We already know that non-commutativity of observables leads to uncertainty in quantum mechanics cf. e.g. this and this Phys.SE post. What about the opposite: Does uncertainty imply noncommutativity?

If the statistical version of an uncertainty principle (i.e. in terms of standard deviation) is assumed to hold in an algebra of observables, can we ever deduce from this that the ambient operator algebra is necessarily noncommutative? (Say we give the notion of standard deviation as in the Robertson uncertainty principle.)

EDIT: In light of the first answer, I should clarify: If there is a universal lower bound for the measurement of the standard deviations in question, must there be noncommutativity?

Let me rephrase the question:

Question: Why should the presence of a statistical uncertainty relation imply that we need a noncommutative quantum theory?

I guess the point is that Heisenberg's "matrix" explanation for Rydberg-Ritz shockingly implies the uncertainty relation. Particularly, imagining that the classical phase space is a circle, the set of real-valued functions on the space are the observables, and we assume that these functions can be expanded in Fourier series...and then by the Fourier transform the algebra of observables is identified with the convolution algebra of the integers. In the quantum case, Rydberg-Ritz gives that we should replace the above abelian convolution algebra with the convolution algebra of a groupoid...hence a noncommutative matrix algebra. From this noncommutativity, the uncertainty relation can be deduced. Is it possible that a fundamental uncertainty relation exist between commuting observables? No reason exists experimentally to think so, except for the fact that the noncommutative theory predicts it for noncommuting observables...I should be a bit ashamed of this question, as there is no physical reason to think it would be true as of now. Yet I ask anyhow...

I think that the first answer does the trick, and that the answer is no, but I'd be interested in any further thoughts on the degree to which uncertainty should force a noncommutative quantum theory.


4 Answers 4


I) OP wrote(v5):

Does uncertainty imply non-commutativity?

In practice, No, not necessarily. In many realistic situations$^1$, it will be impossible get the variance

$${\rm Var}(\hat x) ~:=~\langle (\hat x-\langle \hat x\rangle)^2\rangle $$

of a Hermitian observable, say $\hat x$, below some positive bound $\epsilon>0$ because of various experimental (rather than fundamental) constraints. But we would not conclude that the operator $\hat x$ does not commute$^2$ with itself!

II) One may wonder how Heisenberg deduced non-commutativity between $\hat q$ and $\hat p$ in his matrix mechanics version of quantum mechanics? He did in fact not deduced this from direct experimental measurements of variance of $\hat q$ and $\hat p$. Rather he used a combination of (i) known observed behaviors of quantum systems and (ii) theoretical arguments based on Fourier representations, see e.g. Wikipedia and this Phys.SE post for further details.

III) To counter any misunderstandings, perhaps we should here emphasize the fact that the uncertainty in Heisenberg's uncertainty relations is a fundamental property of quantum systems, and is not a statement about the observational success of current technology.


$^1$ Especially if the observable $\hat x$ does not commute with the Hamiltonian $\hat H$.

$^2$Any (Grassmann-even) operator commutes with itself.

  • $\begingroup$ This is what I expected to hear...it makes me wonder about why canonical commutation relations are canonical. $\endgroup$
    – Jon Bannon
    Nov 9, 2012 at 20:51
  • $\begingroup$ The uncertainty principle can be generalized to any two Hermitian observables $\hat A$ and $\hat B$, whether or not they are canonical variables. $\endgroup$
    – Qmechanic
    Nov 9, 2012 at 21:06
  • $\begingroup$ My prior comment is not what I wanted to say, precisely. If we don't accept wave mechanics etc. as a starting point for QM, an experimentally observed uncertainty relation is no evidence that we should need a noncommutative theory, right? $\endgroup$
    – Jon Bannon
    Nov 9, 2012 at 22:13
  • $\begingroup$ @JonBannon: but noncommunativity is also checkable--project onto $\hat A$, and then project onto $\hat B$, and then do it in the other order. If the end state is not the same, then the theory is not communative. $\endgroup$ Nov 9, 2012 at 23:27
  • $\begingroup$ @JerrySchirmer: I guess what surprises me (and shouldn't) is that the uncertainty principle doesn't necessarily imply noncommutativity...it could be that some fundamental lower bound for commuting observables is still possible. $\endgroup$
    – Jon Bannon
    Nov 10, 2012 at 0:09

No. Non-commutativity implies uncertainty but the converse is not true. E.g. in classical theory we find statistical uncertainties such as $\Delta x \Delta k \gt 0$ on positions and wave-vectors but the underlying algebra is commutative. Those uncertainties reflect ignorance, (e.g. a small uncertainty in the momentum of a particle generates at latter times a uncertainty in the position of the particle). In quantum mechanics uncertainty is ontological not epistemic. Quantum uncertainty reflects the lack of common eigenstate for non-commuting observables.


Well. Let us formulate the uncertainty principle: If A and B are two observables such that [A,B]=C, then we have that $$ \Delta A\Delta B\geq \frac{1}{2}|\langle[A,B]\rangle|\equiv \frac{1}{2} |\langle C\rangle|. $$ This means that, if C differs from 0, if we find a state such that $\langle C\rangle=0$, then we have that $\Delta A\Delta B\geq 0$. This is, for instance, the case of the spin (Pauli matrices): suppose $A\equiv\sigma_x$ and $B\equiv\sigma_y$, then $C=i\sigma_z$, and if we consider one eigenstate of $\sigma_x$, we have that $\langle C\rangle=i\langle\sigma_z\rangle=0$.

@Qmechanic IMPORTANT! The uncertainty principle is only in part related to the variance or the error that we can get in an experiment! This is actually what Heisenberg thought when he formulated the principle, but he was wrong. The uncertainty principle is, instead, a fundamental property of quantum mechanics. An experimental proof of this can be found in http://arxiv.org/pdf/1201.1833.pdf.


If two compact observables commute they have a joint normalized eigenvector $\psi$, in which both observables have zero variance.

In case of noncompact observables we only have unnormalizable eigenvectors, but these can be arbitrarily well approximated by normalizable states, in which both observables have arbitrarily small variance.

Thus a lower bound on the product of the variaces implies commutativity.

  • $\begingroup$ If I can verify these statements, this is exactly what I hoped would be the case. $\endgroup$
    – Jon Bannon
    Nov 12, 2012 at 18:19

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