Spectral projector

Question

I have a question if we say "dw comes from the spectral projectors, $E_x$ $x\in\mathbb{R}$ of an self adjoint operator Q in the sens $dw(x)=\left\langle \phi_0|E_x\phi_0\right\rangle$" I need know what its means. Thanks for your help

Answer 1

The $E_x$ comes from the Spectral Theorem for a selfadjoint operator $A$ on a Hilbert space $H$. For $-\infty < x < \infty$, the function $E_x$ is an orthogonal projection, which is to say $$ E_x^* = E_x = E_x^2. $$ The $E_x$ have the property that the are increasing with $x$, and all of the $E_x$ commute: $$ E_x E_y = E_y E_x = E_x,\;\;\; x \le y. $$ All of the $E_x$ commute with $A$ as well. Finally, you have the vector continuity property $$ \lim_{x\downarrow -\infty} E_x f = 0,\;\;\; \lim_{x\uparrow\infty} E_xf =f. $$ The selfadjoint operator $A$ can then be written as $$ Af = \int_{-\infty}^{\infty}\lambda d_{\lambda}E_{\lambda}f $$ Weakly, $$ (Af,f) = \int_{-\infty}^{\infty}\lambda d_{\lambda}(E_{\lambda}f,f). $$ The properties of $E$ show that the following function is a non-decreasing function of $\lambda$ for a fixed $\lambda$: $$ \lambda \mapsto (E_{\lambda}f,f)= (E_{\lambda}f,E_{\lambda}f)=\|E_{\lambda}f\|^2 $$ And $\lim_{\lambda\downarrow -\infty}\|E_{\lambda}f\|^2 = 0$, $\lim_{\lambda\uparrow\infty}\|E_{\lambda}f\|^2 = \|f\|^2$. Furthermore, the domain of $A$ is fully characterized by the condition $$ \|Af\|^2 = \int_{-\infty}^{\infty}\lambda^2 d\|E_{\lambda}f\|^2 < \infty. $$