Normalization of the infinite square well.

  • Thread starter Jdraper
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  • #1
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Main Question or Discussion Point

I have been going through my textbook deriving equations in preparation for my test on QM tomorrow. I noticed in the infinite square well that i was unable to complete the normalization.

My textbook, Griffiths reads :

(integral from 0 to a) ∫|A|^2 * (sin(kx))^2 =|A|^2 * (a/2) =1 Therefore, |A|^2 = 2/a

When i do this integral i do not get this result, furthermore when i put this integral into wolfram alpha i get this result:

http://www.wolframalpha.com/input/?i=integrate+sin^2(kx)+dx+from+0+to+a

Which is also not the required result.

It's probably something stupid but help would be appreciated none the less,

Thanks in advance, John.
 

Answers and Replies

  • #2
jtbell
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Using the result from Wolfram Alpha, your equation becomes
$$|A|^2 \left( \frac{a}{2} - \frac{\sin(2ak)}{4k} \right) = 1$$
Recall that k is restricted to certain values because of the boundary conditions. What are those values?
 
  • #3
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I derived the wave equation to be:

[itex]\Psi[/itex](x) = Asin(kx), So this means k takes integer values of n?

So k's values are restricted to (n*pi/a) where n=1,2,3

Just as i was writing that out i understood, k can only take integer values so sin(2kx) will always equal 0 meaning the expression is just a/2.

Thanks a bunch :)
 
  • #4
jtbell
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Right, ##\sin(ka) = \sin(2n\pi)## which is always zero for integer n. It's nice when "extra" terms in an equation disappear like that. :smile:
 

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