- 153
- 1
I know that hilbert space is infinite dimension space whereas eucledian is Finite n dimensional space, but what are all other differences between them?
I guess he's taking a QM course. I have seen QM courses being taught by defining Hilbert space as infinite dimensional. Some even define Hilbert space as ##L^2(\mathbb{R})## or ##\ell^2##. I don't know why they don't teach it the right way though.Hilbert spaces are not necessarily infinite dimensional, I don't know where you heard that.
Euclidean space is a specific example of a Hilbert space. Euclidean space is the set of all n lists of numbers (n is is the dimension, and may be infinite dimensional), over real numbers, with the dot product as the inner product.If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?
Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.
I don't know who would describe a hilbert space as infinite dimensional only as one of the most important hilbert spaces in QM is two-dimensional. It's hard to talk about Hilbert spaces properly when a very good portion of the students won't even know what completeness means.I guess he's taking a QM course. I have seen QM courses being taught by defining Hilbert space as infinite dimensional. Some even define Hilbert space as ##L^2(\mathbb{R})## or ##\ell^2##. I don't know why they don't teach it the right way though.
You need to specify an inner product.An example space which is hilbert but not eucledian space can be set of all continuous function in a given interval..correct ?
Yes, Euclidean space is the same as a finite-dimensional Hilbert space. So the Euclidean space does not have infinite dimension by definition. A Hilbert space can be finite or infinite dimensional.If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?
Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.
Sakurai:I don't know who would describe a hilbert space as infinite dimensional only as one of the most important hilbert spaces in QM is two-dimensional. It's hard to talk about Hilbert spaces properly when a very good portion of the students won't even know what completeness means.
I got to say that this is one of the absolute worst descriptions of Hilbert space that I've ever seen. I'm the last one to say that QM books should do rigorous math, but if they do mention things like Hilbert spaces, it should at the very least be approximately correct. /rantLater, in section 1.6, we consider the case of continuous spectra - for example, the position (coordinate) or momentum of a particle - where the number of alternatives is nondenumerably infinite, in which case the vector space in question is known as a Hilbert space after D. Hilbert, who studied vector spaces in infinite dimensions.
I think there's some out-of-date terminology that agrees with this. While you and I know a Hilbert space as an inner-product space whose induced metric space structure is complete, I've been told they used to refer to (the) Hilbert space, which is what we would call a separable, infinite-dimensional Hilbert space. Of course the word "the" is validated by there being only one such space (up to unitary equivalence).Some even define Hilbert space as ##L^2(\mathbb{R})## or ##\ell^2##. I don't know why they don't teach it the right way though.
Yeah. Originally, Hilbert only worked with the ##\ell^2## or ##L^2## space. He didn't consider more general spaces.Sorry for the OT, but...
I think there's some out-of-date terminology that agrees with this. While you and I know a Hilbert space as an inner-product space whose induced metric space structure is complete, I've been told they used to refer to (the) Hilbert space, which is what we would call a separable, infinite-dimensional Hilbert space. Of course the word "the" is validated by there being only one such space (up to unitary equivalence).
That's a very surprising description.I got to say that this is one of the absolute worst descriptions of Hilbert space that I've ever seen. I'm the last one to say that QM books should do rigorous math, but if they do mention things like Hilbert spaces, it should at the very least be approximately correct. /rant
A similar story, about Shizuo Kakutani. He proved a useful generalization of Brouwer's fixed point theorem (to set-valued functions), which is used all over the place in economic theory. Once, he presented at a conference and had a bunch of economists in the audience, so he asked a game theorist buddy why. When the game theorist told him that it was likely because of the Kakutani fixed point theorem, Kakutani replied, "What is the Kakutani fixed point theorem?"Probably apocryphal, but: once, a mathematician gave a lecture in front of other mathematicians. He said "consider a Hilbert space H and...". He was interrupted as somebody in the audience asked "What exactly is a Hilbert space?" That somebody was David Hilbert.
And again, with Banach, when he was reading a paper asked what a Banach space was (he called them B-spaces IIRC).A similar story, about Shizuo Kakutani. He proved a useful generalization of Brouwer's fixed point theorem (to set-valued functions), which is used all over the place in economic theory. Once, he presented at a conference and had a bunch of economists in the audience, so he asked a game theorist buddy why. When the game theorist told him that it was likely because of the Kakutani fixed point theorem, Kakutani replied, "What is the Kakutani fixed point theorem?"
Uhh, the question is a question about definitions. I don't know what else he's asking for if not for clarification of the definitions.It appears that this question has degenerated into a quibble about definition. The original question was about the difference between (finite dimensional) Euclidean space and (infinite dimensional) Hilbert space. I suggest that replies should try to address that question and not get hung up on definitions.
I think you made a typo here. Of course it should be that any bounded sequence has a convergent subsequence.- Every bounded sequence converges to something.
Hah! Thank you. Yes.I think you made a typo here. Of course it should be that any bounded sequence has a convergent subsequence.
Isn't this true in any metric space? Consider the bounded sequence as a metric (sub)space S. Then S is a compact (sub)space so that every sequence has a convergent subsequence.Hah! Thank you. Yes.
Every bounded sequence has a subsequence that converges to something.
No, it's not true in any metric space. One example is an infinite-dimensional Hilbert space. Or the discrete metric space.Isn't this true in any metric space? Consider the bounded sequence as a metric (sub)space S. Then S is a compact (sub)space so that every sequence has a convergent subsequence.