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Homework Help: Normalising wave functions.

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle of unit mass moving in an infinite square well, V=0 for |x|<= a , V=∞ for |x|>a , is described by a wave function u(x) = Asin(3πx/a).

    (i) If I normalise the wave function, what is A?

    (ii) And what is the energy of state described by this wave function?

    3. The attempt at a solution

    (i) Normalised the wave function by saying it is only valid from 0<x<=a and every where else is 0.

    So I square the wave function, integrate it and I get:

    A²a = 1 --> 1/√a

    (ii) And the energy of state decribed by this wave function would be 9h²/8ma² , since we know its on n=3.

    What Im puzzled about is if the width of the well is 2a or a? I know it is 'a' because the denominator of the wave function tells me it is. But when I use the general formula of:

    A= √(2/L) I know L = a so I instead get:

    A= √(2/a) which isn't true?

  2. jcsd
  3. Dec 20, 2009 #2


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    Homework Helper

    Why only [itex]0 < x \le a[/itex]? Take another close look at that question to figure out the range over which the potential is non-infinite.

    The form of the wave function tells you nothing, in general, about the width of the well. You have to get that from other information in the problem - the width of the region in which the potential is zero.
  4. Dec 20, 2009 #3
    Your result is close but awkward on the first part. Go back and redo the integral. You should get the square root of 2 divided by a, not 1 divided by a. The bounds of integration are 0 to a. Given what you've presented here, the width of the well is a.

    You have the right idea for the second part, but it should be:

    [tex]E = 9\pi^2h^2/(2ma^2)[/tex]
  5. Dec 20, 2009 #4
    It's actually from -a< x <a which will give a well of width 2a.

    If I integrate now this will become 2a(A²) = 1 --> 1/√2a

    But If I use A= √(2/L) I know L = 2a so I instead get:

    A = √(1/a)

    Im clearly going wrong somewhere
  6. Dec 21, 2009 #5
    Probably posting your integration helps others to spot your mistakes.

    if you really do not see it on your own.

    Well you probably remember cos^2 + sin^2 = 1, so integrating over full peroids of a sine squared (or cosine squared) gives you half of the peroid length.
  7. Dec 22, 2009 #6
    For the first part the answer is A=1/√a
  8. Dec 24, 2009 #7
    Yea it was a mistake I made with the integration, I finally got the correct answer. Is the energy state for this wave function simply:


    since we know n=3 from the wave function given?
  9. Dec 24, 2009 #8
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