Resolution of the identity (basic questions)

Question

There are a few advanced discussions about the title:

1. Resolution of Identity
2. Resolution of Identity
3. Resolution of identity of a selfadjoint

I just want to ask the most fundamental part of it. (I am not a student in math.)

From the definition in Wiki: https://en.wikipedia.org/wiki/Borel_functional_calculus

Let $T$ be a self-adjoint operator. If $E$ is a Borel subset of $\mathbb{R}$, and $\mathbf{1}_E$ is the indicator function, then $\mathbf{1}_E(T)$ is a self-adjoint projection on $H$. Then mapping $$\Omega: E\mapsto \mathbf{1}_E(T)$$ is a projection-valued measure called the resolution of the identity for the self-adjoint operator $T$ ($H$ is a Hilbert space).

1. I am still confused about how does $\mathbf{1}_E(T)$ really work. $E\subset\mathbb{R}$; however, $T$ is an operator defined on $H$. So $T\notin E$
2. How does $\mathbf{1}_E(T)$ act on vectors in $H$?
3. "resolution of the identity": where does "resolution" and "identity" come from this definition.

Could anyone please let me know these (much better if a concrete example provided)

Answer 1

Let me elaborate a bit more.

Diagonal self-adjoint operators

Let $\mathsf{H}$ be a (separable) Hilbert space with an orthonormal basis $(e_n)$. Take a non-empty compact set $K \subset \mathbb{R}$ and a sequence $(\lambda_n)$ of distinct points from $K$ such that the closure of the set $\{ \lambda_n \colon n \in \mathbb{N}\}$ is $K$. We construct a self-adjoint operator on $\mathsf{H}$ with spectrum $\sigma(A)=K$ by setting $$Ax = \sum_{n \in \mathbb{N}}\lambda_n \left<x, e_n\right>e_n$$ for all $x \in \mathsf{H}$.

The function $E \colon \mathcal{B}(K) \rightarrow B(\mathsf{H})$, where $\mathcal{B}(K)$ stands for the set of all bounded Borel function on $K$ and $B(\mathsf{H})$ is the space of linear bounded operators on $\mathsf{H}$, defined by $$E(B)x = \sum_{n \in \mathbb{N}} \textbf{1}_{B}(\lambda_n) \left<x,e_n\right>e_n$$ is the resolution of identity (spectral measure, link) for $A$ and it is often denoted by $\mathbf{1}_B(A):=E(B)$ (this notation is consistent with the application of functional calculus, discussed further in the answer) for all $B \in \mathcal{B}(K)$. Furthermore the following holds $$(\star) \ \ \qquad Ax = \int_{\sigma(A)} \lambda \ E(\mathrm{d}\lambda)x = \sum_{n \in \mathbb{N}}\lambda_n \left<x, e_n\right>e_n,$$ where $\int_{\sigma(A)} \lambda \ E(\mathrm{d}\lambda)$ is an integral with respect to the spectral measure $E$ of $A$ (I'm not going to define this integral here, check this link, $(\star)$ supposed to give the intuition).

Notation

In Physics people often use the Dirac notation, that is, since the above $A$ can be written by $\sum_{n \in \mathbb{N}} \lambda_n \left|e_n\right>\left<e_n\right|$ or even $\sum_{i} \lambda_i \left|i\right> \left<i\right|$ (if we use the notation $(\left|i\right>)$ to denote the orthonormal basis) then the following notation can be often seen $$A = \int \lambda \ \mathrm{d}\left|i\right>\left<i\right|$$ for any self-adjoint operators (not-necessarily diagonal).

General case

Now take an arbitrary self-adjoint operator on $\mathsf{H}$ (not-necessarily diagonal). The spectral theorem states that $$A = \int_{\sigma(A)} \lambda \ E(\mathrm{d}\lambda),$$ where $E$ is the corresponding resolution of identity (spectral measure) for $A$.

In particular, one may want to consider the Borel functional calculus. Given a bounded borel function $f \in \mathcal{B}(\sigma(A))$ we have that $$ f(A) = \int_{\sigma(A)} f(\lambda) \ E(\mathrm{d} \lambda).$$

Name

The term identity in the name probably comes from the fact that $$\mathbf{1}_{\sigma(A)}(A) = I_{\mathsf{H}},$$ and I think that resolution corresponds to the spectral form of the operator.

Spectral theorem - different version

There is also a multiplication operator version of the spectral theorem which says that any self-adjoint operator is unitarily equivalent to the operator of multiplication on some space $L^2(\mu)$ for some measure $\mu$. Recall that a multiplication operator with symbol $f\in L^{\infty}(\mu)$ is $M_f \in B(L^2(\mu))$ given by $(M_f g)(x) = f(x)g(x)$. One may want to check this link for further information.

In particular, the spectral measure for multiplication operator $M_f$ is given by $E(B) = \mathbf{1}_{f^{-1}(B)}$ for $B \in \mathcal{B}(\sigma(M_f))$.