Infinite or undefined standard deviation in HUP

[nomadreid]

If we measure one conjugate variable in an uncertainty relation precisely , i.e., so its standard deviation is zero, then by the HUP the sd of the other one is either infinite or undefined. But what about the cases when the other conjugate variable has limits: e.g., there cannot be an infinite spread for momentum or energy? Is it then better to just say that the spread is undefined in this case?

 

[A. Neumaier]

Canonically conjugate observables always have all reals in the spectrum, hence are unbounded and have no limits.

 

[nomadreid]

Canonically conjugate observables always have all reals in the spectrum, hence are unbounded and have no limits.

OK. This raises the (certainly naïve) question: suppose we refer to the momentum of particle at a determined point. Since the point is precise, you have the infinite spectrum of momentum. However, beyond a certain momentum the particle has enough mass-energy at that point you end up with a black hole, making the point lose its exact value, which would mean that the spectrum of the momentum would not be able to go beyond that limit. Hence not infinite. There is obviously a basic flaw in my reasoning here, but what?

 

[A. Neumaier]

OK. This raises the (certainly naïve) question: suppose we refer to the momentum of particle at a determined point. Since the point is precise, you have the infinite spectrum of momentum. However, beyond a certain momentum the particle has enough mass-energy at that point you end up with a black hole, making the point lose its exact value, which would mean that the spectrum of the momentum would not be able to go beyond that limit. Hence not infinite. There is obviously a basic flaw in my reasoning here, but what?

Canonical commutation rules for interacting particles are valid in flat space-time only. This excludes black holes.

 

[nomadreid]

Ah, I overlooked that. Excellent point, thank you, A.Neumaier. That probably takes care of my other counterexamples. No superluminary virtual particles, then?

 

[A. Neumaier]

Virtual particles don't have a dynamics, hence the velocities arbitrarily ascribed to them mean nothing.

 

[nomadreid]

Thank you, A.Neumaier, for that answer and the excellent link; my delay in answering was due to my reading through it. (I had a feeling I shouldn't have used the concept "virtual particles"...) Very interesting and well written, and there are a few things there that I intend to try to understand more thoroughly in the course of time.
(Wikipedia is a suitable target -- even in secondary school assignments, Wikipedia is not accepted as a reliable source due to the fact that the articles are not signed.) I was also trying to find out (online) what corresponds to the Uncertainty Relations (i.e., the canonical commutation rules) in curved space-time, but the discussions I found exceed my present level in the subject.