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Calculate the RMSD for the odd solutions

  1. Aug 2, 2007 #1
    Calculate the RMSD for the odd solutions of the infinite symmetric well problem.

    The odd solutions have the form:

    [tex]u_n(x) = 1/\sqrt{a} \cdot sin(n \pi x/a)[/tex]

    So, I should multiply u_n by x, then take the integral over the bottom of the well to get <x>. Then I should multiply u_n by x^2, take the integral over the bottom of the well to get <x^2>. From this I can get delta x.

    Is all that correct?
     
  2. jcsd
  3. Aug 2, 2007 #2

    Gokul43201

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    How is the expectation value of an observable defined?
     
  4. Aug 2, 2007 #3
    I should take the integral over x times the modulus squared of u_n to get <x>. That better.
     
  5. Aug 3, 2007 #4

    Gokul43201

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    Better, but it doesn't look like your wavefunctions have been normalized.
     
  6. Aug 3, 2007 #5
    They have been. Remember this it represents a probability amplitude not a probability density and your region of integration should be -a to a.
     
  7. Aug 3, 2007 #6

    Gokul43201

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    In that case, the wavefunction looks wrong to me.
     
  8. Aug 4, 2007 #7
    Its my book that does it oddly. Instead of breaking it up into cases of n = 0,2,4,.. and n =1,3,5,... they used different arguments for sine and cosine so that they can use n = 1,2,3,... for both cases.
     
  9. Aug 4, 2007 #8

    Gokul43201

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    I don't think that's particularly unusual.

    If your well width is 2a I'd imagine your wavefunctions to be [itex]\sqrt{1/a}~cos(n \pi x/2a) [/itex], and if the well goes from 0 to a then you'd have [itex]\phi_n(x) = \sqrt{2/a}~sin(n \pi x/a) [/itex]. The wavefunction you've written is neither of these.
     
    Last edited: Aug 4, 2007
  10. Aug 4, 2007 #9
    It goes from -a to a. The even solutions are of the form B*cos(k*x) , where k = sqrt(2mE/hbar^2). The boundary conditions give us that cosk(k*a) = 0, so set k_n*a = (n - 1/2)pi to get k_n = (n-1/2)pi/a. Plug that back into cosine and normalize.
     
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