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Normalization factor in wave equation

  1. Aug 18, 2011 #1
    (Note: although arising in QM, this is essentially a calculus question)

    Ѱ (x) = A sin (n╥x/a)

    1 = ∫ l Ѱ (x) l^2 dx with limits of integration a to 0

    1 = ∫ A^2 sin^2 (n╥x/a) dx with limits of integration a to 0

    Indefinite integral ∫ sin^2 x dx = x/2 - sin2x/4

    I know this integral should reduce to 1 = A^2 a/2

    But what I get is

    I = A^2 [a/2 - sin (2n╥a/4a) - 0/2 - sin (2n╥0/4a)]

    Third and fourth term within bracket are 0, but second term is 1

    i.e., sin (n╥/2) = 1

    I assume I'm missing something, but not sure what it is.

  2. jcsd
  3. Aug 19, 2011 #2


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    Homework Helper
    Gold Member

    Your integral is not right. In the second term, the 4 is not inside the argument of the sine function:

    [tex]\int A^2 \sin^2(x) dx = \frac{x}{2}-\frac{\sin(2x)}{4} + C[/tex]
  4. Aug 20, 2011 #3

    That was my problem.

    Correct value now.
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