- 637
- 0
WE all know that:
dE dt >= h/2pi
or
dS dt >= h/2pi
But... mathematically, what does that mean?
dE dt >= h/2pi
or
dS dt >= h/2pi
But... mathematically, what does that mean?
While this may be true, it is not the point of the HUP, in fact to use the Δ of the HUP to a differential is going against the entire meaning the HUP. The point is that neither of those quanities can be zero. To form the integral you must be have ture differentials, this means that they can be arbitrally small.Originally posted by PrudensOptimus
so this means ∫x dp >= h/4pi??
To understand the uncertainty relation one has to first understand what an uncertainty is. For that seeOriginally posted by PrudensOptimus
WE all know that:
dE dt >= h/2pi
or
dS dt >= h/2pi
But... mathematically, what does that mean?
That is incorrect. There is no reason that an uncertainty can't be zero. If dx is zero that means that there is zero knowledge of p.The point is that neither of those quanities can be zero.
Originally posted by meteor
It's not the HUP dx*dp>=~(h/(2*pi))? At least is how it appears in my book "Quantum theory" of David Bohm
h or h bar??Originally posted by jcsd
Most of the time it can be assumed that:
ΔxΔp_{x} ≈ h/2π
Setting either of those quanities to zero violates the HUP. Isn't this a good protion of the argument for zero point energy? The fact that 0 energy is forbidden by the HUP?Originally posted by pmb
That is incorrect. There is no reason that an uncertainty can't be zero. If dx is zero that means that there is zero knowledge of p.
Of course dx is not an infinitesimal x - its an uncertainty in x. Big difference.
Pete
Well ... if we have zero uncertainty in position, then the wavefunction isn't really a function, it's a Dirac delta distribution. The Fourier transform in momentum space is a constant function of infinite uncertainty. So the left-hand side of the uncertainty relation is an ill-defined zero times infinity. But you can define it using limits: take the wavefunction in position space to be an increasingly sharply peaked Gaussian, and thus in momentum space to be an increasingly broad Gaussian. In the limit as the position Gaussian becomes infinitely narrow (a Dirac delta), we can say that the limit of the product of the uncertainties is well-defined, finite, and does not violate the HUP.Originally posted by Integral
Setting either of those quanities to zero violates the HUP.
Setting either does not violate HUP. E.g. if the uncertainty in x is zero and the uncertainty in p_{x} is infinite then the equality is satisfied. Why do you think 0 energy is forbidden?Originally posted by Integral
Setting either of those quanities to zero violates the HUP. Isn't this a good protion of the argument for zero point energy? The fact that 0 energy is forbidden by the HUP?