# Exponential of Hermitian operator

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?

• 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.
– Karl
Commented Mar 30, 2019 at 0:36

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 $$\delta\ge K/(t+\log(1/\epsilon))$$.
• Does this apply also if $H$ is an infinite dimensional (and still bounded, of course) Hermitian operator?
• Yes of course. What matters is $\|T\|\le r$. Commented Apr 1, 2019 at 6:11