How does $(\hat H + C)^2$ act onto a ket wavefunction?

Question

Equation:

$$\Delta E = \sqrt {\langle \Psi | (\hat H + \bar E)^2 | \Psi \rangle}$$

Does the squared term expand normally into:

$$ \langle \Psi |(\hat H + \bar E)^2 | \Psi \rangle = \langle \Psi | \hat H^2 | \Psi \rangle + \langle \Psi | 2\bar E \hat H | \Psi \rangle + \langle \Psi | \bar E^2 | \Psi \rangle$$

And if so, does $\hat H^2$ correspond to the eigenvalue $E^2$ in this equation, ie.

$$\hat H^2 \Psi = E^2 \Psi $$

I'm haven't encountered this form of equation before and am leery of presuming it would be this straightforward.

Answer 1

For your first question: yes the squared term does expand like any other squared term: \begin{align} \langle\Psi|(\hat{H}+E)^2|\Psi\rangle & =\langle\Psi|(\hat{H}^2+2E\hat{H}+E^2)|\Psi\rangle \\ & =\langle\Psi|\hat{H}^2|\Psi\rangle+\langle\Psi|2E\hat{H}|\Psi\rangle+ \langle\Psi|E^2|\Psi\rangle \end{align} For your second question: yes the eigenvalue of $\hat{H}^2$ is $E^2$: \begin{align} \hat{H}\Psi & = E\Psi \\ \hat{H}^2\Psi & =\hat{H}\hat{H}\Psi=\hat{H}E\Psi=E\hat{H}\Psi=E^2\Psi \end{align}