Question:
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What is the coefficient $ N $ for the equation $ \psi_n(x) = N\sin(\frac{n \pi x}{L}) $.
Solution:
We will do that by normalizing the wavefunction.
$ \begin{eqnarray*} 1 &=& \int\limits_{0}^{L}{(N\sin(\frac{n \pi x}{L}))^2dx} \\ &=& N^2\int\limits_{0}^{L}{\sin^2(\frac{n \pi x}{L})dx} \\ &=& N^2\int\limits_{0}^{L}{\frac{1 - \cos(\frac{2 n \pi x}{L})}{2}dx} \\ &=& N^2\frac{L}{2} \\ N &=& \sqrt{\frac{2}{L}} \end{eqnarray*} $
So the answer is $ \sqrt{\frac{2}{L}} $.
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What is the coefficient $ N $ for the equation $ \psi_n(x) = N\sin(\frac{n \pi x}{L}) $.
Solution:
We will do that by normalizing the wavefunction.
$ \begin{eqnarray*} 1 &=& \int\limits_{0}^{L}{(N\sin(\frac{n \pi x}{L}))^2dx} \\ &=& N^2\int\limits_{0}^{L}{\sin^2(\frac{n \pi x}{L})dx} \\ &=& N^2\int\limits_{0}^{L}{\frac{1 - \cos(\frac{2 n \pi x}{L})}{2}dx} \\ &=& N^2\frac{L}{2} \\ N &=& \sqrt{\frac{2}{L}} \end{eqnarray*} $
So the answer is $ \sqrt{\frac{2}{L}} $.
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