Solve Integral of sin*sin: Quantum Eigenfunctions Orthogonal

  • Thread starter Thread starter Mary
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
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.
Mary
Messages
11
Reaction score
0
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 :)
 

Attachments

  • integral.png
    integral.png
    3.6 KB · Views: 459
Physics news on Phys.org
$$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\
\int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$
 
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 :)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K