Question on correctly interpreting a bra-ket equation

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blaisem
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I am trying to solve for the uncertainty in energy ##\Delta E## in the following exercise:
$$\Delta E = \sqrt{\langle \Phi | (\hat H - \bar E )^2 | \Phi \rangle}$$
Questions
  1. What does ##(\hat H - \bar E )^2## mean? Is it a simple binomial expansion into ##\hat H^2 - 2 \bar E \hat H + \bar E^2##, which I distribute over ##|\Phi \rangle##?
  2. If that is the case, does ##\hat H^2 \psi = E^2 \psi##?

I'm not inquiring for the full solution. I simply do not understand the notation in order to begin.

Background
In case a broader context is important, here is the full info available to me:
$$\begin{align} \Phi = a_1 \psi _1 &+ a_2 \psi _2 + a_3 \psi _3 \nonumber \\ \hat H &\psi _i = E_i \psi _i \nonumber \\ \langle \psi _i &| \psi _j \rangle = \delta _{ij} \nonumber \end{align}$$
I also normalized ##\Phi## as ##N=\left( \sum_{1}^3 a_{i}^2 \right)^{-1/2}## and solved ##\bar E = \langle \Phi | \hat H | \Phi \rangle = N^2 \sum_i \left( a_i^2 E_i \right) ## in previous exercises.

Thank you!
 
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1. Yes
2. Consider an operator ##\hat{A}## with an eigenvector-eigenvalue pair ##v, A##. Then ##\hat{A}^{2}v = \hat{A}(\hat{A}\psi) = \hat{A}(Av) = A(\hat{A}v) = A^{2}v##
 
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Perfect! I should be good from here on out. Thanks for the help getting started.
 
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