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I am struggling to formally derive the expression found in the Stone's theorem for one-parameter unitary groups. I am aware that this can be done by using the spectral theorem. I am mostly interested in the discrete 'version' of the spectral theorem.

Here's a short statement of Stone's theorem:

If ${\cal H}$ is a Hilbert space and $U(t)$ is a strongly-continuous, one-parameter unitary group, then $U(t)=\exp\bigl(-itH\bigr)$, where $H$ is self-adjoint.

Can anyone help me out or point me to some reference where this is explicitly done?

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  • $\begingroup$ What do you mean by "discrete version" of the spectral theorem? $\endgroup$ – MaoWao Nov 1 '19 at 10:41
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    $\begingroup$ By discrete I mean the case in which an operator in question has a purely discrete spectrum $\endgroup$ – omsorg Nov 1 '19 at 11:05
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A standard reference is Reed, Simon. Methods of Modern Mathematical Physics I. Stone's theorem is treated in Section VIII.4.

The rough idea is as follows. Define a (possibly unbounded) operator $H$ by $$ D(H)=\left\{\xi\in \mathcal{H}\mid\lim_{t\to 0}\frac{\xi-U(t)\xi}{it}\phantom{x}\text{ exists }\right\},\ \ \ H\xi=\lim_{t\to 0}\frac{\xi-U(t)\xi}{it}. $$

It needs some effort, but one can show that $H$ is self-adjoint. Then one can check that if $\xi\in D(H)$, both $u_1(t)=U(t)\xi$ (by the group property of $U$) and $u_2(t)=\exp(-itH)\xi$ (by the spectral theorem) solve the initial-value problem \begin{align*} \dot u(t)&=-iHu(t),\\ u(0)&=\xi. \end{align*} In particular, $$ \|u_1(t)-u_2(t)\|^2=\int_0^t\frac{d}{ds}\|u_1(s)-u_2(s)\|^2\,ds=-2\int_0^t\operatorname{Re}i\langle u_1(s)-u_2(s),H(u_1(s)-u_2(s)\rangle\,ds=0. $$ Thus $U(t)\xi=\exp(-itH)\xi$. The equality for arbitrary $\xi\in\mathcal{H}$ follows by continuity.

I do not know how purely discrete spectrum for the generator can be detected from the unitary group.

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  • $\begingroup$ thanks a lot! I think i can work it out for my specific case using this! $\endgroup$ – omsorg Nov 1 '19 at 11:49
  • $\begingroup$ @omsorg if this answers your question sufficiently, you should accept it to prevent it from being bumped up to the front page unnecessarily in the future. $\endgroup$ – Cameron Williams Nov 1 '19 at 18:21
  • $\begingroup$ thanks for the headsup, just done! $\endgroup$ – omsorg Nov 4 '19 at 8:55

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