Question on correctly interpreting a bra-ket equation

[blaisem]

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!

 

[GwtBc]

1. Yes
2. Consider an operator $\hat{A}$ with an eigenvector-eigenvalue pair $v, A$. Then $\hat{A}^{2}v = \hat{A}(\hat{A}\psi) = \hat{A}(Av) = A(\hat{A}v) = A^{2}v$

 

[blaisem]

Perfect! I should be good from here on out. Thanks for the help getting started.