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How does Uncertainty Principle work?

  1. Nov 8, 2003 #1
    WE all know that:

    dE dt >= h/2pi

    or

    dS dt >= h/2pi


    But... mathematically, what does that mean?
     
  2. jcsd
  3. Nov 8, 2003 #2

    jcsd

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    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.
     
    Last edited: Nov 8, 2003
  4. Nov 8, 2003 #3
    so this means ∫x dp >= h/4pi??
     
  5. Nov 8, 2003 #4

    jcsd

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    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
     
  6. Nov 8, 2003 #5

    Integral

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    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.
     
    Last edited: Nov 8, 2003
  7. Nov 8, 2003 #6

    jcsd

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    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' instaed of a 'd'.
     
  8. Nov 8, 2003 #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)
     
  9. Nov 8, 2003 #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
     
    Last edited: Nov 8, 2003
  10. Nov 8, 2003 #9

    pmb

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    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
     
  11. Nov 8, 2003 #10

    pmb

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    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
     
  12. Nov 8, 2003 #11

    jcsd

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    Most of the time it can be assumed that:

    &Delta;x&Delta;px &asymp; h/2&pi;
     
    Last edited: Nov 8, 2003
  13. Nov 8, 2003 #12
    h or h bar??
     
  14. Nov 8, 2003 #13
  15. Nov 8, 2003 #14

    Integral

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    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?
     
  16. Nov 8, 2003 #15
    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.
     
  17. Nov 8, 2003 #16

    jcsd

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    Prudens Optimus: In that equation h, but remember h-bar = h/2pi, so I could of simply said "h-bar".
     
    Last edited: Nov 8, 2003
  18. Nov 8, 2003 #17

    pmb

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    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
     
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