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This thread is only Q&A!

Given a Hilbert space $\mathcal{H}$.

Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$

Regard the domain: $$A\in\mathcal{B}(\mathcal{H}):\quad\mathcal{D}(s_A):=\mathcal{D}(H)\times\mathcal{D}(H)$$

Construct the form: $$s_A(\varphi,\psi):=\langle iA\varphi,H\psi\rangle-\langle iAH\varphi,\psi\rangle$$

Suppose it is bounded: $$|s_A(\varphi,\varphi)|\leq\|s\|\cdot\|\varphi\|^2\implies|\overline{s}_A(\varphi,\varphi)|\leq\|\overline{s}_A\|\cdot\|\varphi\|^2$$

By Lax-Milgram one has: $$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H}):\quad \overline{s}_A(\varphi,\psi)=\langle\mathrm{ad}(A)\varphi,\psi\rangle$$

Introduce the evolution: $$A\in\mathcal{B}(\mathcal{H}):\quad\tau^t[A]:=e^{itH}Ae^{-itH}$$

Then equivalence holds: $$\mathrm{ad}^n(A)\in\mathcal{B}(\mathcal{H})\iff\tau[A]\varphi,\tau[A^*]\varphi\in\mathcal{C}^n(\mathbb{R},\mathcal{H})^\quad(\varphi\in\mathcal{H})$$

Especially one has then: $$\frac{\mathrm{d}^n}{\mathrm{d}t^n}\tau^t[A]\varphi=\tau^t[\mathrm{ad}^n(A)]\varphi\quad(\varphi\in\mathcal{H})$$

How can I prove this?

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Evolution

Regard dense elements: $$\overline{\mathcal{D}(H)}=\mathcal{H}:\quad\varphi\in\mathcal{D}(H)$$

By the previous thread: $$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H})\implies\frac{\mathrm{d}}{\mathrm{d}t}\tau^t[A]\varphi=\tau^t[\mathrm{ad}(A)]\varphi$$

That gives the identity: $$\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi\mathrm{d}s=\frac{1}{h}\int_0^h\frac{\mathrm{d}}{\mathrm{d}s}\tau^{t+s}[A]\varphi\mathrm{d}s=\frac{1}{h}\{\tau^{t+h}[A]-\tau^t[A]\}\varphi$$

Now regard elements $$\varphi\in\mathcal{H}:\quad\varphi=\lim_n\varphi_n\quad(\varphi_n\in\mathcal{D}(H))$$

One has the dominant: $$\|\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\|\leq\|\mathrm{ad}(A)\|(1+\|\varphi\|)$$

So by dominated convergence: $$\frac{1}{h}\{\tau^{t+h}[A]-\tau^t[A]\}\varphi=\lim_n\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\mathrm{d}s\\ =\frac{1}{h}\int_0^h\lim_n\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\mathrm{d}s=\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi\mathrm{d}s\stackrel{h\to0}{\to}\tau^{t}[\mathrm{ad}(A)]\varphi$$

Concluding one direction.

Commutator

Denote the derivative: $$\varphi\in\mathcal{H}:\quad D(A)\varphi:=\left(\frac{\mathrm{d}}{\mathrm{d}t}\tau^t[A]\varphi\right)_{t=0}$$

Regard elements: $$\varphi,\psi\in\mathcal{D}(H)=\mathcal{D}(H^*)$$

Then invariance follows: $$\langle\varphi,D(A)\psi\rangle=\langle H\varphi,iA\psi\rangle-\langle\varphi,iAH\psi\rangle\implies iA\psi\in\mathcal{D}(H^*)=\mathcal{D}(H)$$

So one obtains: $$\varphi\in\mathcal{D}(H):\quad D(A)\varphi=\{iHA-iAH\}\varphi$$

For arbitrary elements: $$\langle D(A)\varphi,\psi\rangle=\lim_{h\to0}\langle\frac{1}{h}\{\tau^h[A]-\tau^0[A]\}\varphi,\psi\rangle\\ =\lim_{h\to0}\langle\varphi,\frac{1}{h}\{\tau^h[A^*]-\tau^0[A^*]\}\psi\rangle=\langle\varphi,D(A^*)\psi\rangle$$

So one obtains: $$D(A^*)=D(A)^*\implies D(A)=\overline{D(A)}\implies D(A)\in\mathcal{B}(\mathcal{H})$$

Concluding other direction

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