Stone's Theorem and Functional Calculus

Question

I have been reading up on Stone's theorem for unitary groups, and going through the proof of this theorem. The theorem basically states that for every one parameter unitary group $U(t)$ on a Hilbert space, there exists a self-adjoint operator $H$ such that $e^{-itH}$ generates it. All the proofs I've read use the rigorous definition of $e^{-itH}$ using functional calculus, but none actually explain what this explicitly looks like.

For example, what does $e^{-itH}$ look like when $H=-i\partial_x$ is the momentum operator? I've been able to find the spectral measure for $H$, i.e., $P(\Lambda)=\mathcal{F}^{-1}\chi_\Lambda\mathcal{F}$, where $\chi_\Lambda$ is the characteristic function on the Borel set $\Lambda$ and $\mathcal{F}$ is the Fourier transform, so we have $$e^{-itH}=\int_{\mathbb{R}}e^{-it\lambda}dP(\lambda)$$ using our spectral measure. Is this the best we can do for a definition of our operator? Can we explicitly compute this integral?

Answer 1

It depends a lot on what you call "explicit". In the case of the exponential, you have $$ e^{-itH}=\sum_{k=0}^\infty\frac{(-i)^kt^k}{k!}\,H^k=\sum_{k=0}^\infty\frac{(-1)^kt^k}{k!}\,\partial_x^k. $$

Answer 2

The spectral integral reduces to $$ (e^{-itA}f)(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-its}e^{isx}\int_{-\infty}^{\infty}e^{-isv}f(v)\,dv \\ = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{is(x-t)}\int_{-\infty}^{\infty}e^{-isv}f(v)\,dv = f(x-t). $$ This is translation semigroup. You can verify: $$ \frac{d}{dt}f(x-t)|_{t=0}=-f'(x) = (-iAf)(x) $$ This derivative $\frac{d}{dt}$ is taken in the $L^{2}$ sense for $t \mapsto f(\cdot -t)\in L^{2}(\mathbb{R})$ for any $f \in \mathcal{D}(A)$.