Question:
Consider the time evolution operator $ \hat{U}(\tau) = e^{\frac{−i}{\hbar} \hat{H} \tau} $. Given an eigenstate of the Hamiltonian, $ \hat{H} |E\rangle = E | E\rangle $, at $ t = 0 $, what is $ \hat{U}(\tau) | E \rangle $, where $ \tau $ is some time, $ \tau > 0 $.
Solution:
To solve this problem, we expand the exponential operator using its Taylor series definition as follows:
$ \begin{eqnarray*} \hat{U}(\tau) |E\rangle &=& e^{\frac{−i}{\hbar} \hat{H} \tau} |E\rangle \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k}{k!}}) |E\rangle \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k \hat{H}^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k E^k |E\rangle}{k!}} \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau E)^k }{k!}}) |E\rangle \\ &=& e^{\frac{−i}{\hbar} \tau E} |E\rangle \end{eqnarray*} $
There you go!
Consider the time evolution operator $ \hat{U}(\tau) = e^{\frac{−i}{\hbar} \hat{H} \tau} $. Given an eigenstate of the Hamiltonian, $ \hat{H} |E\rangle = E | E\rangle $, at $ t = 0 $, what is $ \hat{U}(\tau) | E \rangle $, where $ \tau $ is some time, $ \tau > 0 $.
Solution:
To solve this problem, we expand the exponential operator using its Taylor series definition as follows:
$ \begin{eqnarray*} \hat{U}(\tau) |E\rangle &=& e^{\frac{−i}{\hbar} \hat{H} \tau} |E\rangle \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k}{k!}}) |E\rangle \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k \hat{H}^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k E^k |E\rangle}{k!}} \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau E)^k }{k!}}) |E\rangle \\ &=& e^{\frac{−i}{\hbar} \tau E} |E\rangle \end{eqnarray*} $
There you go!
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