Hi
I think common quantum measurements are based on electromagnetic interaction and each measurement has at least one EM interaction. Is it true?
If we can perform a quantum measurement based on weak interaction, does the range of uncertainty principle change? in other word, can right hand side of delta_x*delta_p>h/2 be reduced?
In a sense every particle decay event is a measurement. All the nuclei in a block of C-14 have some decay flux but at any given time we have some probability of observing a decay even at some location. This is mediated by the weak interaction but, of course, the follow up processes for us to see it are electromagnetic. I think this is what you were saying. If you have only a weak process then I think you would not have enough of an effect to driven the selection to a localized site.
The decoherence advocates would probably give some long speculative bound on the time for a measurement to occur. I'm not in their camp.
Principle of uncertainty can not be violated. but yes, we can squeeze uncertainty in one of the quadrature beyond Standard quantum limit (SQL) in non commuting observable. The Squeezing in one quadrature or observable will be compensated by enhanced fluctuation in other quadrature or observable.
But for measurement, we cant go beyond Compton wavelength. The position measurement of particle Beyond Compton wavelength will increase its kinetic energy equals to rest mass energy and pair production will be created (Rev. Mod. Phys. 52, 341 – Published 1 April 1980, page no 354 section E).
Please check these references...
Rev. Mod. Phys. 52, 341 – Published 1 April 1980
Quantum Non-demolition Measurements
DOI:10.1126/science.209.4456.547
Hi,
You are right in noting that QM was based on em phenomena and that its extension to other situations is not trivial. But QM extends well beyond any specific interaction. It is considered now as a general theory and is fully applicable too weak interactions (and is consistent with week interaction measurements),
Actually Quantum Field Theory (QFT) provides today an exceptionally successful of all physics - but Gravity.
Well, you can find pictures of the sun, observed at night via neutrinos. If one could detect neutrinos more efficiently, they would be fantastic new channel for observing the universe.
Dear Care,
this picture is astonshing. I already knew it. Neutrinos penetrate the earth and also light years thick lead without interaction. In this case they interact with a small sensor or photographic film. Furthermore how they are focused?
Best regards
Hans
Research Theory and reality on the experiment of Reines/Cowan 1956
The Planck constant is just a unit of measure of the action. One can choose a system of units in which it is equal to 1. It's the coefficient in front of the Planck constant which may be questioned.
Some authors take it as 1, others as 1/2, but this is not important.
The uncertainty relation has nothing to do with the type of field involved, but with the quantum nature of object (or, in non-rigorous terms, wave-like nature). It comes from the signal theory or from the wave-mechanics. Using Fourier transforms it is proved that the narrower is a signal in time, the wider is its frequency range, and the relation is Δω Δt ≥ 1. Multiply this on both sides by ħ and you get the uncertainty relation for energy and time. Alternatively, replace ω = 2π\nu = 2πV/λ, where λ is the de Broglie wavelength and V the (group) velocity of movement of the object, and you'll get
2π Δk Δ(Vt) ≥ 1
where k is the wave-number. Multiply on both sides by ħ and you get
hΔk Δx ≥ ħ.
Recall that hk = p, and you get
Δp Δx ≥ ħ.
As I said, the uncertainty relations come from the wave-like nature of the quantum objects.
The uncertainty relations have nothing to do with the nature of the fields, but with the wave-like nature of the quantum objects. In the wave mechanics it is proved, using Fourier transforms, that the shorter is a wave, the wider is the range of frequences in its spectrum. The relation is
(1) 2πΔ\nu Δt ≥ 1.
Multiply this by ħ and you'll obtain the energy-time uncertainty relation.
Now, let your wave move with the group-velocity V, and use the de Broglie wavelength λ = V/ \nu, or k = \nu /V, where k is the wave-number. Then do a simple manipulation
(2) 2πΔk Δ(Vt) ≥ 1.
Multiply now by ħv on both sides and recall that 2πħk = p, the linear momentum. You get the uncertainty relation for linear momentum and position.
So, again, it's the wave-nature which generates the uncertainty relations, not the type of field.
Squeezed quantum states in lasers have been claimed to change the right hand side of your equation, making it considerably smaller. Weak interactions tend to be statistical and spontaneous, not providing a method to squeezed quantum states at realistic energies.
Indeed the em interaction is usually shadowing the weak one.. However, the direct detection of neutrinos (e.g. by Davis, by super Kamiyokande etc) as well as the observation of Z-> nu nu^bar (at LEP using ISR radiation that gives rise to single photon events) are examples of weak interaction measurements.
There is no relation between this and the uncertainty principle that does not assume any interaction.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Quantum Physics