Spectral projector 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
 A: 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
$$
                \|Af\|^2 = \int_{-\infty}^{\infty}\lambda^2 d\|E_{\lambda}f\|^2 < \infty.
$$
