Intuition for the spectral representation of an operator

Question

I have been trying to build an intuitive idea/example of the spectral representation of an operator, namely a "more visual" idea of the formula $$A=\int_{\sigma (A)} \lambda dP(\lambda).$$

For instance, I am considering the operator $A:L^2[0,1]\to L^2[0,1]$ such that $(A\varphi)(x)=x\varphi(x)$, as i believe that is rather intuitive to understand the action of this operator.

The spectrum of $A$ is the set $[0,1]$ and for each $E\subset\sigma(A)$, the projection valued spectral measure is given as $$(P(E)\varphi)(x)=\chi_E(x)\varphi(x).$$

But I am failing in build the bridge between the spectral measure and the spectral representation. Moreover, I would like give visual representations of what's going on but I fail completly.

Any help is appreciated.

Thank you for your time.

Answer 1

The intuition is exactly the same as for regular integration. Note that if you partition $[0,1]$ into small almost disjoint intervals of length $\Delta x=\dfrac{1}{n}$. Then $$\int_{0}^{1} f d\mu \approx \sum_{i} f(x_i)\mu (\Delta x) $$ The spectral decomposition of a multiplication operator similarly could be represented as:

$$\hat{x} \approx \sum_i x_iP(\Delta x) $$

You must intuitively think of $P(\Delta x)$ as an infinitesimal projection operator. Notice that this agrees with your explicit example where $P(E)=1_E$ since in that case $(P(\Delta x)\phi)(x)=\psi(x)$ where $\psi$ is only supported on that small $\Delta x$ interval (and is equal to $\phi$ there) but zero otherwise. Then notice that you can easily see that $$(\hat{x}\psi)(x) =x\psi(x) \approx \sum_i x_iP(\Delta x)\psi(x_i) $$

More rigorously we have the following result: Given a domain $\Omega$, a bounded measurable function $f$ and a PVM $\Pi$ then one can partition $\Omega$ into disjoint sets $\Omega_i$ (i=1,2,...) such that for any choice of $w_i \in \Omega_i$ the partial sum

$$\sum_{i=1}^{n}f(w_i)\Pi(\Omega_i)$$

converges to $$\int_\Omega f(w)d\Pi(w)$$

in the operator norm

where $\int_\Omega f(w)d\Pi(w)$ is defined by its action as follows:

$$\bigg<\int_\Omega f(w)d\Pi(w)\psi, \phi\bigg>= \int_\Omega f(w)d\Pi_{\psi,\phi}(w)$$

and $$\Pi_{\psi,\phi}(E)=\big<\Pi(E) \psi, \phi \big>$$

See Bogachev (real and functional analysis) theorem 7.9.3 for proof. The significance of the above result is that it says that the integral with respect to a projection valued measure (initially defined by its action on vectors as above) can be seen as a partial sum of projection operators converging in the operator norm which is closer to our intuition of what integration should be.

Answer 2

Let $X=\{x_1,\dots,x_n\}$ and $C(X)$ the algebra of complex-valued functions on it. Take $f\in C(X)$ to be real-valued. Then, where $\delta_x$ is the Kronecker delta function: $$\delta_x(y)=\begin{cases}1&\text{ if }y=x\\ 0&\text{ otherwise}\end{cases},$$ we can write $f$ in the form: $$f=\sum_{x_i\in X}f(x_i)\,\delta_{x_i}.$$ This is a spectral decomposition.

Note that we can embed $\pi(C(X))\subset B(\mathbb{C}^n)$ with the function $f$ represented by the diagonal matrix $$\pi(f)=\begin{pmatrix}f(x_1) & 0 & \cdots& 0 \\0 & f(x_2) & \cdots & 0 \\ \vdots & & \ddots & \\ 0 & 0 & \cdots & f(x_n)\end{pmatrix},$$ and $\pi(\delta_{x_i})$ equal to the projection $E_{ii}$.

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Consider a finite spectrum $\sigma(f)=\{\lambda_1,\dots,\lambda_n\}$ self-adjoint operator $f\in B(\mathsf{H})$. I use $f$ rather than $A$ because for intuition I really want you to think of as an operator as a function.

When $f$ 'looks' at $\mathsf{H}$ it sees a set: it sees $\{\mathsf{\Lambda}_1,\dots,\mathsf{\Lambda}_n\}$, the set of eigenspaces of $f$. The operator/function $f$ attaches a label, which we call an eigenvalue, $\lambda_i$ to each of these elements, and where $p_i\in B(\mathsf{H})$ is the orthogonal projection $p_i(\mathsf{H})=\mathsf{\Lambda_i}$, we write, as above, $$f=\sum_{\lambda_i\in\sigma(f)}\lambda_i\,p_i.$$ The comparison should be clear.

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The difference between $C(X)$ and $B(\mathsf{H})$ however is that given two distinct self-adjoint $f,g\in C(X)$, the operators $\pi(f),\,\pi(g)\in B(\mathbb{C}^n)$ (or just two commuting self-adjoint operators in $B(\mathsf{H})$) always see the same set when they look at $\mathbb{C}^n$, two non-commuting self-adjoint operators $f,g\in B(\mathsf{H})$ will disagree on what set they see when they look at $\mathsf{H}$. This is the moral of why the product of non-commuting observables is not observable.

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Now, if you have this intuition in the finite case, the infinite case might be a little easier to understand.