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What is the basic postulate on which QM rests. Is it that the position of a particle can only be described only in the probabilistic sense given by the state function $\psi(r)$ ? We can even go ahead and abandon the particle formalism as well. So what is the QM all about ? A probabilistic description of the physical world ? and nothing more ?

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Related: physics.stackexchange.com/questions/6738/… –  Marek Aug 16 '11 at 17:41
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There isn't one basic postulate of QM. There are several postulates that fit together into a surprising and beautiful theory. For some reason, people are upvoting one of these postulates and downvoting another, and haven't even mentioned others (e.g., the superposition principle). –  Peter Shor Aug 17 '11 at 11:13
    
@Peter: that's only half-true. There are many other theories that share lots of properties of QM (e.g. any theory given by linear PDE will have superpositions). Similarly, classical logic and quantum logic are basically the same except for one axiom. Therefore if one is really after one thing that makes QM special, one is inevitably lead to non-commutativity. After all, if there was none of it but everything else was kept untouched (formally, $\hbar \to 0$), you'd get back your plain old boring Poisson algebra on the phase space. –  Marek Aug 17 '11 at 21:10

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up vote 5 down vote accepted

Existence of non-compatible observables: measuring one of them (say, coordinate) leads to an unavoidable uncertainty in the result of a subsequent measurement of the other (say, momentum). This is the essence of the Heisenberg uncertainty principle in the kinematics of your system. There is a detailed discussion along these lines in the beginning of the Quantum Mechanics volume (volume III) in the Course of Theoretical Physics by Landau and Lifshitz. Any measurable (physical) system, be it particle, atom or anything else, is quantum only if you can identify a manifestation of Heisenberg uncertainty principle (non-commutativity of observables).

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"Existence of non-compatible observables", great to know this term..+1. –  Rajesh D Aug 16 '11 at 17:29
    
This on-compatibility of physical observables is the empirical reason why non-commutative objects (operators or matrices) need to be assigned to them in the quantum formalism... –  Slaviks Aug 16 '11 at 17:34
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+1, this is precisely what sets QM apart from other theories. In mathematics the term quantization and deformation (in the sense of deforming commutative algebras to non-commutative ones) are basically equivalent. –  Marek Aug 16 '11 at 17:41
    
"deformation of algebras", great to know the term. +1 –  Slaviks Aug 16 '11 at 17:46

To me, the most basic postulate is that energy comes in discrete packages of $h \nu$. Based on this assumption, you get much of the rest of basic quantum mechanics.

In fact, the Schrödinger equation is related to the Hamilton-Jacobi formulation of classical mechanics with the added assumption of quantized energy, and the Heisenberg picture follows directly from the Poisson bracket formulation of classical mechanics, assuming quantized energy.

Wave-particle duality is also a very important assumption- this gives us a way to interpret exactly what these equations express, which is the probability of locating a particle per some (small) volume of position- or momentum-space.

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Discreteness is not really that important (although it had been a century ago when QM was born; it's also where the term quantum comes from). There are many systems for which all important observables have continuous spectrum. –  Marek Aug 16 '11 at 17:44
    
That's true! I guess it's not the best way to think about deriving all of QM. (Non-commuting observers probably are) In my mind, I like to put everything in a historical context. –  specterhunter Aug 16 '11 at 18:02

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