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--- hamburgminicourse2017:time-dependent_potentials [2017/04/02 20:30] – markus
+++ hamburgminicourse2017:time-dependent_potentials [2017/04/02 22:06] (current) – markus
@@ Line 18: / Line 18: @@
 \]
 **Proof sketch.** We show that potentials $V \in \mathrm{Lip}([0,T], K)$ give a well defined Schrödinger solutions but the proof omits a few important steps that would take up too much space here.[(The full proof and many things more on the subject can be found in my PhD thesis on [[https://arxiv.org/abs/1610.05552|arXiv]], §3.7.)]
+. Take the time interval $[0,t]$ with $t\leq T$ divided into $k$ subintervals $[t_j,t_{j+1}]$ with equal length $t/k$. We define the stepwise static approximation by using [[Stone's theorem]] for $k$ different static Hamiltonians:
+\[
+U_k(t,0) = \exp(-iH(t_{k-1})t/k) \ldots \exp(-iH(t_{0})t/k) =: U_k^{(k)} \ldots U_k^{(1)}.
+\]
+We show now that for $k\rightarrow \infty$ this so-defined evolution operator converges uniformly in $t$ in the strong topology of the Hilbert space $\mathcal{H}$.
+. So we show the Cauchy property of this sequence with a similar trick as in the proof of [[Stone's theorem]]:
+\[
+(U_k(t,0)-U_l(t,0))\psi_0 = i\int_0^t U_l(t,s) (V(s_{(l)}) - V(s_{(k)})) U_k(s,0)\psi_0 \,ds
+\]
+Here the time $s_{(k)}$ is the largest time step $t_k$ in the partitioning belonging to $U_k$ smaller than $s$. Clearly it holds $s_{(l)} - s_{(k)} \rightarrow 0$ if both indices increase. So it seems plausible that $V(s_{(l)}) - V(s_{(k)})$ in the $K$-norm, but we need also $U_k(s,0)\psi_0$ converging strongly in $D(H)$ to be able to even multiply with the potential difference and use the norm estimate ([[classes_of_static_potentials|here bottom]]), this we show in 4. onward.
+. By adding a big enough constant one first makes all $H(t) \geq 1$ (i.e. $\langle \varphi,H(t)\varphi \rangle \geq  1$ for all $\varphi \in D(H)$) and thus has a well-defined inverse $H(t)^{-1}:\mathcal{H} \rightarrow D(H)$ bounded by 1.
+. We add $D(H)$-identities $H(t)^{-1}H(t)$ to the definition of $U_k$:
+\begin{align*}
+U_k(t,0) &= U_k^{(k)} \ldots U_k^{(1)} \\
+&= H(t_{k-1})^{-1}H(t_{k-1}) U_k^{(k)} H(t_{k-2})^{-1} \ldots H(t_{1})H(t_{0})^{-1}H(t_{0}) U_k^{(1)} \\
+&= H(t_{k-1})^{-1} U_k^{(k)} H(t_{k-1})H(t_{k-2})^{-1} \ldots H(t_{1})H(t_{0})^{-1} U_k^{(1)} H(t_{0}).
+\end{align*}
+We used the fact that the Hamitonians commute with the short-stime evolution operators at the same time-step in this.
+. Now we rewrite the $H(t)H(s)^{-1}$ encounters as $K(t,s) + \mathrm{id} : \mathcal{H} \rightarrow \mathcal{H}$ where the operator $K(t,s)$ is bounded by $L |t-s|$ (not shown here), $L$ the Lipschitz constant of $V$ in the Sobolev--Kato--Lipschitz space. We have now:
+\[
+U_k(t,0) = H(t_{k-1})^{-1} U_k^{(k)} \prod_{j=1}^{k-1} (K(t_j,t_{j-1}) + \mathrm{id}) U_k^{(j)} H(0)
+\]
+with a time-ordered product.
+. The leading $H(t_{k-1})^{-1}$ is $\mathcal{H} \rightarrow D(H)$ with bound 1. So an estimate in the $D(H) = H^2$ Sobolev norm gives:
+\begin{align*}
+\|U_k(t,0)\psi_0\|_{H^2} &\leq \prod_{j=1}^{k-1} \left( \frac{Lt}{k} + 1 \right) \|H(0)\psi_0\|_\mathcal{H} \\
+&= \left( \frac{Lt}{k} + 1 \right)^{k-1} \|H(0)\psi_0\|_\mathcal{H} \longrightarrow e^{Lt} \|H(0)\psi_0\|_\mathcal{H}.
+\end{align*}
+This shows convergence for $\psi_0 \in D(H)$ and also the $H^2$ regularity of such solutions.
+. Finally we show the Cauchy property from (2).
+\begin{align*}
+\|(U_k(t,0)-U_l(t,0))\psi_0\|_\mathcal{H} &\leq \int_0^t \|V(s_{(l)}) - V(s_{(k)})\|_K \|U_k(s,0)\psi_0\|_{H^2} ds\\
+&\leq \max_{s \in [0,t]} \|V(s_{(l)}) - V(s_{(k)})\|_K  \|H(0)\psi_0\|_\mathcal{H} \int_0^t e^{Ls} ds \\
+&= \max_{s \in [0,t]} \|V(s_{(l)}) - V(s_{(k)})\|_K  \|H(0)\psi_0\|_\mathcal{H} \left( e^{Lt} -1 \right) \longrightarrow 0
+\end{align*}
+This convergence is uniform because $t$ is from a bounded set $[0,T]$. Finally already (6) showed that the evolution operators are uniformly bounded on $D(H)$ and thus can extended to the whole Hilbert space.
+ >> [[Regularity of Schrödinger solutions and functional differentiability]]