# Bound on the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix?

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.

• so the off-diagonals are set to the min and max of the corresponding row, right? – Suvrit Jun 20 '11 at 0:13
• unless I am mistaken, there seems to be some problem here---it seems that the matrix $M\uparrow$ might be indefinite! – Suvrit Jun 20 '11 at 0:22
• @Suvrit : No, they are set to the max/min over all off-diagonal elements. This is important, as maximizing or minimizing over rows would not necessarily result in a (physical) positive matrix, e.g. M = {{1,0,1},{0,1,1},{1,1,1}} – Jess Riedel Jun 20 '11 at 0:28
• Ah, I see that the confusion is due to my lazy min and max notation. I'll make it more explicit – Jess Riedel Jun 20 '11 at 0:32
• but even with the global max being substituted for all the off-diagonals, it seems that the resulting matrix can be indefinite (i am experimenting only with density matrices $M$ that satisfy all your assumptions, except $m_{ii}=1/d$. – Suvrit Jun 20 '11 at 2:48