# Why quantum mechanics looks so different from classical one?

1. Apr 8, 2015

### Nixom

In many textbooks, the non-commutativity for the canonical pair is considered to lead to the major variaty from classical mechanism(CM) to quantum mechanism(QM), and change the Possion bracket into quantum commutator is a standard procedure called as canonical quantization. But in fact the Possion bracket of x and p in classical mechanism are also non-vanishing. In my opinion both the QM and CM have the same algebraic structure, they ought not to appear much different in formalism, but they do indeed, why?

2. Apr 8, 2015

### naima

The name of the physicist is Poisson (not Possion) It is the french for Fisch!

3. Apr 8, 2015

### bhobba

That is a very deep question and here is the book that answers it:

Be warned however - its what mathematicians call non trivial - meaning its hard.

I have a copy and every now and then feel emboldened to study it - it doesn't last long - it is tough going. To be fair though if you have something like a math degree - its doable with effort - its just if you want to spend the effort. Despite having the proper background having studied analysis at uni my willingness to put in the effort waxes and wanes.

Thanks
Bill

Last edited by a moderator: May 7, 2017
4. Apr 8, 2015

### JK423

Why do you think that they have similar algebraic structure?

First of all, the mathematical space that the two theories live in is different, and the 'states' of the particles in the two theories (quantum states for QM and position for Classical M.) are different mathematical objects that behave in a different way. These mathematical differences also lead to different experimental predictions. I say this because a theory may be formulated in two completely different ways that look different to each other, but since the two ways describe the same thing they must be equivalent and lead to same predictions. That's not the case with QM and CM, they are different both mathematically/algebraically-wise and prediction-wise.

5. Apr 8, 2015

### ddd123

In classical mechanics the state of a system having f degrees of freedom is described by any unique vector in phase space, a 2f-dimensional vector space spanned by Lagrangian coordinates and the conjugated momenta. The solution of the equations of motions is a trajectory.

In quantum mechanics the state of a system is described by a ray (or orbit, when considering Gauge transformations) in the Hilbert space, so its vectors are not unique and cannot be null, also the space can be infinite-dimensional (and generally is when you want to consider position and momentum) while phase space generally is not unless you consider literally infinite number of particles.. In quantum mechanics even a single particle has nonzero entropy since it has spin, the particles are indistinguishable at close range and this solves the Gibbs’ paradox, and so on. Many physical differences which stem from the algebraic differences.

6. Apr 8, 2015

### bhobba

Get the book I mentioned - you are the exact person its aimed at

Thanks
Bill

7. Apr 8, 2015

### aleazk

They look different because they are very different!

Classical physics is based on classical, standard probability theory, in which the propositions about the system form a sigma Boolean algebra (a type of commutative lattice) which is isomorphic to the Borel algebra of sets of phase space. States are probability measures on this algebra of sets, i.e., standard probability.

In QM, the propositions about the system form a non-distributive lattice (because of the existence of incompatible propositions) and states are generalized measures on this lattice. Thus, QM is based on a generalization of classical probability theory. The change is so drastic that the lattice of propositions cannot be realized anymore as a Borel algebra of sets. Instead, one has to use an identification between propositions and orthogonal projectors in a Hilbert space. With this, observables become operators rather than functions from phase space to the real numbers.

You say "But in fact the Possion bracket of x and p in classical mechanism are also non-vanishing". Yes, but in classical mechanics we have another product of observables too! the usual pointwise product of functions (observables). With this product, the space of functions can be made a commutative C*-algebra. In analogy to the algebraic formulation of quantum theory, one can formulate classical physics in terms of this algebra. The abelian character of the algebra leads to standard probability. In quantum theories, both products can only be realized in terms of the usual commutators of operators; so, the CCR give, at the same time, a non-commutative character to the basic C*-algebra of observables too, and thus to a non-classical probability.

8. Apr 8, 2015

9. Apr 9, 2015

### Demystifier

10. Apr 9, 2015

### Nixom

In textbooks, the analog between Poisson bracket and commutator is considered as the bridge connecting CM and QM formalism, converting Poisson bracket into commutator is always the common procedure for quantization. After the convert, the energy of harmonic oscillator system become quantized due to the noncommunity of x and p. So what I'm wondering is if the same procedure could be applied to CM either, although the variables have very different connotation in these two mechanism. Could we image a case in which the canomical variables in CM could be taken as some operators of the C function(the analog object to the state in QM)? Or what's the crucial difference occurs after the canoical quantization and leads to the differenct formalism between two theories?

11. Apr 9, 2015

### Nixom

Actually the appearance of the concept of QM state is also indigested to me, it shows up in an "unusal" way in the QM axiom as a new concept. In my opnion, the state of CM defined in phase space by (x,p) is not a comparable thing even in mathematical sense. Is is necessary to create such entity for the new(respective to CM) theory? It is known that the QM state is a vector in Hilbert space, so the question becomes why the theory must be established on the Hilbert space?

12. Apr 9, 2015

### Nixom

Last edited by a moderator: May 7, 2017
13. Apr 9, 2015

### Nixom

Do you mean CM can be formulated in the same formalism with QM?

So the difference is due to the definition of products. Do I get it?

14. Apr 9, 2015

### ddd123

The bottom line physical explanation is that the probability distribution which is measured show signs of interference between phases associated to the particles (e.g. the double slit or the Aharanov-Bohm effect etc), so you need a complex functional space to handle the amplitudes.

15. Apr 9, 2015

### bhobba

It isn't. The real basis of both QM and Classical Mechanics is symmetry which explains why the Poisson bracket method works.

If you haven't seen it before read the following:
Landau - Classical Mechanics
Ballentine - Quantum Mechanics - A Modern Development - especially the first three chapters

The situation is this - the Galilean transformations + the Principle Of least Action (PLA) leads to classical mechanics. The Galilean transformations + the 2 axioms of QM from Ballentine leads to standard QM. Those two axioms also lead to the PLA.

When one moves to Quantum Field Theory symmetry is basically what determines the equations.

Thanks
Bill

Last edited: Apr 9, 2015
16. Apr 9, 2015

### bhobba

The textbook I gave at the beginning gives a sophisticated answer that has been further bolstered by a new theorem called Soler's theorem:
https://golem.ph.utexas.edu/category/2010/12/solers_theorem.html

A weaker version is what's called Pirons theorem you may also like to look into.

Its deep stuff - like I said - definitely non trivial.

Personally though I like the idea its an extension of probability theory that allows continuous transformations between pure states:
http://arxiv.org/pdf/quant-ph/0101012v4.pdf

QM is the simplest most reasonable generalised probability model that allows that.

Thanks
Bill

17. Apr 9, 2015

### Demystifier

Almost. See the paper in post #9 above.

18. Apr 9, 2015

### naima

the link in post #8 by Dr Du is also interesting
there is an equivalence between QM and phase space with Wigner quasiprobabilities defined on it.
It is written that this formulation avoids the operators on Hilbert space.
Is it possible to describe a particle with the phase space and the Bloch sphere for its spin?

Last edited: Apr 9, 2015
19. Apr 10, 2015

### naima

20. May 21, 2015

### andresB

The commutator of classical position and the operator and classical momentum commute unlike the quantum ones
See

http://en.wikipedia.org/wiki/Koopman–von_Neumann_classical_mechanics