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Schrodinger equation normalization to find A -Griffiths

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data
    In David Griffiths Introduction to Quantum Mechanics (2nd ed.), page 32 he normalizes a time independent wave function to get the coefficient A. He dropped the sine part of the integration with no explanation. What is the justification.

    2. Relevant equations
    The time independant wave function given is [itex]\varphi[/itex] = Asin(kx)
    Griffiths gets :
    [itex]\int^{a}_{0}|A|^{2}sin^{2}(kx) dx = |A|^{2}\frac{a}{2} = 1[/itex]
    But:
    cos(2kx) = 1 - 2sin[itex]^{2}(kx)[/itex]
    and:
    sin[itex]^{2}(kx) = \frac{1}{2} - \frac{1}{2}cos(2kx)[/itex]
    This means that the integral should really give:
    [itex]|A|^{2}(\frac{a}{2} - \frac{1}{4k}sin(2ka))[/itex]

    3. The attempt at a solution
    What is the justification for dropping the sine term.
     
  2. jcsd
  3. Nov 23, 2011 #2

    BruceW

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    Should this be in the homework section? Anyway, with the information given, there is no justification. But I am guessing there is more information given in the book. Does it say anything about what the value of the wave function is when x=a?
     
  4. Nov 23, 2011 #3

    vela

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    The spatial frequency k takes on only certain values in order to satisfy the boundary conditions. For those values, you have that sin(2ka)=0.
     
  5. Nov 23, 2011 #4
    Yes at x=a, x=0
    Also not sure where this should go. It is a clarification of the textbook not a homework problem. At least this was my reasoning. I've been wrong before.

    Gary
     
  6. Nov 23, 2011 #5
    Thank you Vela.
    I had fogotten that. That makes sense. It's too bad that Griffiths didn't point this out. Generally the text is great, especially for self study but he sloughs thing sometimes.
     
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