This post follows this answer of Ben Crowell.

The problem with the time-energy uncertainty relation $$[H,T]=i\hbar$$ is that it implies that the spectrum of the energy operator $H$ cannot be bounded below, which generates grave problem in quantum mechanics.

But this implication occurs on $L^2(\mathbb{R})$ where the spectrum of the position operator (analogous of above energy operator, dixit Stone-von Neumann uniqueness theorem) is $\mathbb{R}$. Now, on $L^2(\mathbb{S}^1)$, the position operator is just a bilateral shift operator on $\ell^2$ and its spectrum is just $\mathbb{S}^1$, so bounded.

Question: Does the time-energy uncertainty relation work well on $\mathbb{S}^1$ or are there other obstructions? Is there a reference for this case?

Remark: If I am not mistaken, for any compact space $X$, the spectrum of the position operator on $L^2(X)$ should also be bounded. Then should we conclude that the time-energy uncertainty relation implies that the universe is compact (for example, $SU(2)$)?

  • $\begingroup$ You have skipped or critically eluded the heart of your question: how do you define B of the mainstream Mandelshtam-Tamm definition in this compact space, precisely and consistently? $\endgroup$ Dec 13, 2019 at 12:54
  • $\begingroup$ @CosmasZachos: Let us consider $X=\mathbb{S}^1$ first. Is there a problem in this case? $\endgroup$ Dec 13, 2019 at 15:19
  • $\begingroup$ @CosmasZachos: question is: Does the time-energy uncertainty relation work well on $\mathbb{S}^1$ or are there other obstructions? Is there a reference for this case? $\endgroup$ Dec 13, 2019 at 15:31
  • $\begingroup$ @CosmasZachos: Sorry, I am not physicist (but mathematician). For me the time-energy uncertainty relation is just about finding a self-adjoint time operator $T$ such that $[H,T]=i\hbar$. For $\mathbb{R}$ such a relation leads to grave problem, I just want to know whether everything is fine for $\mathbb{S}^1$. $\endgroup$ Dec 13, 2019 at 15:44
  • $\begingroup$ I see.. you are seeking a meaningful operator B normalized by the norm of its time derivative... $\endgroup$ Dec 13, 2019 at 16:13

2 Answers 2


The relation $[A,B]=i\hbar$ for self-adjoint operators $A,B$ with a common dense domain implies (in the physically relevant case where the operators generate a Heisenberg group) by the Stone-von Neumann theorem that both operators $A$ and $B$ have the whole real line as spectrum. Thus such a relation is impossible with $A$ or $B$ being the translation generator $H$ on the circle, which has a discrete spectrum only.

A fairly complete treatment of the problems associated with a time operator is given in

  • P. Busch, The time-energy uncertainty relation, pp. 73-105 in: Time in quantum mechanics (J. Muga et al., eds.), 2nd ed., Lecture Notes in Physics, Vol. 734, Springer, Berlin 2008. quant-ph/0105049.
  • $\begingroup$ The Stone–von Neumann theorem is formulated in terms of the Weyl relations, and the Weyl relations are not strictly equivalent to the canonical commutation relation, counterexamples exist satisfying the canonical commutation relations but not the Weyl relations, on the circle (counterexample) $\endgroup$ Jan 5, 2020 at 17:25
  • $\begingroup$ @SebastienPalcoux: This counterexample is not defined on a circle since $A$ is ill-defined at $\theta=0=2\pi$. $\endgroup$ Jan 6, 2020 at 13:43
  • $\begingroup$ Moreover, for an adequate interpretation in terms of the physics of time and energy, one would need the Weyl relations (i.e., the Heisenberg group) and not only the commutation relations (the Heisenberg Lie algebra). $\endgroup$ Jan 6, 2020 at 14:23

This is an interesting, fundamental question! You most propably have read this compact writing from Baez. Which take's the step closer to more generel case; $p$ and $q$;

The reason is this: by the Stone-von Neumann uniqueness theorem, any pair of operators satisfying the canonical commutation relations [H,T] = i ℏ can only be a slightly disguised version of the familiar operators p and q. These operators p and q are unbounded below - i.e., their spectra extend all the way down to negative infinity. But a physically realistic Hamiltonian must be bounded below!

This physical reality actually should just be accepted; The realistic physics is bounded from below. The linked wiki-page related to the Stone-von Neumann phrases this;
the left-hand side is zero, the right-hand side is non-zero. Further analysis2 shows that, in fact, any two self-adjoint operators satisfying the above commutation relation cannot be both bounded. For notational convenience, the nonvanishing square root of ℏ may be absorbed into the normalization of $p$ and $x$, so that, effectively, it is replaced by 1.

It should also not be forgotten that this $px=1$ is describing a vector which has a wave character, and these $p$ and $x$ are thus Conjugate variables, also pairs of variables mathematically defined in such a way that they become Fourier transform duals.

So the $p$ is describing a momentum of a wave, while $x$ is describing the position of that same wave. Now, measuring a position of a wave is impossible, if the exact timing of the cycle is not defined. This is the "uncertainty principle". For impulse, the whole cycle must be measured, which makes impossible to measure some timing of that cycle, leaving position "undefined" within this cycle.

Further, it can be noted that though $xp \not= px$ is also purely a character of matrix calculations of these both vectors, and thus basically $xp=-px$ where this minus simply means an opposite direction. This further concludes that the Canonical commutation relation $xp-px=i \hbar=1$ must mean that a one full cycle is described from one single point heading on opposite directions.

So this all just concludes that this Realistic physic is (must be) bounded below. I have approached this idea with this question. And I agree with the original Question, that this provides insights to Quantum Gravity, though they might be quite different than expected today, as it turns out that the very first idea of Gravity might have been correct.

So, to conclude your Question; I consider the Uncertainty principle not only works well when bounded, it also defines that these bounds must exist. I don't have link for proper references, but here is more of my own original ideas about this.

  • $\begingroup$ Downvoted because I don't think it addresses the question "Does the time-energy uncertainty relation work well on $\mathbb S^1$ or are there other obstructions? Is there a reference for this case?" at all. $\endgroup$
    – Gonenc
    Jan 5, 2020 at 0:34
  • $\begingroup$ @Gonenc But $\mathbb{S}^1$ does describe bounded set ? en.wikipedia.org/wiki/Bounded_set $\endgroup$
    – Jokela
    Jan 5, 2020 at 1:49

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