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Normalizing Wave Function of A Ring

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \psi_{n}(\theta)=A_{n} \exp(\imath n \theta)
    [/tex]
    where n is an integer

    Calculate the factor [tex]A_{n}[/tex] if the wave function is normalized between
    [tex]\theta = 0[/tex] and [tex]\theta = 2\pi[/tex].

    2. Relevant equations
    NA

    3. The attempt at a solution

    [tex]
    1=\int_0^{2\pi} |\psi_{n}(\theta)|^2 d\theta
    [/tex]

    [tex]
    1=|A_{n}|^2\int_0^{2\pi} \exp(2\imath n \theta) d\theta
    [/tex]

    [tex]
    1=|A_{n}|^2 [\frac{.5sin(2n\theta)}{n} - \frac{.5cos(2n\theta)}{n} \imath]_0^{2\pi}
    [/tex]

    [tex]
    1=|A_{n}|^2 [0]
    [/tex]

    [tex]
    1=0
    [/tex]

    Obviously 1 does not equal 0 :) . Am I missing something?
     
  2. jcsd
  3. Nov 22, 2008 #2

    George Jones

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    What does [itex]\| \exp(\imath n \theta) \|^2[/itex] equal?
     
  4. Nov 22, 2008 #3
    [itex]
    \| \exp(\imath n \theta) \|^2 = \| \exp(2\imath n \theta) \| = \|cos(2n\theta)+sin(2n\theta)\imath \|
    [/itex]

    correct?
     
  5. Nov 22, 2008 #4
    You just took the square of the wavefunction, instead of the absolute value squared.
     
  6. Nov 22, 2008 #5

    George Jones

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    If [itex]z = a + \imath b[/itex], where [itex]a[/itex] and [itex]b[/itex] are both real, what does [itex]|z|^2[/itex] equal?
     
  7. Nov 22, 2008 #6
    [itex]
    \| \exp(\imath n \theta) \|^2 = 1
    [/itex]

    Man, I can't believe I missed that!

    [tex]
    1=|A_{n}|^2\int_0^{2\pi} \|\exp(\imath n \theta)\|^2 d\theta
    [/tex]

    [tex]
    1=|A_{n}|^2\int_0^{2\pi} 1 d\theta
    [/tex]

    [tex]
    1=|A_{n}|^2[\theta]_0^{2\pi}
    [/tex]

    [tex]
    1=|A_{n}|^2*2\pi
    [/tex]

    [tex]
    A_{n}=\frac{1}{\sqrt{2\pi}}
    [/tex]

    This is the correct solution, right?
     
  8. Nov 22, 2008 #7

    George Jones

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    This looks good and is probably the expected answer, but note that multiplying your [itex]A_n[/itex] by any phase factor would give something that also works.
     
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