Quantum mechanics and classical physics

1. Nov 6, 2005

jhe1984

Hi,

I was wondering if there is a standard "margin of error" concerning the level of divergence from classical physics we see when looking at atoms and particles in a quantum context. That is, I know it is the case that we can't pinpoint the exact next location of a particle, but is it the case that there is a general "range" that our predictions can become relatively accurate?

This might be a very fundamental point but I'm a newcomer.

Thanks.

JHE

2. Nov 6, 2005

masudr

Whenever the action of the system you are considering is of the order of Planck's constant, you will generally get quantum behaviour. Although there are several quantum effects which have literally no classical counterpart, such as entanglement, spin etc.

edit: in fact the fact that the operators commute isn't that interesting - we have the canonical commutation relations in classical mechanics too. The most important point about QM is that we have amplitudes for systems which are complex numbers, and we add them and then only use their absolute value. This is, if you think about it, a very silly thing to do, but it works beautifully.

3. Nov 10, 2005

Careful

Indeed, entanglement has no classical counterpart, but spin has (I know Feynman said it didn't but he has proven to be wrong there). The edit is a bit crazy: non commuting observables are at the hart of quantum mechanics (the classical Poisson bracket has nothing to do with observation´´, while the commutator obviously has). This issue gave rise to the famous 1935 EPR paper where Einstein Podolsky and Rosen argued the incompletess of quantum mechanics by providing a setup in which one could know the values of two non commuting observables at the *same* time through retrodiction (a possiblity Heisenberg himself was also aware of).

4. Nov 18, 2005

masudr

The principle of superposition is at the heart of quantum mechanics. That's what gives it the marked difference from classical mechanics. The value of $\hbar$ does define the scale at which quantum effects are important, but it doesn't modify the quantum effects in any way. The commutator is not what gave rise to the EPR paper; instead it was the principle of superposition which lead to that.

5. Nov 18, 2005

Galileo

What's classical about spin? Can you show me this proof?

6. Nov 18, 2005

Careful

To be exact, it was the existence of irreducible superpositions of productstates which, when applied in a rather special situation, led to the debate about the reality of values of non commuting observables versus the Bohr complementarity principle (with its instantaneous wave function collapse). My comment was mainly about your claim that noncommuting operators aren't very interesting. The commutator did indeed not lead to the EPR paper, but was seriously involved in its outcome.

7. Nov 18, 2005

Careful

There exist some classical theories of spin which have very close *resemblance* to the properties of quantummechanical spin (as for example the correct gyromagnetic factor for the electron/Feynman claimed that no classical theory would be able to produce the number of two here). In classical GR, you might look at rotating Kerr charged shell models (see Rosquist 2004 on the Arxiv and references therein). If you want to, I dig up more, but to my knowledge there does not exist yet a full classical theory of spin which is known to reproduce ALL the EXPERIMENTAL results. However, the reasons why a classical theory of spin would not exist are shown to be incorrect. In the same way, it is not known to my knowledge that all the Feynman GEDANKEN experiments (with three SG apparatus in a row) do actually give the correct result (if someone would know the contrary and could provide me with references, I would be glad to recieve those). It is however known that all evidence for spin is INDIRECT (ie SG experiments with electrons give no indication whatsoever) and one might conjecture that spin only occurs due to EM interactions in a multi particle bound state (actually some classical approaches are based upon this idea - I seem to remember that Bohr himself held this idea for quite some time).

8. Nov 18, 2005

ahrkron

Staff Emeritus
I'm pretty sure that is incorrect. There is a very nice article by David Mermin in which he shows a couple of configurations that simply cannot be described by classical physics. I'll try to find the reference.

Another, older and somewhat more complicated demonstration of the incompatibility of classical and QM is the Kochen-Specker theorem. I seem to remember that it was formulated in terms of spins, but that it has a more general applicability.

Maybe the exact experiments he depicts have not been performed, but *all* experiments in which people have applied the same rules he used come out as predicted by QM.

9. Nov 19, 2005

akhmeteli

The situation may be less clear-cut, and QM does not necessarily require complex amplitudes (or complex wavefunctions). Example (Shroedinger, Nature (1952), v.169, p.538): for any solution of the equations of the Klein-Gordon-Maxwell electrodynamics (a scalar charged field \psi interacting with electromagnetic field) there exists a physically equivalent solution with a real (not complex) field, which can be obtained from the original solution by a gauge transform. Thus, the entire range of physical phenomena described by the Klein-Gordon-Maxwell electrodynamics may be described using real fields only. Shroedinger's comment: "That the wave function ... can be made real by a change of gauge is but a truism, though it contradicts the widespread belief about 'charged' fields requiring complex representation."

10. Nov 19, 2005

Careful

11. Nov 19, 2005

Careful

Good remark. It is clear that you can do that locally, but I assume you have to be very careful doing this when considering the Bohm Aharonov effect (where you get topological winding numbers). Your real wave equation would contain a term with a dirac function somewhere I guess.

Last edited: Nov 19, 2005
12. Nov 19, 2005

akhmeteli

I agree, extra care never hurts:-), but I believe that if we can do it locally, we can do it globally, provided we are talking about single-valued wave functions. In fact, what are we doing? If we have a wave function \psi=a*\exp(i*b), where a and b are real functions, then we replace \psi with a and add the 4-gradient of b times a constant to the 4-potential of electromagnetic field. If the gradient of b is singular somewhere, it is singular locally there as well. If \psi is continuous, problems may arise in the points where \psi=0. For example, if \psi approximately equals \rho*\exp(i*(\phi+\omega*t)) in the vicinity of such point, where \rho and \phi are the polar coordinates with the center in this point, and t is time, then the 4-gradient of b equals (\omega, 0, \frac{1}{\rho},0), where the last three components are spacial cylindrical coordinates. So we do have a singularity in the 4-potential, however its contribution to electromagnetic fields vanishes. Perhaps more serious problems can arise, say, if you have a singular distribution of charge or current, which is often considered in the context of the Aharonov-Bohm effect.

13. Nov 21, 2005

Careful

No, it is a bit more subltle than that: *psi* is gobally twice differentiable (that is no problem) but your *b* is not necessarily single valued (if the problem were in 2D - then I would say there is a branch were b down = b up + 2*n*pi with n a natural number). Gauging b away would involve dirac distributions...

14. Nov 23, 2005

akhmeteli

What I tried to say is the following: if the situation you describe does have place, there will be points in the infinitesimal vicinity of which this problem arises as well (e.g., if there are branches in 2D, then there is at least one branching point). That is why I said: if we can do it locally, we can do it globally. Because if we cannot do it globally, there must be (I believe) a point where it cannot be done locally. Where I was wrong - in the example that I gave I overlooked that the fields are zero only outside of the zero of the coordinat system (so the fields do represent distributions). I stand corrected.