Spectral Measures: Riemann-Lebesgue

Question

Problem

Given a Hilbert space the Lebesgue measure.

Consider a selfadjoint Hamiltonian: $$H:\mathcal{D}\to\mathcal{H}$$

Denote its associated Borel spectral measure by: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$ its corresponding evaluated measures by: $$\mu_{\varphi\psi}(A):=\langle\varphi,E(A)\psi\rangle$$ $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$ and its related functional calculus by: $$\varphi\in\mathcal{D}f(H):\quad\langle f(H)\varphi,\eta\rangle:=\int fd\mu_{\varphi\eta}\quad(\eta\in\mathcal{H})$$

Then it follows by Riemann-Lebesgue: $$e^{itH}\varphi\stackrel{|t|\to\infty}{\rightharpoonup}0\quad(\varphi\in\mathcal{H}_{ac})$$

How to prove this from scratch?

Caution

Be warned that the above does not hold strongly: $$\|e^{itH}\varphi\|\equiv\|\varphi\|\nrightarrow0$$ (The problem lies in missing pointwise convergence.)

Application

For compact operators this turns into a strong version: $$C\in\mathcal{C}(\mathcal{H}):\quad Ce^{itH}\varphi\stackrel{|t|\to\infty}{\to}0\quad(\varphi\in\mathcal{H})$$ (This applies for example in scattering theory.)

Answer 1

The probability measures are absolutely continuous: $$\varphi\in\mathcal{H}_{ac}\iff\nu_\varphi\ll\lambda$$

It holds the estimate: $$|\mu_{\varphi\eta}(A)|=|\langle E(A)\varphi,\eta\rangle|\leq\|E(A)\varphi\|\cdot\|\eta\|=\sqrt{\nu_\varphi(A)}\cdot\|\eta\|$$

So one has absolute continuity: $$\nu_\varphi\ll\lambda\implies|\mu_{\varphi\eta}|\ll\lambda\implies\mu_{\varphi\eta}\ll\lambda\implies\mu_{\varphi\eta}\ll\lambda$$

One obtains an integrable derivative: $$h\in\mathcal{L}(\lambda):\quad|\mu|=\int h\mathrm{d}\lambda$$

Arriving at the operator analogue of Riemann Lebesgue: $$\langle e^{itH}\varphi,\eta\rangle=\int_{-\infty}^{+\infty}e^{itx}\mathrm{d}\mu_{\varphi\eta}(x)=\int_{-\infty}^{+\infty}e^{itx}u(x)\mathrm{d}|\mu_{\varphi\eta}|(x)=\int_{-\infty}^{+\infty}e^{itx}u(x)h(x)\mathrm{d}\lambda(x)\stackrel{|t|\to\infty}{\longrightarrow}0$$

Answer 2

Remember that $E(A)^{2}=E(A)=E(A)^{\star}$. So $$ \mu_{x,x}=(E(A)x,x)=(E(A)^{2}x,x)=(E(A)x,E(A)^{\star}x)=(E(A),E(A)x)=\nu_{x}. $$ It is interesting that $\mu_{x,y} << \mu_{x,x}$ and $\mu_{x,y} << \mu_{y,y}$ because $$ |\mu_{x,y}(A)| \le \mu_{x,x}(A)^{1/2}\mu_{y,y}(A)^{1/2}. $$ Radon-Nikodym gives you something interesting, and it can help define and study Fourier transforms in an abstract way. Pick a vector $x$ and stick to it. Then $y \mapsto f_{x,y}$ has some nice properties, where $d\mu_{y,x}=f_{x,y}d\mu_{x,x}$. This gives a representation of the cyclic subspace spanned by $x$ under the operators $T=\int \lambda dE(\lambda)$ and $T^{\star}$ as an $L^{2}_{\mu_{x}}$ space. On this subspace $T$ becomes a multiplication operator by the spectral parameter $\lambda$, which is what classical Fourier transforms do. This offers a type of generalized Fourier transform, and is a subset of the GNS theory of $C^{\star}$ algebra representation. Quantum Mechanics benefits from this point-of-view.