A What separates Hilbert space from other spaces?

[chiro]

For those that haven't done mathematics - you either have finite dimensions or not. Can't be both.

Take a variable - call it n for the number of dimensions.

Can n be finite or infinite dimensional at the same time? It can't. You have to remember that it is a variable and you can't assign infinity to a natural number [which is often the case for things like dimension].

Also for quantum mechanics, read Von Neumanns treatise on it and you'll find that they use Hilbert spaces because of the infinite dimensional aspect.

In quantum mechanics we have infinitely many states [a continuum has to]. What do you think the consequences are for this when they use Hilbert spaces?

Also - don't read Von Neumanns work on finite linear spaces since that is completely different - read the one on quantum mechanics just in case you get confused.

 

[SemM]

Thanks Chiro

 

[TeethWhitener]

Hilbert spaces have a dot product which induces a norm and Banach spaces just a norm, no inner product required.

I'm a bit curious about this. One of the examples that's usually trotted out for Hilbert vs. Banach is $L_p$ spaces, where $p \neq 2$. It's clear that the Hilbert space inner product induces a natural choice for a norm, but is it required that such a choice actually be the norm? In other words, can you take $L_3$ space, with a norm of
$$||f|| = \left(\int (f(x))^3 dx\right)^{1/3}$$
and invent an inner product
$$\langle f,g\rangle = \left(\int (f(x))^{3/2}(g(x))^{3/2} dx\right)^{1/3}$$
or something similar which satisfies the requirements of being an inner product? Basically, can you choose independent (unrelated) functions to be inner product and norm, or does the inner product necessarily have to induce the norm for it to be a Hilbert space?

 

[FactChecker]

Also for quantum mechanics, read Von Neumanns treatise on it and you'll find that they use Hilbert spaces because of the infinite dimensional aspect.

I find this entire argument to be silly. A space (with a choice of a norm) is either a Hilbert space or it is not. If you want to use R², R³, or Rⁿ, you are "using a Hilbert space". You have no choice. Who in their right mind would use Rⁿ and ignore the concepts of the inner product and angles?

 

[fresh_42]

Basically, can you choose independent (unrelated) functions to be inner product and norm, or does the inner product necessarily have to induce the norm for it to be a Hilbert space?

Yes, for the first part, no, for the second. However, in cases the two are different, it should be made very clear, because given a dot product, people usually assume the norm to be chosen accordingly.

I've gathered a few examples at the end of: https://www.physicsforums.com/insights/hilbert-spaces-relatives/
and especially 5.2 and 5.3 show that there is no unique way. One of the things physicist use to forget if they talk about the Hilbert space, or almost always assume the square integral norm to be given without ever mentioning it.

 

[FactChecker]

Yes, for the first part, no, for the second. However, in cases the two are different, it should be made very clear, because given a dot product, people usually assume the norm to be chosen accordingly.

I've gathered a few examples at the end of: https://www.physicsforums.com/insights/hilbert-spaces-relatives/
and especially 5.2 and 5.3 show that there is no unique way. One of the things physicist use to forget if they talk about the Hilbert space, or almost always assume the square integral norm to be given without ever mentioning it.

Good point. After reading this, I just had to edit my post #54 to say "A space (with a choice of a norm) is either a Hilbert space or it is not."

 

[fresh_42]

Good point. After reading this, I just had to edit my post #54 to say "A space (with a choice of a norm) is either a Hilbert space or it is not."

I liked your formulation

If you want to use $\mathbb{R}^2$, $\mathbb{R}^3$, or $\mathbb{R}^n$, you are "using a Hilbert space". You have no choice.

You saved my day!

 

[aabottom]

Hi, I have the impression that the special thing about Hilbert space for Quantum Mechanics is that it is simply an infinite space, which allows for infinitively integration and derivation of its elements, f(x), g(x), their linear combination, or any other complex function, given that the main operators in QM, which are differentiation, multiplication and addition, are all linear self-adjoint operator which thus act orthogonally on orthogonal elements, and thus operate in infinitively many dimensions. This is not possible in $\mathbb{C}$ and $\mathbb{R}$, where both include coordinates as their elements.

What about Banach spaces however? Is my explanation given above on the Hilbert space sufficient to explain to anyone why that makes it ideal for QM?

Thanks!

SemM
I'm late to the game, but I the found fresh_42 article, "Hilbert Spaces and Their Relatives" [1], to be very informative. In that article, fresh_42 outlines the difference between Hilbert spaces and Banach spaces and other spaces. He also has a flow chart of spaces that outlines their differences.

The article states that a Banach space does NOT require the existence of an inner product, but a Hilbert space does. Thus all Hilbert spaces are Banach spaces. But not all Banach spaces are Hilbert spaces.

Thanks fresh_42, for that article.

[1] https://www.physicsforums.com/insights/hilbert-spaces-relatives/

 

[chiro]

FactChecker - Did you read the treatise or not?

Simple question - and it's relevant because the whole hilbert space connection is defined in that treatise.

You do realize that the definitions are in that treatise and if you don't agree with that treatise [and the assumption that definitions can't change in mathematics] then it means you don't agree with the mathematicians that developed and introduced it.

If you see an infinity sign then it means infinity - not a natural number.

Again - read the actual treatise because it is used to take mathematics [including hilbert spaces] and apply it to physics and I know that you are associating this with quantum mechanics.

What exactly do you have a problem with when it comes to the treatise and definitions used within? [Remember - this was done by Von Neumann himself who was the one who started to make quantum mechanics mathematically consistent - this is not some random author].

 

[fresh_42]

FactChecker - Did you read the treatise or not?

Simple question - and it's relevant because the whole hilbert space connection is defined in that treatise.

I very much doubt that we need to quote John von Neumann to define a Hilbert space, nor to decide whether they can be of finite dimension or not. $\mathbb{R}^n$ is a Hilbert space with respect to the ordinary scalar product, because it fulfills the requirements. End of plea.

Before this threads becomes even more personal based on ridiculous assumptions and wrong statements, I'll close it now.