Heisenberg famously introduced and illustrated the uncertainty principle with a hypothetical gamma-ray microscope experiment. The diffraction limit played a central role in this derivation. Meanwhile the diffraction limit has been shown to be no fundamental limitation at all and "super-resolution" is achieved by a number of novel imaging techniques. Does this imply trouble for the uncertainty principle?
What you mean by super-resolution?
Heisenberg's rationale for his microscope, was criticised as wrong - see criticism in
https://en.wikipedia.org/wiki/Heisenberg%27s_microscope#Problems_with_the_argument
On the other hand, the uncertainty principle is a firm law of the nature, not typical only to quantum mechanics but also to wave-machanics. For instance we encounter this principle when we learn of representation and transformation of signals from the time domain to frequency domain and vice-versa i.e. when we learn "Fourier transforms", (but we don't call it by the name "uncertainty principle"). The longer is a signal as time duration, the sharper is it frequency band, and vice-versa. People working in telecommunications have to do a lot with this principle.
As to your question, nothing affects the uncertainty principle.
Hi
Uncertainty principle include all measurements of micro-particles.
Hope you will understand the this,
B.Rath
Uncertainty principle is a mathematical fact realized already in classical physics for Fourier transform.
A derivation of the uncertainty relation based on the commutation relation of position and momentum can be found in
https://de.wikipedia.org/wiki/Heisenbergsche_Unschärferelation
Dear Walter,
I am glad that you mentioned the commutation relations possibility to prove the uncertainty relations.
With this occasion I want to say that I saw in some place the wrong opinion that the uncertainty principle is correct only if proved in base of the commutation relations. This is not true. The time-energy uncertainty derives from wave-mechanics, and is widely known in signal processing. As I said, the shorter is the time duration of a signal (e.g. of an electric pulse of current, or of voltage), the broader is its frequency spectrum. The Continuous to Continuous Fourier Transform shows that immediately.
About position and linear momentum, we also don't need a proof based on commutation relations,one can prove the uncertainty absolutely analogously with the time-energy proof.
Dear Sofia,
You are absolutely right. Instead of using the commutation relations, you may, alternatively, use a 'representation' of the commutation relations by a 'wave function' to derive uncertainty relations. Specific to quantum mechanics, however, is the occurrence of the Planck constant. You will not find this constant in the context of signal processing.
Dear All,
many thanks for your remarks. Indeed, I do not expect any serious trouble for the uncertainty principle and know of the derivations which are based on more solid ground than Heisenbergs Gedankenexperiment involving the gamma-ray microscope. With respect to the commutator issue: Uffink and Hilgevoord have done marvelous research on this. To view the Robertson relation as the embodidment of the uncertainty relation is critisised by them very conclusivly.
However, I wonder if something interesting could be learned here nevertheless... and if only about the shortcomings of Heisenbergs early reasoning. What provokes this idea further: Recently it came to a reappraisal of measurments-disturbance relation by Busch, Lahti, Werner et al. (e.g. Phys. Rev. Lett. 111, 160405) They view their work as the formal justification for Heisenbergs gamma ray microscope experiment...
Dear Christian,
What you say makes sense to ME. But with some refinements. (Leaving aside that Heisenberg's microscope which is not a rigorous example) you can improve, at least theoretically, the resolution of one of the non-compatible observables until its uncertainty becomes zero. But you pay in the resolution of the other observable. This one increases. Thus, if you are able to measure one of the observables with very high precision, the measurement of the other observable will tell you almost nothing.
Well, I am sure that you know these things.
Dear Walter,
You pose interesting problems.
I'll try to explain what happens in the wave-mechanics, i.e. why we don't have to do there with Planck's constant.
In the QM we have the uncertainty relation
(1) ΔE Δt ≥ ħ.
But the energy of a quantum system is defined as
(2) E = ħω = h\nu.
where ω = 2π\nu. Substituting (2) in (1) and dividing on both sides by ħ, one gets the wave-mechanics relation
(3) Δω Δt ≥E ≥ 1.
The same with the uncertainty relation between position and linear momentum. In QM we have
(4) Δp Δx ≥ ħ,
together with the relation
(5) p = ħҟ,
where ҟ = 2π/λ and λ is the wavelength. Introducing in (5) in (4) and dividing on both sides by ħ, we get the wave-mechanics relation
(6) Δҟ Δx ≥ 1.
The appearence of Planck constant in uncertainty relation is due to normalisation and of course reflects a new representation of p and x. In Fourier analysis the Planck constant is replaced with unity since one has k and x (than than p= hbar*k and x)
My dear Christian,
Thank you for the article, it is interesting indeed. Let me say what it says: that we can do quantum mechanics without using Planck's constant, by working with frequency instead of energy, with the phase of the wave-function without writing it as S/ħ, and other similar replacements.
It is indeed as I showed in my previous comment.
But, there is a problem: Planck's constant has a deeper meaning, as stressed by Feynman when explaining the evolution of a quantum system vis-a-vis a classical system which evolves along a well-defined trajectory. Planck's constant establishes the border between classical and quantum systems, in the sense that for classical systems ħ can be approximated to zero. Here is a citation from "Quantum Mechanics and Path Integral" by Feynman & Hibbs:
"The classical approximation, however, corresponds to the case that the dimensions, masses, times, etc., are so large that S is enormous in relation to ħ (= 1.05×10–27erg∙sec). Then the phase of the contribution S/ħ is some very, very large angle. . . . . . Therefore, no path really needs to be considered if the neighboring path has a different action; for the paths in the neighborhood cancel out the contribution. But for the special path x̄(t), for which S is an extremum, a small change in the path produces, in the first order at least, no change in S. All the contributions from the paths in this region are nearly in phase, . . . , and do not cancel out. . . . . But at an atomic level, S may be comparable to ħ, and then all the trajectories must be added . . . in detail."
Best regards, and thanks for posing the problem,
Sofia
here's a simple example
to what accuracy can one track the centre of mass of a particle?
the point spread function of a microscope can be related to the uncertainty principle which means that nay particle no matter how small will appear to a a finite size.
however, if the image is of high quality and the image itself contains many many photons then the centre of the particle can be estimated with a arbitrarily high precision depending on the single to noise in the image. ie the accuracy is much better than a simple interpretation of the uncertainty principle might suggest.
key here is that the image contains many many photons. in essence one is making an estimate of the mean based on many measurements of the same distribution, one might imaging that the precision of the estimate of the mean can be reduced below the point spread function of the microscope in proportion to the square root of the photon number.
the uncertainty principle escapes :)
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