Question:
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Given $ \phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right) $, what must be true?
Solution:
This is the Parseval's theorem. Since $ \phi(x, t)^* \phi(x, t) $ represents a probability, we have $ \int\limits_{-\infty}{^\infty}{\phi(x, t)^* \phi(x, t) dx} = 1 $. Substitute in the definition, we have:
$ \begin{eqnarray*} 1 &=& \int\limits_{-\infty}^{\infty}{\phi(x, t)^* \phi(x, t) dx} \\ &=& \int\limits_{-\infty}^{\infty}{(\sum_{n=1}^{\infty} a_n^* \psi_n(x) \exp \left(iE_n t/\hbar \right))(\sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)) dx} \\ &=& \int\limits_{-\infty}^{\infty}{\sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \psi_p(x) \psi_q(x)}} dx} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \int\limits_{-\infty}^{\infty}{\psi_p(x) \psi_q(x) dx}}} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \delta_{pq}}} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2 \exp \left(iE_p t/\hbar \right) \exp \left(-iE_p t/\hbar \right)} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2} \\ \end{eqnarray*} $
Therefore the answer is $ \sum_{p=1}^{\infty}{|a_p|^2} = 1 $.
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Given $ \phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right) $, what must be true?
Solution:
This is the Parseval's theorem. Since $ \phi(x, t)^* \phi(x, t) $ represents a probability, we have $ \int\limits_{-\infty}{^\infty}{\phi(x, t)^* \phi(x, t) dx} = 1 $. Substitute in the definition, we have:
$ \begin{eqnarray*} 1 &=& \int\limits_{-\infty}^{\infty}{\phi(x, t)^* \phi(x, t) dx} \\ &=& \int\limits_{-\infty}^{\infty}{(\sum_{n=1}^{\infty} a_n^* \psi_n(x) \exp \left(iE_n t/\hbar \right))(\sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)) dx} \\ &=& \int\limits_{-\infty}^{\infty}{\sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \psi_p(x) \psi_q(x)}} dx} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \int\limits_{-\infty}^{\infty}{\psi_p(x) \psi_q(x) dx}}} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \delta_{pq}}} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2 \exp \left(iE_p t/\hbar \right) \exp \left(-iE_p t/\hbar \right)} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2} \\ \end{eqnarray*} $
Therefore the answer is $ \sum_{p=1}^{\infty}{|a_p|^2} = 1 $.
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