Solve Integral of sin*sin: Quantum Eigenfunctions Orthogonal

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Homework Help Overview

The discussion revolves around demonstrating the orthogonality of quantum eigenfunctions, specifically involving the integral of the product of sine functions. Participants are exploring the mathematical properties and identities related to this integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for trigonometric identities to simplify the integral. There is confusion about whether the eigenfunctions in question are orthogonal to themselves or to each other, particularly between different quantum states.

Discussion Status

Some participants have offered insights into the use of trigonometric identities, while others are clarifying the specific eigenfunctions involved. There is an acknowledgment of a mistake in the notation of the eigenfunctions, which has led to further questions about the problem's requirements.

Contextual Notes

Participants mention the importance of having a reference for trigonometric identities and express uncertainty about the implications of the notation error on the problem's setup.

Mary
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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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