What, exactly, does it take to make an infinite-dimensional Hilbert Space?

Question

I was trained as an engineer, where the beauty of the foundational math is often brushed away - or interpreted in a 'useful' way. I am now learning about quantum computation, which means learning more of how physicists see it.

Suppose we look at the space of well-behaved, continuous curves on the interval $[0,1]$. As an engineer, you are told this is an infinte-dimensional space, precisely because any such function has a Fourier transform, and the infinite number of different Fourier terms form the basis.

A physics/quantum text I looked at argued thus: think of the value at any point $x$ that is in $[0,1]$. In fact, just as we usually give subscripts for bases ($i, j,k$), imagine $x$ being the subscript - and then each point $x$ is a 'dimension' and the space is infinite-dimensional.

Maybe there is something wrong with that, but I note that it meets the basics we are looking for, including the fact that the inner product, $<f,g>=\int f(x)\cdot g(x) dx$ sure looks like the limiting case of the usual dot product, but extended to infinite dimensions. I also note that if you want every function to be expressible as the infinite sum of projections/inner products, it seems to work so long as you allow $\delta (x)$ to be used as the bases.

What is the more rigorous approach to what this infinite-dimensional (Hilbert) space really is based upon? In what way is it infinite-dimensional? Are both or either of these views right, and why or why not?

I note, by the way, that if you take the view the book I am reading takes (what I will call the 'axis approach') then that space always has a dimension that is uncountably infinite, because the points on the real line are all indexes. The same would be true for the Fourier transform view...except for this asymmetry: for functions that are periodic on $[0,1]$, the 'axis' view is still uncontably infinite in dimension, but the Fourier view is now countably infinite.

Answer 1

The big thing that's different in the infinite vs finite dimensional cases is the topology. If you're not experienced with math, topology may be a completely new concept. It's the common way the idea of a continuous function is formalized. Certain things (Like "A function you can draw with a pencil without picking up the tip") work ok for definitions, but have large numbers of failings.

One big failing is that this works in spaces where we have points, and functions that map points to points. What about spaces with functions and abstract functions between functions. The Fourier Transform is precisely this concept, which is makes it important to treat the topology rigorously in infinite dimensions.

The biggest reason to worry about topology in infinite dimensions is that it's "automatic" in finite dimensions. By this, I mean that it's not difficult to show that any linear transformation of finite dimension vector spaces is continuous. Because of this, it's something you don't have to worry about at all. In infinite dimensions, we can have cases where the identity operator's inverse isn't bounded. A consequence of this is that the identity's inverse isn't continuous (it can be proven that an operator having bounded operator norm is equivalent to it being continuous). So, functions we might think are "obviously continuous" are no longer so.

For $L^2([0,1])$, this is the space of all functions $f:[0,1]\to \mathbb{R}$ that satisfy some technical property called being measurable (that's similar to being continuous, but distinct) instead of being continuous. These functions also need to have finite norm, so $\langle f,f\rangle = \int_0^1 |f(x)|^2dx <\infty$. It can be shown that these functions have a vector space structure. It's known that $\{e^{2\pi ik}\}_{k = -\infty}^\infty$ is an orthonormal basis for this space, and writing some function $f$ in terms of this basis is precisely expressing it in terms of it's fourier series.

Note that this is a countable basis, so it has as many elements as $\mathbb{Z}$ (or $\mathbb{N}$, or your favorite countable set). The basis you suggest would be $\{\delta_r(x)\}_{r\in\mathbb{R}}$, which is a much larger set. Summing over this set is also not super well-defined (if you could define some sum that matches your intuition with $\sum_{k = 1}^\infty$, it'd require giving an enumeration of $\mathbb{R}$, which would imply $\mathbb{R}$ is countable). It does turn out that this countable basis is good enough for the aforementioned "measurable functions" (and for continuous functions, which are all measurable in this context).

Answer 2

Every vector space has a basis.

Every basis of a given vector space has the same cardinality.

The dimension of a vector space is the cardinality of a basis for it.

Hence a vector space is infinite dimensional iff it has a basis which is infinite.

If your infinite dimensional space has an inner product and is complete with respect to the induced norm then it is an infinite dimensional Hilbert space.

That's all it takes to make an infinite dimensional Hilbert space.