Hamiltonian: Derivative Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Regard the evolution:
$$A=A^*:\quad A(t):=e^{-itH}Ae^{itH}$$
Suppose invariance:
$$e^{ihH}\mathcal{D}(A)\subseteq\mathcal{D}(A)$$
And uniform bound:
$$\varphi\in\mathcal{D}(A):\quad\|Ae^{ihH}\varphi\|_{|h|<\varepsilon}<\infty$$
For the common domain:
$$\varphi,\psi\in\mathcal{D}(A)\cap\mathcal{D}(H)$$
Then the derivative writes:
$$\frac{\mathrm{d}}{\mathrm{d}t}\langle A(t)\varphi,\psi\rangle=\langle A(t)\varphi,iH\psi\rangle+\langle iH\varphi,A(t)\psi\rangle$$
How can I prove this from scratch??
 A: Meanwhile I got it...
The derivative writes as:
$$\langle\tfrac{1}{h}\{A(t+h)-A(t)\}\varphi,\psi\rangle=\langle\tfrac{1}{h}\{A(h)-A(0)\}e^{itH}\varphi,e^{itH}\psi\rangle$$
Expand the expression:
$$|\langle\tfrac{1}{h}\{A(h)-A(0)\}\varphi,\psi\rangle-\langle A\varphi,iH\psi\rangle-\langle iH\varphi,A\psi\rangle|\\
\leq|\langle Ae^{ihH}\varphi,\tfrac{1}{h}e^{ihH}\psi\rangle-\langle Ae^{ihH}\varphi,\tfrac{1}{h}1\psi\rangle-\langle Ae^{ihH}\varphi,iH\psi\rangle|\\
+|\langle\tfrac{1}{h}e^{ihH}\varphi,A\psi\rangle-\langle\tfrac{1}{h}1\varphi,A\psi\rangle-\langle iH\varphi,A\psi\rangle|\\
+|\langle Ae^{ihH}\varphi,iH\psi\rangle-\langle A\varphi,iH\psi\rangle|$$
Control the first term by:
$$\|Ae^{ihH}\varphi\|_{|h|<\varepsilon}\cdot\|\tfrac{1}{h}\{e^{ihH}-1\}\psi-iH\psi\|\stackrel{h\to0}{\to}0$$
And the second term by:
$$\|\tfrac{1}{h}\{e^{ihH}-1\}\varphi-iH\varphi\|\cdot\|A\psi\|\stackrel{h\to0}{\to}0$$
One has the bound:
$$\|Ae^{ihH}\varphi\|_{|h|<\varepsilon}<\infty$$
So for the third term:*
$$\langle Ae^{ihH}\varphi,iH\psi\rangle\to\langle A\varphi,iH\psi\rangle$$
Concluding the assertion.
*See the thread: Weak Convergence
