How does Uncertainty Principle work?

In summary: I'm sorry, I was wrong. I was thinking of x and px as both being uncertain (my mistake). But the uncertainty in x can be zero only if the uncertainty in px is infinite. Is that correct?PeteYes, that is correct. The HUP states that the product of the uncertainties in position and momentum must be greater than or equal to h/2pi. So if one of the uncertainties is zero, the other must be infinite. This is why 0 energy is not forbidden by the HUP, as the uncertainty in energy can be zero if the uncertainty in time is infinite.In summary, the Heisenberg Uncertainty Principle (HUP) states that the product of the uncertainties in position and momentum
  • #1
PrudensOptimus
641
0
WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi


But... mathematically, what does that mean?
 
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  • #2
What's dS? That to me is change in entropy. Also the uncertainty is gretae than or equal to h/4pi

There are several sets of complementary pairs, for example:

ΔxΔpx ≥ ħ/2

ΔEΔt ≥ ħ/2

ΔωxΔωz ≥ ħ/2

etc.

What this does is place a definte limit on how accurately the results of an experiment can be predicted, note: this is not due experimental error, the uncertainity is intrinsic.
 
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  • #3
so this means ∫x dp >= h/4pi??
 
  • #4
No, that's why using d instead of Δ may lead to confusion.

ΔxΔpx ≥ ħ/2

means that, for an elecrton (for example), uncertainty in postion along the x axis(Δx) mutplied by uncertainty in momentum along the x-axis (Δpx) is always greater than or equal to the rationalised Planck's constant (syn. with Dirac's constant and h/2pi) over 2
 
  • #5
Originally posted by PrudensOptimus
so this means ∫x dp >= h/4pi??

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.

Edit, the more I think about this the less true is seems. Where is you differential on the RHS? Your relationship simply does not work.
 
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  • #6
Infact is worth noting that when a 'd' appears infront of a quantity in physics it does not mean you should neccesarily start thniking about intergration. For example in the formula for a change in entropy dS = dq/T, the dS and dq are actually non-integral and to avoid such confusion are sometimes annoted with a 'd-bar' instead of a 'd'.
 
  • #7
The Δ in the HUP actually refers to a standard deviation, i.e. if A is an observable, then

&Delta;A = &radic;(<A2>-<A>2)
 
  • #8
It's not the HUP dx*dp>=~(h/(2*pi))? At least is how it appears in my book "Quantum theory" of David Bohm
 
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  • #9
Originally posted by PrudensOptimus
WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi


But... mathematically, what does that mean?

To understand the uncertainty relation one has to first understand what an uncertainty is. For that see

http://www.geocities.com/physics_world/qm/probability.htm

But note that dt above is not an uncertainty, in fact dE dt >= h/2pi is not a true uncertainty relation.

Pete
 
  • #10
The point is that neither of those quanities can be zero.
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
 
  • #11
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


Most of the time it can be assumed that:

&Delta;x&Delta;px &asymp; h/2&pi;
 
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  • #12
Originally posted by jcsd
Most of the time it can be assumed that:

&Delta;x&Delta;px &asymp; h/2&pi;

h or h bar??
 
  • #14
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

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?
 
  • #15
Originally posted by Integral
Setting either of those quanities to zero violates the HUP.

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.
 
  • #16
Prudens Optimus: In that equation h, but remember h-bar = h/2pi, so I could of simply said "h-bar".
 
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  • #17
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?
Setting either does not violate HUP. E.g. if the uncertainty in x is zero and the uncertainty in px is infinite then the equality is satisfied. Why do you think 0 energy is forbidden?

Pete
 

1. What is the Uncertainty Principle?

The Uncertainty Principle, also known as Heisenberg's Uncertainty Principle, is a fundamental concept in quantum mechanics that states it is impossible to precisely measure certain pairs of physical quantities, such as position and momentum, at the same time.

2. How does the Uncertainty Principle work?

The Uncertainty Principle works by establishing a limit to the precision with which certain physical quantities can be measured simultaneously. This is due to the wave-like nature of particles at the quantum level, where the act of measuring one quantity will inevitably disturb the other quantity being measured.

3. Is the Uncertainty Principle a universal law?

Yes, the Uncertainty Principle is considered a universal law in quantum mechanics. It applies to all particles, regardless of their size or energy level, and has been proven to hold true in numerous experiments.

4. What are the implications of the Uncertainty Principle?

The Uncertainty Principle has several implications, including the inability to precisely predict the behavior of particles at the quantum level and the limitations it places on our ability to measure and understand the fundamental nature of the universe.

5. Can the Uncertainty Principle be overcome?

No, the Uncertainty Principle cannot be overcome. It is a fundamental principle in quantum mechanics that is deeply rooted in the nature of our universe. However, scientists have developed techniques, such as quantum entanglement, to mitigate the effects of the Uncertainty Principle and improve the precision of measurements.

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