Exploring Quantum Physics - Week 2 Question 3

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!