Solve Integral of sin*sin: Quantum Eigenfunctions Orthogonal

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In summary, the author is having trouble with one of the examples in their quantum book. They are suppose to be showing that two eigenfunctions are orthogonal and have solved the integral attached, but they do not understand the steps. They are looking for help and are grateful for any help that is offered.
  • #1
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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  • #2
$$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\
\int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$
 
  • #3
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.
 
  • #4
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[itex]\pi[/itex]x /a for both eigen functions. One should be l[itex]\pi[/itex]x /a. Thank you.
 
  • #5
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[itex]\pi[/itex]x/a. So, should the problem be different then?
 
  • #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!
 
  • #7
No worries, it's all good stuff :)
 

Related to Solve Integral of sin*sin: Quantum Eigenfunctions Orthogonal

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value of a function over a given interval.

2. What is the sin*sin function?

The sin*sin function is a mathematical function that multiplies two sine functions together. In this case, the two sine functions are sin(x) and sin(x). This function is commonly used in quantum mechanics to describe the behavior of particles in a system.

3. What are quantum eigenfunctions?

Quantum eigenfunctions are mathematical functions that represent the possible states of a quantum system. These functions are solutions to the Schrödinger equation and are used to describe the behavior and properties of particles in a quantum system.

4. What does it mean for quantum eigenfunctions to be orthogonal?

Two quantum eigenfunctions are said to be orthogonal if their inner product is equal to zero. This means that they are perpendicular to each other and have no overlap. Orthogonality is an important concept in quantum mechanics as it allows for the description of different states of a system as independent from each other.

5. How do you solve the integral of sin*sin?

To solve the integral of sin*sin, you can use the trigonometric identity sin(x)*sin(y) = (1/2) * (cos(x-y)-cos(x+y)). This identity can be used to simplify the integral and make it easier to solve. Once the integral is simplified, you can use integration techniques such as substitution or integration by parts to find the solution.

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