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EDIT: sorry for all the crappy formulas, I tried to add the more context I can, I hope it is better.


I am currently trying to analytically bound the following term: $$ \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert $$ with $$ S_2(\lambda) = e^{\frac{\lambda}{2}H_0}e^{\lambda H_1}e^{\frac{\lambda}{2}H_0} $$ for $H_0$ and $H_1$ two non-commuting matrices with the following properties:

  • $H_0^2 = H_1^2 = I$
  • $\Lambda = \max \left\{ \vert\vert H_0\vert\vert, \vert\vert H_1\vert\vert \right\} = \vert\vert H_0\vert\vert = \vert\vert H_1\vert\vert = 1$.

My final goal is to bound the following error $$ \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)^r\right\vert\right\vert \leqslant g\left(\lambda, \Lambda, r\right) \leqslant \epsilon $$ by a function $g$ and to solve $g\left( \lambda, \Lambda, r \right) \leqslant \epsilon$ for $r$ to obtain an expression of $r$ as: $$ r = h\left( \lambda, \Lambda, \epsilon \right). $$

I currently have 2 bounds that are far from optimal from [1] (Section F.1, page 24-26): $$ r^{\text{analytic}} = \left\lceil \max \left\{ 4 \Lambda \vert\lambda\vert, \sqrt{\frac{\left(4 \Lambda \vert\lambda\vert\right)^3}{3\epsilon}} \right\} \right\rceil $$ and $$ r^{\text{minimized}} = \min \left\{ r \in \mathbb{N}^* : \frac{\left(4 \Lambda \vert\lambda\vert\right)^3}{3r^2} \exp\left(\frac{4 \Lambda \vert\lambda\vert}{r}\right) \leqslant \epsilon \right\}. $$

During the mathematical development of the error term I stumbled upon a mathematical expression that I failed to major finely enough to obtain a usable error bound: $$ \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert \leqslant \sum_{n=0}^{+\infty} \vert\lambda\vert^n \Lambda^n \left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right] $$ with $$ f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left\vert \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right\vert. $$

Using the triangle inequality, I ended up with

$$ f(n) \leqslant \frac{n(n+1)}{2 n!} + \frac{2^n - 1}{n!} $$ by using the identity $$ \sum_{i=0}^n \frac{n!}{i! (n-i)!} = 2^n $$ which leads to $$ \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert \leqslant 2 \left[ e^{2\vert\lambda\vert\Lambda} - e^{\vert\lambda\vert \Lambda} \right]. $$ and finally to $$ \begin{align} \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)^r\right\vert\right\vert &\leqslant r \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)\right\vert\right\vert\\ &\leqslant g(\lambda, \Lambda, r) = 2r e^{\vert\lambda\vert\Lambda / r} \left( e^{\vert\lambda\vert\Lambda / r} - 1 \right). \\ \end{align} $$

The main problem with $g(\lambda, \Lambda, r)$ is that its limit when $r \to +\infty$ is not $0$ (it is $2\Lambda\vert\lambda\vert$), so I will not be able to solve $g\left( \lambda, \Lambda, r \right) \leqslant \epsilon$ for $r$ as this equation might have no solution.

My question is then, do you have an idea on how I could compute a better bound?

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    $\begingroup$ That expression goes to $0$ very rapidly as $n$ increases, so perhaps you ought to tell us what values of $n$ you are interested in, and how small a bound you are hoping to achieve. $\endgroup$ – saulspatz Sep 3 '19 at 15:39
  • $\begingroup$ I will add the whole context in the next few hours, sorry for the missing information. $\endgroup$ – Nelimee Sep 3 '19 at 16:19
  • $\begingroup$ I added a few material based on quick and dirty regressions for your small numbers. $\endgroup$ – Claude Leibovici Sep 6 '19 at 4:06
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This is too long for a comment.

Reading the post, I have been wondering about the meaning of the vertical bars in the definition of $f(n)$.

If it is $$f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left( \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right)$$ we have $$\sum_{i=0}^{n-j} \left( \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right)=\frac{n+1-j}{n!}-\frac{1}{j! (n-j)!}$$ which makes $$f(n)=\frac{n^2+n-2^{n+1}+2}{2 n!}$$ and then $$\left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right]=0$$

Edit

If we consider $$f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left\vert \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right\vert$$ and compute the value of $f(n)$ for the range $3 \leq n \leq 250$, a quick and dirty regression for the model $$\log[f(n)]=a+b n^c +d n^{2c}$$ which is equivalent to the minimization of $$SSQ=\sum _{n=3}^{250} \left| \frac{\Delta f(n)}{f(n)}\right|^2$$ we have (with $R^2=0.999998$) $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +6.666622 & 0.199625 & \{+6.273405,+7.059838\} \\ b & -0.671761 & 0.006089 & \{-0.683754,-0.659767\} \\ c & +1.329758 & 0.001918 & \{+1.325979,+1.333537\} \\ d & +0.000030 & \approx 0 & \{+0.000029,+0.000030\} \\ \end{array}$$ Similarly, using the same model for $$g(n)=\left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right]$$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +4.126499 & 0.156637 & \{+3.817960,+4.435038\} \\ b & -0.732346 & 0.005281 & \{-0.742749,-0.721943\} \\ c & +1.313991 & 0.001539 & \{+1.310961,+1.317022\} \\ d & +0.000032 & \approx 0 & \{+0.000031,+0.000032\} \\ \end{array}$$

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  • $\begingroup$ Thank you for your answer! The vertical bars are absolute values, that is why I tried the triangle inequality, without success. The absolute value comes from the fact that I normed the whole sum and used: $||A+B||\leqslant ||A||+||B||$, $||AB|| \leqslant ||A||||B||$ (for the spectral norm I am using) and $||\lambda A|| = |\lambda| ||A||$ (the coefficient in the sum is the $\lambda$ here). $\endgroup$ – Nelimee Sep 4 '19 at 7:47
  • $\begingroup$ @Nelimee. I am sorry ! $\endgroup$ – Claude Leibovici Sep 4 '19 at 8:09
  • $\begingroup$ Your comment gave me an idea! I tested it, and I will edit my question to include this idea. The result is not better than the 2 bounds from the paper linked in the question, but is way better than my previous bound! $\endgroup$ – Nelimee Sep 9 '19 at 11:11

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