---
# System prepended metadata

title: Extremal Density Matrices
tags: [Entanglement Bootstrap, ' Quantum Marginal Problem']

---

---
tags: Entanglement Bootstrap, Quantum Marginal Problem
---

# Extremal Density Matrices

:::danger
**Last update: 8/30/2021, Isaac Kim**

Unfortunately, this line of approach seems to be too restrictive; see the last section for the underlying reason. Therefore, the notion of extremality, at least the one defined in this document, does not seem like a promising approach.

Nevertheless, this document is left as is because there are some interesting concepts introduced here. Some of those concepts may be migrated to a separate note in the future.
:::

Consider a quantum spin system arranged on $\mathbb{R}^2$. We can partition $\mathbb{R}^2$ into a set of $2$-cells, $1$-cells, and $0$-cells. Each $2$-cell and $1$-cell is homeomorphic to a disk and an open interval, respectively. The $0$-cells are points. In the figure below, a $2$-cell is represented by a red cell, a $1$-cell is represented by a blue line, and a $0$-cell is represented by a green dot. 

![](https://i.imgur.com/JRh1CCn.png)
 
(Physically, we can think of a $2$-cell as a coarse-grained degree of freedom representing a collection of microscopic degrees of freedom.)

## Primer on two-dimensional topology

The set of $0$-, $1$-, and the $2$-cells form a CW-complex, *complex* for short. We shall denote the set of $k$-cells as $\mathcal{C}_k$. A subset $\mathcal{C}'\subset \mathcal{C}_2$ of $2$-cells induces a complex defined by 
\begin{equation}
\mathcal{C}_k' := \{c: c\in \mathcal{C}_k, c\in \bar{c}'\,\, \text{for some } c'\in \mathcal{C}'\}
\end{equation}
for $k<2$. (Here $\bar{c}'$ means a closure of $c'$ in $\mathbb{R}^2$.)

:::success
**Definition 1.** 
Let $c\in\mathcal{C}'$. A *minimal neighborhood* of $c$ is
\begin{equation}
N_{\text{min}}^{\mathcal{C}'}(c) := \{c'\in \mathcal{C}': \bar{c}' \cap \bar{c} \neq \emptyset \}.
\end{equation}
Similarly, a minimal neighborhood of $\mathcal{C}''\subseteq \mathcal{C}'$ is
\begin{equation}
N_{\text{min}}^{\mathcal{C}'}(\mathcal{C}'') = \cup_{c\in \mathcal{C}''} N_{\text{min}}^{\mathcal{C}'}(c)
\end{equation}
:::

:::success
**Definition 2.** 
A *path* from $\mathcal{C}'\subseteq \mathcal{C}_2$ to $\mathcal{C}'' \subseteq \mathcal{C}_2$ is a sequence of sub complexes induced by a sequence of $2$-cells $(\mathcal{C}_2)_{i=1}^n$, $(\mathcal{C}_2)_1= \mathcal{C}'$, $(\mathcal{C}_2)_n = \mathcal{C}''$ s.t.
\begin{equation}
|(\mathcal{C}_2)_{i+1} \setminus (\mathcal{C}_2)_{i}|=1 \,\, \text{or} \,\, |(\mathcal{C}_2)_{i} \setminus (\mathcal{C}_2)_{i+1}|=1
\end{equation}
:::

:::success
**Definition 3.** 
A path is *increasing* if $|(\mathcal{C}_2)_{i+1} \setminus (\mathcal{C}_2)_{i}|=1$ for all $i$. A path is *decreasing* if $|(\mathcal{C}_2)_{i} \setminus (\mathcal{C}_2)_{i+1}|=1$ for all i.
:::


:::success
**Definition 4.** 
Let $\mathcal{C}'\subseteq \mathcal{C}_2$. $\overline{\mathcal{C}'}:= \cup_{c\in \mathcal{C}'} \bar{c}$.
:::

:::success
**Definition 5.**
A path $(\mathcal{C}_2)_{i=1}^n$ is *smooth* if $\overline{(\mathcal{C}_2)_{i}} \cong \overline{(\mathcal{C}_2)_{i+1}}$ for all $i$.
:::

:::danger
**Lemma 1.** 
Let $\mathcal{C}' \subset \mathcal{C}'' \subseteq \mathcal{C}_2$ and $\overline{\mathcal{C}'} \cong \overline{\mathcal{C}''} \cong D^2$. There exists a $c\in \mathcal{C}'' \setminus \mathcal{C}'$ s.t. $\overline{\mathcal{C}'} \cong \overline{\mathcal{C}'\cup \{c \}}$.
:::
**Proof of Lemma 1.**
Let us prove the contrapositive statement. First, we show that there must exist a $c\in \mathcal{C}'' \setminus \mathcal{C}'$ such that $c\in N_{\text{min}}^{\mathcal{C}''}(\mathcal{C}')$. If this is not the case, one can partition $\overline{\mathcal{C}''}$ into $\overline{\mathcal{C}'}\cup \overline{\mathcal{C}''\setminus \mathcal{C}'}$ so that $\overline{\mathcal{C}''\setminus \mathcal{C}'}$ has no points in common with  $\overline{\mathcal{C}'}$. Therefore, $\overline{\mathcal{C}''}$ is not connected and cannot be homeomorphic to $D^2$. Thus, $c\in N_{\text{min}}^{\mathcal{C}''}(\mathcal{C}')$.

Next, we show that there exists $c\in N_{\text{min}}^{\mathcal{C}''}(\mathcal{C}')$ s.t. $\overline{\mathcal{C}'} \cong \overline{\mathcal{C}'\cup \{c \}}$. If not, for all $c\in N_{\text{min}}^{\mathcal{C}''}(\mathcal{C}')$, $\overline{\mathcal{C}'\cup \{\mathcal{c}\}}$ is not homeomorphic to $D^2$. Since $\overline{\mathcal{C}'\cup \{c \}}$ is a connected subset of $\mathbb{R}^2$, $\overline{\mathcal{C}'\cup \{c \}}$ must be homeomorphic to a closed disk with $k>0$ open punctures. In particular, the interior of at least one of the punctures is not a subset of $\overline{\mathcal{C}''}$. Since $\overline{\mathcal{C}'\cup \{c \}}\subset \overline{\mathcal{C}''}$, $\mathcal{C}''$ must have a puncture and thus cannot be homeomorphic to $\overline{D^2}$. Thus, there exists $c\in N_{\text{min}}^{\mathcal{C}''}(\mathcal{C}')$ s.t. $\overline{\mathcal{C}'} \cong \overline{\mathcal{C}'\cup \{c \}}$. $\blacksquare$


:::info
**Theorem 1.**
Let $\mathcal{C}'\subset \mathcal{C}'' \subseteq \mathcal{C}_2$ and $\overline{\mathcal{C}'} \cong \overline{\mathcal{C}''} \cong \overline{D^2}$. There is a smooth increasing path from $\mathcal{C}'$ to $\mathcal{C}''$. 
:::
The proof follows straightforwardly from Lemma 1.

Let us also prove an intuitive fact which will prove to be useful later.

:::warning
**Proposition 1.**
Let $c\in \mathcal{C}_2$. Let $\mathcal{C}'$ be a subset of $N_{\text{min}}(c)$ which includes $c$. Then $\overline{\mathcal{C}'} \cong \overline{D^2}$.
:::

**Proof of Proposition 1.**

If $\mathcal{C}' = N_{\text{min}}(c)$, $\overline{\mathcal{C}'}\cong \overline{D^2}$ by definition. Note that, for $A,B \cong D^2$, $B\subset A$, s.t. $\partial A \cap \partial B \cong D^1$, $A\setminus B \cong D^2$. Thus, if $\mathcal{C}'$ is a strict subset of $N_{\text{min}}(c)$ s.t. $|\mathcal{C}'| = |N_{\text{min}}(c)|+1$, since  $c\in \mathcal{C}' \ \setminus N_{\text{min}}(c)$, $\overline{\mathcal{C}'} \cong \overline{D^2}$. This argument can be used recursively to prove $\overline{\mathcal{C}'}\cong \overline{D^2}$ for $|\mathcal{C}'| = |N_{\text{min}}(c)|+k$ for any $k>1$.

## Extremal density matrices

Now we move onto the realm of quantum mechanics. Consider a Hilbert space which has a tensor product structure over the $2$-cells:
\begin{equation}
\mathcal{H} = \bigotimes_{c\in\mathcal{C}_2} \mathcal{H}_c,
\end{equation}
where $\dim(\mathcal{H}_c) < \infty$.

:::success
**Definition 6.** 
Let $(\mathcal{C}_2)_{i=1}^n$ is a smooth increasing path and $\rho$ be a density matrix supported on $(\mathcal{C}_2)_n$, where $|(\mathcal{C}_2)_{1}|=1$.
\begin{equation}
MED^{(\mathcal{C}_2)_{i=1}^n}(\rho)= MED^{(\mathcal{C}_2)_{i=1}^{n-1}}(\rho) + 
S(\delta_n| (\mathcal{C}_2)_{n-1} \cap N_{\text{min}}(\delta_n)),
\end{equation}
where $\delta_n = (\mathcal{C}_2)_{n}\setminus (\mathcal{C}_2)_{n-1}$.
:::

:::success
**Definition 7.**
A density matrix $\rho$ over $\mathcal{C}'$ is *extremal* if for all smooth increasing path $P$ from a $2$-cell to $\mathcal{C}'$,
\begin{equation}
MED^P(\rho) = S(\rho).
\end{equation}
:::

* Remark: By SSA, $MED^{P}(\rho) \geq S(\rho)$ for general quantum states.

:::success
**Definition 8.**
A density matrix $\rho$ over $\mathcal{C}'$ is *locally extremal* if for all $c\in \mathcal{C}'$, the reduced density matrix of $\rho$ on $\mathcal{C}'\cap N_{\text{min}}(c)$ is extremal.
:::

:::danger
**Lemma 2.**
Let $\rho_A$ be extremal. If $B\subseteq A$ and $\overline{B}\cong \overline{A}$, $\rho_B$ is extremal. **Note: This is wrong. See the k-point function section.**
:::
**Proof or Lemma 2.**
Since $\overline{B}\cong \overline{A}$, there is a smooth increasing path from $B$ to $A$. For any smooth increasing path  from a single $2$-cell to $B$, we can concatenate this path with a path from $B$ to $A$, yielding a path from a single $2$-cell to $A$. Let us denote this path as $(P_1, P_2)$, where $P_1$ is a path from a $2$-cell to $B$ and $P_2$ is a path from $B$ to $A$.

\begin{equation}
MED^{(P_1, P_2)}(\rho_A) - S(\rho_A) = MED^{P_1}(\rho_B) - S(\rho_B) + \ldots,
\end{equation}
where the ellipsis represents a sum of $S(\rho_B)$ conditional entropies which altogether is nonnegative because of SSA. The left-hand-side is $0$ by definition and the right-hand-side is a sum of nonnegative terms. Therefore, both those terms must be zero. Thus, $MED^{P_1}(\rho_B) = S(\rho_B)$ for any path from a $2$-cell to $B$. $\blacksquare$

## BK Entropy: Key to efficient characterization

Naively, verifying the extremality condition seems like a formidable task because the number of paths grows very quickly with $|\mathcal{C}'|$. Note that, for a connected set of size, its boundary size is $\Omega(\sqrt{n})$. So, for the first $k= O\left(\sqrt{|\mathcal{C}'|}\right)$ steps, the number of distinct paths is $\Omega(\sqrt{k!})$. This naive counting argument thus leads to a number of conditons that scales *faster* than exponentially with $|\mathcal{C}'|$. 

Remarkably, there is a sufficient condition for verifying the extremality which reduces this cost dramatically. This is the main subject of this section. The main protagnist of this story is the *Bethe-Kikuchi entropy* ---BK entropy for short --- which we shall denote as $S_{\text{BK}}(\rho_A)$. The BK entropy depends implicitly on the notion of *elementary clusters*. In our context, this is simply a family of sets of $2$-cells, wherein each set is representing a local patch of the underlying physical system supported on a ball of bounded radius. Let $\mathcal{EC}$ be an ordered set of elementary clusters. Define $\mathcal{EC}_A$, where $A$ is an ordered set of $2$-cells, as 
\begin{equation}
\mathcal{EC}_A = (C_i: C_i\in \mathcal{EC} \text{ where } \forall c\in C,  c\in A).
\end{equation}
The BK entropy is defined as
\begin{equation}
S_{BK}^{\mathcal{EC}}(\rho_A) = \sum_{1\leq i \leq n} S(\rho_{C_i}) - \sum_{1\leq i<j\leq n} S(\rho_{C_i\cap C_j}) + \sum_{1\leq i<j<k\leq n}S(\rho_{C_i\cap C_j \cap C_k}) - \ldots.
\end{equation}


BK entropy satisfies the following *removal lemma*.
:::danger
**Lemma 3.** (Removal lemma) Let $\mathcal{EC}' = \mathcal{EC} \cup \{C' \}$ where $C' \subset C$ for some $C\in \mathcal{EC}$. Then
\begin{equation}
S_{\text{BK}}^{\mathcal{EC}'}(\rho_A) = S_{\text{BK}}^{\mathcal{EC}}(\rho_A).
\end{equation}
:::

**Proof of Lemma 3.**

If $C'\not\in \mathcal{EC}_A$, the statement is trivial. Thus, without loss of generality, suppose there are $n$ $C_i\in \mathcal{EC}_A$ s.t. $C' \subset C_i$. Define two sets:
\begin{equation}
\begin{aligned}
\mathcal{I}_A &= \{C: C\in \mathcal{EC}_A, C' \subset C\}, \\
O_A &= \mathcal{EC}_A \setminus \mathcal{I}_A.
\end{aligned}
\end{equation}
Note that $|\mathcal{I}_A|=n$.

Consider the decomposition of $S_{\text{BK}}^{\mathcal{EC}'}(\rho_A) - S_{\text{BK}}^{\mathcal{EC}}(\rho_A)$ in terms of $S(\rho_{C'})$, $S(\rho_{C'\cap C_1})$, $S(\rho_{C'\cap C_1\cap \ldots C_k})$, etc, where $C_i \in \mathcal{O}_A$. Without loss of generality, consider the term $S(\rho_{C'\cap C_1\cap \ldots C_k})$. This term appears in the entropy of an intersection of $C_1,\ldots, C_k\in \mathcal{O}_A$, $C'$ and elements in $\mathcal{I}_A$. Let $\mathcal{I}_{A,i}$ be the set of elements in $\mathcal{I}_A$ lying between $C_i$ and $C_{i+1}$; similarly, let $\mathcal{I}_{A,0}$ and $\mathcal{I}_{A,k}$ be the set of elements in $\mathcal{I}_A$ appearing before and after $C_1$ and $C_k$, respectively. The integer coefficient of $S(\rho_{C'\cap C_1\cap \ldots C_k})$ can be determined by summing over all possible subsets of $\mathcal{I}_{A,i}$ --- denoted as $\tilde{\mathcal{I}}_{A,i}$ --- and weighing them by $(-1)^{\sum_{i=0}^k |\tilde{\mathcal{I}}_{A,i}|}$.

This sum is exactly zero. To see why, fix all but one $\tilde{\mathcal{I}}_{A,i}$ and take the sum. Let $\tilde{\mathcal{I}}_{A,k}$ be the variable subset. The corresponding sum is proportional to
\begin{equation}
\sum_{j=0}^{m_k} \binom{m_k}{i}(-1)^j = 0,
\end{equation}
where $m_k = |\tilde{\mathcal{I}}_{A,k}|$ because the sum is equal to $(1+ (-1))^{m_k}$ by the binomial expansion. $\blacksquare$


Moreover,
:::danger
**Lemma 4.** (Inclusion-Exclusion Principle) Suppose every element in $\mathcal{EC}$ is contained in $AB, BC,$ or $AC$. Then
\begin{equation}
S_{\text{BK}}^{\mathcal{EC}}(\rho_{ABC})= S_{\text{BK}}^{\mathcal{EC}}(\rho_{AB})+
S_{\text{BK}}^{\mathcal{EC}}(\rho_{BC})+
S_{\text{BK}}^{\mathcal{EC}}(\rho_{AC})-
S_{\text{BK}}^{\mathcal{EC}}(\rho_{A})-S_{\text{BK}}^{\mathcal{EC}}(\rho_{B})-S_{\text{BK}}^{\mathcal{EC}}(\rho_{C}).
\end{equation}
:::

The proof follows straightforwardly from the assumption. On a similar note, if every element of $\mathcal{EC}$ is contained in $A$ or $B$,
\begin{equation}
S_{\text{BK}}^{\mathcal{EC}}(\rho_{AB})=
S_{\text{BK}}^{\mathcal{EC}}(\rho_{A})+
S_{\text{BK}}^{\mathcal{EC}}(\rho_{B}).
\end{equation}

## Extremality and BK entropy.

Now we establish an equivalence of extremal density matrices and locally extremal density matrices whose entropies are equal to their respective BK entropies. In this discussion, we shall choose $\mathcal{EC}$ using the following rule. Consider a graph obtained by assigning $2$-cells to the vertices and an edge between $c$ and $c'$ if $\overline{c} \cap \overline{c'}\neq \emptyset$. This is a planar graph, so we can naturally define a notion of face. We choose $\mathcal{EC}$ to be the union of the set of vertices, edges, and faces **on the newly defined planar graph**. (Note that this is not exactly a dual graph of the original planar graph.) Lastly, we will focus on a hexagonal lattice. (Its dual lattice is thus triangular.) While we suspect the proofs in this section will apply to more general planar graphs, we leave this as an open problem.

![](https://i.imgur.com/ylZjqHB.png)
In the figure above, the dual $0$-, $1$-, and $2$-cell is colored in red, green, and blue, respectively. Since $\mathcal{EC}$ will be fixed throughout this section, we drop it in $S_{\text{BK}}^{\mathcal{EC}}$, instead using $S_{\text{BK}}$ as a short-hand notation.

:::danger
**Lemma 5.**
Let $A\subset N_{\text{min}}(c)$ for some $c\in \mathcal{C}_2$. A density matrix $\rho_A$  is extremal iff it is (i) locally extremal and (ii) $S(\rho_A) = S_{\text{BK}}(\rho_A)$.
:::

**Proof of Lemma 5.**

Simply consider all possibilities and use induction. $\blacksquare$


:::danger
**Lemma 6.**
Suppose a density matrix $\rho_A$ is locally extremal and $\overline{A} \cong D^2$. Then $MED^P(\rho_A)  = S_{\text{BK}}(\rho_A)$ for any path $P$ from a $2$-cell to $A$.
:::
**Proof of Lemma 6.**
The proof is based on induction. Suppose we have proven this statement for all disks $B\subset A$. 
\begin{equation}
MED^{P}(\rho_A) = MED^{P'}(\rho_B) + S(A\setminus B| N_{\text{min}}(A\setminus B)\cap B),
\end{equation}
where $P'$ is a subsequence of $P$ obtained by removing the last element. By our hypothesis,
\begin{equation}
\begin{aligned}
MED^{P}(\rho_A) &= S_{\text{BK}}(\rho_B)  + S(A\setminus B| N_{\text{min}}(A\setminus B)\cap B). 
\end{aligned}
\end{equation}

Now we relate the conditional entropy to the BK entropy. Note that, the set we are focused on is a subset of a minimal neighborhood. By the local extremality, the conditional entropy can be further decomposed into a BK-entropy. Further, using the inclusion-exclusion principle, we get $MED^{P}(\rho_A) = S_{\text{BK}}(\rho_A)$. $\blacksquare$




:::info
**Theorem 2.** 
A density matrix $\rho_A$ s.t. $\overline{A} \cong D^2$  is extremal iff it is (i) locally extremal and (ii) $S(\rho_A) = S_{\text{BK}}(\rho_A)$.
:::
**Proof of Theorem 2.**

If $\rho_A$ is extremal, by Lemma 2, $\rho_B$ is extremal for every $\overline{B}\cong \overline{A}$ which is a subset of $A$. Since $\overline{A \cap N_{\text{min}}(c)}\cong \overline{D^2}$ for every $c\in A$ (see Proposition 1), $\rho_A$ is locally extremal. 

Next, we show that extremality of $\rho_A$ implies $S(\rho_A) = S_{\text{BK}}(\rho_A)$, by induction. Suppose we have proved this statement for any $B\subset A$ s.t. $\overline{B}\cong \overline{A}$. By the extremality of $\rho_A$, 
\begin{equation}
\begin{aligned}
S(\rho_A) &= MED^P(\rho_B) + S(A\setminus B| N_{\text{min}}(A\setminus B)\cap B) \\
&= S_{\text{BK}}(\rho_B) + S(A\setminus B| N_{\text{min}}(A\setminus B)\cap B)
\end{aligned}
\end{equation}
for any path $P$ from a $2$-cell to $B$. By the local extremality and our hypothesis,
\begin{equation}
\begin{aligned}
S(\rho_A) &= S_{\text{BK}}(\rho_B) + S_{\text{BK}}(\rho_{N_{\text{min}}(A\setminus B)\cap A}) + S_{\text{BK}}(\rho_{N_{\text{min}}(A\setminus B)\setminus B}).
\end{aligned}
\end{equation}

Now we are in a position to use the inclusion-exclusion principle of the BK entropy; see Lemma 4 and a remark below it.  Note that every element in $\mathcal{EC}^A$ is either in $B$ or $N_{\text{min}}(A\setminus B)\cap A$. Moreover, since there is no cell with nontrivial overlap with an element in $B\setminus N_{\text{min}}(A\setminus B)$ and an element in $A\setminus B$, we get, using the inclusion-exclusion principle,
\begin{equation}
S(\rho_A)=S_{\text{BK}}(\rho_A).
\end{equation}
Note that the base case of our induction argument follow from Lemma 5. Thus we have proved the "only if" part of the statement.

Now let us prove the converse statement. Given $A$, pick a $B\subset A$ s.t. $|B| = |A|-1$ and $\overline{B} \cong \overline{A}$. Suppose we have proven that, for all those $B$, that if $\rho_B$ is locally extremal and  $S_{\text{BK}}(\rho_B)= S(\rho_B)$ then $\rho_B$ is extremal. If $S(\rho_A)= S_{\text{BK}}(\rho_A)$. Then by the local extremality and the inclusion-exclusion principle of the BK entropy,
\begin{equation}
S(\rho_A) - S_{\text{BK}}(\rho_B)= S(A\setminus B| N_{\text{min}}(A\setminus B)\cap B).
\end{equation}
By the local extremality, $S_{\text{BK}}(\rho_B) = MED^P(\rho_B)$ for any path $P$ from a $2$-cell to $B$ (Lemma 6). Because any path from a $2$-cell to $A$ is a path from the same $2$-cell to some disk $B\subset A$ followed by $A\setminus B$, $\rho_A$ is extremal. $\blacksquare$

Thus, now we have a better way of certifying extremality. Simply check the local extremality and calculate the BK entropy and check if it is equal to its global entropy. Alas, while this reduces the number of constraints that need to be checked to something that scales linearly with the number of $2$-cells, this is still a difficult condition to verify. Caculating the von Neumann entropy of a large system is not easy!

## Merging extremal density matrices

Now we show that extremal density matrices form a "nice" set. What we mean by this is that given two extremal density matrices, we can "merge" them together form a new extremal density matrix. The only extra condition needed is local consistency. That is, the two density matrices must yield the same reduced density matrix on their overlapping support. Obviously, if this condition is not satisfied, no such density matrix can exist (let alone something extremal).

:::danger
**Lemma 7.**
(Merging lemma) Let $\rho_{ABC}$ and $\sigma_{BCD}$ be density matrices such that $\rho_{ABC} \stackrel{c}{=} \sigma_{BCD}$ and $I(A:C|B)_{\rho} = I(B:D|C)_{\sigma} =0$. Then there exists a density matrix $\tau_{ABCD}$ such that $\tau_{ABC}=\rho_{ABC}$, $\tau_{BCD}=\sigma_{BCD}$, and $I(AB:D|C) = I(A:CD|B)=0$.
:::

By the uniqueness of the maximal entropy state, the state $\tau_{ABCD}$ that satisfies the above-mentioned conditions is unique. We shall represent this as $\tau_{ABCD} = \rho_{ABC} \Join \sigma_{BCD}$ and $\tau_{ABCD}$ as the *merged density matrix*.


:::danger
**Lemma 8.**
(Extremal merging lemma) Let $\rho_A$ and $\sigma_B$ be extremal density matrices and $\rho_{A} \stackrel{c}{=}\sigma_B$. If there is a partition of $A\cap B = C \cup D$ for disjoint sets $C$ and $D$ such that $A\setminus D$, $B\setminus C$, $C\cup D$ are disks and $d(A\setminus B,B\setminus C), d(B\setminus A, A\setminus C)>0$, there exists an extremal density matrix $\tau_{A\cup B} \stackrel{c}{=}\rho_A, \sigma_B$.
:::
Again, by the uniquenes of the maximal entropy state, $\tau_{A\cup B}$ is unique. So we can simply say that, under the conditions described above, $\tau_{A\cup B}= \rho_A \Join \sigma_B$ is extremal.

## Applications

### Solution to the quantum marginal problem 

Using the following merging sequence, given a set of locally consistent extremal density matrics on $3\times 3$ patches, we can infer the existence of a global state consistent with those density matrices. Note that at all stage all the density matrices remain extremal. For each diagram, the the red and the blue regions are the supports of the respective density matrices.

* Step 1: Merging the first three rows.
![](https://i.imgur.com/AoxTGZt.png)
* Step 2: Merging the fourth row.
![](https://i.imgur.com/spdTMZ0.png)
![](https://i.imgur.com/OXlvxBJ.png)
* Step 3: Repeat Step 2.




### Free energy bound

Using the exrtremality of the merged density matrix, we can write down an exact expression for the global entropy; this is precisely the BK-entropy.

### $k$-point functions
$k$-point functions can be computed efficiently as long as the $k$ points all lie on a single line.
![](https://i.imgur.com/2B3Qcoc.png)
![](https://i.imgur.com/hE8Zioj.png)
Here the colored region forms a one-dimensional Markov chain. These don't work because there is no path from them to a "regular" large disk, technically speaking. However, this idea does work if we thicken the chain sufficiently.

This kind of flexibility shows that the notion of extremal density matrix is too restrictive. One can argue that, based on the consistency of the $2$-point correlation functions over two different paths (specifically, one which is short and another which is long), that correlation ought to vanish strictly outside some bounded region. Let's discuss more about this in detail below.


## Why extremality is too much.

Extremality is quite restrictive because it requires $S(\rho_A) = MED^{P}(\rho_A)$ for all path from a $2$-cell to $A$. For example, consider a "trivial" example which has a Markov chain along the $x$-direction on a single $y$ coordinate and a trivial product state on the rest. Markov chains can have nonzero correlation and one can show that the extremality condition does not apply to this example. The basic reason is that if one considers a path in which, if we consider the subsequence formed by the points with $y$-coordinates on which the Markov chain exists, the ordering is not strictly increasing or decreasing, in order for the extremality condition to hold, the Markov chain must have zero correlation.



