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(Quantum Mechanics) Infinite Square Well

  1. Aug 23, 2011 #1
    Hi, I'm stuck in this Griffiths' Introduction to QM problem (#2.8)

    1. The problem statement, all variables and given/known data
    A particle in the infinite square well has the initial wave function

    [itex]\Psi(x,0) = Ax(a-x)[/itex]

    Normalize [itex]\Psi(x,0)[/itex]

    2. Relevant equations
    [itex] \int_{0}^{a} |\Psi(x)|^2 dx = 1 [/itex]

    3. The attempt at a solution
    Haha, this is supposed to be the least of my problems but... doing

    [itex] A^2 \int_{0}^{a} x^2 (a-x)^2 dx = 1 [/itex]
    gives us [itex] A = \sqrt{\frac{30}{a^5}} [/itex].

    When the correct answer is [itex] A = \sqrt{\frac{2}{a}} [/itex]. I have no clue what I did wrong...
  2. jcsd
  3. Aug 23, 2011 #2


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    The normalization constant A = sqrt(2/a) is appropriate for ψ(x) = A sin(nπx/a). Here you have ψ(x) = Ax(a-x). Different wavefunctions have different normalization constants.
  4. Aug 23, 2011 #3
    Kuruman... indeed [itex]A = \sqrt{\frac{2}{a}}[/itex] is appropriate for [itex]\psi[/itex] as a sin function. But when checking the result i got, i found this solution for the exact same problem:

    http://img714.imageshack.us/img714/6186/64732846.jpg [Broken]

    The integral limit is set to 0,a/2... I'm not sure why, and if this is correct. For now i'll stick with my solution

    By the way... (a correlated problem) trying to prove that: [itex] \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m (x) dx = \delta_{n,m}[/itex] (Kronecker delta).

    For n = m, and [itex]\psi_n (x) = \sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a}),\in \Re[/itex], i'm coming up with

    [itex] \int_{-\infty}^{\infty} \psi_{n}^{2}(x) dx [/itex] = [itex]\frac{-1}{a} cos(\frac{n\pi x}{a}) sin(\frac{n\pi x}{a})|_{-\infty}^{\infty} + \frac{1}{a} \int_{-\infty}^{\infty} dx[/itex]

    The first term goes to 0, but the second would result in 1 for like... [itex] \int_{0}^{a}[/itex], not [itex]\int_{-\infty}^{\infty} [/itex]. Again... where am I messing up?
    Last edited by a moderator: May 5, 2017
  5. Aug 23, 2011 #4


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    It is not the exact same problem. The problem above has yet a third wavefunction that is constant for 0≤x≤a/2 and zero everywhere else. That is the reason why the upper limit is set at a/2. If you extend the integral beyond a/2, you will be adding a whole bunch of zeroes since ψ(x) is zero, so you might as well use a/2 for the upper limit.

    Here I assume that you are using particle-in-the-box wavefunctions. All of them are zero for x≤0 and x≥a. What do you think your limits of integration ought to be?
    Last edited by a moderator: May 5, 2017
  6. Aug 23, 2011 #5
    Hmm, ok! So it's a different problem

    It makes sense now, I forgot the conditions... You're right, it is [itex]\int_{0}^{a}[/itex]!
    Thank you
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