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Rigid Box Particle and Heisenberg Principle

  1. Jul 16, 2013 #1
    1. The problem statement, all variables and given/known data
    For the ground state of a particle in a rigid box, we
    have seen that the momentum has a definite magni-
    tude (h/2π)k but is equally likely to be in either direction.
    This means that the uncertainty in p is Δp≈ (h/2π)k. The
    uncertainty in position is Δx≈ a/2. Verify that these
    uncertainties are consistent with the Heisenberg
    uncertainty principle.

    2. Relevant equations
    Heisenberg Relations
    Δp= h/2pi Δk ...(1)
    ΔxΔk ≥ 1/2
    ΔxΔp≥ (h/2pi)/2

    3. The attempt at a solution
    (a/2)Δp≥h/4pi

    Δp≥h/(2pi*a )

    p=(pi/a)h-bar

    I am not sure, if I am in the right direction. I feel that I am going in circles. Any help?
     
  2. jcsd
  3. Jul 16, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    You're given that Δp ≈ hk/(2π). Use what you know about the ground state of a particle in a box to express k in terms of a.

    Then see what you get for ΔpΔx.
     
  4. Jul 17, 2013 #3
    k=π/a

    a≈(h/2)(1/Δp) ,a=2Δx
    2Δx=(h/2)(1/Δp)

    Δp*Δx=(h/4)

    I still don't see it. Help, please.
     
    Last edited: Jul 17, 2013
  5. Jul 18, 2013 #4

    TSny

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    Yes, k = π/a. I think what you did is ok.

    I would have written it out a little differently:

    You were given in the statement of the problem that you could assume Δp≈hk/(2π). So, when you substitute for k you get

    Δp ≈ h(π/a)(1/(2π)) = h/(2a).

    You were also given that Δx≈a/2.

    So, putting it together, ΔpΔx ≈ h/(2a) (a/2) = h/4. This is the same as your result.

    The question asks you to state whether or not this is consistent with the uncertainty principle. What do you think?
     
  6. Jul 18, 2013 #5
    Thank you, Freunde.

    It agrees with Heisenberg equation.
    ## \Delta p \Delta x \geq \dfrac {\hbar}{2} = 5.272 \times 10^{-35} m^{2} \times \dfrac {kg}{s} ##
    ## \Delta p \Delta x= h/4 = 16.5 \times 10^{-35} \times m^{2} \times \dfrac{kg}{s} ##
     
  7. Jul 18, 2013 #6

    TSny

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    Good!
     
  8. Jul 26, 2013 #7
    I Thought this will be helpful

    I have attached two pdf files.problem 4.2 in the pdf named uncertainty will give you the actual uncertainty relation for your problem and the second pdf file will show you that how this uncertainty results in zero point energy.
     

    Attached Files:

  9. Aug 4, 2013 #8
    Thank you for the information.
    Could you cite your sources?
    Also, can anyone else corroborate that the sources are right?
     
  10. Aug 7, 2013 #9
    oh it is taken from a excellent book named concept and application of quantum mechanics by Nouredine Zettili.this book is easily available you can check it yourself
     
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