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In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble understanding the convexity argument in the proof of Lemma 2. (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecessary to solve the issue.

In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble understanding the convexity argument in the proof of Lemma 2 (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecessary to solve the issue.

In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble understanding the convexity argument in the proof of Lemma 2. (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecesssary to solve the issue.

Let us consider a quantum system for which the Hilbert space can be factorised into $m$ different "sites". That is to say $\mathcal{H} = \bigotimes_{k=1}^{m} \mathcal{H}_k$ where all the individual Hilbert spaces $\mathcal{H}_k$ are $2d$ dimensional for some fixed $d$, i.e. $\mathcal{H}_k = \mathbb{C}^{2d}$ for all $k$. Now let us define a set of localised two outcome observables $M_{k,i}$ with $i=1,2$ and $k=1,2,...,m$ such that they only have $\pm 1$ as their eigenvalues, and are traceless. The authors go on to show that it is possible to choose a basis in which both $M_{k,1}$ and $M_{k,2}$ take the block diagonal form $\oplus_{j=1}^{d}\mathbb{C}^2$ simultanously for all $k$. The quantity of interest is the Bell correlation $$B = \langle{\psi|\mathcal{B}|\psi}\rangle.$$ where $$\mathcal{B} = \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m}\bigotimes_{k=1}^{m}M_{k,i_k}$$ is the Bell operator for some fixed $c_{i_1...i_m}$ coefficients. We're taking $M_{k,0}=1$ as the identity operator. As I understsand it, this is useful in having terms in $\mathcal{B}$ that have operators from, say, only two sites a la the CHSH case. The authors claim that "the maximum quantum value of the Bell inequality is achieved by a state that has support on a qubit at each site". I'm assuming by "value of Bell inequality" they mean value $B$. Moreover I'm assuming by "support over one quibit", they mean when the state $|\psi\rangle$ lies entirely in exaclty one $\mathbb{C}^2$ sector in all $\mathcal{H}_k$'s. Please correct me if either of these assumptions are incorrect or unreasonble.

1. What does the $\rho_{j,k}$ mean? It couldn't possibly mean an actual reduced density matrix obtained by tracing out all the degrees of freedom apart from $k$ because then the formula $\langle{\psi|\mathcal{B}|\psi}\rangle = \text{tr} \mathcal{B} \rho_{j,k}$ wouldn't hold as $\mathcal{B}$ has nontrivial operators on other sites too. As far as I know, this kind of expection value formula holds only when our operator is nontrivial only on one site and identity on the rest.

2. What does the convexity argument mean? I know what a convex sum of vectors is but I don't see a vector, or anything that can be construed as a vector, on either side of the equation. It's scalar on both sides.

3. How does said convexity mean that $B$ is maximised with support only on one sector? What is the corresponding theorem being used here?

In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble understanding the convexity argument in the proof of Lemma 2. (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecessary to solve the issue.

Let us consider a quantum system for which the Hilbert space can be factorised into $m$ different "sites". That is to say $\mathcal{H} = \bigotimes_{k=1}^{m} \mathcal{H}_k$ where all the individual Hilbert spaces $\mathcal{H}_k$ are $2d$ dimensional for some fixed $d$, i.e. $\mathcal{H}_k = \mathbb{C}^{2d}$ for all $k$. Now let us define a set of localised two outcome observables $M_{k,i}$ with $i=1,2$ and $k=1,2,...,m$ such that they only have $\pm 1$ as their eigenvalues, and are traceless. The authors go on to show that it is possible to choose a basis in which both $M_{k,1}$ and $M_{k,2}$ take the block diagonal form $\oplus_{j=1}^{d}\mathbb{C}^2$ simultaneously for all $k$. The quantity of interest is the Bell correlation $$B = \langle{\psi|\mathcal{B}|\psi}\rangle.$$ where $$\mathcal{B} = \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m}\bigotimes_{k=1}^{m}M_{k,i_k}$$ is the Bell operator for some fixed $c_{i_1...i_m}$ coefficients. We're taking $M_{k,0}=1$ as the identity operator. As I understand it, this is useful in having terms in $\mathcal{B}$ that have operators from, say, only two sites a la the CHSH case. The authors claim that "the maximum quantum value of the Bell inequality is achieved by a state that has support on a qubit at each site". I'm assuming by "value of Bell inequality" they mean value $B$. Moreover I'm assuming by "support over one qubit", they mean when the state $|\psi\rangle$ lies entirely in exactly one $\mathbb{C}^2$ sector in all $\mathcal{H}_k$'s. Please correct me if either of these assumptions are incorrect or unreasonable.

1. What does the $\rho_{j,k}$ mean? It couldn't possibly mean an actual reduced density matrix obtained by tracing out all the degrees of freedom apart from $k$ because then the formula $\langle{\psi|\mathcal{B}|\psi}\rangle = \text{tr} \mathcal{B} \rho_{j,k}$ wouldn't hold as $\mathcal{B}$ has nontrivial operators on other sites too. As far as I know, this kind of expectation value formula holds only when our operator is nontrivial only on one site and identity on the rest.

2. What does the convexity argument mean? I know what a convex sum of vectors is but I don't see a vector, or anything that can be construed as a vector, on either side of the equation. It's scalar on both sides.

3. How does said convexity mean that $B$ is maximised with support only on one sector? What is the corresponding theorem being used here?

In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble undrstanding the convexity argument in the proof of Lemma 2. (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecesssary to solve the issue.

Let us consider a qunatum system for which the Hilbert space can be factorised into $m$ different "sites". That is to say $\mathcal{H} = \bigotimes_{k=1}^{m} \mathcal{H}_k$ where all the individual Hilbert spaces $\mathcal{H}_k$ are $2d$ dimensional for some fixed $d$, i.e. $\mathcal{H}_k = \mathbb{C}^{2d}$ for all $k$. Now let us define a set of localised two outcome observables $M_{k,i}$ with $i=1,2$ and $k=1,2,...,m$ such they only have $\pm 1$ as their eigenvalues, and are traceless. The authors go on to show that it is possible to choose a basis in which both $M_{k,1}$ and $M_{k,2}$ take the block diagonal form $\oplus_{j=1}^{d}\mathbb{C}^2$ simultanously for all $k$. The quantity of interest is the Bell correlation $$B = \langle{\psi|\mathcal{B}|\psi}\rangle.$$ where $$\mathcal{B} = \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m}\bigotimes_{k=1}^{m}M_{k,i_k}$$ is the Bell operator for some fixed $c_{i_1...i_m}$ coefficients. We're taking $M_{k,0}=1$ as the identity operator. As I understsand it, this is useful in having terms in $\mathcal{B}$ that have operators from, say, only two sites a la the CHSH case. The authors claim that "the maximum quantum value of the Bell inequality is achieved by a state that has support on a qubit at each site". I'm assuming by "value of Bell inequality" they mean value $B$. Moreover I'm assuming by "support over one quibit", they mean when the state $|\psi\rangle$ lies entirely in exaclty one $\mathbb{C}^2$ sector in all $\mathcal{H}_k$'s. Please correct me if either of these assumptions are incorrect or unreasonble.

In this paper, the authors give a proof of the monogamy principle in quantum physics. I'm having trouble understanding the convexity argument in the proof of Lemma 2. (penultimate paragraph, page 2). I suggest the reader read the problem first because the background may well be unnecesssary to solve the issue.

Let us consider a quantum system for which the Hilbert space can be factorised into $m$ different "sites". That is to say $\mathcal{H} = \bigotimes_{k=1}^{m} \mathcal{H}_k$ where all the individual Hilbert spaces $\mathcal{H}_k$ are $2d$ dimensional for some fixed $d$, i.e. $\mathcal{H}_k = \mathbb{C}^{2d}$ for all $k$. Now let us define a set of localised two outcome observables $M_{k,i}$ with $i=1,2$ and $k=1,2,...,m$ such that they only have $\pm 1$ as their eigenvalues, and are traceless. The authors go on to show that it is possible to choose a basis in which both $M_{k,1}$ and $M_{k,2}$ take the block diagonal form $\oplus_{j=1}^{d}\mathbb{C}^2$ simultanously for all $k$. The quantity of interest is the Bell correlation $$B = \langle{\psi|\mathcal{B}|\psi}\rangle.$$ where $$\mathcal{B} = \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m}\bigotimes_{k=1}^{m}M_{k,i_k}$$ is the Bell operator for some fixed $c_{i_1...i_m}$ coefficients. We're taking $M_{k,0}=1$ as the identity operator. As I understsand it, this is useful in having terms in $\mathcal{B}$ that have operators from, say, only two sites a la the CHSH case. The authors claim that "the maximum quantum value of the Bell inequality is achieved by a state that has support on a qubit at each site". I'm assuming by "value of Bell inequality" they mean value $B$. Moreover I'm assuming by "support over one quibit", they mean when the state $|\psi\rangle$ lies entirely in exaclty one $\mathbb{C}^2$ sector in all $\mathcal{H}_k$'s. Please correct me if either of these assumptions are incorrect or unreasonble.