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Someone asked in a deleted comment, something along the lines of "Which Hilbert space?", pointing out that a Hilbert space is just a vector space which has an i
by drdeca 6y ago
Someone asked in a deleted comment, something along the lines of "Which Hilbert space?", pointing out that a Hilbert space is just a vector space which has an inner product, and is complete under the metric induced by the norm.
I wrote a response to this, and while the initial comment was deleted (presumably by the author of the comment), I think my response might still be informative to others who have the same question. It is as follows:
All countably infinite dimensional Hilbert spaces are isomorphically isometric.
Pick an orthonormal basis for it, and that gives you an isomorphism with \ell^2(\mathbb{N}) (which preserves the metric).
As such, some people will sometimes just refer to "the Hilbert space", to refer to any countably infinite dimensional Hilbert space over the complex numbers.
If the Hilbert space in question is countably dimensional, then this is "the" one they mean.
I think there are some reasons that some people say one might have to use one with an uncountable dimension? I don't understand these reasons. I think they are connected to quantum field theory? My impression is that it is somewhat controversial whether one with uncountably infinite dimension is ever necessary to model physics. Wikipedia has more to say about this question. Because I don't understand it, you would be better off reading it on wikipedia than reading me trying to summarize something written on Wikipedia which I don't understand.
edit : upon actually reading more of the link, I might see why the comment I meant to reply to was deleted : the paper answers their question, saying that the Hilbert space is determined by the dimension.
So, my comment is a bit redundant.
edit2: Oh, and, furthermore, the article says that in a bounded region, we have reason to expect the dimension of the space to be finite dimensional? This surprises me.
- nonameiguess 6y agoThe paper itself says the dimension of the Hilbert space is e^(e^123), so very big but not infinite, countably or otherwise. That said, I also do not remotely understand this paper. I think I at least grasp the basics of quantum theory itself modeling the very small world as a Hilbert space that collapses to a single real-valued vector upon being measured and the math checks out according to what it predicts and what we actually measure, but after reading this I cannot tell why this should have any consequences for ontology. I think I understand the basic desire physicists have to include an ontology in their theories. Old-school billiard ball physics and even general relativity have very obvious and intuitive analogies to basic geometric objects we can envision. We definitely get a map with the math and at least a clear allusion to the territory by reifying these geometric analogies. With quantum physics, nothing at all comes out as some clear candidate for a territory. So this guy is just going galaxy brain and proposing maybe there is no territory at all and the most ontologically basic thing from which all observable phenomena we actually experience arises is just the math itself. I don't really see a justification in the paper, but that's kind of the nature of philosophy. At the most fundamental level, nothing can be justified.
- bjornsing 6y ago> Oh, and, furthermore, the article says that in a bounded region, we have reason to expect the dimension of the space to be finite dimensional? This surprises me. Does it? Iirc a classic basis for a countably infinite Hilbert space is a Fourier series. If you have a finite space to describe, and a maximum frequency you need to represent (which I would say is reasonable to assume), then you only need a finite dimensional space.
- drdeca 6y agoI guess the maximum frequency is the surprising thing to me? Idk, I guess I thought maybe the limited information was, not really limiting the dimension? Or, maybe I thought because of no clean objective separation between the observable universe, and the rest of it, that that would cause an issue? Not really sure what I was thinking. I guess it makes sense that it would be finite dimensional. Hm, actually, I am confused about what counts as the observable universe. Is it that which is in our past light cone, or, that which is in a future light cone of our past light cone, or? Also, there’s the issue of time in different reference frames. If the state of the observable universe, is supposed to be given by a vector in the same Hilbert space at each time t, uh, well, that appears to depend on the reference frame. Like, the points in space time that are “simultaneous” depends on the reference frame. So, a vector which describes the state of the universe in a given time slice in a given reference frame, I don’t see how it would also describe it for a different reference frame.
- vlmutolo 6y agoThis question is addressed directly by Griffiths in his “Introduction to Quantum Mechanics”. It’s in a footnote at the beginning of chapter 3, “Formalisms”. Basically he says that when physicists say “Hilbert space”, they’re actually referring to a specific Hilbert space, namely the set of all square integrable functions, sometimes referred to as “L2”.