Why denote the spectral decomposition of a bounded operator as an integral?

Question

In the spectral theorem for bounded self-adjoint operators we get, for each self-adjoint (bounded) operator $A$ in a Hilbert space $\mathcal{H}$, and each bounded Borel-measurable function $f \in \mathcal{B}_0(\sigma(A))$ (where I denote by $\mathcal{B}_0(\sigma (A))$ the set of all Borel-measurable functions in the spectrum $\sigma(A)$ of $A$) a bounded operator $f(A)$ defined as

$$\langle \psi, f(A) \psi \rangle = \int_{\sigma(A)} f(\lambda) d\mu_{\psi}(\lambda) \ ,$$

where $\mu_{\psi}$ is a measure defined by a family of "projection-valued measures" $(P_B)_{B \in \mathcal{B}(\sigma(A))}$ which are just $\chi_B(A)$ in the sense of the functional calculus, for each Borel subset $B$ of $\sigma(A)$. Given all this information, the operator $A$ is denoted as

$$ A = \int_{\sigma(A)} \lambda dP_\lambda \ ,$$

and this decomposition is unique in the sense that any other projection-valued measure satisfying this equality should be equal to the first. My questions are: Is there some other reason for this notation for the operator $A$? How this integral should be interpreted, as it isn't actually constructed as a Lebesgue integral? The only way I know to obtain the operator in this case is to use the somewhat more general expression above, and use the polarization identity + Riesz lemma, as usual.

Sorry for the long question, my doubt is quite simple but there was a lot of context before I could express it.

Answer 1

I'll answer the second question first, then the first question.

The interpretation of the integral comes the following result:

If $(P_B)_{B \in \Omega}$ is a projection-valued measure for $(X,\Omega,H)$ and $f:X\to\mathbb C$ is a bounded $\Omega$-measurable function, then there is a unique operator $T$ in $\mathcal B(H)$ such that if $\varepsilon>0$ and $\{B_1,\ldots,B_n\}$ is an $\Omega$-partition of $X$ with $\sup\{|f(x_0)-f(x_1)|:x_0,x_1\in B_k\}<\varepsilon$ for $1\leq k\leq n$, then for any $x_k\in B_k$, $$\|A-\sum_{k=1}^nf(x_k)P_{B_k}\|<\varepsilon.$$

For a proof of this result, see Proposition IX.1.10 of Conway's functional analysis book. Given the result, we define $T=\int f\ dE$. In the course of the proof, it is seen that the relation you state, namely $\langle \psi, T \psi \rangle = \int_{\sigma(A)} f(\lambda) d\mu_{\psi}(\lambda)$, is shown to hold.

The reason for the notation is because this is (part of) an infinite-dimensional analogue of the spectral theorem of linear algebra. In the finite-dimensional case, self-adjoint operators are sums of projections on eigenspaces. In infinite-dimensions, self-adjoint operators are integrals ("limits" of sums) of projections associated with the spectrum.

Answer 2

For a continuous function $f$, you can treat the integral as a Riemann-Stieltjes type of vector integral $$ \int_a^b f(\lambda)dP(\lambda)x = \lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{j=1}^{n}f(\lambda_j)P[\lambda_{j-1},\lambda_j)x. $$

The notation is very suggestive in that form. As for notation, if you think of $dP(\lambda)$ as an "infinitestimal" projection onto the eigenspace associated with $\lambda$, then $$ A dP(\lambda) = \lambda dP(\lambda). $$ The projections sum to the identity $$ I = \int dP(\lambda) $$ which is another way of saying that you have a "continuous" basis of eigenvectors which may include discrete components as well. Hence, $$ Ax = A\int dP(\lambda)x=\int AdP(\lambda)x=\int \lambda dP(\lambda)x. $$ And the eigenspaces are mututally orthogonal, represented symbolically as $$ dP(\lambda)dP(\mu) = 0,\;\; \lambda\ne \mu, \\ dP(\lambda)dP(\lambda)=dP(\lambda). $$ which makes integrals easy to multiply $$ \int f(\lambda)dP(\lambda)\int g(\mu)dP(\mu)x = \int f(\lambda)g(\lambda)dP(\lambda)x. $$ And there is a Parseval identity, $$ \|\int f(\lambda)dP(\lambda)x\|^2 = \int |f(\lambda)|^2 d\|P(\lambda)x\|^2, \\ \| x\|^2 = \|\int dP(\lambda)x\|^2 = \int d\|P(\lambda)x\|^2. $$ It makes sense to write $f(A)=\int f(\lambda)dP(\lambda)$ because $$ A^nx = \int \lambda^n dP(\lambda)x. $$ More generally, $(fg)(A)x=f(A)g(A)x$. It's effort to make all of this rigorous, but that was not your question.

This is John von Neumann's integral notation. He used $dE(\lambda)$ where $E$ and $\lambda$ were used to be suggestive to the Quantum Physicist.