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Monday, 5 September 2011

Physics GRE - # 34

Eigenfunctions for a rigid dumbbell rotating about its centre have a \phi dependence of the form \psi(\phi)=Ae^{im\phi}, where m is a quantum number and A is a constant. Which of the following values of A will properly normalize the eigenfunction?

  1. \sqrt{2\pi}

  2. 2\pi

  3. (2\pi)^2

  4. \dfrac{1}{\sqrt{2\pi}}

  5. \dfrac{1}{2\pi}

Solution :

Choice 4 is the answer.

For \psi(\phi) to be normalized on a to b, recall the normalization condition: \int_a^b{\left|\psi(\phi)\right|^2}d\phi=1. Since \phi ranges from 0 to 2\pi, we have: \int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1 where the second equality holds as |e^{im\phi}|=1 for any m and any \phi.

Thus, we know that: \int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.

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