A What separates Hilbert space from other spaces?

[chiro]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

 

[fresh_42]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

This is a debatable point of view. In any case these examples have still to hold, if theorems on arbitrary Hilbert spaces are proven. They are a good test object.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

... since you can have infinite dimension. Whether a geometric intuition is needed is a very personal decision. I think no intuition and merely formulas are still better than a wrong intuition. And to draw a line with a pencil without interruption is wrong since the days of Cantor.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

I work with the definitions and finite dimensional examples. A couple of things are different in finite and infinite dimensional vector spaces, but this wan't the question here. The answer to the OP's question is: inner product and completeness, not dimensionality.

 

[FactChecker]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

It may be better to say that our knowledge of Rⁿ or Cⁿ is abstracted by identifying their Banach and Hilbert properties and applying them to other spaces. Then we can have Rⁿ or Cⁿ motivated insight in those spaces.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

The Dirac delta function, associated comb functions and other generalized functions are very important in Fourier analysis. Continuity is not required for a function to be measurable and Lebesgue integration and measure can handle very strange functions.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

Integration or convolution with an appropriate kernel function is the infinite-dimensional equivalent of matrices.

 

[bhobba]

Just elaborating a bit on what FactChecker said.

Hilbert spaces were invented by David Hilbert and others to help in areas like Fourier Analysis, Partial Differential Equations and eventually QM.

Nowadays, especially for applications to QM and Fourier Transforms, IMHO the easiest way to naturally motivate infinite Hilbert spaces is via what's called Rigged Hilbert spaces of great value in understanding mathematically difficult/unusual things like the Dirac Delta Function - which is not part of a Hilbert space.

What you do is take the set of all complex (just to be definite row) vectors of finite length then consider its dual. Its in fact any infinite sequence at all. Well that leads you naturally to look at dual's of subspaces of that huge and very general space. What you notice is sometimes that dual contains the original space and sometimes its the other way around. There is one very special space whose dual is basically itself - that is called the Hilbert space. This is expressed in the so called Gelfland Triple:

S* ⊃ H ⊃ S - where H is a Hilbert space (ie the space such that the sum of the squares of the sequences absolute value exists) - S is some subspace of H and S* is dual.

This view is especially useful in studying distribution theory that makes Fourier transform's a snap - otherwise you quickly run into issues of convergence and functions that do not belong to a Hilbert space eg that dreaded Dirac Delta Function mentioned before. It bedeviled the greats like Von-Neumann who I believe actually invented the name - Hilbert Space - in honor of David Hilbert who he was assistant to (wow what a pair that would be - like Wheeler and Feynman).

That's just a way of motivating why its interesting to study it - you can of course treat it purely mathematically with no motivation at all.

Added Later:
In light of Marks response - note the original space of finite vectors. Finite vectors are Hilbert spaces ie they all can be placed in one to one correspondence with its dual. We see in that light infinite Hilbert spaces are simply a generalization of that interesting property to infinite spaces. The Dirac Delta function belongs to the dual of a subset of a Hilbert space - the space of what are called good functions is usually used because it has the interesting property of the Fourier transform of a good function is a good function. Its dual is called a Schwartz space - and its easy to define the Fourier Transform of any member of that space because of the Fourier Transform property of good functions. The Dirac Delta function is a member of the Schwartz space. It took quite a while for mathematicians to figure it all out and solve what the great Von-Neumann couldn't.

Thanks
Bill

 

[Mark44]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

A Hilbert space can be of infinite dimension, but doesn't have to be.

Emphasis added in both quotes.
From Wikipedia (https://en.wikipedia.org/wiki/Hilbert_space):

(A Hilbert space) extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions.

From Wolfram MathWorld (http://mathworld.wolfram.com/HilbertSpace.html)

Examples of finite-dimensional Hilbert spaces include
1. The real numbers $\mathbb R^n$ with <v, u> the dot product of v and u.
2. The complex numbers $\mathbb C^n$ with <v, u> the dot product of v and the complex conjugate of u.

 

[SemM]

Please forgive me if I repeat things that have been said before.
A Banach space is complete and has a norm, so distance and convergence is defined. Every Hilbert space is a Banach space that additionally has a dot product which defines the norm and angles. So the extra things that you get with a Hilbert space are angles, orthogonal decomposition, Pythagorean Theorem, parallelogram law, etc. If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.

Does this mean that Banach spaces can be more suitable for classical mechanics, diffusion problems etc?

 

[Mark44]

If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.

Does this mean that Banach spaces can be more suitable for classical mechanics, diffusion problems etc?

I don't believe it means that at all.
It appears to me that you are committing a logical error by starting from an implication that is assumed true, and then negating both the hypothesis and conclusion.
FactChecker: If (you're working in QM) then (a Hilbert space is preferred)
SemM: If (you're not working in QM) then (you should not use a Hilbert space (i.e., use a Banach space))
The second implication is faulty, because nerely negating the hypothesis and conclusion isn't valid.

Here's a simple mathematical example that shows why this negation is invalid.
If x = 2, then x² = 4
This is clearly a true statement.

The implication "If x ≠ 2, then x² ≠ 4" is untrue, as is easily seen by the counterexample x = -2.

 

[SemM]

Take for instance curvilinear phenomena as gravity and relativistic time/space asymmetry phenomena, having no orthogonality and having no need for inner product, would Banach space be more suitable than Hilbert space , for these ?

 

[PeroK]

Take for instance curvilinear phenomena as gravity and relativistic time/space asymmetry phenomena, having no orthogonality and having no need for inner product, would Banach space be more suitable than Hilbert space , for these ?

No need for inner product in GR?

 

[SemM]

No need for inner product in GR?

That is what I am saying, its curvilinear, and Banach space has no inner product...so Banach space OK for GR?

 

[Orodruin]

That is what I am saying, its curvilinear, and Banach space has no inner product...so Banach space OK for GR?

No. Even asking that question shows a fundamental misunderstanding of how GR works and the central role played by the metric.

 

[SemM]

No. Even asking that question shows a fundamental misunderstanding of how GR works and the central role played by the metric.

Please explain, that would be the meaning of this thread... when Banach? When Hilbert (QM).. When RIenman? When others? That may not be the original question, but it helps in understanding and answering the original question. Let's be pedagogic here shall we, so we can make Physicsforum pleasant and informative!

 

[Orodruin]

Please explain, that would be the meaning of this thread... when Banach? When Hilbert (QM).. When RIenman? When others? That may not be the original question, but it helps in understanding and answering the original question. Let's be pedagogic here shall we, so we can make Physicsforum pleasant and informative!

GR is based upon differential geometry and the description of space-time as a Lorentzian manifold, i.e., a manifold equipped with a non-degenerate symmetric bilinear form with signature (1,3). This defines a pseudo-inner product on all tangent spaces.

 

[FactChecker]

Curvature alone does not rule out an inner product. A space can have curvature but when you zoom into a tiny neighborhood of any point, then there are inner products, angles, right angles and orthogonal vectors. They are local to that point. As long as you have them, you should (judiciously) use them where they apply. Because they are defined differently at every point, they are trickier to use. Moving continuously from one point to another, the definitions of the angles and coordinates also change continuously.

 

[SemM]

Curvature alone does not rule out an inner product. A space can have curvature but when you zoom into a tiny neighborhood of any point, then there are inner products, angles, right angles and orthogonal vectors. They are local to that point. As long as you have them, you should (judiciously) use them where they apply. Because they are defined differently at every point, they are trickier to use. Moving continuously from one point to another, the definitions of the angles and coordinates also change continuously.

Thanks, I thought curvature implied non-orthogonality of it elements.

 

[PeroK]

Thanks, I thought curvature implied non-orthogonality of it elements.

If you don't have an inner product, what does "orthogonal" mean?

 

[SemM]

Nothing, that is what I am saying here:

"I thought curvature implied non-orthogonality of it elements." No orthogonality, no inner product.

 

[fresh_42]

Nothing, that is what I am saying here:

"I thought curvature implied non-orthogonality of it elements." No orthogonality, no inner product.

The curvature of a space is usually defined by its behavior of tangents (second derivative), i.e. linear approximations. Usually here means, I'm not sure whether there are any exotic possibilities to do it otherwise. But with tangents of different directions at a point, we have an angle between them, an angle between intersecting lines. This is the reason why manifolds are considered. They behave locally, i.e. in small open neighborhoods of points like a Euclidean space, which means for arbitrary close approximations we have all geometric advantages.

 

[Orodruin]

To add to what has been said, curvature is a priori completely unrelated to the inner product as it determines the failure of parallel transport around a closed loop to return the same tangent vector. This depends only on the affine connection that is placed on the manifold and relates nearby tangent spaces. It does not even require a metric tensor on the manifold. It is related to the metric only if the connection is assumed to be metric compatible, which is the case, e.g., for the Levi-Civita connection (which is the unique torsion free metric compatible connection).

 

[FactChecker]

A familiar example is the surface of the Earth. At any point, p, on Earth (except the North and South poles), one can talk about the angles between lines of latitude and lines of longitude. They are at right angles. An inner product, orthogonal vectors, and angles are defined at that point. Yet the surface of the Earth has curvature at that point.

 

[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.