2.6.7 So­lu­tion herm-g

Ques­tion:

Show that if $A$ is a Her­mit­ian op­er­a­tor, then so is $A^2$. As a re­sult, un­der the con­di­tions of the pre­vi­ous ques­tion, $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$ is a Her­mit­ian op­er­a­tor too. (And so is just ${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$, of course, but $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$ is the one with the pos­i­tive eigen­val­ues, the squares of the eigen­val­ues of ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$.)

An­swer:

To show that $A^2$ is Her­mit­ian, just move the two op­er­a­tors $A$ to the other side of the in­ner prod­uct one by one. As far as the eigen­val­ues are con­cerned, each ap­pli­ca­tion of $A$ to one of its eigen­func­tions mul­ti­plies by the eigen­value, so two ap­pli­ca­tions of $A$ mul­ti­plies by the square eigen­value.