Solve Integral of sin*sin: Quantum Eigenfunctions Orthogonal

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The discussion revolves around solving an integral to demonstrate the orthogonality of quantum eigenfunctions. The original poster struggled with the integral involving sine functions and suspected a need for trigonometric substitution. Clarification was sought on whether to show orthogonality between the same eigenfunction or different ones. After realizing a mistake in labeling the eigenfunctions, the poster found clarity by reviewing trigonometric identities. Ultimately, the issue was resolved with the help of the forum members.
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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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$$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\
\int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$
 
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.
 
Simon Bridge said:
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.

Ah, yes, I accidently put n\pix /a for both eigen functions. One should be l\pix /a. Thank you.
 
lurflurf said:
$$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\
\int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$

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\pix/a. So, should the problem be different then?
 
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!
 
No worries, it's all good stuff :)
 

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