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Consider the following inequality of Lemma 1 arising in The law of large numbers for quantum stochastic filtering and control of many-particle systems :

$$\Big|tr(L\gamma LB) - \frac{1}{2}tr(B(L\gamma + \gamma L))tr(BL + \gamma L) + tr(B\gamma)tr(BL)tr(\gamma L) \Big| \leq 5||L||^2 tr\big((Id-\gamma)B\big),$$

where $\gamma$ is a ($n\times n$) density matrix of rank one (see https://en.wikipedia.org/wiki/Density_matrix for the definition of density matrix), $B$ is a density matrix and $L$ a hermitian matrix. My question is whether the above inequality holds (up to some constant in front of $tr\big((Id-\gamma)B\big)$) by assuming only that $\gamma$ is a density matrix (with the assumptions on $\gamma, L$ unchanged)?

PS : Without loss of generality we may assume that $\gamma$ is diagonal with its diagonal elements $\gamma_1,\ldots, \gamma_n\in\mathbb R_+$ s.t. $\sum_{i=1}^N \gamma_i=1$.

PS2 : Denote respectively the l.h.s. and r.h.s. by

$$f(\gamma):=\Big|tr(L\gamma LB) - \frac{1}{2}tr(B(L\gamma + \gamma L))tr(BL + \gamma L) + tr(B\gamma)tr(BL)tr(\gamma L) \Big|$$

and

$$g(\gamma):=5||L||^2 tr\big((Id-\gamma)B\big).$$

Following the hint of Narutaka OZAWA, I try to reduce the general case to the case where $\gamma$ has rank one. The difficulty for me is that $g$ is affine while $f$ is not (it is even not trivial for me to show $f$ is convex).

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  • $\begingroup$ @JosephVanName That's a really unexpected observation. Could you please give more details how computation yields the inverse inequality? I know very few on computer experiments. Many thanks $\endgroup$
    – Fawen90
    Commented Jan 27, 2023 at 7:53
  • $\begingroup$ What you had is right. I was getting the inequality in the wrong direction. $\endgroup$ Commented Jan 27, 2023 at 17:04
  • $\begingroup$ Using computer experiments with gradient descent (I used the Flux library in the language Julia). I conjecture the inequality $$|\text{tr}(L\gamma LB)−0.5\cdot \text{tr}(B(L\gamma+\gamma L))\text{tr}(BL+\gamma L)+\text{tr}(B\gamma)\text{tr}(BL)\text{tr}(\gamma L)|\leq \|L\|_p^2\cdot\text{tr}((Id−\gamma)B)⋅4^{1/p}$$ for $1\leq p\leq\infty$ where $\|\cdot\|_p$ denotes the Schatten $p$-norm. $\endgroup$ Commented Jan 27, 2023 at 17:06
  • $\begingroup$ This is true, but for an uninteresting reason. The $\mathrm{tr}$-norm distance from $\gamma$ (or $B$) to a rank-one projection is dominated by a constant (perhaps $2$) multiple of $\mathrm{tr}((1-\gamma)B)$. Hence the problem reduces to the case where $\gamma$ has rank one (modulo changing the constant in front of $\mathrm{tr}((1-\gamma)B)$ as you suggested). $\endgroup$ Commented Jan 30, 2023 at 6:58
  • $\begingroup$ @NarutakaOZAWA Thanks so much for pointing out this. Could you please detail how to prove what you have mentioned? I'm not very familiar with the tricks in matrix analysis and I'm interested in this inequality. Thank you very kindly for your consideration $\endgroup$
    – Fawen90
    Commented Jan 30, 2023 at 13:34

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