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text:
Nestle has a new plant-based burger, and it is Awesome
Aimee Levitt
Filed to:fake meat
Photo: Sweet Earth (PR Newswire)
In the beginning, there was the veggie burger. And it was... not so good. Then came Beyond Meat. And it was better. And then came Impossible Foods, followed by the Impossible Whopper. And Big Food saw that all this was generating lots of buzz and said, "Lo, we should get in on that." So earlier this month, Kellogg announced its own fake meat, which it improbably decided to call Incogmeato. And now, at last, Nestlé, the world's biggest food company, has joined the fray with the Awesome Burger, which has already begun shipping to grocery stores. Surely the world's biggest food company has the resources to come up with a better name?
This in-depth piece from CNN doesn't delve into this issue. Instead it talks about how, since Nestlé is so enormous (308,000 employees worldwide), it's likely to avoid the steep losses and staffing and supply-chain issues that have made expansion so difficult for startups like Impossible and Beyond. (It helps that Nestlé isn't building its fake meat division from the ground up but instead taking over Sweet Earth, a California-based company.) It will also be good for Nestlé's PR department: Earlier this year, the company had to deal with allegations of animal cruelty at one of its suppliers and it continues to face criticism from activists for producing its bottled water.
The Awesome Burger is made of pea protein, coconut oil, wheat gluten and canola oil. It has more protein and fiber and less fat than actual beef or its fake-beef competitors. I saw the Awesome Burger for myself at Foodscape, a food innovation expo I attended yesterday afternoon. I didn't have a chance to taste it, but by sight, I would not confuse it with an actual patty of ground beef. But that's the great thing about burgers: Put enough stuff on them, and the meat doesn't matter as much.
Burger King's Impossible Whopper changes the game
Kellogg's thinks we'll want to eat something called Incogmeato
To dream the Impossible dream… at home
Whole Foods CEO: Plant-based "meat" not all that good for you
Aimee Levitt is associate editor of The Takeout.

redpajama_set_name: RedPajamaCommonCrawl


text:
Thai cave boys: Members of the Wild Boars soccer team recollects their ordeal after survival
Members of a Thai youth soccer team and their coach have described their rescue from a flooded cave as a miracle, thanked the experts who saved them and discussed how the experience will affect the rest of their lives.
In their first public remarks since emerging from their two-week ordeal last week, the boys recounted their side of an extraordinary story that captured the imagination of the world.
Dressed in matching team shirts, the boys and their coach appeared happy and relaxed as they faced the world's media after being discharged from the hospital in Chiang Rai on Wednesday.
The boys, all members of the Wild Boars junior soccer team, introduced themselves to the media, shared their nicknames and told the audience what position they played on the team.
Sitting beside the boys were the Thai Navy SEALs who stayed inside the cave with them after they were found, as well as members of the medical team who looked after them after the rescue. In a carefully arranged press conference, for which questions were pre-screened, the boys told of the moment they realized they were trapped, how they adapted to their surroundings and their eventual joy at being found, ten days later.
Authorities said that more than 100 questions were sent in from members of the media, though only a handful were selected. All 12 players and their coach had been under close supervision at Chiang Rai Prachanukroh hospital, near the border with Myanmar, since they were rescued from the cave on July 10.
Why go in?
Until Wednesday, the question of why the boys and their coach had decided to go into the Tham Luang cave on June 23 has been a point of speculation. It had been suggested the boys had been engaged in an initiation rite, or had been celebrating a team-member's birthday.
In fact, 25-year-old coach Ekapol Chantawong explained, the boys were merely curious to look inside as some of them had never visited it before. The coach, whose nickname is Ake, said it was not unusual for the group to participate in group activities after soccer practice on Saturday afternoons.
They explored the underground tunnels for about an hour, before deciding to turn back. But by this time the cave had become partially flooded and their exit was blocked. "Someone said are we lost?" said Ake, who reassured the group that help would come.
At his point, the realization dawned that they were trapped. With the entrance flooded and no obvious way out, the group retreated further into the cave to find somewhere to to rest for the night. "We moved further in for about 200 meters," Ake said. "There we found a bit of slope and there was a small water source inside the cave." Ake knew that the water dripping from the roof of the cave would be purer than the dirty floodwater on the floor. "I told them it's better to be near a water source," Ake said.
"Before we slept, I told them, 'Let's say a prayer.' So we said a prayer that night."
The team were not scared, Ake told the enraptured audience, explaining that he hoped the water level would drop the next day, and that help would arrive.
Surviving members of the Thai Wild Boars team
Waters rising
The waters did not subside, however. Instead, Ake described the moment that he heard the sound of flowing water — and saw the levels rising fast. In response, he ordered the group to find higher ground. Concerned that they might soon be submerged, he instructed the boys to start digging and look for a potential exit.
Having eaten after soccer practice, the boys had no food during their ordeal. Instead, they filled themselves with water from the cave. "I tried not to think about food because it would make me hungry," said the youngest of the boys, 11-year-old Chanin "Titan" Wibrunrungrueang.
The moment they were found
Adun Sam-On, the 14-year-old boy who became famous after responding in English to the first diver to reach the group, spoke of his shock on realizing they had been discovered.
Adun, like other members of the group, was busy digging — looking for a possible way out — when some of the boys thought they heard the sound of people talking.
Coach Ake told the group to stay quiet. He asked one of the boys to move closer to the ledge and shine a flashlight on the water, but the boy was too scared, said Adun, who volunteered instead.
When the British divers breached the surface, Adun said he was so shocked, all he could think to say was "hello!"
"I thought this was really a miracle. I didn't know how to respond," Adun said.
The boys described how they formed a bond with the Thai Navy SEALs who remained with them in the cave while rescuers worked out a plan to free them. Titan described how they played checkers — and that one of the Navy SEALs sitting alongside them at the press conference always won. "He was king of the cave," Titan said.
When the decision was made to extract the boys through the floodwaters, coach Ake joked he and the boys made the decision on who should go first based on who lived the furthest away. Ake thought the rescued kids would go straight home and those who got out first could spread the word, not realizing the global media had descended on the cave.
When asked about the lessons they've learned from the incident, Ake said he was going to live life more carefully.
Ardun said though people can't predict the future, the experience had taught him about the consequences of acting carelessly.
Other boys said though they still dreamed of becoming soccer players, some said they now wanted to become Navy SEALs.
Many of the boys apologized to their parents for not telling them they went to the cave.
Now that the boys' ordeal is over, there are concerns over their long-term psychological health. "We don't know what wounds the kids are carrying in their hearts," said Tawatchai Thaikaew, an official at the Thai justice ministry. He urged the media to respect the boys' privacy in the future, out of concern for their health, Reuters said.
Some of the boys are stateless, and the process of granting them Thai citizenship is under way, officials confirmed.
The largely joyous mood of the press conference was tempered, however, when the boys and their coach discussed the loss of Saman Kunan, the former Thai Navy SEAL who had died during the rescue effort. Coach Ake said the team were shocked to learn of Saman Kunan's death, called him a hero and said he had sacrificed his life for theirs.
In memory of the navy diver, Ake said the boys would spend time as novice Buddhist monks — a practice considered a high honor in Thailand.
Source: Rebecca Wright|| CNN
Ekapol Chantawong
Saman Kunan
Tawatchai Thaikaew
Thai youth soccer team
Tham Luang
Wild Boars
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text:
\section{Explicit Formula for FI MeNoS}
\label{App:Explicit}
In this Section, we perform the maximization from \eqref{maxsimp}  in order to derive \eqref{chian}. We have
\begin{equation}
\label{max2}
    \underset{0 \le \tilde N_1 \le \mathbb{1}}{\textrm{max}} \textrm{Tr}\left[A_1 \tilde N_1 + A_K \left(\mathbb{1}-\tilde N_1 \right) \right] = l_K^2 + \underset{0 \le \tilde N_1 \le \mathbb{1}}{\textrm{max}} \textrm{Tr}\left[ (A_1-A_K) \tilde N_1 \right]
\end{equation}
Let $\ket{e_i}$ be orthonormal diagonalizing basis of $A_1-A_K$, we can express both $A_1-A_K$ and $\tilde N_1$ in this basis:
\begin{equation}
    A_1 - A_K = \sum_{i=1}^R \lambda_i \ket{e_i}\!\!\bra{e_i},~~ \tilde N_1 = \sum_{i,j=1}^R \tilde N_{ij} \ket{i}\!\!\bra{j},
\end{equation}
$R=\textrm{dim}\mathcal{H}_\textrm{S}$ is the size of the matrices. We choose the ordering of $\ket{e_i}$ such that for some $P \in \{1,...,R\}$, $\lambda_1,....,\lambda_P \ge 0$ and $\lambda_{P+1},...,\lambda_R <0$. The maximization problem from the RHS of \eqref{max2} transforms to
\begin{equation}
    \underset{0 \le \tilde N_1 \le \mathbb{1}}{\textrm{max}} \textrm{Tr}\left[ (A_1-A_K) \tilde N_1 \right] = \underset{\tilde N_{ii}}{\textrm{max}} \sum_i \tilde N_{ii} \lambda_i
\end{equation}
From positivity of $\tilde N_1$ and of $\mathbb{1}- \tilde N_1$, $\tilde N_{ii} \ge 0$ and $\tilde N_{ii} \le 1$, therefore
\begin{equation}
    \underset{0 \le \tilde N_1 \le \mathbb{1}}{\textrm{max}} \textrm{Tr}\left[ (A_1-A_K) \tilde N_1 \right] = \sum_{i=1}^P \lambda_i,
\end{equation}
which is the sum of positive eigenvalues of $A_1-A_K$. Moreover, we have
$
    \textrm{Tr} (A_1 - A_K) = \sum_{i=1}^P \lambda_i + \sum_{i=P+1}^R \lambda_i$,  $\|A_1-A_K\|_1 = \sum_{i=1}^P \lambda_i - \sum_{i=P+1}^R \lambda_i$,
which means that
\begin{equation}
     \sum_{i=1}^P \lambda_i = \frac{1}{2} \left(\textrm{Tr}(A_1-A_K) + \|A_1-A_K\|_1 \right) = \frac{1}{2}(l_1^2-l_K^2+ \|A_1-A_K\|_1).
\end{equation}
Taking all this results together, we obtain
\begin{equation}
\label{maxG}
   \underset{\bm{N} \in \mathcal{M}}{\textrm{max}} G[\bm{N}] = \underset{0 \le \tilde N_1 \le \mathbb{1}}{\textrm{max}} \textrm{Tr}\left[A_1 \tilde N_1 + A_K \left(\mathbb{1}-\tilde N_1 \right) \right]  = \frac{1}{2}(l_1^2+l_K^2+ \|A_1-A_K\|_1)
\end{equation}
which, after inserting into \eqref{chiMN2} leads to \eqref{chian}.
\section{The most robust among the most informative measurements: Theorems}
\label{App:Theorems}
In this Section, we show how to approach a problem of finding a QCRB-sat. measurement with a minimal FI MeNoS. Before showing solutions for examples mentioned in the main text, let us introduce some theorems that help to systematically describe potential candidates for such a measurement. Let us start with quoting a condition for QCRB-saturability, firstly introduced in \cite{Braunstein1994}, then reformulated in \cite{Zhou2020} in a following form:

\vspace{0.5\baselineskip}
\noindent
\textbf{Theorem 1.}
POVM $\bm{M} = \{M_i\}_i$ is QCRB-sat. for the family of states $\rho_\theta$ around $\theta=\theta_0$ iff
\begin{equation}
\label{sat1}
     M_i^{1/2} L_{jk} M_i^{1/2} = 0,~ \forall i,j,k
\end{equation}
and
\begin{equation}
\label{sat2}
    \forall i~ \textrm{s.t.}~ \textrm{Tr}\left( M_i \rho_{\theta_0} \right)=0,~~ M_i^{1/2} \Lambda_{\theta_0} \ket{\psi_{{\theta_0},j}}=0  ,~\forall j
\end{equation}
Here $\rho_\theta = \sum_j p_{\theta,j} \ket{\psi_{\theta,j}} \!\! \bra{\psi_{\theta,j}}$ is the diagonalization of $\rho_\theta$, $p_{\theta,j} > 0$, $\Lambda_\theta$ is the symmetric logarithmic derivative matrix, and $L_{jk} =  \ket{\psi_{\theta_0,j}} \!\! \bra{\psi_{\theta_0,k}} \Lambda_{\theta_0}- \Lambda_{\theta_0} \ket{\psi_{\theta_0,j}} \!\! \bra{\psi_{\theta_0,k}} $.

\vspace{0.5\baselineskip}
\noindent
\emph{Proof.} See Ref.\cite{Zhou2020} for a proof. $\blacksquare$
\vspace{0.5\baselineskip}

Next two lemmas will help us to reduce the set of potential candidates for minimal MeNoS QCRB-sat. measurements to those containing only rank-one elements $M_i$ (see Theorem 2).

\vspace{0.5\baselineskip}
\noindent
\textbf{Lemma 1} Let $p_\theta(i)$ be the family of probability distributions indexed by $\theta$, $i \in \{1,...,K\}$. Let $ q_\theta(j) = \sum_i p_\theta(i) t(j|i)$ be another probability distribution, $j \in \{1,...,L\}$, $t(j|i) \ge 0$, $\sum_j t(j|i)=1$. Then
\begin{align}
        1.&~F_\textrm{C}[p_\theta(i)] \ge F_\textrm{C}[q_\theta(j)], \\
        2.&~F_\textrm{C}[p_\theta(i)] = F_\textrm{C}[q_\theta(j)] ~ \textrm{iff}  ~\forall_{j,i_1,i_2} \left( t(j|i_1) \ne 0 \land t(j|i_2) \ne 0 \right) \Rightarrow l_{i_1} = l_{i_2},
    \end{align}
where $l_i = \frac{ \dot p_\theta(i)}{p_\theta(i)}$. Point 2. means, that random mixing of results does not decrease the CFI iff only results with the same logarithmic derivatives are mixed with each other.

\vspace{0.5\baselineskip}
\noindent
\emph{Proof.}
Let us adopt a short-hand notation $p_i \equiv p_\theta(i)$, $q_j \equiv q_\theta(j)$. The CFI associated with the distribution $q_\theta(j)$ can be written as
\begin{equation}
\label{FcFj}
    F_\textrm{C} \left[q_\theta(j) \right] = \sum_{j=1}^L F_j,~\textrm{where}~ F_j = \frac{\dot q_j^2}{q_j}.
\end{equation}

Now we want to prove, that
\begin{equation}
\label{Fjin}
    F_j = \frac{\left(\sum_{i=1}^K t(j|i) \dot p_i \right)^2}{q_j} \le \sum_{i=1}^K t(j|i) \frac{\dot p_i^2}{p_i}.
\end{equation}
For a fixed $j$, let us introduce the notation $\alpha_i = \frac{t(j|i) p_i}{q_j}$. From the definition of $q_j$ follows that $\sum_i \alpha_i =1$, moreover $\alpha_i \ge 0$. Taking this into account, we have
\begin{equation}
\label{ineq12}
   \left|\sum_{i=1}^K t(j|i) \dot p_i \right| = \left| \sum_{i=1}^K \alpha_i\frac{t(j|i) \dot p_i}{ \alpha_i} \right| \overset{(1)}{\le} \sum_{i=1}^K \alpha_i \left|\frac{t(j|i) \dot p_i}{ \alpha_i} \right| \overset{(2)}{\le}  \sqrt{\sum_{i=1}^K \alpha_i \left( \frac{t(j|i) \dot p_i}{ \alpha_i} \right)^2},
\end{equation}
where (1) is triangle inequality, and (2) follows from the weighted power mean inequality (between arithmetic and quadratic mean, AM-QM in short).
After squaring both sides of \ref{ineq12}, substituting the definition of $\alpha_i$ in the RHS, and dividing both sides by $q_j$ we obtain inequality \ref{Fjin}. Subsequently, using \ref{Fjin} and \ref{FcFj} , we obtain
\begin{equation} \label{FciFcj}
    F_\textrm{C} \left[q_\theta(j) \right] \le \sum_{j=1}^L \sum_{i=1}^K t(j|i) \frac{\dot p_i^2}{p_i} = \sum_{i=1}^K \frac{\dot p_i^2}{p_i} = F_\textrm{C} \left[p_\theta(i)\right],
\end{equation}
which is exactly part 1. of our Lemma. In order to prove part 2., let us notice, that inequality \ref{FciFcj} is saturated iff inequalities (1) and (2) in \ref{ineq12} are saturated for all $j$. For a given $j$, part (1) of \ref{ineq12} becomes equality iff for all $i$ s.t. $t(j|i) \ne 0$, terms $\frac{t(j|i) \dot p_i}{ \alpha_i}= q_j \frac{\dot p_i}{p_i}$ have the same sign. Part (2), which is AM-QM inequality becomes saturated, when all terms in the mean with non-zero weights are equal, which implies that for a given $j$, for all all $i$ s.t. $t(j|i) \ne 0$ terms $\left|q_j \frac{\dot p_i}{p_i}\right|$ are equal. From these two saturability conditions, we obtain part (2) of Lemma 1. $\blacksquare$
\vspace{0.5\baselineskip}

The next lemma states that random mixing of the elements of POVM either decreases the CFI or remains both the CFI and FI MeNoS unaffected.

\vspace{0.5\baselineskip}
\noindent
\textbf{Lemma 2.}
Let $\bm{M}, \bm{M}' \in \mathcal{M}$, $\bm{M}=\{M_i\}_{i \in \{1,...,K\}}$, $\bm{M}'=\{M_j'\}_{j \in \{1,...,L\}}$, $M_j' = \sum_i t(j|i) M_i$, where $t(j|i) \ge 0$, $\sum_j t(j|i) = 1$. The family of quantum states $\rho_\theta$ is fixed. Then
\begin{align}
        1.&~F_\textrm{C}[\bm{M}] \ge F_\textrm{C}[\bm{M}'], \\
        2.&~F_\textrm{C}[\bm{M}] = F_\textrm{C}[\bm{M}'] \Rightarrow \chi \left[ \bm{M} \right] = \chi \left[ \bm{M}' \right].
    \end{align}

\noindent
\emph{Proof.}
The classical probability distributions associated with measurements $\bm{M}$ and $\bm{M}'$ are $p_\theta(i)= \textrm{Tr} (\rho_\theta M_i)$, $q_\theta(j) = \textrm{Tr} (\rho_\theta M_j) $. From the definition of $\bm{M}'$, we have $q_\theta(j) = \sum_i t(j|i) p_\theta(i)$. Therefore, part 1. of Lemma 2 is a direct consequence of part 1. of Lemma 1.

The logarithmic derivative of $q_j$ is
\begin{equation}
    l'_j = \frac{\dot q_j}{q_j} = \frac{\sum_i t(j|i) \dot p_i}{\sum_i t(j|i) p_i} = \frac{\sum_i t(j|i) l_i p_i}{\sum_i t(j|i) p_i}
\end{equation}
When $F_\textrm{C} [\bm{M}] = F_\textrm{C} [\bm{M}'] $, then, according to part 2. of Lemma 1.,
\begin{equation}
     l_{j_1}=l_{j_2}=...=l_{j_{K_j}}=l_j',
\end{equation}
where $j_1,j_2,...j_{K_j}$ are the indices satisfying $t(j|j_i) \ne 0$. That means, that the sequence of logarithmic derivatives $l'_j$ consist on exactly the same elements as the sequence $l_i$ (some elements may repeat), and the minimal and maximal logarithmic derivatives for $\bm{M}$ and $\bm{M}'$ are the same, so according to \eqref{chian}, $\chi \left[ \bm{M} \right] = \chi \left[ \bm{M}' \right]$. $\blacksquare$

\vspace{0.5\baselineskip}
\noindent
\textbf{Theorem 2.}
For any family of quantum states $\rho_\theta$, there exists a QCRB-sat. measurement with a minimal FI MeNoS whose all elements are rank-one matrices.

\vspace{0.5\baselineskip}
\noindent
\emph{Proof.} Let $\bm{M}' = \{M'_i\}_i $ be any QCRB-sat. POVM with a minimal FI MeNoS, $F_\textrm{C} \left[ \bm{M}' \right] = F_\textrm{Q}$, $\chi [\bm{M}'] = \chi_\textrm{Q}$. In a diagonal form, $M'_i = \sum_j \lambda_{i,j} \ket{\phi_{i,j}}\!\!\bra{\phi_{i,j}}$, $\lambda_{i,j} \ge 0$ because $M'_i$ are positive-semidefinite. Let us consider a POVM $\bm{M} = \{\lambda_{i,j}\ket{\phi_{i,j}}\!\!\bra{\phi_{i,j}}\}_{i,j}$, whose elements are all rank-one matrices. According to Lemma 2, $F_\textrm{C} \left[\bm{M}\right] \ge F_\textrm{C} \left[\bm{M}'\right] = F_\textrm{Q} $, but on the other hand, the CFI cannot be greater then the QFI, so $F_\textrm{C} \left[\bm{M}\right]=F_\textrm{C} \left[\bm{M}'\right]= F_\textrm{Q} $. Therefore, from part 2. of Lemma 2., $\chi[\bm{M}] = \chi[\bm{M}']= \chi_\textrm{Q}$, which means, that $\bm{M}$ is a QCRB-sat. POVM with a minimal MeNoS consisting of rank-one matrices only. $\blacksquare$
\vspace{0.5\baselineskip}

The next theorem is helpful for problems involving states with block-diagonal structure, it will be used in the next section to study super-resolution imaging.

\vspace{0.5\baselineskip}
\noindent
\textbf{Theorem 3.}
Let $\rho_\theta, \dot \rho_\theta \in \mathcal{L}(\mathcal{H})$ be a density matrix and its derivative, which can be decomposed as
\begin{align}
    &\rho_{\theta} = \rho_{\theta}^{(1)} + \rho_{\theta}^{(2)} + ... + \rho_{\theta}^{(L)}, \\
    & \dot \rho_{\theta} = \dot \rho_{\theta}^{(1)} + \dot \rho_{\theta}^{(2)} + ... + \dot \rho_{\theta}^{(L)},
\end{align}
where $\rho_{\theta}^{(l)}, \dot \rho_{\theta}^{(l)}$ act on a subspace $\mathcal{H}^{(l)} \subset \mathcal{H}$ (it has zeros outside this subspace), $\mathcal{H} = \mathcal{H}^{(1)} \oplus \mathcal{H}^{(2)} \oplus ... \oplus \mathcal{H}^{(L)}$, subspaces $\mathcal{H}^{(l)}$ are orthogonal to each other. Then, there exists a QCRB-sat. measurement with a minimal MeNoS, which can be written as \begin{equation}
\label{POVMsep}
\bm{M} = \left\{M_1^{(1)},...,M_{K_1}^{(1)},M_1^{(2)},...,M_{K_2}^{(2)},...,M_1^{(L)},...,M_{K_L}^{(L)}, \right\} \end{equation}
 where for each $l \in \{1,...,L\}$ matrices $M_1^{(l)},...,M_{K_l}^{(l)}$ act on   $\mathcal{H}^{(l)} $ and form a QCRB-sat. POVM for a family of normalized states $\tilde \rho_{\theta}^{(l)} = \rho_{\theta}^{(l)}/\textrm{Tr} \left( \rho_{\theta}^{(l)} \right)$.

\vspace{0.5\baselineskip}
\noindent
\emph{Proof.}
Let us denote an orthogonal projector on $\mathcal{H}^{(l)}$ by $P^{(l)}$. Let $\bm{M}' = \{M'_i\}_{i \in \{1,...,K\}} $ be any QCRB-sat. POVM with a minimal FI MeNoS. Let us construct another POVM $\bm{M}'' = \{M''_i\}_{i \in \{1,...,K\}}$, where
$
     M''_i = P^{(1)} M_i' P^{(1)}+P^{(2)} M_i' P^{(2)}+...+P^{(L)} M_i' P^{(L)}.
$

We have
\begin{equation}
    \textrm{Tr}(\rho_{\theta} M_i') = \textrm{Tr}\left(\left(\rho_{\theta}^{(1)} + \rho_{\theta}^{(2)}+...+\rho_{\theta}^{(L)} \right) M_i' \right) = \textrm{Tr} \left( \rho_{\theta} \left( P^{(1)} M_i' P^{(1)} + ... +P^{(L)} M_i' P^{(L)} \right) \right) = \textrm{Tr} \left( \rho_{\theta} M_i'' \right),
\end{equation}
where we used the fact that $\rho_{\theta}^{(l)} = P^{(l)} \rho_{\theta} P^{(l)} $.
Similarly,
\begin{equation}
     \textrm{Tr}(\dot \rho_{\theta} M_i') = \textrm{Tr}\left(\left(\dot \rho_{\theta}^{(1)} + \dot \rho_{\theta}^{(2)}+...+\dot \rho_{\theta}^{(L)} \right) M_i' \right) =\textrm{Tr} \left( \dot \rho_{\theta} M_i'' \right).
\end{equation}
Therefore, POVMs $\bm{M}'$ and $\bm{M}''$ are fully equivalent---probabilities of different outcomes $p_{\theta}(i)$ and their derivatives $\dot p_{\theta}(i)$ are the same for $\bm{M}'$ and $\bm{M}''$. Consequently, $F_\textrm{C} [ \bm{M}']=F_\textrm{C} [ \bm{M}'']$ and $\chi [ \bm{M}']=\chi [ \bm{M}'']$.

Let us finally construct a POVM
\begin{equation}
    \bm{M} = \left\{ P^{(1)} M_1' P^{(1)}, P^{(1)} M_2' P^{(1)},...,P^{(1)} M_K' P^{(1)},..., P^{(L)} M_1' P^{(L)},P^{(L)} M_2' P^{(L)},...,P^{(L)} M_K' P^{(L)}\right\},
\end{equation}
which is clearly of the form from \eqref{POVMsep} because $P^{(l)} M_i' P^{(l)}$ act on $  \mathcal{H}^{(l)}$. Moreover, elements of $\bm{M}''$ are linear combinations of the elements of $\bm{M}$, so, according to Lemma 2, $F_\textrm{C} \left[ \bm{M} \right] = F_\textrm{Q}$ and $\chi [\bm{M}] = \chi_\textrm{Q}$. Therefore, we constructed a QCRB-sat. POVM with a minimal MeNoS which has a form defined in \eqref{POVMsep}. Let us now prove, that when $\bm{M}$ from \eqref{POVMsep} is QCRB-sat., then for any $l \in \{1,...,L\}$, POVM $\bm{M}^{(l)} = \{ P^{(l)} M_1' P^{(l)}, P^{(l)} M_2' P^{(l)},...,P^{(l)} M_K' P^{(l)}  \}$ is QCRB-sat. for $\tilde \rho_{\theta}^{(l)}$. It is straightforward to show that
\begin{equation}
\label{Fcmany}
    F_\textrm{C} \left[ \rho_\theta, \bm{M} \right] = \sum_{l=1}^L F_\textrm{C} \left[  \tilde \rho_\theta^{(l)}, \bm{M}^{(l)} \right] + F_\textrm{C} \left[ P_\theta(l) \right],
\end{equation}
where $P_\theta(l) = \textrm{Tr} \left( \rho_\theta^{(l)} \right)$ is the probability of measuring $\rho_\theta$ in a subspace $\mathcal{H}^{(l)}$.
When for some $l$, $\bm{M}^{(l)}$ is not QCRB-sat., then there exists a POVM $\bm{M}_2^{(l)}$ for which $F_\textrm{C} \left[  \tilde \rho_\theta^{(l)}, \bm{M}_2^{(l)} \right] > F_\textrm{C} \left[  \tilde \rho_\theta^{(l)}, \bm{M}^{(l)} \right]$. Then, according to \eqref{Fcmany}, $F_\textrm{C} \left[ \rho_\theta, \bm{M}_2 \right]>F_\textrm{C} \left[ \rho_\theta, \bm{M} \right]$, where $\bm{M}_2 = \left\{ \bm{M}^{(1)},...,\bm{M}_2^{(l)},...,\bm{M}^{(K)} \right\}$. This is a contradiction with the assumption that $\bm{M}$ is QCRB-sat.---therefore $\bm{M}^{(l)}$ is QCRB-sat. for $\tilde \rho_\theta^{(l)}$. $\blacksquare$

\vspace{0.5 \baselineskip}
\noindent
\textbf{Theorem 4.}
Function $\chi[\bm{M}]$ defined in \eqref{chian} is  non-decreasing with $l_K$ and  non-increasing with $l_1$ for fixed $F_\textrm{C}[\bm{M}]$ (when $l_1$, $l_K$, $F_\textrm{C}[\bm{M}]$ are treated as independent variables).

\vspace{0.5 \baselineskip}
\noindent
\emph{Proof.}
Let us introduce a notation
\begin{equation}
    G[\bm{N},l_1,...,l_K] = \sum_{i=1}^K \textrm{Tr}(A_i N_i),~~A_i=l_i^2 \rho_\theta - 2 l_i \dot \rho_\theta,
\end{equation}
which is a more precise version of \eqref{Gdef}. We assume that $\rho_\theta$ and $\dot \rho_\theta$ are fixed, so we do not write them down explicitly as arguments of $G$. Let  $l_1 \le ...\le l_K \le l_{K+1} $ be an arbitrary ascending sequence of real numbers. For any $K$-element POVM $\bm{N}=\{N_1,...,N_K\}$ it is possible to construct a $K+1$-element POVM $\bm{\tilde N}$ satisfying
\begin{equation}
    G\left[ \bm{N},l_1,...,l_K \right] = G \left[ \bm{\tilde N},l_1,...,l_K,l_{K+1} \right],
\end{equation}
just by defining $\bm{\tilde N} = \{N_1,...,N_K,0\}$. Consequently,
\begin{equation}
    \underset{\bm{N} \in \mathcal{M}}{\textrm{max}} G\left[ \bm{N},l_1,...,l_K \right] \le \underset{\bm{\tilde N} \in \mathcal{M}}{\textrm{max}} G\left[ \bm{\tilde N},l_1,...,l_K,l_{K+1} \right].
\end{equation}
After applying \eqref{maxG} to both sides of this inequality, we obtain
\begin{equation}
\frac{1}{2}(l_1^2+l_K^2+ \|A_1-A_K\|_1) \le \frac{1}{2}(l_1^2+l_{K+1}^2+ \|A_1-A_{K+1}\|_1),
\end{equation}
which means that for a fixed $F_\textrm{C} [\bm{M}]$, $\chi[\bm{M}]$ is non-decreasing with $l_{K+1}$ because real number $l_K \le l_{K+1}$ can be arbitrary. Analogously, we can prove that for any pair $l_0 \le l_1$,
\begin{equation}
    \frac{1}{2}(l_1^2+l_K^2+ \|A_1-A_K\|_1) \le \frac{1}{2}(l_0^2+l_{K}^2+ \|A_0-A_{K}\|_1),
\end{equation}
from which follows the 2nd part of our thesis. $\blacksquare$

\section{The most robust among the most informative measurements: Examples}
\subsection{Pure state models}
\label{Sec:ApPure}
Let $\rho_\theta = \ket{\psi_\theta} \!\! \bra{\psi_\theta}$, $\dot \rho_\theta = \ket{\psi_\theta}\!\! \bra{\dot \psi_\theta} + \ket{\dot \psi_\theta}\!\! \bra{\psi_\theta}$. Then, the symmetric logarithmic derivative matrix is $\Lambda_\theta = 2 \left( \ket{\psi_\theta}\!\! \bra{\dot \psi_\theta} + \ket{ \dot \psi_\theta}\!\! \bra{ \psi_\theta} \right) $, and the QFI is
\begin{equation}
    F_\textrm{Q} = \textrm{Tr} (\rho_\theta \Lambda_\theta^2) = 4 \left( \braket{\dot \psi_\theta | \dot \psi_\theta} - \left| \braket{\dot \psi_\theta | \psi_\theta} \right|^2  \right).
\end{equation}
For a fixed value of a parameter $\theta=\theta_0$ we define
\begin{align}
\label{canb1}
    &\ket{0} = \frac{1}{\sqrt{2}} \ket{\psi_{\theta_0}} + i \sqrt{2 F_\textrm{Q}^{-1}} \left( \ket{ \dot \psi_{\theta_0}} - \braket{\psi_{\theta_0}|\dot \psi_{\theta_0}} \ket{\psi_{\theta_0}} \right) \\
    \label{canb2}
    &\ket{1} =\frac{1}{\sqrt{2}} \ket{\psi_{\theta_0}} - i \sqrt{2 F_\textrm{Q}^{-1}} \left( \ket{ \dot \psi_{\theta_0}} - \braket{\psi_{\theta_0}|\dot \psi_{\theta_0}} \ket{\psi_{\theta_0}} \right).
\end{align}
It is straightforward to check that $\ket{0}$ and $\ket{1}$ form o.-n. basis of $\textrm{span} \left\{ \ket{\psi_{\theta_0}}, \ket{\dot \psi_{\theta_0}} \right\}$,  the representations of $\rho_{\theta_0}$ , $\dot \rho_{\theta_0}$ and $\Lambda_{\theta_0}$ in this basis are:
\begin{equation}
\label{rhocan}
    \rho_{\theta_0} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \ket{+}\!\! \bra{+}, ~~
    \dot \rho_{\theta_0} = \frac{1}{2} \sqrt{F_\textrm{Q}} \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \frac{1}{2} \sqrt{F_\textrm{Q}} \sigma_y, ~~ \Lambda_{\theta_0} = \sqrt{F_\textrm{Q}} \sigma_y
\end{equation}
The condition \eqref{sat1} from Theorem 1 says, that all elements $M_i$ of any QCRB-sat. POVM must satisfy
\begin{equation}
\label{condQCRB}
    M_i^{1/2} L_{11} M_i^{1/2} = 0,
\end{equation}
where
\begin{equation}
    L_{11} = \ket{\psi_{\theta_0}}\!\! \bra{\psi_{\theta_0}} \Lambda_{\theta_0} - \Lambda_{\theta_0} \ket{\psi_{\theta_0}}\!\! \bra{\psi_{\theta_0}}   = i \sqrt{F_\textrm{Q}} \sigma_z
\end{equation}
in the basis $\ket{0},\ket{1}$.
After taking the trace of $n$-th power of both sides of \ref{condQCRB}, we obtain
\begin{equation}
\label{condQCRB2}
    \textrm{Tr} \left( (M_i P)^n \right)=0.
\end{equation}
The most general parametrization of $M_i$ is
\begin{equation}
    M_i = \begin{bmatrix} \alpha_i & \beta_i - i \gamma_i \\ \beta_i + i \gamma_i & \delta_i \end{bmatrix},
\end{equation}
where $\alpha_i, \beta_i, \gamma_i, \delta_i \in \mathbb{R}$. Using \ref{condQCRB2} with $n=1$ and $n=2$, we obtain the following conditions:
\begin{equation}
    n=1: \alpha_i = \delta_i,~~
    n=2:  \det M_i = 0
\end{equation}
From the 2nd condition, one eigenvalue of $M_i$ must be $0$, so $M_i$ is rank-one, and can be written as
\begin{equation}
\label{POVMproj}
    M_i = \lambda_i \ket{\phi_i}\!\! \bra{\phi_i},~~
    \ket{\phi_i} = \begin{bmatrix} a_i \\ b_i \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ e^{i \varphi_i} \end{bmatrix} ,
\end{equation}
 $0 \le \lambda_i \le 1$, $\left|a_i\right| = \left|b_i\right|=1/\sqrt{2}$ because $\alpha_i =\delta_i$ and because of normalization of $\ket{\phi_i}$. We obtained necessary conditions for QCRB saturation, but it turns out, that they are also sufficient---all POVMs, whose elements are of the described form, saturate QCRB. To show it, let us compute the CFI explicitly:
\begin{equation}
    F_\textrm{C} = \sum_i \lambda_i \frac{\left(\braket{\phi_i|\dot \rho_{\theta_0} |\phi_i} \right)^2}{\braket{\phi_i| \rho_{\theta_0} |\phi_i}} = \frac{F_\textrm{Q}}{2} \sum_i \lambda_i \left( 1 -a_i \bar b_i - \bar a_i b_i \right) = F_\textrm{Q},
\end{equation}
we used the identity $\sum_i M_i = \mathbb{1}$, from which follows that $\sum_i \lambda_i = \textrm{Tr} (\mathbb{1}) = 2$, and $\sum_i \lambda_i \bar a_i b_i = \sum_i \lambda_i a_i \bar b_i =0$. To conclude, for a given parametrization of pure states family, a POVM is QCRB-sat. iff it contains only elements proportional to projectors on states from a Bloch sphere equator.

At this point, we are ready to compare different QCRB-sat. measurements according to their FI MeNoS. Let us start with projective measurements, i.e. those for which $K=2$, $\lambda_1 = \lambda_2 =1$, $\varphi_1 = -\varphi$, $\varphi_2 = \pi - \varphi$, we refer to notation from \eqref{POVMproj}, $\varphi \in [0, 2 \pi ]$ corresponds to phase $\varphi$ which has to be added to the lower arm of Mach-Zender interferometer from Fig.~\ref{Fig:interferometer} in order to implement a described projective measurement. Logarithmic derivatives associated with this measurement are
\begin{equation}
    l_1 = \frac{\textrm{Tr} (\dot \rho _{\theta_0} M_1)}{\textrm{Tr} ( \rho _{\theta_0} M_1)} = \sqrt{F_\textrm{Q}} \tan (\varphi/2),~~l_2 = \frac{\textrm{Tr} (\dot \rho _{\theta_0} M_2)}{\textrm{Tr} ( \rho _{\theta_0} M_2)} = -\sqrt{F_\textrm{Q}} \tan^{-1}(\varphi/2).
\end{equation}
There are only two logarithmic derivatives, so one of them is maximal, and another is minimal. After inserting the values of $l_1$, $l_2$ into $\eqref{chian}$ and computing eigenavalues of of $A_1-A_2$ we obtain \eqref{chiint}, from which follows that $\chi[\bm{M}] \ge 4$ for all QCRB-sat. projective measurements, and the inequality is saturated only for $\varphi = \pi/2$. Let us now show, that $\chi[\bm{M}] \ge 4$ for all QCRB-sat. measurements (not only projective). For $\bm{M}$ with $K$ elements of the form defined in \eqref{POVMproj}, we construct a noise POVM
\begin{equation}
    \bm{N} = \left\{ N_1,...,N_k\right\},~~N_i = \lambda_i \left( \mathbb{1} - \ket{\phi_i}\!\!\bra{\phi_i} \right).
\end{equation}
Then
\begin{equation}
    G[\bm{N}] = \sum_i \textrm{Tr}(A_i N_i) = \sum_i \lambda_i l_i^2 - p_i l_i^2 + 2 \dot p_i l_i = F_\textrm{C} + \sum_i \lambda_i l_i^2,
\end{equation}
and therefore
\begin{equation}
    \chi[\bm{M}, \bm{N}] = 2+ F_\textrm{Q}^{-1} \sum_i \lambda_i l_i^2,
\end{equation}
where we substituted $F_\textrm{C}$ with $F_\textrm{Q}$ because $\bm{M}$ is QCRB-sat. In our case, $l_i = - \sqrt{F_\textrm{Q}} \tan (\varphi_i/2)$, so
\begin{equation}
\label{bineq2m2}
    \chi[\bm{M}, \bm{N}] = 2 - \sum_i \lambda_i + \sum_i \lambda_i \cos^{-2}(\varphi_i/2) = 2 \sum_i \frac{\lambda_i}{2} \left| \cos^{-1} (\varphi_i/2) \right|^2,
\end{equation}
where we used the identity $\sum_i \lambda_i = \textrm{Tr}(\mathbb{1})=2$. Using power mean inequality (between powers $2$ and $-2$, $\lambda_i/2$ are weights), we obtain
\begin{equation}
    \label{ineq2m2}
    \sqrt{\sum_i \frac{\lambda_i}{2} \left| \cos^{-1} (\varphi_i/2)  \right|^2} \ge \left(\sum_i \frac{\lambda_i}{2} \left| \cos^{-1} (\varphi_i/2)  \right|^{-2} \right)^{-1/2} = \sqrt{2},
\end{equation}
the last equality follows from the fact that $\sum_i \lambda_i \cos^2(\varphi_i/2) = \frac{1}{4} \sum_i \lambda_i (2+e^{i\varphi_i} + e^{- i \varphi_i})=1$ because the sum over $i$ of anti-diagonal terms of $M_i$, which are $\lambda_i e^{\pm i \varphi_i}$, is equal to $0$. After inserting \eqref{ineq2m2} into \eqref{bineq2m2}, and using an inequality $\chi[\bm{M}] \ge \chi[\bm{M}, \bm{N}] $, we obtain that $\chi[\bm{M}] \ge 4$ for any QCRB-sat. $\bm{M}$. This inequality can only be saturated if \eqref{ineq2m2} is saturated, which means that all terms in the average are equal, and therefore $\cos^2(\varphi_1/2) = ... = \cos^2(\varphi_K/2)$. Then we have $1 = \sum_i \lambda_i \cos^2(\varphi_i/2) = \cos^2(\varphi_1/2) \sum_i \lambda_i  = 2 \cos^2(\varphi_1/2) $, so $\cos^2(\varphi_1/2) = ... = \cos^2(\varphi_K/2)=1/2$, and consequently $\varphi_i = \pm \pi/2$ for each $i$. Therefore, $\chi[\bm{M}]=4$ iff the measurement is a projective measurement on eigenstates of $\sigma_y$.
\subsection{Super-resolution optical imaging}
\label{App:Superres}
Let us firstly follow the construction from \cite{Tsang2016} of vectors $\ket{0}_\textrm{s}$,$\ket{1}_\textrm{s}$,$\ket{0}_\textrm{a}$,$\ket{0}_\textrm{a}$, where $\ket{0/1}_\textrm{s}$ form o.-n. basis of $\mathcal{H}_\textrm{s}$, $\ket{0/1}_\textrm{a}$ form o.-n. basis of $\mathcal{H}_\textrm{a}$, $\mathcal{H}^{(\theta)} = \mathcal{H}_\textrm{s} \oplus \mathcal{H}_\textrm{a}$, for a fixed value of $\theta$, $\rho_\theta, \dot \rho_\theta \in \mathcal{L} \left( \mathcal{H}^{(\theta)} \right)$. We define
\begin{align}
    \ket{0}_\textrm{s} &= \frac{1}{\sqrt{2 (1+\delta)}} \left( \ket{u_{+,\theta}} + \ket{u_{-,\theta}} \right) \\
    \ket{1}_\textrm{s} &= \frac{1}{c_4} \left[ -\sqrt{2} \left( \partial_\theta \ket{u_{+,\theta}}+\partial_\theta \ket{u_{-,\theta}}  \right) + \frac{\gamma}{\sqrt{1+\delta}} \ket{0}_\textrm{s}\right] \\
    \ket{0}_\textrm{a} &= \frac{1}{\sqrt{2 (1-\delta)}} \left( \ket{u_{+,\theta}} - \ket{u_{-,\theta}} \right)\\
    \ket{1}_\textrm{a} &= \frac{1}{c_3} \left[ -\sqrt{2} \left( \partial_\theta \ket{u_{+,\theta}}-\partial_\theta \ket{u_{-,\theta}}  \right) - \frac{\gamma}{\sqrt{1-\delta}} \ket{0}_\textrm{a}\right],
\end{align}
where
\begin{align}
    \delta &= \braket{u_{+,\theta}|u_{-,\theta}}= e^{-\theta^2/8 \sigma^2},\\
    \gamma &= \bra{u_{+,\theta}} \partial_\theta \ket{u_{+,\theta}} = -\frac{\theta  e^{-\frac{\theta ^2}{8 \sigma ^2}}}{4 \sigma ^2} \\
    c_3 &= \frac{1}{4} \sqrt{\frac{8 \sigma ^2 \sinh \left(\frac{\theta ^2}{8 \sigma ^2}\right)-\theta ^2}{\sigma ^4 \left(e^{\frac{\theta ^2}{8 \sigma ^2}}-1\right)}},\\
    c_4 &= \frac{1}{4} \sqrt{\frac{8 \sigma ^2 \sinh \left(\frac{\theta ^2}{8 \sigma ^2}\right)+\theta ^2}{\sigma ^4 \left(e^{\frac{\theta ^2}{8 \sigma ^2}}+1\right)}}.
\end{align}
Direct computations show, that $\ket{0}_\textrm{s}$, $\ket{1}_\textrm{s}$,$\ket{0}_\textrm{a}$,$\ket{1}_\textrm{a}$ indeed form o.-n. basis, and $\rho_\theta$, $\dot \rho_\theta$ can be expressed in this basis as
\begin{equation}
\rho_\theta = \frac{1}{2}\begin{bmatrix}1+\delta & 0 &0&0 \\
0 & 0 &0&0 \\
0 & 0 &1-\delta&0 \\
0 & 0 &0&0 \\
\end{bmatrix}, ~ \dot \rho_\theta = \frac{1}{2} \begin{bmatrix}\gamma & -\frac{c_4}{2} \sqrt{1+\delta} & 0 & 0 \\
 -\frac{c_4}{2} \sqrt{1+\delta} & 0  & 0 & 0 \\
0 & 0 & -\gamma &  -\frac{c_3}{2} \sqrt{1-\delta } \\
 0 & 0   &  -\frac{c_3}{2} \sqrt{1-\delta } & 0 \\
\end{bmatrix}
\end{equation}
which is equivalent to \eqref{rhosup}, \eqref{drhosup} after substituting
\begin{equation}
    \alpha = \frac{\gamma}{2},~ \beta_\textrm{s} = -\frac{c_4}{4} \sqrt{1+\delta},~ \beta_\textrm{a} = -\frac{c_3}{4} \sqrt{1-\delta }.
\end{equation}
Both $\rho_\theta$ and $\dot \rho_\theta$ have block-diagonal structure (as in Theorem 3):
\begin{equation}
    \rho_\theta = \rho_\theta^{(\textrm{s})} +\rho_\theta^{(\textrm{a})},~ \dot \rho_\theta =\dot \rho_\theta^{(\textrm{s})} +\dot \rho_\theta^{(\textrm{a})},
\end{equation}
normalized components of $\rho_\theta$ are $\tilde \rho_\theta^{(\textrm{s/a})} = \ket{0}_\textrm{s/a}\!\!\bra{0}$,  their derivatives: $\dot{ \tilde{\rho}}_\theta^{(\textrm{s})}= \frac{2 \beta_{\textrm{s}}}{1+\delta} \sigma_x^{\textrm{(s)}} $, $\dot{ \tilde{\rho}}_\theta^{(\textrm{a})}= \frac{2 \beta_{\textrm{a}}}{1-\delta} \sigma_x^{\textrm{(a)}}$. Therefore, the evolution of the normalized density matrix in each subspace is locally equivalent to rotation of a pure state $\ket{0}$ around $y$ axis of a Bloch sphere. Consequently, taking into account Theorem 3 and the characterization of QCRB-sat. measurements for pure states from Section \ref{Sec:ApPure}, we conclude that potential candidates for minimal MeNoS QCRB-sat. measurements are of the form
\begin{equation}
    \bm{M} = \left\{\bm{M}^\textrm{(s)}, \bm{M}^\textrm{(a)}  \right\},~
    \bm{M}^\textrm{(s)} = \left\{  M_1^\textrm{(s)},...,M_{K_1}^\textrm{(s)} \right\},~\bm{M}^\textrm{(a)} = \left\{  M_1^\textrm{(a)},...,M_{K_2}^\textrm{(a)} \right\}
\end{equation}
where $M_i^\textrm{(s/a)} = \lambda_{i,\textrm{s/a}} \ket{\varphi_{i,\textrm{s/a}}}_\textrm{s/a} \!\!\bra{\varphi_{i,\textrm{s/a}}}$ , $ \ket{\varphi}_\textrm{s/a}=\cos(\varphi/2)\ket{0}_{s/a} + \sin(\varphi/2) \ket{1}_{s/a}$. In this case, QCRB-sat. measurements in each subspace consist on projectors on states from a meridian of a Bloch sphere, not from equator, because parametrization is different from the one used in Section \ref{Sec:ApPure}. As the last step, let us prove that $\bm{M}$ with a minimal MeNoS is of the form \eqref{supresoptM}, which means that $K_1=K_2=2$.  We will prove,that if $K_1 \ge 3$, then it is possible to construct a QCRB-sat. measurement $ \bm{ \tilde M} = \left\{ \bm{\tilde M}^\textrm{(s)}, \bm{ M}^\textrm{(a)} \right\}$ such that $\chi[\bm{\tilde M}] \le \chi[\bm{M}]$ and $\bm{\tilde M}^\textrm{(s)}$ contains $K_1-1$ elements. Let us start with choosing two indices $i,j \in \{1,...,K_1\}$ such that
\begin{equation}
\label{ijconds}
    \varphi_{i,\textrm{s}} \le \varphi_{j,\textrm{s}}~ \land~ \left( \varphi_{i,\textrm{s}},\varphi_{j,\textrm{s}} \in \left[0,\pi\right] \lor \varphi_{i,\textrm{s}},\varphi_{j,\textrm{s}} \in \left[\pi,2\pi\right]  \right).
\end{equation} Such a choice is always possible for $K_1 \ge 3$ because of pigeonhole principle. After some algebra, we obtain the following identity
\begin{equation}
\label{MiMjsum}
    M_i^\textrm{(s)} + M_j^\textrm{(s)} = A \mathbb{1}_\textrm{s} + B \ket{\tilde \varphi}_\textrm{s}\!\!\bra{\tilde \varphi},
\end{equation}
where
\begin{equation}
    B=\sqrt{\lambda_{i,\textrm{s}}^2+\lambda _{j,\textrm{s}}^2+2 \lambda_{i,\textrm{s}} \lambda_{j,\textrm{s}} \cos \left(\varphi_{j,\textrm{s}}-\varphi_{i,\textrm{s}} \right)},~~ A= \frac{1}{2} \left(\lambda_{i,\textrm{s}}+\lambda_{j,\textrm{s}}-B\right)
\end{equation}
are positive constants, $A < 1$ because $\lambda_{i,\textrm{s}}+\lambda_{j,\textrm{s}} < \textrm{Tr} (\mathbb{1}_\textrm{s}) = 2$, and
\begin{equation}
    \tan\left( \tilde \varphi \right) = \frac{C \tan \left(\varphi _{i,\textrm{s}}\right)+D \tan \left(\varphi _{j,\textrm{s}}\right)}{C+D},~~C=\lambda _{i,\textrm{s}} \cos^{-1} \left(\varphi _{j,\textrm{s}}\right),~~D=\lambda _{j,\textrm{s}} \cos^{-1} \left(\varphi _{i,\textrm{s}}\right).
\end{equation}
From the above formula it is clear that  $\tan(\tilde \varphi)$ lies between $\tan(\varphi_{i,\textrm{s}})$ and $\tan(\varphi_{j,\textrm{s}})$, so using \eqref{ijconds}, we obtain $\tilde \varphi \in [\varphi_{i,\textrm{s}},\varphi_{j,\textrm{s}} ]$. After removing  $M_i^\textrm{(s)}$ and $M_j^\textrm{(s)}$ from $\bm{M}^\textrm{(s)}$, and replacing them with $B \ket{\tilde \varphi}_\textrm{s}\!\!\bra{\tilde \varphi}$, the sum of all elements is decreased from $\mathbb{1}_\textrm{s}$ to $(1-A) \mathbb{1}_\textrm{s} $, according to \eqref{MiMjsum}. Therefore, to obtain a valid POVM we need to rescale all elements after such a replacement, and then we end up with
\begin{equation}
    \bm{\tilde M}^\textrm{(s)} =  \left\{ \left.(1-A)^{-1} M_k^\textrm{(s)} \right| _{k \in \{1,...,K_1 \} \setminus \{i,j\} } , (1-A)^{-1} B \ket{\tilde \varphi}_\textrm{s}\!\!\bra{\tilde \varphi}  \right\},
\end{equation}
which is a POVM containing $K_1-1$ elements. From the construction, $\bm{\tilde M}^\textrm{(s)}$ contains only elements proportional to projectors on a Bloch sphere meridian, so it is QCRB-sat.. Let us now prove, that for a given construction, $\chi[\bm{\tilde M}] \le \chi[\bm{M}]$.
The logarithmic derivative of $p_\theta(k,\textrm{s}) = \textrm{Tr} \left( \rho_\theta M_k^{\textrm{(s)}} \right)$ is
\begin{equation}
    l_{k,\textrm{s}} = \frac{\textrm{Tr} \left( \dot \rho_\theta M_k^{\textrm{(s)}} \right)}{\textrm{Tr} \left( \rho_\theta M_k^{\textrm{(s)}} \right)} = f(\varphi_{k,\textrm{s}}),~\textrm{where}~f(\varphi) = \frac{2}{1+\delta} \left( \alpha + 2 \beta_\textrm{s} \tan (\varphi/2) \right),
\end{equation}
notice that $l_{k, \textrm{s}}$ does not depend on $\lambda_{k,\textrm{s}}$, and that $f(\varphi)$ is monotonic in range $\left[0,\pi\right]$ and in range $\left[\pi,2\pi\right]$. That means, that after replacing $\bm{M}^\textrm{(s)}$ with $\bm{\tilde M}^\textrm{(s)}$, logarithmic derivatives of outcomes with indices $k \in \{1,...,K_1 \} \setminus \{i,j\}$ are not affected, and  $l_{i, \textrm{s}}$, $l_{j, \textrm{s}}$ are replaced with a logarithmic derivative $\tilde l = f(\tilde \varphi)$. From inclusion $\tilde \varphi \in [\varphi_{i,\textrm{s}},\varphi_{j,\textrm{s}} ]$, we conclude that $\tilde l$ lies between $l_{i, \textrm{s}}$ and $l_{j, \textrm{s}}$. This means, that the described modification of $\bm{M}$ neither increased the largest logarithmic derivative, nor decreased the smallest one. Therefore, according to Theorem 4, $\chi[\bm{\tilde M}] \le \chi[\bm{M}]$. It means, that we can lower the value of $K_1$ without affecting the CFI, and without increasing MeNoS, as long as $K_1>2$. The same reasoning can be applied to the assymetric subspace, so we can also lower $K_2$. Therefore, we can always find a QCRB-sat. measurement with a minimal MeNoS with $K_1=K_2=2$, and such a measurement must be of the form defined in \eqref{supresoptM}.

In the main text, we also considered the family of measurements in Hermite-Gaussian modes, $M_i= \ket{\phi_i}\!\!\bra{\phi_i}$ for $i \in \{1,2,...,K-1\}$, $M_K = \mathbb{1} - M_1-...-M_{K-1}$, where the representation of $\ket{\phi_i}$ in a position basis is
\begin{equation}
    \braket{x|\phi_i}= \left( \frac{1}{2 \pi \sigma^2} \right)^{1/4} \frac{1}{\sqrt{2^i i!}} H_q \left( \frac{x}{\sqrt 2 \sigma} \right) \exp \left( - \frac{x^2}{4 \sigma^2} \right).
\end{equation}
According to Ref.\cite{Tsang2016}, the probability of obtaining $i$-th outcome for an input state defined in \eqref{imagingstate} is
\begin{equation}
    p_\theta(i) = \braket{\phi_i|\rho_\theta|\phi_i} = \exp \left(- \theta^2/16\sigma^2 \right) \frac{\left( \theta^2/16\sigma^2 \right)^i}{i!}~~\textrm{for}~i \in \{1,...,K-1\},~~ p_\theta(K) = 1-p_1-...-p_{K-1},
\end{equation}
 corresponding logarithmic derivatives are
\begin{equation}
    l_i = -\frac{\theta}{8 \sigma^2} + \frac{2i}{\theta} ~~\textrm{for}~i \in \{1,...,K-1\},~~l_K = -\frac{l_1 p_1 + ... + l_{K-1} p_{K-1}}{p_K}.
\end{equation}
Both CFI and FI MeNoS plotted in Fig.~\ref{Fig:superresolution} are computed directly using above relations and formulas \eqref{CFIdef} and \eqref{chian}.
\end{document}

redpajama_set_name: RedPajamaArXiv


The Messermeister Four Seasons 12" Kullenschliff Flexible Fillet Knife has a stamped, long, narrow and extremely flexible blade that cuts extremely close to a spine, bone or other hard surfaces. An excellent blade for the day-to-day demands of a commercial kitchen but useful for any fishing enthusiast. The alternating kullenschliffs (a.k.a. kullens, hollows or grantons) on each side of the knife reduce drag and prevent food from sticking to the blade.


For chic, adventure-driven women, the TechnoMarine MoonSun watch with Nato Strap may be the perfect accessory. It projects toughness while exuding classic sophistication. The watch has a silver dial encased in a gold-tone 40mm stainless steel and flame fusion crystal. Made with the precision of Quartz movement and water resistant up to 50 meters, it has a durable blue and white Nato band with leather lining, and may be worn for short periods of recreational swimming. With the new TechnoMarine MoonSunNato watch, there is no time to wait but to enjoy life's daily adventures without compromising style.
To know more, visit TechnoMarine stores in Greenbelt 3, Robinson's Galleria, Robinson's Magnolia, Shangri-La Plaza Mall, SM Mall of Asia, SM Megamall, SM North EDSA, SM City Cebu, SM Lanang Premier Davao, TriNoma, and Uptown Mall, or follow TechnoMarinePH on Facebook and Instagram.


text:
Canada's imminent trademark law changes attract brand trolls
MarkRubens / iStockphoto.com
Second Circuit rules safe harbour does apply to pre-1972 recordings
UK court dismisses ISP appeal in Richemont counterfeits case
Developing a moving target
Canada has decided to become fully integrated with the international trademark regime, but trolls are already taking advantage of the rule changes, writes Ashlee Froese of Froese Law.
In 2014, Canada ratified the adoption of three international trademark treaties that will impact aspects of domestic Canadian trademarks law: the Nice Agreement, the Singapore Treaty and the Madrid Protocol.
These are the most significant changes made to Canadian trademark law in 50 years.
The implementation of these treaties into domestic law has been slow. Four years later, we are still waiting for the full integration of these new laws—but we are advised that they are expected to come into force in 2019.
Brand trolls are already taking advantage of the anticipated changes, to the detriment of legitimate brand owners.
Domestic trademark law is shaped by international trademark treaties, should countries ratify those treaties.
For better or worse, to date, Canada has been an outlier to the most significant international trademark treaties. In 2014, Canada finally decided to become fully integrated with the international trademark regime, by ratifying these international treaties.
Significant changes
Some of the imminent changes are substantive; others are administrative. Below is a summary of the significant changes:
Use will no longer be required as a basis of registration, nor a requirement for registration.
The definition of what constitutes a trademark will be widely expanded to include scent, texture, taste, and hologram.
Similar to the rules in the US, it will be possible to divide trademark applications.
The registration term will be reduced from 15 years to 10 years.
Third parties will be able to file letters of protest during the examination stage of a trademark application.
Incremental government fees per class of wares/services will be incurred.
What's the impact?
Canada has always been a 'use it or lose it' jurisdiction. Currently, you can obtain a trademark registration only if you are able to attest to use of the trademark, either domestically or internationally.
This requirement acted as a safeguard to ensure (for the most part) that those trademarks that register in Canada are actually being used in Canada.
With the upcoming removal of the use requirement, this means that brand trolls can now register trademarks in Canada with no threshold to demonstrate use of the trademark.
Essentially, it enables brand trolls to seek to register trademarks that are owned by legitimate trademark owners and hold those trademarks hostage, blocking the brand owner out of registering the trademarks.
We are already seeing some brand trolls file Canadian trademark applications in anticipation of the full phase-in in 2019, when there will be no requirement to file a declaration of use at the final stage of the registration process.
This will place a greater importance on trademark monitoring, trademark watch services and trademark enforcement proceedings than before.
"After the 2019 phase-in, the office will charge additional government fees for each class of good and/or service."
Through your trademark lawyer, it is possible to place watches to see if any third parties are filing identical or confusingly similar trademarks as your brand.
Similarly, greater emphasis will be placed on expungement proceedings and opposition proceedings that are trademark enforcement proceedings which are embedded within the trademark registration process.
Currently, the Canadian Intellectual Property Office charges one government fee, regardless of how many classes are included in the application.
After the 2019 phase-in, the office will charge additional government fees for each class of good and/or service. This means it is cheaper to file trademark applications now, especially if you are filing multi-class applications.
With the 2019 phase-out of the use basis, brand trolls are targeting Canada. File your trademark applications sooner than later, to ensure that you are staking your claim and squeezing out opportunities for brand trolls to take your trademarks hostage.
Work with your trademark lawyer to strategise trademark watching and monitoring services to ensure that you are guarding against brand trolls.
Ashlee Froese is a branding and fashion lawyer. She is the founder of Toronto-based boutique branding firm Froese Law. Her scope of practice includes trademarks, copyright, domain name, social media, marketing, advertising, packaging and labelling, licensing and commercial agreements. She can be contacted at ashlee@froeselaw.com and followed at @froese_law
Canada, trademark, trademark law, trolls, Ashlee Froese, Froese Law, Nice Agreement, Singapore Treaty, Madrid Protocol, brand trolls


i just sent this email to sjobs@mac.com and sjobs@pixar.com.
i seriously doubt he'll even see it, but maybe if we all flood his email with complaints, he might have to make good on his wrong. it's worth the minute it will take you to write it and it just MIGHT make a difference.
I for one, know the penalty of early adoption, but a HUGE price break in just 2 months! ridiculous. mark my word, come end of september or at least by xmas, 16gb iPhone for $499.
"i have been an Apple customer for 25 years, preferring the mac OS over windows any day of the week.
when the first mac came out, bought it.
when the powerbook came out, bought it.
when the imac came out, bought it.
when the g3 came out, bought it.
when the g4 came out, bought it.
when the g5 came out, bought it.
when the ipod came out bought it.
WHEN THE iPhone CAME OUT, STOOD IN LINE FOR 8 HOURS, BOUGHT IT.
2 months later you reduce the price by $200. 2 MONTHS!
way to completely screw over the Apple faithful.
you need to make good on this immediately! maybe a $200 gift certificate to iTunes store, or something. i don't know. i'm sure you can come up with something as innovative as you are.
i never considered another company for my future computing needs but boy am i now.
However, I don't think I'll email Steve. I sent it to Apple via the iPhone story part on their website. I'm really loyal to Apple, but this puts a huge dent in my loyalty. Sorry Apple.
i just sent this on "share your story:"
"bought iPhone on opening day. stood in line for 8 hrs. got the 8gb for $630.00 with tax. have loved it until today. Apple faithful for 25 years. spent roughly $20,000.00 on your products since then.
just saw ipod keynote. iPhone price drop of $200.00 in 2 months! i know there's a price for adopting a product early and supporting you, but THIS IS RIDICULOUS! sure prices go down as technology gets better. i understand. But this is the way you reward the Apple faithful?!?
now every time i look at or use my wonderful iPhone, it's going to remind me of this day and how the company whom i have supported for years, took complete advantage of me.
thanks Apple. you may have just lost a customer."
I am very upset. Like so many other early adopters, I bought the iPhone on launch day. I spent two hours waiting in line outside in the Texas summer heat. I forgave the lack of real Exchange server support (that I hoped would be temporary). I forgave the lack of 3rd party apps. I believed that many great apps would be coming through software updates. I bought the iPhone.
Now, two months later I have no new applications, no new features, no corporate exchange support, and you have depreciated my *very* expensive purchase by 1/3 for those 2 months of waiting. I even went without it for a week while it was shipped off for repair. This is outrageous. This is insulting to those that supported the iPhone. You've made the very people who enthusiastically hyped the product look like chumps, and I for one don't like to be humiliated. This is a tipping-point issue with Apple for me, among others including quality. Out of the 5 Apple products that I have owned, 4 of them have had quality issues or needed repair - including the iPhone. After each issue, I was disappointed but amicable. I cannot continue to be after this.
This price drop is a slap in the face to early adopters, and will be the last straw for me personally with Apple. If there is no compensation at all for early adopters to come, then I am completely done with Apple and any future products. Not only that, but Apple will have earned at least $200 in negative word-of-mouth advertising to anyone who will listen (and at my advertising rates, this is should last for years).


The Castine Historical Society offers guided Walking Tours of Castine during the summer season. Tours begin every Saturday at 10 AM in front of the Abbott School at 17 School Street or by appointment, and are offered at no cost.
Meet your guide in front of the Abbott School, and take in the sights and sounds of Castine while learning about our beautiful town and its historical significance.


text:
Uotilansaari ou Vasikkasalo est une île du lac Päijänne à Kuhmoinen en Finlande.

Géographie
Uotilansaari mesure 3,7 kilomètres de long, 2 kilomètres de large et une superficie de 2,43 kilomètres carrés.
Uotilansaari est située dans la partie ouest du Tehinselkä, dans une baie juste en face du village de Kuhmoinen, où, avec Rekisalo, remplit presque la baie avec sa grande taille.

Entre Uotilansaari et Rekisalo se trouvent les détroits Kessalmi, Tanskansalmi et Kiimasalmi, qui forment un ensemble de détroits de 40 à 600 mètres de large.
À l'ouest des îles se trouve un détroit de cinq kilomètres de long qui s'élargit à plus de 200 mètres à proximité d'Uotilansaari.

Au nord de l'île, pointent Varrasniemi et Rautsaarenkärki.
La péninsule Varrasniemi n'est qu'à 140 mètres du continent et Rautsaarenkärki à 180 mètres.

Au nord d'Uotilansaari se trouvent les îles Ansiossaari à l'embouchure du Kukkostensalmi, Laviosari au large de Lavioslahti et Harakkasaaret et Lehtosaaret à l'est d'Uotilansaari.

Uotilansaari est relativement plate, sauf dans sa partie sud-ouest, où se trouvent les collines Pitkävuori, Huhkainvuori et la colline qui s'élève au nord de la baie Telkänlahti.
Pitkävuori culmine à 49 mètres au dessus du lac Päijänne.

Entre les collines se trouve un marécage. À l'extrémité nord de l'île se trouvent les baies   Torvelanlahti, Vasainlahti, Syvälahti et Lavioslahti.

L'île est située à proximité du continent et a donc été habitée en permanence.
Les derniers champs sont maintenant des prairies et les habitants permanents se sont déplacés vers le continent, de sorte que l'île n'a plus qu'une cinquantaine de maisons de vacances.

Références

Liens externes

 Geonames

Liens internes
 Liste des îles du lac Päijänne

Île du Päijänne
Kuhmoinen

redpajama_set_name: RedPajamaWikipedia


text:
E-Waste – All You Need To Know
Post MY GP Team August 7, 2022 August 7, 2022 no commentNo tags
Electronics that have reached or are at the end of their "useful life" and are undesired, broken, or inefficient are referred to as "e-waste." Electronic devices used on a daily basis include computers, televisions, VCRs, stereos, copiers, and fax machines. It is not a new problem to try to figure out the best way to get rid of old and useless equipment. However, a lot has changed since then, most notably the volume of devices thrown away today. We are going to talk about this in detail in – e waste collection Singapore.
What is E-waste?
Any electrical or electronic device that has been thrown out is considered e-waste. This covers products that are both in functional and broken condition that are discarded in the trash or given to a charity retailer like Goodwill. The item will frequently be thrown away if it is unsold at the store. E-waste is particularly hazardous because it contains poisonous compounds that, when buried, naturally leach from the metals inside.
But more and more e-waste these days isn't regarded as being composed of inoperative or outmoded products. Many electronic equipment that still function well are now regarded as being antiquated since technological advancements are occurring at such a breakneck pace. E-waste is produced when someone decides to redesign a device that uses electricity and believes they can do it better. We are concerned about this since obsolete electronics have been clogging up landfills all around the world for a long time.
E-waste Covers Up Toxic Materials
Modern technology is safe to use and be around while it is above ground. Beryllium, cadmium, mercury, and lead are only a few of the dangerous substances found in most electronics, though, and they all represent significant threats to the environment, including harm to our land, water, air, and wildlife. E-waste that is dumped in a landfill may disintegrate into the foul sludge that permeates the landfill in minute traces. These harmful material traces eventually gather in the ground beneath the waste. Leaching is what's happening here. These trace hazardous substances are more prevalent in the groundwater as the dump receives more metals and e-waste.
Leeching Poisons Nearby Water
The issue is that there is so much electronic waste that over time, the trace levels have increased dramatically. The contaminated water continues underneath the landfill and continues to flow. It goes on to reach the groundwater and the origins of all the freshwater in the neighborhood. This is not only detrimental to anyone using a natural well, but it also harms the local animals. Due to the high concentration of these minerals, this in turn leads to lead, arsenic, cadmium, and other metal poisonings in wildlife.
E-waste – A Big Problems
It's likely that e-waste will continue to have a broad meaning. As technology develops quickly, more and more highly advanced electronic products are being developed and produced. Consider the idea of a "smart home" for a moment. It's obvious how many electronic gadgets are now capable of doing a wide range of tasks, including security, lighting control, and brewing fresh coffee before we wake up. Sadly, owners are dismissing an alarmingly large proportion of e-waste as rubbish. Computers, laptops, and cellphones are the most notable examples of that. Even as the present model seems to be operating without a hitch, new models are introduced. Despite this, the most recent edition always offers extra features that make it look irresistible.
Electric gadgets designed to simplify and improve our lives in every way imaginable are still being produced by technological innovators. However, we appear to be all too prone to swiftly discarding the machines we already have. It makes no difference how happy we've been with them so far.
What led to this?
The e-waste recycling sector is one area that has seen substantial growth since. This sector has the most effective method for preventing e-waste from ending up in landfills or being burned in incinerators. For the benefit of nearby manufacturers and businesses, the recycling industry has been committed to recovering usable components from obsolete electronic gadgets. As e-waste production has increased, so too has this business, which today generates hundreds of thousands of jobs globally by recycling the devices we no longer need.
We hope you found our blog – e waste collection Singapore, helpful.
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