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Uncertainty relation

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    The (classical) energy of one-dimensional linear oscillator is
    2.gif
    a) show, using the uncertainty relation, that the energy can be written as
    gif.gif
    b) Show that the minimum energy of the oscillator is 2.gif
    Where m%29.gif


    2. Relevant equations
    Δp Δx >= ħ/2
    p ≈ ħ/2x

    3. The attempt at a solution
    I'm really stuck on this problem, I could say that the p_x = ħ^2 / (2x)^2 then substitute in the equation, but still where did the π and the 32 come from? I couldn't find any other relevant equations
    Edit: Oh I just realized ħ = h/2π , then p^2 = h^2/(4πx)^2 , substituting in the equation would give the answer.
    But now what about b)? I believe to find the minimum energy we set Δx >= ħ/2Δp , but it isn't clear to me how the frequency and angular velocity come in
     
    Last edited: Oct 25, 2015
  2. jcsd
  3. Oct 25, 2015 #2

    Orodruin

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    Are you familiar with the difference between h and ħ?
     
  4. Oct 25, 2015 #3
    Oh yes, I just realized that a minute ago and edited my post, but now I'm still having trouble with part b)
     
  5. Oct 25, 2015 #4

    Orodruin

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    Are you familiar with how to find the minimum of a function?
     
  6. Oct 25, 2015 #5
    Hmm I believe I have to take the derivative with respect to Δp, minimum energy dE/dΔp = 0
     
  7. Oct 25, 2015 #6

    Orodruin

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    Δp is no longer part of your expression for the energy ... You have just expressed it as a function of x only.
     
  8. Oct 25, 2015 #7
    Then the derivative of p_x = ħ^2 / (2x)^2 ?
     
  9. Oct 25, 2015 #8

    Orodruin

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    No, it is the energy you wish to minimise.
     
  10. Oct 25, 2015 #9
    use dE/dx = 0 ?
     
  11. Oct 25, 2015 #10

    Orodruin

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    Why dont you try it and see what you get?
     
  12. Oct 25, 2015 #11
    Would get
    E = -2h^2/32mx^3π^2 + 2Cx
    Doesn't make sense because I still don't see where the omega and f come from
     
    Last edited: Oct 25, 2015
  13. Oct 25, 2015 #12

    Orodruin

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    You care far too much about the omega and f. You are just supposed to find the minimum energy, which is a certain given expression in terms of omega where omega is an expression in other variables which arethe actual variables you have.
     
  14. Oct 25, 2015 #13
    So you mean E = -2h^2/32mx^3π^2 + 2Cx can be simplified into 2.gif ? how?
     
  15. Oct 25, 2015 #14

    Orodruin

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    No, that is your energy derivative. You need to put it to zero to find the minimum.
     
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