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Normilzation of a wavefunction

  • Thread starter Nezva
  • Start date
  • #1
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Normalize this wavefunction
psi(x) = Acos(2x) from -pi/2 to pi/2

I'm unsure exactly hoe to 'normalize' the equation. I've heard a few conflicted explanations.
 

Answers and Replies

  • #2
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Ok, so the square of the wave function set equal to 1 (i.e. 100% probability) will give me the euqation I need to normalize the wavefunction, I believe.

So for this one,
int[A^2 * cos^2(2x)]=1, insert trig ident. for cos^2
int(A^2 * [1+cos(2x)/2])=1

set u = 2x and thus dx = 0.5du
rewrite:
1 = A^2/2 * int[1 + cos(u)*1/2du]

I get mathematically lost at this point. My calculus is not that strong. I'm continuing to work it and may be back soon if I figure it out.
 
  • #3
585
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Just integrate and you are done.
 
  • #4
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Ugh that was simple to figure out, I just separated the int[1+cos(2x) into two integrations prior to u substitution. This left me with this (Please check my work for errors):

1 = (A^2)/2 * (x) + ([A^2]/2 * int[cos(2x)])

note: x is the integral of 1 from the 1+cos(2x)

NOW I do the u-substitution to find the integral of cos(2x):

[A^2](x)/2 * [A^2]/2int[cos(2x)] = [A^2](x)/2 * -sin(2x)] = 1

Solving for A, A = sqrt[2/(x-0.5[sin(2x)])] from the -pi/2 to pi/2. So to find normalization A I subtract the large A from the small A?
 
  • #5
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I'm sorry but that is just really really super hard for me to read. Please try using latex. Click the capital sigma all the way to the right of the toolbar above the reply box.

I think you are right but I'm not sure what you mean by large A and small A. I did the problem, what is your final answer?
 
  • #6
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I've gotten to A=[tex]\sqrt{2/[x-0.5*sin(2x)]}[/tex]. I've not yet run it for the interval -[tex]\Pi/2[/tex] to [tex]\Pi/2[/tex]

edit: I was wondering if subtract [tex]x=-\Pi/2[/tex] from the value of [tex]x=\Pi/2[/tex] or vice versa?
 
  • #7
585
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The fundamental theorem of calculus states:

[tex]\int^{b}_{a}f(x)dx=F(b)-F(a)[/tex]
 
Last edited:
  • #8
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Trust me I'm working on it, I'm already in the class. Thank you for your time. I was simply unclear how that applied to a "normalization constant" since it's not an integral. Since it's a constant it's never actually integrated, yet it's subject to the A and B boundaries. Does that make any nonsense? Maybe I'm thinking on it too hard, need to take a step back.
 
  • #9
585
2
Hmm I don't really understand, but I think you may be confusing yourself with the way you solved for A before evaluating the integral.

[tex]\int^{\pi/2}_{-\pi/2}\psi*\psi dx=1[/tex]

[tex]A^{2}\int^{\pi/2}_{-\pi/2}cos^{2}(2x)dx=1[/tex]

[tex]=A^{2}\int^{\pi/2}_{-\pi/2}(\frac{1}{2}+\frac{cos2x}{2})[/tex]

[tex]=\frac{A^{2}}{2}(x+\frac{sin2x}{2})|^{\pi/2}_{-\pi/2}[/tex]

[tex]=\frac{A^{2}}{2}(\frac{\pi}{2}-\frac{-\pi}{2})[/tex]

[tex]=A^{2}\frac{\pi}{2}=1[/tex]
 
  • #10
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Oh, I did make a math error on top of it all :S. Thank you! This was very enlightening.
 
  • #11
585
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I screwed up! When you reduce the power of the cosine, you should get cos(4x) right! :redface: Anyway I'm sure you can work that out.
 
  • #12
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Oh yah I noticed that when I was working it. I probably assisted your mistake in the way I wrote the problem. Thanks for teaching me this problem and how to use the forum boards more correctly.
 

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