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Integral of sin*sin

  1. Apr 13, 2014 #1
    Hey, I'm having trouble with one of the examples in my quantum book. I'm suppose to be showing that two eigenfunctions are orthogonal and in order to do that I have to solve the integral I have attached to this forum. I have the solution but I don't understand the steps! I believe it may be a trig substitution that I just can't remember. I would really appreciate any help. Thank you :)
     

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  2. jcsd
  3. Apr 13, 2014 #2

    lurflurf

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    $$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\
    \int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$
     
  4. Apr 13, 2014 #3

    Simon Bridge

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    Puzzled...
    Is ##\psi_n=\sqrt{\frac{2}{a}}\sin (n\pi x/a)## orthogonal to ##\psi_n## ?
    Or are you supposed to show that ##\psi_n## is orthogonal to ##\psi_m## where ##m\neq n##?

    Anyway - I'd add that you will do well to arm yourself with a table of trig identities:
    http://en.wikipedia.org/wiki/List_of_trigonometric_identities
    ... after a while you'll just remember the ones you use all the time.
     
  5. Apr 13, 2014 #4
    Ah, yes, I accidently put n[itex]\pi[/itex]x /a for both eigen functions. One should be l[itex]\pi[/itex]x /a. Thank you.
     
  6. Apr 13, 2014 #5
    Thank you for your quick response. I just realized that I accidently put n for both eigenfunctions. One of the sin's should be l[itex]\pi[/itex]x/a. So, should the problem be different then?
     
  7. Apr 13, 2014 #6
    Ah, nevermind, I see now. After reviewing the trigometric identities on wikipedia I see how simple it is now. Thanks a bunch for all of your all's help!
     
  8. Apr 14, 2014 #7

    Simon Bridge

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    No worries, it's all good stuff :)
     
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