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Let $|n \rangle$ denote the $n^{th}$ stationary state of the harmonic oscillator, with energy $E_n = \hbar \omega(n+\frac{1}{2})$

How would I find $\langle x\rangle$ and $\sigma_x$

I know that

$$\langle x\rangle = \langle \alpha|x\alpha \rangle = \sqrt{\frac{\hbar}{2m\omega}}\langle \alpha|(a_+ + a_-)\alpha \rangle=\sqrt{\frac{\hbar}{2m\omega}}(\alpha + \alpha^*)$$

but that is just from the definition of ladder operator, what does it have to do with energy eigenstate?

Any help would be appreciated

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closed as off-topic by user10851, John Rennie, Emilio Pisanty, Dan, Waffle's Crazy Peanut Nov 8 '13 at 3:14

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Assuming you are attempting to determine $\langle n|x|n\rangle$, you're on the right track. Here are some

Hints:

  1. Determine what the ladder operators do the the energy eigenstates; \begin{align} a|n\rangle = ?, \qquad a^\dagger |n\rangle = ? \end{align}
  2. Write the position operator in terms of ladder operators as you have done.

  3. What are the inner products of the energy eigenstates: \begin{align} \langle m|n\rangle = ? \end{align}

  4. Combine this all together to obtain the desired expressions for $\langle x\rangle = \langle n|x|n\rangle$.

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