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? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:30:30Z http://mathoverflow.net/feeds/question/68249 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68249/can-the-von-neumann-entropy-of-a-positive-positive-semi-definite-and-unit-trace 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? Jess Riedel 2011-06-19T23:37:41Z 2011-08-16T08:59:53Z <h2>Statement of problem</h2> <p>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:</p> <p>$\quad M^\dagger = M, \quad 0 \le M \le 1, \quad \mathrm{tr} \, \, \, M = 1 .$ </p> <p>For now, we also assume that $M$ has equal on-diagonal elements: $m_{i,i} = 1/d$.</p> <p>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:</p> <p>$\quad m_{i,j}^{\uparrow} = \max_{k \neq l} (m_{k,l}) , \qquad i \neq j$</p> <p>$\quad m_{i,j}^{\downarrow} = \min_{k \neq l} (m_{k,l}) , \qquad i \neq j$</p> <p>$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i}$</p> <p>(Above, the extremizations are taken over <em>all</em> 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 </p> <p>$\quad S[A]= \mathrm{tr} (-A \ln A),$</p> <p>can we bound</p> <p>$\quad S[M^\uparrow] \le S[M] \le S[M^\downarrow] \quad ?$</p> <h2>Physics Intuition</h2> <p>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 $v_i$:</p> <p>$\quad m_{i,j} = (v_i , v_j )$</p> <p>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 $\{p_i = (v_i , v_i ) \}$.</p> <p>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</p> <p>$\quad \vert s_i \rangle \otimes \vert e_0 \rangle \to \vert s_i \rangle \otimes \vert e_i \rangle$</p> <p>where $\langle e_i \vert e_j \rangle \equiv (e_i,e_j) = m_{i,j} \sqrt{p_i p_j}$. Then </p> <p>$\quad \mathrm{tr}_\mathcal{E} [ U ( \vert \psi \rangle \langle \psi \vert \otimes \vert e_0 \rangle \langle e_0 \vert ) U^\dagger ] = \rho_{\mathcal{S}} \equiv M$</p> <p>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 <em>all</em> the $\vert e_0 \rangle$ have all positive components.</p> <p>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 <em>increasing</em> the decoherence. Therefore, the entropy should go up as we pass from $M$ to $M^{\downarrow}$. Likewise, $M^{\uparrow}$ has <em>less</em> decoherence, and should have a lower entropy.</p> <h2>Numerical Evidence</h2> <p>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 </p> <p>$M \succ M^{\downarrow}$</p> <p>but</p> <p>$M^{\uparrow} \nsucc M \quad \mathrm{and} \quad M^{\uparrow} \nprec M$</p> <p>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, <em>not</em> just that it's Shur-concave.</p> <h2>Generalizing to unequal diagonal elements</h2> <p>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 $p_i$:</p> <p>$\quad m_{i,i} = (v_i,v_i) = p_i$</p> <p>$\quad m_{i,j} = (v_i,v_j) = \sqrt{p_i p_j} (e_i,e_j), \quad i \neq j .$</p> <p>Additional decoherence will correspond to less overlap (more distinguishability) between the environmental states $e_i$, which remain normalized. This means we define </p> <p>$\quad \gamma^{\uparrow} = \max_{k \neq l} (e_k,e_l) = \max_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$</p> <p>$\quad \gamma^{\downarrow} = \min_{k \neq l} (e_k,e_l) = \min_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$</p> <p>$\quad m_{i,j}^{\uparrow} = \gamma^{\uparrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$</p> <p>$\quad m_{i,j}^{\downarrow} = \gamma^{\downarrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$</p> <p>$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i} = p_i$</p> <p>(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.</p> http://mathoverflow.net/questions/68249/can-the-von-neumann-entropy-of-a-positive-positive-semi-definite-and-unit-trace/72968#72968 Answer by KoenraadA for 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? KoenraadA 2011-08-16T08:59:53Z 2011-08-16T08:59:53Z <p>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.</p> <p>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.</p>