Let $\mathcal{H}$ be an Hilbert space.
Firstly, I shall define some notions as their definitions may vary:
A spectral resolution is a function $E:\mathbb{R}\to\mathcal{L}(\mathcal{H})$ (the space of continuous operators from and to $\mathcal{H}$) iff the following holds:
- $E(t)$ is an orthogonal projection for any $t\in\mathbb{R}$
- $\text{Im}(E(s))\subset\text{Im}(E(t))\forall s<t$, which is denoted by $E(s)<E(t)$
- $\lim_{s>0,s\to 0}E(t+s)=E(t)$ (right continuity), i.e. $\cap_{s>t}\text{Im}(E(s))=\text{Im}(E(t))$
- $\lim_{t\to -\infty}E(t)x=0$ and $\lim_{t\to\infty}E(t)x=x$ for all $x\in\mathcal{H}$
Define $E(t^{-}):=\lim_{\epsilon>0,\epsilon\to 0}E(t-\epsilon)$. It can be seen as the orthogonal projection onto $\cup_{s<t}\text{Im}(E(s))$.
As $s\mapsto (E(s)f,f)$ is positive, increasing and right-continuous for any fixed $f\in\mathcal{H}$, we can define the following Lebesgue-Stieltjes integral ($\varphi$ measurable and bounded):
$$\int_{\mathbb{R}}\varphi(s)\text{d}(E(s)f,f)$$
This naturaly defines a space $L^{2}(\mathbb{R},(E(\cdot)f,f))$. Now, we define
$$\begin{align*} E[a,b]&=E(b)-E(a^{-})\\ E(a,b)&=E(b^{-})-E(a)\\ E[a,b)&=E(b^{-})-E(a^{-})\\ E(a,b]&=E(b)-E(a) \end{align*}$$
Let $\phi=\sum_{i=1}^{m}c_{i}\mathbb{I}(I_{i})$ where $I_{i}$ is a real interval (i.e. let $\phi$ be a simple function) and where $I_{i}\cap I_{j}=\emptyset$ for all $i\neq j$. We define: $$\int_{\mathbb{R}}\phi(s)\text{d}E(s):=\sum_{i=1}^{m}c_{i}E(I_{i})$$
(this integral is an operator, not a scalar).
Now, for any $\varphi\in L^{2}(\mathbb{R},(E(\cdot)f,f))$, the following is well-defined:
$$\left(\int_{\mathbb{R}}\varphi(s)\text{d}E(s)\right)f=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_{n}(s)\text{d}(E(s)f,f)$$
where $\phi_{n}$ are simple functions and
$$\phi_{n}\overset{L^{2}(\mathbb{R},(E(\cdot)f,f))}{\longrightarrow}\varphi$$
A version of the spectral theorem may be expressed as follows: for any self-adjoint operator $T$ over $\mathcal{H}$, there exists a spectral resolution $E_{T}$ such that
$$T=\int_{\mathbb{R}}s\text{d}E(s)$$
and the domain of $T$ can be expressed as $D(T)=\left\{f\in\mathcal{H}\mid\int_{\mathbb{R}}s^{2}\text{d}(E(s)f,f)<\infty\right\}$
Now, if $\varphi:\mathbb{R}\to\mathbb{C}$ is measurable, bounded over the spectrum $\sigma(T)$ of $T$, we define
$$\varphi(T):=\int_{\mathbb{R}}\phi(s)\text{d}E(s)$$
and we have the following properties:
- $\varphi(T)$ is bounded;
- $\varphi(T)+\psi(T)=(\varphi+\psi)(T)$;
- $\varphi(T)\circ\psi(T)=(\varphi\psi)(T)$;
- $\varphi(T)^{\ast}=\bar{\varphi}(T)$ where $U^{\ast}$ denotes the adjoint of $U$ for any operator $U$ and $\bar{\varphi}:s\mapsto \overline{\varphi(s)}$ maps $s$ to the complex conjugate of $\varphi(s)$;
- Moreover, if $\varphi$ is continuous, $\varphi(\sigma(T))=\sigma(\varphi(T))$.
Now comes my question:
We know that $s\mapsto e^{-its}$ ($t\in\mathbb{R}$) is a bounded, measurable function, so that $e^{-itU}$ is a well-defined operator for any self-adjoint operator $U$. We want to prove that if $f_{0}$ is an eigenvector of $U$ with associated eigenvalue $\lambda$, then $f_{0}$ is an eigenvector of $e^{-itU}$ with eigenvalue $e^{-it\lambda}$. From the properties mentioned above, it is clear that $e^{-it\lambda}$ is an eigenvalue. However, how can I prove that $f_{0}$ is an eigenvector?
As $e^{-itU}$ is bounded, I am tempted to say that
$$e^{-itU}=\sum_{n=0}^{\infty}\frac{(-it)^{n}U^{n}}{n!}$$
but I am not sure how to prove this. If this is true, we can easily deduce that $e^{-it\lambda}$ is an eigenvalue (which we already know) associated with the eigenvector $f_{0}$.
Any help is appreciated.
The eigenvector $f_0$ is in the range of the spectral projection $E_[\lambda, \lambda] = E(\lambda) - E(\lambda^-)$, while $E(\lambda^-) f_0 = 0$ and $(E(x) - E(\lambda)) f_0 = 0$ for $x > \lambda$. Thus for any $f \in \mathcal H$, the measure $\langle f, dE(s)\; f_0 \rangle = d \langle f, E(s) f_0\rangle$ is a point mass of $\langle f, f_0\rangle$ at $s=\lambda$ and $$ \langle f, \phi(T) f_0 \rangle = \int_{\mathbb R} \phi(s) \;\langle f, dE(s) f_0 \rangle = \phi(\lambda) \langle f, f_0 \rangle = \langle f, \phi(\lambda) f_0 \rangle$$ which implies $\phi(T) f_0 = \phi(\lambda) f_0$.