# 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??

• What is $R(i)$? – Robert Israel May 2 '15 at 0:26
• @RobertIsrael: Ah I forgot: $R(i)=(i-H)^{-1}$ (I wrote it out now.) – C-Star-W-Star May 2 '15 at 0:27
• Please avoid over-editing your post. Regards, – Pedro Tamaroff May 12 '15 at 17:58

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.