Intuition behind the direct integral of a family of Hilbert spaces

Question

In order to understand better the mathematically rigorous version of Dirac's formalism in Quantum Mechanics I've been reading about direct integrals of Hilbert spaces, projector-valued measures and so on. Although the definition of the direct integrals of Hilbert spaces I've seem is quite clear, I simply couldn't get the idea behind it.

The definition I have is the following:

Definition 1: Let $\mu$ be a Radon measure on the $\sigma$-compact locally compact Hausdorff space $X$. A measurable field of Hilbert spaces is a collection $\{H_x : x \in X\}$ of Hilbert spaces together with a linear subspace $\mathcal{S}$ of $\prod_{x\in X}H_x$, whose elements are called measurable sections satisfying the following axioms:

• If $\eta\in \prod_{x\in X}H_x$, then $\eta\in \mathcal{S}$ if and only if $x\mapsto \langle \xi(x),\eta(x)\rangle$ is measurable for all $\xi \in \mathcal{S}$.
• There is a sequence $(\xi_n)$ in $\mathcal{S}$ such that for almost every $x\in X$, the closed linear span of $\xi_n(x)$ is $H_x$.

Definition 2: Let $\mu$ be a Radon measure on the $\sigma$-compact locally compact Hausdorff space $X$, and $\{H_x: x\in X\}$ a measurable field of Hilbert spaces over $X$. We define the direct integral of the $H_x$ to be the set of all $\xi \in \mathcal{S}$ (modulo agreeing on measure zero sets) such that $\int_X |\xi(x)|^2d\mu(x) < \infty$ and we denote this direct integral by

$$\int_X^{\oplus}H_x d\mu(x).$$

On the direct integral we have the inner product

$$\langle \xi,\eta\rangle = \int_X \langle \xi(x),\eta(x)\rangle_{H_x}d\mu(x),$$

which turns the direct integral into a Hilbert space.

Now, although the definition by itself is clear, I can't get the intuition behind it. We have a collection of Hilbert spaces indexed by one topological space with a Radon measure. We then consider a linear subspace of the product, satisfing two properties.

This is my first question: what is the intuition behind the defining properties of a measurable section? What we are really defining there?

After that, one simply picks all the measurable sections with the property that the integral of the square of the norm of the vectors at each point converge.

This is my second question: what is the intuition behind this definition? What we are really trying to define with this construction of the direct integral?

The direct sum, for example, can be understood in a intuitive way as the space of sums of elements of each space alone. In the case of the direct integral I'm not seeing one simple inution like that.

Answer 1

Disclaimer: I'm new to the subject too and thus don't understand these things on a very deep level.

One thing that direct integrals makes rigorous is the concept of eigenvectors of operators that, in a quantum mechanical terminology, belong to different space than the physical states $\psi$. In quantum mechanics books the eigenvalues of the position operator are often said to be the delta functions $d_{x_0}(x) = \delta(x-x_0)$. If the state space is for example $L^2$ (as a set of equivalence classes of regular functions), then of course it doesn't contain delta distributions and thus the position operator $P: L^2 \to L^2$ has eigenvectors outside of $L^2$.

It may help to consider a less general case of the direct integral first. Also, the measurability details can be ignored at first. Set $X = \mathbb{R}$ and let $H_x$ be some Hilbert space for all $x \in \mathbb{R}$. Now we would like to glue all the spaces $H_x$ together to obtain some space (the direct integral) which consists of of objects $s$ which are functions from $\mathbb{R}$ to the collection $\bigcup_{x \in \mathbb{R}} H_x$ in some sense. From $s$ we also require that for each $x \in \mathbb{R}$ we have $s(x) \in H_x$. Now if $H_x = \mathbb{R}$ for each $x$, then $s$ would be a function $\mathbb{R} \to \mathbb{R}$.

To further restrict what kind of elements belong to the direct integral, we assume that $s$ satisfies $$ \int_{\mathbb{R}} \langle s(x), s(x) \rangle_{H_x} d \mu(x) < \infty\,, $$ which really is just a more abstract square integrability condition. Again, if $H_x = \mathbb{R}$ for each $x$, we get the space $L^2$.

Now if we have a position operator on the direct integral, it works like this: $$ P: \int_{\mathbb{R}}^{\oplus} H_x \, d \mu(x) \to \int_{\mathbb{R}}^{\oplus} H_x \, d \mu(x) \,, \quad [P(s)](x) = x s(x)\,. $$ Remember that $[P(s)](x) \in H_x$. Now if $P$ had a eigenvalue $\lambda$ with the eigenvector $s_\lambda$, we get $P(s_\lambda) = \lambda s_\lambda$. Pointwise this is $$ [P(s_\lambda)](x) = [\lambda s_\lambda](x) = \lambda s_\lambda(x)\,, $$ where $s_\lambda(x) \in H_x$. Now, again, if $H_x = \mathbb{R}$ this would be a regular eigenvector on $L^2$. Of the operator $[Pf](x) = x f(x)$.

Now, in quantum mechanics, we often claim that the functions $\delta(x-x_0)$ are eigenfunctions of the position operator on $L^2$ since $x \delta(x-x_0) = x_0 \delta(x-x_0)$. If on the direct integral we set $H_x$ to be some space of distributions containing the delta functions, then the section $d(x) = 0$ for $x \neq x_0$ and $d(x_0) = \delta(\cdot - x_0)$ makes sense. Now $$ [P(d)](x) = [xd](x) = x d(x) = \begin{cases} x_0 \delta( \cdot - x_0) \,, \quad x = x_0 \\  0\,, \quad x \neq x_0  \end{cases} $$ which is the same as the section $x \mapsto x_0 d(x)$. Thus the "delta-section" is an eigenvector of the position operator. Remember that our position operator acts on the direct integral. Thus if the direct integral is our quantum mechanical state space, then we can have all the regular $L^2$ functions in the direct integral, but also the delta-distributions which are the eigenvectors of the position operator.

The measurability details are not really relevant to the intuition; they are there just to make sure that all the integrals norms and inner products on the direct integral make sense. Also the "simultaneous basis" $(\xi_n)$ is a technical detail to make everything work nicely.

So in conclusion, the direct integral is a more general version of the normal $L^2$ space which also makes the notion of eigenvectors more flexible.