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Normalization of the infinite square well.

  1. Jan 16, 2014 #1
    I have been going through my textbook deriving equations in preparation for my test on QM tomorrow. I noticed in the infinite square well that i was unable to complete the normalization.

    My textbook, Griffiths reads :

    (integral from 0 to a) ∫|A|^2 * (sin(kx))^2 =|A|^2 * (a/2) =1 Therefore, |A|^2 = 2/a

    When i do this integral i do not get this result, furthermore when i put this integral into wolfram alpha i get this result:

    http://www.wolframalpha.com/input/?i=integrate+sin^2(kx)+dx+from+0+to+a

    Which is also not the required result.

    It's probably something stupid but help would be appreciated none the less,

    Thanks in advance, John.
     
  2. jcsd
  3. Jan 16, 2014 #2

    jtbell

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    Staff: Mentor

    Using the result from Wolfram Alpha, your equation becomes
    $$|A|^2 \left( \frac{a}{2} - \frac{\sin(2ak)}{4k} \right) = 1$$
    Recall that k is restricted to certain values because of the boundary conditions. What are those values?
     
  4. Jan 16, 2014 #3
    I derived the wave equation to be:

    [itex]\Psi[/itex](x) = Asin(kx), So this means k takes integer values of n?

    So k's values are restricted to (n*pi/a) where n=1,2,3

    Just as i was writing that out i understood, k can only take integer values so sin(2kx) will always equal 0 meaning the expression is just a/2.

    Thanks a bunch :)
     
  5. Jan 16, 2014 #4

    jtbell

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    Staff: Mentor

    Right, ##\sin(ka) = \sin(2n\pi)## which is always zero for integer n. It's nice when "extra" terms in an equation disappear like that. :smile:
     
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