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.



Bound on the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix?
Indefiniteness is not an issue. If M is positive semidefinite, then M-uparrow and M-downarrow are positive semidefinite too; that's easy to prove.


However, I'm afraid one can find counterexamples, at least for the lower bound. Take two random numbers between 0 and 1/d, and allow all off-diagonal elements of M to be one or the other (while preserving symmetry and checking for positive semidefiniteness of M). The entropy of M can then be below the entropy of M-uparrow.
The following matrix, for example:
M=[1 a b b; a 1 b b;b b 1 b;b b b 1]/4
with a=0.01 and b=0.6
has entropy 0.934, while its M-uparrow has entropy 0.940.


