(1-w)\sigma \preceq _{{\mathcal {C}}}\rho ,&\end{aligned}$$

(A2)

$$\begin{aligned}&\sigma \in {\mathcal {F}}.&\end{aligned}$$

(A3)

Let ${\tilde{\sigma }}=(1-w)\sigma$, then we have

$$\begin{aligned} {\text {minimize}} \quad&1-\langle U, {\tilde{\sigma }}\rangle ,&\end{aligned}$$

(A4)

$$\begin{aligned} \text{ subject } \text{ to }\quad&{\tilde{\sigma }}\preceq _{{\mathcal {C}}}\rho ,&\end{aligned}$$

(A5)

$$\begin{aligned}&{\tilde{\sigma }}\in cone({\mathcal {F}}).&\end{aligned}$$

(A6)

The Lagrangian is given by

$$\begin{aligned} L({\tilde{\sigma }}; X,Y)&=1-\langle U, {\tilde{\sigma }}\rangle -\langle X, \rho -{\tilde{\sigma }}\rangle -\langle Y,{\tilde{\sigma }}\rangle \\&=1-\langle X, \rho \rangle -\langle U-X+Y, {\tilde{\sigma }}\rangle , \end{aligned}$$

where $X\in {\mathcal {C}}^*$ and $Y\in {\mathcal {F}}^*$.

Noticing that

$$\begin{aligned} \mathop {\min }\limits _{{\tilde{\sigma }}} L({\tilde{\sigma }}; X,Y) =\left\{ \begin{array}{ll} {1-\langle X, \rho \rangle ,} &{} { \text{ if } U-X+Y=\mathbf {0}, Y\in {\mathcal {F}}^*}, \\ {-\infty ,} &{} {\text{ otherwise } }. \\ \end{array}\right. \end{aligned}$$

Optimizing over the Lagrange multipliers $X\in {\mathcal {C}}^*$ and $Y\in {\mathcal {F}}^*$, we obtain the following dual form:

$$\begin{aligned} {\text {maximize}} \quad&1-\langle X, \rho \rangle ,&\end{aligned}$$

(A7)

$$\begin{aligned} \text{ subject } \text{ to }\quad&X\in {\mathcal {C}}^*,&\end{aligned}$$

(A8)

$$\begin{aligned}&\langle X,\sigma \rangle \ge 1, \forall \sigma \in {\mathcal {F}},&\end{aligned}$$

(A9)

where Eq. (A9) follows from $Y=X-U$ and $\langle Y, \sigma \rangle \ge 0$ for any $\sigma \in {\mathcal {F}}$.

Appendix B: The proof of Lemma 2

Let $N'_i=wN_i$, then Eq. (18) can be rewritten as the following form:

$$\begin{aligned} {\text {minimize}} \quad&w,&\end{aligned}$$

(B1)

$$\begin{aligned} \text{ subject } \text{ to }\quad&M_i-N'_i\succeq _\mathcal {E_F}\mathbf {0},&\end{aligned}$$

(B2)

$$\begin{aligned}&N'_i\in {\mathcal {C}}^*, \forall i&\end{aligned}$$

(B3)

$$\begin{aligned}&\sum \limits _iN'_i=wU.&\end{aligned}$$

(B4)

Then we can rewrite it as

$$\begin{aligned} \begin{aligned} \mathcal {W_{E_F}}({\mathbb {M}})=min\{w\in {\mathbb {R}}_+ \mid M_i-N'_i\in \mathcal {E_F}, \forall i,N'_i\in {\mathcal {C}}^*,\forall i, wU-\sum _iN'_i=0_{{\mathcal {V}}^*}\}. \end{aligned} \end{aligned}$$

The Lagrangian is given by

$$\begin{aligned} \begin{aligned} L(\{N_i\}, w;\{\sigma _i\},\{\delta _i\},\eta )&=w-\sum _i\langle M_i-N'_i, \sigma _i\rangle -\sum _i\langle N_i', \delta _i\rangle -\langle wU-\sum _iN'_i,\eta \rangle \\&=w(1-\langle U, \eta \rangle )+\sum _i\langle N'_i, \eta +\sigma _i-\delta _i\rangle -\sum _i\langle M_i, \sigma _i\rangle . \end{aligned} \end{aligned}$$

Optimizing over the Lagrange multipliers $\sigma _i\in \mathcal {E_F}^*$, $\delta _i\in {\mathcal {C}}$ and $\eta \in {\mathcal {V}}$, we obtain the following dual form:

$$\begin{aligned} {\text {maximize}} \quad&-\sum _i\langle M_i, \sigma _i\rangle ,&\end{aligned}$$

(B5)

$$\begin{aligned} \text{ subject } \text{ to }\quad&-\sigma _i\preceq _{\mathcal {C}} \eta , \forall i&\end{aligned}$$

(B6)

$$\begin{aligned}&\eta \in {\mathcal {V}},&\end{aligned}$$

(B7)

$$\begin{aligned}&\sigma _i\in \mathcal {E_F}^*, \forall i.&\end{aligned}$$

(B8)

$$\begin{aligned}&\langle U, \eta \rangle =1.&\end{aligned}$$

(B9)

Defining $\rho _i=\eta +\sigma _i$, we have

$$\begin{aligned} {\text {maximize}} \quad&1-\sum _i\langle M_i, \rho _i\rangle ,&\end{aligned}$$

(B10)

$$\begin{aligned} \text{ subject } \text{ to }\quad&\rho _i\in {\mathcal {C}}, \forall i&\end{aligned}$$

(B11)

$$\begin{aligned}&\eta \in {\mathcal {V}},&\end{aligned}$$

(B12)

$$\begin{aligned}&\langle F,\eta \rangle \le \langle F,\rho _i\rangle , \forall i, \forall F\in \mathcal {E_F},.&\end{aligned}$$

(B13)

$$\begin{aligned}&\langle U, \eta \rangle =1,&\end{aligned}$$

(B14)

where Eq. (B10) follows from $\sum _i\langle M_i, \eta \rangle =\langle U, \eta \rangle =1$ and Eq. (B13) follows from $\rho _i-\eta \in \mathcal {E_F}^*$.

Appendix C: The proof of Proposition 2

We only give the proof of condition (i), the proof of condition (ii) is similar with that. First, we prove the “only if” part. Suppose that there exists ${\tilde{\Lambda }}\in {\mathcal {O}}$ such that $\rho '={\tilde{\Lambda }}(\rho )$. Then for any ${\mathbb {M}}\in {\mathcal {M}}$, we have

$$\begin{aligned} \begin{aligned} {\tilde{p}}_\mathrm{err}'({\mathbb {M}},\rho ')&=\mathop {\min }\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}} \sum _i p_i\langle M_i, \Lambda _i(\rho ') \rangle \\&=\mathop {\min }\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}} \sum _i p_i\langle M_i, \Lambda _i\circ {\tilde{\Lambda }}(\rho ) \rangle \\&\ge \mathop {\min }\limits _{\Lambda '_i \in {\mathcal {O}},\{p_i, \Lambda '_i\}} \sum _i p_i\langle M_i, \Lambda '_i(\rho ) \rangle \\&={\tilde{p}}_\mathrm{err}'({\mathbb {M}},\rho ), \end{aligned} \end{aligned}$$

(C1)

where the inequality is due to the closedness of ${\mathcal {O}}$ under concatenation.

Then, we prove the “if” part. Suppose that $\forall {\mathbb {M}}\in {\mathcal {M}}$, we have ${\tilde{p}}_\mathrm{err}'({\mathbb {M}},\rho )\le {\tilde{p}}_\mathrm{err}'({\mathbb {M}},\rho ')$. This signifies that

$$\begin{aligned} \begin{aligned} 0&\le \mathop {inf}\limits _{\{M_i\}\in {\mathcal {M}}}\bigg [ \mathop {\min }\limits _{\Theta _i \in {\mathcal {O}},\{q_i, \Theta _i\}}\sum _i q_i\langle M_i, \Theta _i(\rho ') \rangle -\mathop {\min }\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}}\sum _i p_i\langle M_i, \Lambda _i(\rho ) \rangle \bigg ]\\&\le \mathop {inf}\limits _{\{M_i\}\in {\mathcal {M}}}\bigg [\sum _i p_i\langle M_i, \rho ' \rangle -\mathop {min}\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}}\sum _i p_i\langle M_i, \Lambda _i(\rho ) \rangle \bigg ]\\&=\mathop {inf}\limits _{\{M_i\}\in {\mathcal {M}}}\mathop {\max }\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}}\sum _i p_i\langle M_i, \rho '-\Lambda _i(\rho ) \rangle \\&\le \mathop {\min }\limits _{\{M_i\}_{i=0}^{N}\in {\mathcal {M}}}\mathop {max}\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}_{i=0}^{N-1}}\sum _i p_i\langle M_i, \rho '-\Lambda _i(\rho ) \rangle \\&=\mathop {\max }\limits _{\Lambda _i \in {\mathcal {O}},\{p_i, \Lambda _i\}_{i=0}^{N-1}}\mathop {\min }\limits _{\{M_i\}_{i=0}^{N}\in {\mathcal {M}}}\sum _i p_i\langle M_i, \rho '-\Lambda _i(\rho ) \rangle ,\\ \end{aligned} \end{aligned}$$

(C2)

where the second inequality follows from setting each $q_i=p_i$ and any $\Theta _i=id$, the third inequality is because we restricted the minimization over $N+1$-outcome measurements where $N\ge 2$, the last equality follows from Sion’s minimax theorem [71].

To prove that $\exists \Lambda \in {\mathcal {O}}$ such that $\rho '=\Lambda (\rho )$, suppose that $\forall \Lambda \in {\mathcal {O}}$, we have $\rho '\ne \Lambda (\rho )$. Since any $\Lambda _i$ is a physical channel and thus normalization preserving, we have

$$\begin{aligned} \langle U, \rho '-\Lambda _i(\rho )\rangle =0, \forall i. \end{aligned}$$

(C3)

In fact, since $\rho '-\Lambda _i(\rho )\ne \mathbf {0}$ by assumption, we have $\rho '-\Lambda _i(\rho )\notin C$ for all i. If there exists i such that $\rho '-\Lambda _i(\rho )\in C$, then we have $\langle U, \rho '-\Lambda _i(\rho )\rangle >0$, which is contradictory with Eq. (C3). Therefore, by the hyperplane separation theorem [72], for every $\Lambda _i$ there exists an effect $E_i\in C^*$ such that $\langle E_i, \rho '-\Lambda _i(\rho )\rangle <0$. We now construct an incomplete measurement $\{M_i\}_{i=0}^{N-1}$ by

$$\begin{aligned} M_i=\frac{E_i}{\parallel \sum _j E_j\parallel _{\Omega }^{\circ }}, \end{aligned}$$

(C4)

such that $\sum _i M_i\preceq _{C^*} U$, and so we can define a complete measurement $\{M_i\}_{i=0}^{N}$ as

$$\begin{aligned} M_i=\left\{ \begin{array}{ll} \frac{E_i}{\parallel \sum _j E_j\parallel _{\Omega }^{\circ }}, &{} { \text{ if } i\ne N}, \\ {U-\sum _{i=0}^{N-1} M_i,} &{} { \text{ if } i=N}. \end{array}\right. \end{aligned}$$

(C5)

Notably, the effect $M_N$ does not affect the measurement with respect to the ensemble $\{p_i, \Lambda _i\}_{i=0}^{N-1}$. For this measurement $\{M_i\}_{i=0}^{N}$, we have

$$\begin{aligned} \sum _i p_i\langle M_i, \rho '-\Lambda _i(\rho ) \rangle =\sum _i\frac{p_i}{\parallel \sum _j E_j\parallel _{\Omega }^{\circ }}\langle E_i, \rho '-\Lambda _i(\rho ) \rangle <0. \end{aligned}$$

(C6)

That is to say that for any ensemble $\{p_i, \Lambda _i\}_{i=0}^{N-1}$ there exists such a measurement such that Eq. (C6) holds. This is contradictory with Eq. (C2), which says that there exists an ensemble $\{p_i, \Lambda _i\}_{i=0}^{N-1}$ such that any measurement $\{M_i\}_{i=0}^{N}$ gives $\sum _i p_i\langle M_i, \rho '-\Lambda _i(\rho ) \rangle \ge 0$. It suggests that our original assumption must be wrong, hence there exists $\Lambda \in {\mathcal {O}}$ such that $\rho '=\Lambda (\rho )$.

Appendix D: The proof of Proposition 4

We only give the proof of condition (i), the proof of condition (ii) is similar with that. First, we prove the “only if” part. If there exists $\Gamma \in \mathcal {O_E}$ such that ${\mathbb {M}}'=\Gamma ({\mathbb {M}})$, then for any ensemble ${\mathcal {A}}=\{p_i,\sigma _i\}$ we have

$$\begin{aligned} \begin{aligned} {\tilde{p}}_\mathrm{err}({\mathcal {A}}, {\mathbb {M}}')&=\mathop {\min }\limits _{\Lambda ^* \in \mathcal {O_E}} \sum _i p_i\langle M_i', \Lambda (\sigma _i) \rangle \\&=\mathop {\min }\limits _{\Lambda ^* \in \mathcal {O_E}} \sum _i p_i\langle \Gamma (M_i), \Lambda (\sigma _i) \rangle \\&=\mathop {\min }\limits _{\Lambda ^* \in \mathcal {O_E}} \sum _i p_i\langle \Lambda ^*\circ \Gamma (M_i),\sigma _i \rangle \\&\ge \mathop {\min }\limits _{{\tilde{\Lambda }}^* \in \mathcal {O_E}} \sum _i p_i\langle {M_i},{\tilde{\Lambda }}(\sigma _i) \rangle \\&={\tilde{p}}_\mathrm{err}({\mathcal {A}},{\mathbb {M}}), \end{aligned} \end{aligned}$$

(D1)

where the inequality is due to the closedness of $\mathcal {O_E}$ under concatenation.

Then, we prove the “if” part. Suppose $\forall {\mathcal {A}}$, ${\tilde{p}}_\mathrm{err}({\mathcal {A}},{\mathbb {M}})\le {\tilde{p}}_\mathrm{err}({\mathcal {A}},{\mathbb {M}}')$, we have

$$\begin{aligned} \begin{aligned} 0&\le \mathop {\min }\limits _{{\mathcal {A}}}\bigg [ \mathop {\min }\limits _{\Theta ^* \in \mathcal {O_E}}\sum _i p_i\langle M_i', \Theta (\sigma _i) \rangle -\mathop {\min }\limits _{\Lambda ^* \in \mathcal {O_E}}\sum _i p_i\langle M_i, \Lambda (\sigma _i) \rangle \bigg ]\\&\le \mathop {\min }\limits _{{\mathcal {A}}}\bigg [\sum _i p_i\langle M_i', \sigma _i \rangle -\mathop {\min }\limits _{\Lambda ^* \in \mathcal {O_E}}\sum _i p_i\langle M_i, \Lambda (\sigma _i) \rangle \bigg ]\\&=\mathop {\min }\limits _{{\mathcal {A}}}\mathop {\max }\limits _{\Lambda ^* \in \mathcal {O_E}}\sum _i p_i\langle M_i'-\Lambda ^*(M_i), \sigma _i\rangle \\&=\mathop {\max }\limits _{\Lambda ^* \in \mathcal {O_E}}\mathop {\min }\limits _{{\mathcal {A}}}\sum _i p_i\langle M_i'-\Lambda ^*(M_i), \sigma _i\rangle ,\\ \end{aligned} \end{aligned}$$

(D2)

where the second inequality follows from setting $\Theta ^*=id$, and the last equality follows from Sion’s minimax theorem [71].

Next, we will prove that there exists $\Lambda ^*\in \mathcal {O_E}$ such that $M_i'-\Lambda ^*(M_i)=\mathbf {0}$ for all i. To achieve this, we use proof by contradiction. Suppose that $\forall \Lambda ^*\in \mathcal {O_E}$ there exists i such that $M_i'-\Lambda ^*(M_i)\ne \mathbf {0}$. Due to $\Lambda ^*\in \mathcal {O_E}$, for all $\Lambda ^*$ we immediately have

$$\begin{aligned} \sum _i\big [M_i'-\Lambda ^*(M_i)\big ]=U-\Lambda ^*(U)=\mathbf {0}. \end{aligned}$$

(D3)

In particular, it holds that

$$\begin{aligned} \left\langle \sum _i M_i'-\Lambda ^*(M_i), \rho \right\rangle =0, \forall \rho \in C, \end{aligned}$$

(D4)

which means that there exists $i^*$ such that $M_{i^*}'-\Lambda ^*(M_{i^*})\notin C^*$. Otherwise, it implies that $M_i'-\Lambda ^*(M_i)=\mathbf {0}$ for all i, which contradicts with the assumption. Therefore, for any $\Lambda ^*\in \mathcal {O_E}$ there exist an index $i^*$ and a state $\sigma \in \Omega ({\mathcal {V}})$ such that

$$\begin{aligned} \langle M_{i^*}'-\Lambda ^*(M_{i^*}), \sigma \rangle <0. \end{aligned}$$

(D5)

Define the ensemble $\{p_i,\sigma _i\}$ as $p_i=0$ if $i\ne i^*$ and $p_{i^*}=1, \sigma _{i^*}=\sigma$, then we have

$$\begin{aligned} \sum _i p_i \langle M_i'-\Lambda ^*(M_i), \sigma _i\rangle <0. \end{aligned}$$

(D6)

That is to say that for any $\Lambda ^*\in \mathcal {O_E}$ there exists a ensemble ${\mathcal {A}}$ such that Eq. (D6) holds. This is contradictory with Eq. (D2), which says that there exists $\Lambda ^*\in \mathcal {O_E}$ such that any ensemble ${\mathcal {A}}$ gives $\sum _i p_i\langle M_i'-\Lambda _i^*(M_i), \sigma _i\rangle \ge 0$. It suggests that our original assumption must be wrong, hence there exists $\Lambda ^*\in \mathcal {O_E}$ such that ${\mathbb {M}}'=\Lambda ^*({\mathbb {M}})$.