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If we assume superposition and define an Hilbert space with canonical commutation relations we can derive uncertainty relations. So it seems the uncertainty principle isn't required, or should be called the "canonical commutators principle".

On the other direction, if we would like to start assuming the uncertainty relations, it would be tricky even to give them a precise meaning, not having yet a phase space.

So is it safe to assert that uncertainty is called a principle maybe for some historical reason but it's hard to give it that role from a logical point of view?

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  • $\begingroup$ Depends on what you mean by "principle", I suppose. Usually in physics, "principle" does not usually mean "axiom". It sometimes means "a general rule that is useful to order your thinking". $\endgroup$ – knzhou Feb 12 at 11:27
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    $\begingroup$ What does it mean to "assume superposition"? $\endgroup$ – probably_someone Feb 12 at 11:50
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Yes, uncertainty relations are a theorem in nowadays approach to quantum mechanics, and the word principle is the remnant of a different perspective at some stage of the construction of the conceptual framework of QM.

However, for the aim of historical correctness, one should stress that the meaning of the Heisenberg's uncertainty principle was not exactly the same as the present uncertainty relations. The former was a statement about the uncertainty on the value of non commuting observable at the level of an individual measurement, while the latter refers to the statistical spread of values in an ensemble of measurements. The two things are correlated, but are not the same. Some interesting debate about the relations between the two problems has gone on in the last two decades, starting from Ozawa's work.

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  • $\begingroup$ The canonical commutation relations are also derived in some very popular textbooks, so I'm not sure you can say that the canonical commutation relations aren't also "a theorem in nowadays approach to quantum mechanics." Griffiths, for example, starts by defining the position and momentum operators in position space and from there derives the canonical commutation relations. $\endgroup$ – probably_someone Feb 12 at 12:47
  • $\begingroup$ @probably_someone And what is Griffiths derivation if not a theorem? There are some hypothesis and after a chain of deductions one arrives to the uncertainty relations. I would call it a theorem. Or maybe I didn't understand your comment. $\endgroup$ – GiorgioP Feb 12 at 13:29
  • $\begingroup$ My point was that Griffiths derived the canonical commutation relations rather than assuming them. As such, the commutation relations are also a theorem under the set of assumptions that Griffiths chose, and so the fact that the uncertainty principle is a theorem is not something that distinguishes it from the canonical commutation relations. Since the OP was asking about the relative status of the uncertainty principle and the canonical commutation relations, I don't see how you can answer "yes, they enjoy different status" given Griffiths' derivation. $\endgroup$ – probably_someone Feb 12 at 13:32
  • $\begingroup$ @probably_someone I do not understand your point. It is true that one can arrive to the uncertainty relation you write at the beginning of your answer without making explicit the value of the commutator. But after that? $\endgroup$ – GiorgioP Feb 12 at 13:42
  • $\begingroup$ Griffiths conclusion in that there are uncertainty relations for any pair of observable which do not commute. As far as I can see, he never claims that one could derive the value of commutator from the uncertainty relation. $\endgroup$ – GiorgioP Feb 12 at 13:44
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The uncertainty principle states that for any two Hermitian operators $\hat{A}$ and $\hat{B}$, the variances $\sigma_A$ and $\sigma_B$ that is measured in the corresponding observables $A$ and $B$ are related by the following inequality:

$$\sigma_A\sigma_B\geq\frac{1}{2}\left\vert\langle[\hat{A},\hat{B}]\rangle\right\vert$$

The proof of this does not depend on the canonical commutation relations; rather, it only depends on the existence of the Hermitian operators involved, the definition of variance, and the Cauchy-Schwarz inequality. The proof can be found in Griffiths's quantum mechanics textbook, as well as on Wikipedia here: https://en.wikipedia.org/wiki/Uncertainty_principle#Robertson%E2%80%93Schr%C3%B6dinger_uncertainty_relations.

In order to prove the Heisenberg uncertainty principle with this approach (which is a special case of the uncertainty principle where $\hat{A}=\hat{x}$ and $\hat{B}=\hat{p}$), you do need to provide extra information. This information can either be provided by directly assuming the canonical commutation relations, or by defining the action of the position and momentum operators themselves on a wavefunction in any basis, by which the canonical commutation relations can be derived, as is shown, for example, here: Derivation of canonical position-momentum commutator relation.

The uncertainty principle is not an assumption, but it is required; it's a direct mathematical consequence of the definition of operators and the definition of variance. In addition, the Heisenberg uncertainty principle is not necessarily dependent on assuming the canonical commutation relations. There are other choices of assumptions one can make in which the canonical commutation relations are derived from other axioms (I specified one such choice here: defining the position and momentum operators), so it should be clear that even for the Heisenberg uncertainty principle, the canonical commutation relations do not necessarily need to be assumed. They often are in practice, but there is nothing in principle that elevates that set of axioms above any other self-consistent set, except for aesthetic elegance and a later connection to the assumptions of QFT.

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