Should relativity be factored into the uncertainty principle. I cant move faster than c so should that affect the measurement process? Should time dilation and length contraction be considered or maybe some of these factors cancel when 2 effects are considered.
How, exactly?Relativistic quantum mechanics and quantum field theory take care about that.
As opposed to x being...? Please clarify.Although this may be related to the issue of whether x is really the position operator.
Which paradox do you mean?One question .. in the twi split experiment ... do scientist explain the paradox with heisenbergs uncertainity principle .. ?
Hmmm...this alternative position operator X is then a differential operator by including p in its definition, isn't it?Relativistic quantum mechanics is a multiparticle theory. To talk about single-particle quantities you need to restrict it. In particular to exclude the antiparticles you have to use just the positive energy modes.
But it turns out to be impossible, using only positive energy modes, to construct a state which is entirely localized at a point (δ-function). Also, the operator in momentum space that one would normally call the position, i∇p, is not even Hermitian. So for both these reasons, to answer the pertinent question, "what is the probability of finding the particle at a particular point", you can't use x to specify the point.
Newton and Wigner defined a different position operator X to be the Hermitian part of i∇p, namely X = i∇p -ip/(p2 + (mc)2). X obeys the right commutation relations, [X, p] = iħ. And its eigenstates are little Hankel-function-shaped things about a Compton wavelength across.
Sure it is, so is x = iħ∇p!
I know it's backwards from the usual situation where p is represented by a differential operator and x is not. But it works either way, and in quantum field theory the momentum representation is used all the time and for many purposes is a lot simpler.
For the standard x, the eigenvalue equation is iħ∇pφ = x0φ, with solution φ(p) = e-(i/ħ)p·x0. To get the corresponding eigenfunction ψ(x) in coordinate space, Fourier transform it. The Fourier transform of a plane wave of course is δ(x - x0), a wavefunction completely localized at x0.
For the Newton-Wigner X, the same procedure leads to a φ(p) which is just slightly different: φ(p)= √p0 e-(i/ħ)p·x0. But the Fourier transform of this back to x space is no longer a delta function, it's slightly spread out.
I mean if something was moving fast I could just put myself in that inertial frame and use the HUP. But if Im moving do I need to put in a Lorentz factor to make it work.
in that frame it couldn't radiate it doesn't have any energy to lose at least classically. If In another frame it does radiate, this radiation would affect its momentum. how would HUP take care of that.
i mean that in the two slit experiment .. the electrons first behave like a wave .. but when scientist put an detector to see form which slit each electrons go (or both) the electron behaved like a particle ....Which paradox do you mean?
With wave functions of the particle, and one of the many interpretations of quantum physics.i mean that in the two slit experiment .. the electrons first behave like a wave .. but when scientist put an detector to see form which slit each electrons go (or both) the electron behaved like a particle ....
how did they explain this... ?
While mass influences the wavelength<->energy relation, electrons can behave like waves with mass, too. You can perform double-slit experiments with massive electrons (in fact, all electrons are massive).i thought that electrons behave like waves when they dont have any mass..
