Let $H$ be a Hermitian matrix with operator norm $||H|| \leq 1$. I am trying to show that for each $\varepsilon > 0$ I can find a $\delta$ such that

$$\left|\left|e^{iHt}-\sum_{k=0}^{\delta(t + \log(1/\varepsilon))-1} \frac{(iHt)^k}{k!} \right|\right|=\left|\left|\sum_{k=\delta(t + \log(1/\varepsilon))}^{\infty} \frac{(iHt)^k}{k!} \right|\right| \leq \varepsilon $$

What I tried to do was manipulating that sum using the fact that $k! \geq(k/e)^k$ and the triangle inequality which turns it into showing that there is a $\delta$ such that $$\sum_{k=\delta(t + \log(1/\varepsilon))}^{\infty} \left|\left|\left(\frac{iHte}{k}\right)^k \right|\right| \leq \varepsilon $$ How would I proceed?

  • $\begingroup$ What I meant is that if the second equality is proven, the first will be true a fortiori. But it might be a failing strategy, of course. $\endgroup$
    – Karl
    Commented Mar 30, 2019 at 0:36

1 Answer 1


By the Cauchy-Hadamard theorem, the operator series $\sum_k\frac{1}{k!}(itH)^k$ converges since $\|itH\|\le t$ is within the radius of convergence of $\sum_k\frac{z^k}{k!}$, which is infinite. Thus there is some integer $K$ such that $$\left\|\sum_{k=K}^\infty\frac{1}{k!}(itH)^k\right\|<\epsilon$$ Hence let $\delta\ge K/(t+\log(1/\epsilon))$.

  • $\begingroup$ Does this apply also if $H$ is an infinite dimensional (and still bounded, of course) Hermitian operator? $\endgroup$
    – Karl
    Commented Mar 30, 2019 at 18:31
  • 1
    $\begingroup$ It applies to bounded infinite-dimensional operators (Hermitian or not), and to Banach algebras in general. $\endgroup$ Commented Mar 31, 2019 at 6:01
  • $\begingroup$ And also to finite dimensional linear operators, right? $\endgroup$
    – Karl
    Commented Mar 31, 2019 at 17:46
  • $\begingroup$ Yes of course. What matters is $\|T\|\le r$. $\endgroup$ Commented Apr 1, 2019 at 6:11

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