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How did they get 1=A^2(L/4) when integrating?

  1. Oct 12, 2016 #1
    I was looking for questions to practice normalizing a wave function, so I visited the following online pdf, http://people.physics.tamu.edu/syeager/teaching/222/hw1solution.pdf. The first question was to find the normalization constant, A of ψ(x) = A cos (2πx/L) for (−L/4) ≤ x ≤ (L/4). After attempting to solve it on my own I checked my answer verses that of the pdf and was surprised to see they arrived at A^2=4/L. To see their entire solution, visit the pdf.

    Here was my way of solving it:

    1 = ∫A^2 cos^2 (2πx/L)dx
    A^2 ∫cos^2(2πx/L)dx =(A^2)((2πx/2L)+(1/4)(sin(4πx/L))
    = (A^2) ((πx/L)+(sin(4πx/L)/4)
    = (A^2) ((π((L/4)-(-L/4))/L)+sin((4π((L/4)-(-L/4)))/L)/4)
    = (A^2)((π(L/2)/L)+sin((4π(L/2))/L)/4)
    = (A^2)((π/2)+(sin(2π))/4)

    so
    A^2(π/2)=1
    That is what I got, and from there you would get A, so I was wondering how the solution showed A^2=4/L instead of A^2=2/π?
     
    Last edited: Oct 12, 2016
  2. jcsd
  3. Oct 12, 2016 #2

    TSny

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    Welcome to PF!
    You can see something's wrong here. The integral on the left side (not including the A2) has the same dimensions as x. But the right hand side (not including the A2) is dimensionless.

    Also, there needs to be a set of parentheses so that the A2 on the right multiplies the entire right side.
     
  4. Oct 12, 2016 #3
    TSny, I wasn't familiar on how to put in the dimensions for the right hand side with this text editor. The dimensions can be seen in the pdf, but sorry for not being able to include it.
     
  5. Oct 12, 2016 #4

    TSny

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    You don't need to specify the dimensions explicitly. But you can tell that your evaluation of the integral cannot be correct by dimensional analysis.

    How did you get the result
    A^2 ∫cos^2(2πx/L)dx =(A^2)((2πx/2L)+(1/4)(sin(4πx/L)) ?

    It appears to me that you have left out an overall factor that involves ## L## and ##\pi##.
     
  6. Oct 12, 2016 #5
    I merely integrated cos^2(2πx/L), can you elaborate upon what I am missing? If you are correct, then maybe this missing factor could be what I am lacking!
     
  7. Oct 12, 2016 #6

    TSny

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    You'll need to show your steps in the integration process. Then we can identify the specific mistake. (There's a good chance you will catch your mistake yourself while writing out the steps.)
     
  8. Oct 12, 2016 #7
    The part I skipped was showing how I integrated, which is because of the way I learned to integrate cos^2. The following website sums up what I did, http://calc101.com/special_2.html. I basically used this rule to integrate, which is how I got the part you are asking about.
     
  9. Oct 12, 2016 #8
    I also wrote my steps while preforming the problem, and did so at least three times. I caught a mistake once, but that didn't get me the correct answer.
     
  10. Oct 12, 2016 #9
    My question's purpose was for someone to explain how exactly the solution was A^2=4/L? If someone could explain how the problem was integrated to get the following result, I can analyze it and derive what I am missing.
     
  11. Oct 12, 2016 #10

    TSny

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    OK, this is a good way to integrate ##\cos^2 x##. But you nevertheless made a mistake here:
    The only way we can determine where you made the mistake is for you to show the steps in getting to the right side of the above equation. The mistake probably has to do with the fact that the link shows you how to integrate ##\cos^2 x##, but you need to integrate ##\cos^2 \left(2 \pi x/L \right)##. Show us how you handled this.
     
  12. Oct 12, 2016 #11
    I agree that cos^2 (x) must be different than the situation we are facing. My mistake then would be foolishly treating 2πx/L as the x in cos^2(x), which upon thought would be incorrect because you are integrating it with dx. If you look at it closely you can see how I replaced x from the integration of cos^2(x) with 2πx/L in the right side. Seeing this mistake, how would I integrate cos^2(2πx/L)?
     
  13. Oct 12, 2016 #12

    TSny

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    Do you recall the method of "u substitution"?
     
  14. Oct 12, 2016 #13
    I realized that just as I sent that reply, thanks for making me analyze my mistake. I used u substitution and got the correct answer.
     
  15. Oct 12, 2016 #14

    TSny

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    Great! Good work.
     
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