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

Then invariance follows: $$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H})\implies A\mathcal{D}(H)\subseteq\mathcal{D}(H)$$

So for adjoint and product: $$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H})\iff\mathrm{ad}(A^*)\in\mathcal{B}(\mathcal{H})$$ $$\mathrm{ad}(A),\mathrm{ad}(B)\in\mathcal{B}(\mathcal{H})\implies\mathrm{ad}(AB)\in\mathcal{B}(\mathcal{H})$$

Especially for commutator: $$\mathrm{ad}(A^*)=\mathrm{ad}(A)^*$$ $$\mathrm{ad}(AB)=\mathrm{ad}(A)B+A\mathrm{ad}(B)$$

How to prove this from scratch?

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

And the elements: $$\psi\in\mathcal{D}(H)=\mathcal{D}(H^*)$$

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

Thus for the domain: $$A\mathcal{D}(H),B\mathcal{D}(H)\subseteq\mathcal{D}(H)\implies AB\mathcal{D}(H)\subseteq\mathcal{D}(H)$$

So for adjoint and product: $$|\langle iA\varphi,H\varphi\rangle-\langle iAH\varphi,\varphi\rangle|=|\langle iA^*\varphi,H\varphi\rangle-\langle iA^*H\varphi,\varphi\rangle|$$ $$\|i\{HAB-ABH\}\varphi\|\leq\|i\{HA-AH\}B\varphi\|+\|Ai\{HB-BH\}\varphi\|$$

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

By density for commutator: $$\langle\mathrm{ad}(A^*)\varphi,\chi\rangle=\langle i\{HA^*-A^*H\}\varphi,\chi\rangle=\langle\mathrm{ad}(A)^*\varphi,\chi\rangle$$ $$\langle\mathrm{ad}(AB)\varphi,\chi\rangle=\langle i\{HAB-ABH\}\varphi,\chi\rangle=\langle\{\mathrm{ad}(A)B+A\mathrm{ad}(B)\}\varphi,\chi\rangle$$

Concluding the assertions.

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