Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is in continuation to my previous question. It is not a duplicate of the previous one. This question arises because of the answers and discussions in that question.

Can we call a theory, quantum theory, if it is consistent with HUP? For example, suppose there is a finite and self consistent theory of gravity which incorporates the uncertainty principle. Can we at once call this theory a quantum theory of gravity or does it have to satisfy other conditions too?

This question may be too basic but it is intriguing my mind.

share|improve this question
add comment

4 Answers 4

A quantum theory is a theory that obeys the postulates of quantum mechanics, see e.g.

http://en.wikipedia.org/wiki/Postulates_of_quantum_mechanics#Postulates_of_quantum_mechanics

One may choose a slightly different wording of the conditions, add some detailed ones or unify the conditions above to a smaller list, but it must always be effectively the same thing. There are no exceptions from this definition. In particular, a theory of quantum gravity is a theory predicting gravity that is compatible with the postulates of quantum mechanics. (We usually want a theory of quantum gravity to be compatible with the principles of relativity, too.)

share|improve this answer
add comment

In mathematical reasoning, and not only, there are what are defined as necessary conditions, and others that are defined as sufficient conditions.

Necessary means that without this condition the theory cannot be proven/ be consistent. Sufficient means that just from this condition the theory is consistent.

The QM postulates are necessary and sufficient for a quantum mechanical theory to be fully developed. The HUP is a theorem arising from the postulates of the QM theory, so it is a necessary attribute. But it is not sufficient to define the postulates, they cannot be derived from it, and thus not sufficient to define a theory as a quantum theory.

Look at the answer by @Roy Simpson in your previous question. I remember this "uncertainty principle" from my classical electrodynamics course. That it exists, it does not make classical electrodynamics a quantum theory.

share|improve this answer
1  
+1 btw, what was the uncertainty principle in classical electrodynamics? –  Approximist Apr 11 '11 at 7:14
add comment

Apparently, No. To quote SEP

A second point is the question whether the theoretical structure or the quantitative laws of quantum theory can indeed be derived on the basis of the uncertainty principle, as Heisenberg wished. Serious attempts to build up quantum theory as a full-fledged Theory of Principle on the basis of the uncertainty principle have never been carried out. Indeed, the most Heisenberg could and did claim in this respect was that the uncertainty relations created "room" (Heisenberg 1927, p. 180) or "freedom" (Heisenberg, 1931, p. 43) for the introduction of some non-classical mode of description of experimental data, not that they uniquely lead to the formalism of quantum mechanics. A serious proposal to construe quantum mechanics as a theory of principle was provided only recently by Bub (2000). But, remarkably, this proposal does not use the uncertainty relation as one of its fundamental principles.

All the mystique surrounding the Heisenberg Uncertainty principle vanishes if you take into account the deBroglie hypothesis. I remember from the Feynman lectures, '...what follows is something that is familiar to anyone who works with waves'. It might be worthwhile also to look at Terry Tao's post.

share|improve this answer
add comment
  1. .NET is a framework
  2. Software that is built using languages supported by .NET is called .NET software.
  3. That software may or may not break if some part of .NET breaks.
  4. It is necessary to have .NET in place for the software to work.

Now replace .NET with QM and software with theory. :-D

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.