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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?

Qmechanic
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J.Ask
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  • 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". – knzhou Feb 12 '19 at 11:27
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    What does it mean to "assume superposition"? – probably_someone Feb 12 '19 at 11:50

2 Answers2

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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.

probably_someone
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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.

GiorgioP-DoomsdayClockIsAt-90
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  • 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. – probably_someone Feb 12 '19 at 12:47
  • @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. – GiorgioP-DoomsdayClockIsAt-90 Feb 12 '19 at 13:29
  • 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. – probably_someone Feb 12 '19 at 13:32
  • @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? – GiorgioP-DoomsdayClockIsAt-90 Feb 12 '19 at 13:42
  • 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. – GiorgioP-DoomsdayClockIsAt-90 Feb 12 '19 at 13:44
  • The section in Griffiths I am referring to is on pages 109-110 in the first edition, Ch. 3: Formalism, Sec. 3.4: The Uncertainty Principle. He uses the definitions of the position and momentum operators in the position basis to derive the canonical commutation relations by computing the commutator directly on an arbitrary wavefunction. As such, Griffiths does not assume the canonical commutation relations; the canonical commutation relations are a theorem derived from the assumption that is the definition of the position and momentum operators in the position basis. – probably_someone Feb 12 '19 at 14:13
  • I never claimed that the canonical commutation relations could be derived from the uncertainty principle. My claim was that the uncertainty principle does not necessarily have to be derived using the canonical commutation relations. Depending on the set of assumptions one chooses, the two statements are not necessarily logically dependent, which would mean that the answer to the question would be "no." You have answered "yes." In any case, what are you referring to with the question "after that?"? – probably_someone Feb 12 '19 at 14:17
  • @probably_someone I know very well Griffiths derivation. I do not think that the way the value of the commutator is introduced (either via direct calculation or by assuming its value as an axiom) plays any role in the physical content of uncertainty relations or in relation with the original question. On the other hand, the choice of the operator representation of $\hat x$ and $\hat p_x$ is not independent on the choice of the commutator. But, from the point ov view of the uncertainty relations what really matters is the value of commutator on the chosen state. – GiorgioP-DoomsdayClockIsAt-90 Feb 12 '19 at 14:33
  • To reiterate for the third time, if we initially assume the definition of the position and momentum operators in a certain basis, we can derive the Heisenberg uncertainty principle without assuming the canonical commutation relations. The question asked whether the canonical commutation relations and the uncertainty principle enjoyed different logical status (this is what "logical dependence" most likely means, based on the rest of the question; the OP is asking whether you need to assume one to prove the other). Under this set of assumptions, the answer to that question is no. – probably_someone Feb 12 '19 at 14:44
  • You still, of course, eventually need to know the value of the commutator; my only point, this whole time, is that the value of the commutator doesn't need to enjoy any special logical status; in other words, it doesn't need to be an assumption. You can calculate the value of the commutator from other assumptions. I would also like to know how, exactly, you determine "what really matters" for the purposes of answering this question. – probably_someone Feb 12 '19 at 14:48
  • But assuming definition of position and momentum is equivalent to and not independent on the canonical commutation relations. I do not understand your insisting on such a point and I disagree on the negativa answer to the question. – GiorgioP-DoomsdayClockIsAt-90 Feb 12 '19 at 14:48
  • The two sets of axioms are not equivalent for the purposes of this question, which is asking about the logical structure underlying quantum mechanics; you get equivalent physical predictions, certainly, but you do not have an equivalent logical structure, because you have chosen different axioms, and some things that are axioms in one system are theorems in the other. Since the question is asking about the logical relationship of two statements, the logical structure of quantum mechanics is critically important here. – probably_someone Feb 12 '19 at 14:52