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Timeline for Why we use $L_2$ Space In QM?

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Apr 13, 2017 at 12:39 history edited CommunityBot
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Oct 29, 2012 at 16:40 history edited Qmechanic CC BY-SA 3.0
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Oct 29, 2012 at 14:46 comment added Ron Maimon @A.O.Tell: I agree with this--- it would make a fine answer BTW.
Oct 29, 2012 at 12:40 comment added MBN Point 3. is not phrased well. "to make the Hilbert space complete", Hilbert spaces are complete. May be it should say "so that the space of states is complete" or something like that.
Oct 27, 2012 at 15:10 history edited Qmechanic CC BY-SA 3.0
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Oct 27, 2012 at 6:55 comment added A.O.Tell @RonMaimon, I'm very aware of all that, and it's why I called the physical reasons a starting point. And that's all it is. Without the physics you wouldn't have a reason to consider this norm or inner product. I'm not saying that the mathematical constructions entirely follow from physical principles, only that they are motivated.
Oct 26, 2012 at 21:31 comment added Ron Maimon @A.O.Tell: I don't think you realize what kind of functions are contained in L_2. This space includes wavefunctions which are completely fractally supported, discontinuous everywhere, with infinite expected energy, infinite energy variance, ridiculous unphysical propagation, and so on. L2 is an idealization which has very little to do with physics, but with the type of closure operations mathematicians like to make. The idealization physicists make is a lattice, with the limit of finite energy wavefunctions. That's not the same as L2, but so what, the mathematicians can do whatever they want.
Oct 26, 2012 at 15:28 history edited Qmechanic CC BY-SA 3.0
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Oct 26, 2012 at 9:59 comment added A.O.Tell These are all good mathematical reasons, but in my eyes the most important reasons are of physical nature: The Schroedinger equation preserves the 2-norm, and probabilities are related to an inner product that induce this norm. So the mathematical framework for quantum theory is naturally based on these sensible starting points.
Oct 26, 2012 at 0:00 comment added Ron Maimon @TMS: I just mean make space discrete, make a grid of positions, and then there is no difference between the different $L_p$ spaces. All the issues are with what completions you are considering when the grid is eensy teensy, and this is not physics, but pure mathematics. The "right" completion is L_2, but so what, who cares. I don't know a reference, the lattice version is just something you work out on your own, but all physicists imagine a lattice down there anyway, just to regulate things like delta-functions and infinite volume limits, which are only interesting to mathematicians.
Oct 25, 2012 at 23:31 comment added TMS @Ron: can you please explain or provide a reference to understand what you mean by need for limit +1? it's sounds to me something equivalent to the completeness, no?
Oct 25, 2012 at 23:29 vote accept TMS
Oct 25, 2012 at 23:28 history edited Qmechanic CC BY-SA 3.0
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Oct 25, 2012 at 23:28 comment added TMS thx for detailed answer, then we using it just because it feels to be natural, anyway can you please explain why not all $L_p$ spaces Hilbert spaces, and what you mean by "It is true that other Hilbert space (modeled over the position space R3) do exist, but they would typically rely on additional structure"
Oct 25, 2012 at 23:25 comment added Ron Maimon One should add that this is for mathematical convenience in the small lattice limit, which is technically unphysical. For any lattice spacing, it is irrelevant which L space you use, since the topology is the same, but L2 gives the convenient limit, +1 anyway.
Oct 25, 2012 at 23:22 history edited Qmechanic CC BY-SA 3.0
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Oct 25, 2012 at 23:17 history answered Qmechanic CC BY-SA 3.0