Can the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix with equal on-diagonal terms be bounded by equalizing all off-diagonal elements to their highest/lowest value?

## Statement of problem

Consider the density matrix $M = (m_{i,j})$ in $d$-dimensions with all positive elements: $m_{i,j} > 0$. From physics, a density matrix is Hermitian, positive semi-definite, and has unit trace:

$\quad M^\dagger = M, \quad 0 \le M \le 1, \quad \mathrm{tr} \, \, \, M = 1 .$

For now, we also assume that $M$ has equal on-diagonal elements: $m_{i,i} = 1/d$.

Now, consider the two density matrices $M^\uparrow$ and $M^\downarrow$ formed by setting all off-diagonal elements of $M$ to their maximum and minimum value, respectively:

$\quad m_{i,j}^{\uparrow} = \max_{k \neq l} (m_{k,l}) , \qquad i \neq j$

$\quad m_{i,j}^{\downarrow} = \min_{k \neq l} (m_{k,l}) , \qquad i \neq j$

$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i}$

(Above, the extremizations are taken over *all* off-diagonal elements, i.e. all $k$ and all $l$ such that $k \neq l$.) With the von Neumann entropy of a matrix $A$ defined as

$\quad S[A]= \mathrm{tr} (-A \ln A),$

can we bound

$\quad S[M^\uparrow] \le S[M] \le S[M^\downarrow] \quad ?$

## Physics Intuition

Because $M$ is a positive matrix, it can be expressed as a Gram matrix for some set of $d$ vectors $\{v_i\}$. That is, the elements of $M$ are just the inner products of the $v_i$:

$\quad m_{i,j} = (v_i , v_j )$

These $v_i$ are unique up to a global unitary. If we want, we can work with the normalized set $\{e_i = v_i/ \sqrt{p_i} \}$ with probability distribution $\{p_i = (v_i , v_i ) \}$.

The density matrix we describe above can be obtained by starting with a system $\mathcal{S}$ in a pure state $\vert \psi \rangle= \sum_{i = 1}^d \sqrt{p_i} \vert s_i \rangle$ (where $\{{\vert s_i \rangle\}}$ is an orthonormal basis for $\mathcal{S}$) and environment $\mathcal{E}$ in pure state $\vert e_0 \rangle$, and evolving forward under the unitary $U$ which sends

$\quad \vert s_i \rangle \otimes \vert e_0 \rangle \to \vert s_i \rangle \otimes \vert e_i \rangle$

where $\langle e_i \vert e_j \rangle \equiv (e_i,e_j) = m_{i,j} \sqrt{p_i p_j}$. Then

$\quad \mathrm{tr}_\mathcal{E} [ U ( \vert \psi \rangle \langle \psi \vert \otimes \vert e_0 \rangle \langle e_0 \vert ) U^\dagger ] = \rho_{\mathcal{S}} \equiv M $

Having equal on-diagonal elements of $M$ is equivalent to $p_i = 1/d$. $M$ being positive means that all the vectors $\vert e_0 \rangle$ can be fit in the first "orthant", i.e. there is a single basis in which *all* the $\vert e_0 \rangle$ have all positive components.

The physics intuition is that by decreasing the off-diagonal elements of this density matrix to their min value (i.e. we are making more distinguishable the environment states corresponding to distinct system states) we are just *increasing* the decoherence. Therefore, the entropy should go up as we pass from $M$ to $M^{\downarrow}$. Likewise, $M^{\uparrow}$ has *less* decoherence, and should have a lower entropy.

## Numerical Evidence

I've sampled many millions of density matrices of the described form, and have never found a violation of the inequality in question. However, it's also always been true (numerically) that

$\quad M \succ M^{\downarrow}$

but

$\quad M^{\uparrow} \nsucc M \quad \mathrm{and} \quad M^{\uparrow} \nprec M$

where $\succ$ is the majorization partial order on density matrices. (Entropy is a Shur-concave function, and therefore preserves the majorization order.) It was a surprise to me that $M^{\uparrow} \nsucc M$, and this lowers confidence in the physics intuition described above. If the inequality concerning entropies is true, it must make use of the specific properties of the entropy function, *not* just that it's Shur-concave.

## Generalizing to unequal diagonal elements

When the diagonal elements of $M$ are unequal, we go back to the physics motivation to define $M^{\uparrow}$ and $M^{\downarrow}$. This situation corresponds to unequal $p_i$:

$\quad m_{i,i} = (v_i,v_i) = p_i$

$\quad m_{i,j} = (v_i,v_j) = \sqrt{p_i p_j} (e_i,e_j), \quad i \neq j .$

Additional decoherence will correspond to less overlap (more distinguishability) between the environmental states $e_i$, which remain normalized. This means we define

$\quad \gamma^{\uparrow} = \max_{k \neq l} (e_k,e_l) = \max_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$

$\quad \gamma^{\downarrow} = \min_{k \neq l} (e_k,e_l) = \min_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$

$\quad m_{i,j}^{\uparrow} = \gamma^{\uparrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$

$\quad m_{i,j}^{\downarrow} = \gamma^{\downarrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$

$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i} = p_i$

(Here, $\gamma^{\uparrow}$ and $\gamma^{\downarrow}$ are the largest and smallest "decoherence factors".) We can then ask whether the above inequality is true in this more general case.