Role of convexity in proof of Monogamy of Bell correlations

Question

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.

Quick Background:

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.

Problem:

Let $|\psi\rangle$ be some state in $\mathcal{H}$. In the second to last paragraph in page 2, the authors define $\rho_{k,j}$ as the "reduced density matrix obtained by projecting $|\psi\rangle$ onto the $j$’th $\mathbb{C}^2$ factor of the $\bigoplus_{j=1}^{d}\mathbb{C}^2$ subspace induced by $M_{k,1}$ and $M_{k,2}$ at site $k$." Now they state that $B$ can be written as $$B = \sum_{j=1}^{d} \text{tr} (\mathcal{B}\rho_{j,k}).$$ They further argue since this is a convex sum over the $\mathbb{C}^2$ sectors, "it follows that the maximum is achieved by a state with support on a qubit at site $k$." I'm assuming they're talking about projecting $|\psi\rangle$ on the $j$'th $\mathbb{C}^2$ subspace of $k$'th site. Here are my questions.

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?

My Attempt at a solution

Let's define a projector $P_{j,k}$ which projects a state in $\mathcal{H}$ to the $j$'th $\mathbb{C}^2$ sector of the $k$'th site. For a state $|\psi\rangle$, I'm assuming $\rho_{j,k}\equiv P_{j,k}|\psi\rangle \langle \psi | P_{j,k}$. This may not be, and in all possibility isn't, the $\rho_{j,k}$ the paper is talking about, but I couldn't find another reasonable interpretation. Now, since both $M_{k,1}$ and $M_{k,2}$ are block diagonal in the sectors we obtain the required form $B = \sum_{j=1}^{d} \text{tr} (\mathcal{B}\rho_{j,k})$. Now assuming a dot product structure on the space of all operators on $\mathcal{H}$ given by $A\cdot B \equiv \text{tr}(AB)$, we have $B = \sum_{j=1}^{d} \mathcal{B} \cdot \rho_{j,k}$. Since $\rho_{j,k}\cdot \rho_{j',k}=0$ if $j\neq j'$, all the $\rho_{j,k}$'s are orthogonal vectors. I guess now there is some argument with which we can say that the maximum value of $B$ can be achieved when $\mathcal{B}$ is aligned perfectly parallel to one of the $\rho_{j,k}$ and it's only that $\rho_{jk}$ that's nonzero. But this doesn't seem to be a particularly strong argument since it suggests that $\mathcal{B}$ lives is a $d$ dimensional vector space, which is obviously not true because we can arbitrary increase the degrees of freedom in its definition by increasing $k$. Moreover this argument doesn't seem to use convexity anywhere so this is probably not what the authors are getting at.

Any help regarding elucidating this proof will be of great help. Thanks in advance.

Answer 1

This simultaneous block diagonalization is known as Jordan's lemma (not the complex analysis one though). It's a very common technique in the analysis of device-independent protocols as it allows you (under certain conditions) to reduce the analysis from an arbitrary dimension system to qubit systems. You should take a look at Section 4.3.1 in Scarani15 for more details. I'll try to run through the calculation briefly but there will definitely be some gaps that need filling for a rigorous argument.

To start we are considering a two-input, two-output, $m$-party Bell-inequality with outputs $\pm 1$. We'll assume each party measures projectively and each Hilbert space is of the same dimension (otherwise dilate). As you note in your background for each party we can assume that the two observables they have can be simulataneously block diagonalized. That is, there exists a basis of their Hilbert space in which the observables are both block diagonal with blocks of size $2\times 2$. Working from now on in this basis we have that for party $k$, $$ M_{k,1} = \bigoplus_n M_{k,1,n} \\ M_{k,2} = \bigoplus_n M_{k,2,n} $$ where $M_{k,i,n}$ are qubit observables with eigenvalues $\pm 1$.

Now consider a single term in the Bell-operator $\mathcal{B}$ (without the coefficient). This is something of the form $$ M_{1,i_1} \otimes M_{2,i_2} \otimes \dots \otimes M_{m,i_m} $$ which is equal to by the block diagonalization $$ \bigoplus_{n_1} M_{1,i_1,n_1} \otimes \bigoplus_{n_2} M_{2,i_2,n_2} \otimes \dots \otimes \bigoplus_{n_m} M_{m,i_m,n_m}. $$ Now this tensor product of block diagonal operators is itself a block diagonal operator, one block for each tuple $(n_1,\dots, n_m)$, so at this point I think its cleanest to just define some indexing over the tuples $n = (n_1,\dots, n_m)$ and redefine this as something of the form $$ \bigoplus_n M_{1,i_1,n_1} \otimes M_{2,i_2,n_2} \otimes \dots \otimes M_{m,i_m,n_m}. $$ Notice that for each index $n$ the operator in the block is a $\pm 1$ eigenvalued observable on $m$ qubits. As both of the observables for each party are simultaneously block diagonal (and the identity is trivially block diagonal) this block diagonal form is true for each term in the Bell operator and thus we can write $$ \begin{aligned} \mathcal{B} &= \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m} M_{1,i_1} \otimes M_{2,i_2} \otimes \dots \otimes M_{m, i_m} \\ &= \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m} \bigoplus_n M_{1,i_1,n_1} \otimes M_{2,i_2,n_2} \otimes \dots \otimes M_{m,i_m,n_m} \\ &= \bigoplus_n \sum_{i_1=0}^{2}...\sum_{i_m=0}^{2}c_{i_1 ... i_m} M_{1,i_1,n_1} \otimes M_{2,i_2,n_2} \otimes \dots \otimes M_{m,i_m,n_m} \\ &= \sum_n \mathcal{B}_n. \end{aligned} $$ Where on the final line we have defined $\mathcal{B}_n$ to be the operator that is the same as $\mathcal{B}$ on block $n$ but is zero elsewhere. If we let $\Pi_n$ be the projector onto block $n$ then $\Pi_{n_1} \mathcal{B}_r \Pi_{n_2} = \delta_{r n_1} \delta_{n_1 n_2} \mathcal{B}_r$.

Now consider an arbitrary state $\rho$ for the $m$ parties and its projection onto the blocks i.e. $\hat{\rho} = \sum_n \Pi_n \rho \Pi_n$. Note that $\hat \rho$ is also a state. Then the expected Bell-value for the projected state is $$ \begin{aligned} B &= \mathrm{Tr}[\mathcal{B} \hat\rho] \\ &= \sum_n \mathrm{Tr}[ \mathcal{B}_n \hat\rho ] \\ &= \sum_n \mathrm{Tr}[ \mathcal{B}_n \sum_{r} \Pi_{r} \rho \Pi_{r}] \\ &= \sum_n \sum_{r}\mathrm{Tr}[\Pi_{r} \mathcal{B}_n \Pi_{r} \rho] \\ &= \sum_n \mathrm{Tr}[\Pi_n \mathcal{B}_n \Pi_n \rho] \\ &= \mathrm{Tr}[\sum_n \Pi_n \mathcal{B}_n \Pi_n \rho] \\ &= \mathrm{Tr}[\mathcal{B} \rho]. \end{aligned} $$ Thus the expected Bell-value does not change when we consider a state $\rho$ and the projection of $\rho$ onto the blocks. For the purposes of computing the maximum Bell-value it is sufficient to consider states that are also block diagonal.

In the same vein as $\mathcal{B}_n$, let $\rho_n = \Pi_n \rho \Pi_n$. Note that the restriction of $\rho_n$ to its support is a subnormalized $m$-qubit state with a trace $p(n) = \mathrm{Tr}[\Pi_n \rho \Pi_n]$. So now we want to try to compute the maximum violation. Lets begin with the $m$-qubit problem $$ \max \mathrm{Tr}[\mathcal{B}_0 \tau] $$ where we maximize over all possible $m$-qubit measurements and $m$-qubit states $\tau$. Let $\mathcal{B}^*_0$ and $\tau^*$ be the optimal choice for this maximization.

Now back to the full problem. Let $\tilde \rho_n := \frac{\rho_n}{p(n)}$ be the normalized version of $\rho_n$. Taking the maximum over all measurements and states we have $$ \begin{aligned} \max_{\mathcal{B}, \rho} \mathrm{Tr}[\mathcal{B} \rho] &= \max_{\mathcal{B}_n, \rho_n} \sum_n \mathrm{Tr}[\mathcal{B}_n \rho_n] \\ &= \max_{\mathcal{B}_n, \tilde{\rho}_n, p(n)} \sum_n p(n) \mathrm{Tr}[\mathcal{B}_n \tilde\rho_n] \\ &\leq \max_{p(n)} \sum_n p(n) \max_{\mathcal{B}_n, \tilde{\rho}_n} \mathrm{Tr}[\mathcal{B}_n \tilde\rho_n] \\ & = \max_{p(n)} \sum_n p(n) \mathrm{Tr}[\mathcal{B}_0^* \tau^*] \\ & = \mathrm{Tr}[\mathcal{B}_0^* \tau^*] \end{aligned} $$ Where on the third line we split the maximization into a maximization of the subnormalization weights and a maximization over the separate blocks and then used the fact that a maximization of a sum is always smaller than the sum of the maximizations. On the fourth line we have identified that $\max \mathrm{Tr}[\mathcal{B}_n \tilde\rho_n]$ is effectively an $m$-qubit problem as the operators in the trace have support only on some $m$-qubit subspace and that $(\mathcal{B}_0^*, \tau)$ is the optimizer of the $m$-qubit problem. Then on the final line we just have $\sum_n p(n) = 1$.

So we have found that the maximum Bell-value can always be upper bounded by the maximum Bell-value for an $m$-qubit system hence it is sufficient to consider an $m$-qubit system when considering the maximum Bell-value. And moreoever, the maximum Bell-value can always be achieved by an $m$-qubit system.