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Application of Heisenberg Uncertainty Principle

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A parallel beam of electrons traveling in x- direction falls on a slit of width d.
    If after passing the slit, an electron acquires momentum py in the y direction then for a majority of electrons passing through the slit
    A. |py|d ≈ h
    B. A. |py|d > h
    C. |py|d < h
    D. |py|d >> h

    2. Relevant equations
    Δx Δp = h/4π

    3. The attempt at a solution
    I only know Hiesenberg uncertainty equation
    How to deal here with inequalities?
     
  2. jcsd
  3. Apr 20, 2015 #2
    First the Heisenberg uncertainty relation should be written ΔxΔp ≥ h/4π . What does this expression imply?
     
  4. Apr 20, 2015 #3
    This implies that we have uncertainty in position and momentum calculation of an electron.
    How to apply that in this question ?
     
  5. Apr 20, 2015 #4
    When the electron passes through the slit of width d what are you doing? Using the Heisenberg Uncertainly relation what statement can you make about the lateral momentum
     
  6. Apr 20, 2015 #5
    Δp >= h/4πd
    Is that correct?
     
  7. Apr 20, 2015 #6
    Yes, and what does this suggest?
     
  8. Apr 20, 2015 #7
    Sorry, I think it should be Δp>= h/4πΔd ?
    There should be uncertainty involved.
     
  9. Apr 20, 2015 #8
    No. you are localizing the electron in a space of width d which is the significance of Δx.
     
  10. Apr 20, 2015 #9
    That uncertainty in momentum is greater?
     
  11. Apr 20, 2015 #10
    Yes, or the uncertainty is not less than h/4πd.

    I have to do something so I will be away for a few hours sorry.
     
  12. Apr 20, 2015 #11
    How does that relate to options?

    Think you will reply after some hours.
     
  13. Apr 20, 2015 #12
    You should be able to exclude two immediately>
     
  14. Apr 20, 2015 #13
    Do that mean certainty is less then h/4πd ?
    Which is py<h/4πd ?
    Why absolute value is being considered of momentum in options and why 4π term is neglected?
    This is not true , if pyd< h/4π
    Then |py|d< h ?
     
    Last edited: Apr 21, 2015
  15. Apr 21, 2015 #14
    The expression of the uncertainty principle was original suggested by Heisenberg to be ΔpΔx ≥ h. It was later formally derived and found to be ΔpΔx ≥ h/4π.

    It seems to me that the two forms are mixed here. First even is they are mixed A and C are not consistent with either one. But if Δp ≥ h/4πd you cannot say Δp > h/d let alone Δp >> h/d. So to make any sense out of the choices you have to use Heisenberg's original formulation ΔpΔx ≥ h.

    The absolute value is used because momentum and position are vector quantities and only the magnitudes of these quantities are of interest.
     
  16. Apr 21, 2015 #15
    So option C is correct, by original formulation also and also by modern formulation?
    If pyd < h/4π
    Then 4πpyd < h
    Which definitely implies pyd <h
    So no matter what formulation we take ,we only arrive at this answer ?
     
  17. Apr 21, 2015 #16
    If Δpd ≥ h/4π then 4π Δpd ≥ h. But this does not imply universally that Δpd < h.
     
  18. Apr 21, 2015 #17
    Is this correct?
     
  19. Apr 21, 2015 #18
    No. you reversed the inequality sign.
     
  20. Apr 21, 2015 #19
    But I am taking certainty.
    When we remove delta sign , inequality changes?
     
  21. Apr 21, 2015 #20
    It is my understanding that the uncertainty relationship makes no statement about the actual momentum of a particle.
     
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