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

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

Regard a sequence: $$\varphi_n\in\mathcal{D}(H):\quad\varphi_n\to\varphi\in\mathcal{D}(H)$$

Suppose one has: $$\|H\varphi_n\|_{n\in\mathbb{N}}<\infty$$

Then one has weakly: $$\chi\in\mathcal{H}:\quad\langle H\varphi_n,\chi\rangle\to\langle H\varphi,\chi\rangle$$

How can I prove this?

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Since the sequence $(H \varphi_n)_n$ is bounded, it suffices to check the weak convergence on a dense subspace, e.g. for $\chi \in\mathcal{D}(H)=\mathcal{D}(H^\ast )$.

But for those $\chi$, we have

$$ \langle H \varphi_n,\chi \rangle =\langle \varphi_n , H \chi\rangle \to \langle \varphi, H \chi\rangle =\langle H\varphi, \chi\rangle. $$

Here, the last step used $\varphi\in\mathcal{D}(H)$.

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  • $\begingroup$ Got it thanks!! :) $\endgroup$ – C-Star-W-Star May 12 '15 at 17:29

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