Question on correctly interpreting a bra-ket equation

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The discussion focuses on interpreting the bra-ket equation for uncertainty in energy, specifically the term ##(\hat H - \bar E )^2##. It is clarified that this expression can indeed be expanded using binomial expansion, leading to ##\hat H^2 - 2 \bar E \hat H + \bar E^2##, which can be distributed over the state ##|\Phi \rangle##. Additionally, it is confirmed that for an operator ##\hat{H}## acting on its eigenvector, the relation ##\hat H^2 \psi = E^2 \psi## holds true. The user expresses gratitude for the clarification, indicating they can proceed with their calculations. Understanding these notations is crucial for solving quantum mechanics problems related to energy uncertainty.
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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