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Hilbert space formulation for nonquantum mechanics? 
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#1
May912, 05:07 PM

P: 148

Note: I am NOT talking about the classical limit of quantum mechanics, where in the limit of numbers that are large compared to h the average values approach the classical values, nor am I talking about Lagrangin/Hamiltonian mechanics in phase space; I am talking about using vectors with classicalscale values in a Hilbert space INSTEAD of classical mechanics.
For example, take a bead constrained to move on a wire (so its position and momentum can be take values on the x axis), and say for this experiment we are interested if it is to the left of the origin,or at/to the right of the origin, and similarly for the momenta; therefore, our Hilbert Space contains four possible state vectors: (x<0, p<0), (x≥0, p<0), (x<0, p≥0), (x≥0, p≥0), and since it is a classical system, it will ALWAYS be in an eigenstate of the position/momentum operators, with two eigenvalues each. It seems like a formulation like this would have great pedagogical value, since the results obtained can be compared to everyday experience (or to those calculated via the more traditional classical formalistseg F=ma)...that is, supposing it works. I see no reason it shouldn't, but I'm having trouble wrapping my mind around it enough to work out a sample problem to test it out. So  does anyone know of any treatments of this, either online or in the literature? Or perhaps a simple reason why there wouldn't be any such treatments (eg “it won't work because...”)? Thanks  Justin 


#2
May1012, 04:11 PM

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P: 6,066

Hilbert space in mathematics is a well defined concept, essentially an extension of Euclidean space to infinite dimensions. There is no physics involved per se.
Physics uses the concepts of Hilbert space in quantum mechanics because it fits. 


#3
May1012, 05:03 PM

P: 148

But even beyond that, in the context of QM, the "Hilbert space formulation" also implies that observable values a obtained by finding the eigenvalue of a linear operator on the state vector. So that's what I was looking for  a formulation of classical physics using *that* formalism, which happens to use vectors in a Hilbert space. 


#4
May1112, 03:15 PM

Sci Advisor
P: 6,066

Hilbert space formulation for nonquantum mechanics?
My exposure to Hilbert space was through the mathematics. You probably know a lot more about the physics application than I do.



#5
May1112, 05:23 PM

P: 148




#6
May1112, 10:32 PM

P: 446

In classical mechanics there are no superpositions of states. In a Hilbert Space there can be superpositions. I can't state this with certainty, but I would bet that if you could find a Hilbert Space formulation of Classical Mechanics, it would be so messy as to provide no insight.



#7
May1212, 06:50 AM

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P: 7,151

The state vectors you gave in your example seem rather arbitrary to me. The difference between x < 0 and x ≥ 0 depends on the origin of your coordinate system, not on the behaviour of the bead on the wire. This seems fundamentally different from a notion like "the spin of an electron", which doesn't depend on how fast your reference frame is rotating!
In classical physics there are plenty of situations where a differential equation plus its boundary conditions generates an eigenvalue problem, and the possible solutions to the DE are linear combinations of the eigenvectors. The general study of this (independent of particular physical problems) is SturmLiouville theory, which predates quantum mechanics by half a century or more  and in fact the eigenpairs form the basis functions of a Hilbert space .... I'm not sure whether that is where you are heading, but there seems to be an analogy between the mathematical methods used in QM and classical mechanics here, even if there isn't really a physical analogy. 


#8
May1212, 09:34 AM

P: 148

[QUOTE=AlephZero;3909217]The state vectors you gave in your example seem rather arbitrary to me. The difference between x < 0 and x ≥ 0 depends on the origin of your coordinate system, not on the behaviour of the bead on the wire. This seems fundamentally different from a notion like "the spin of an electron", which doesn't depend on how fast your reference frame is rotating![\quote]
If I'm not mistaken, while the total spin is frameindependent, the measured spin about any axis isnt  but even putting that aside, there are position operators in QM that operate exactly like that example  where the value obviously depends on the chosen frame, but would transform appropriately under a change of frame. The more common formulation breaks the axis into an infinity of infinitesimalsized chunks rather than just two chunks, but to get a meaningful probability, you would have to integrate over a finite area....but anyway... 


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