2.5.1 So­lu­tion eigvals-a

Ques­tion:

Show that $e^{{{\rm i}}kx}$, above, is also an eigen­func­tion of ${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$, but with eigen­value $\vphantom{0}\raisebox{1.5pt}{$-$}$$k^2$. In fact, it is easy to see that the square of any op­er­a­tor has the same eigen­func­tions, but with the square eigen­val­ues.

An­swer:

Dif­fer­en­ti­ate the ex­po­nen­tial twice, [1, p. 60]:

$$\frac{{\rm d}}{{\rm d}x}e^{{\rm i}kx} = {\rm i}k e^{{\rm i}k... ...d}x}\left(ik e^{{\rm i}kx}\right) = ({\rm i}k)^2 e^{{\rm i}kx}$$

So ${\rm d}^2$$\raisebox{.5pt}{$/$}$​${\rm d}{x}^2$ turns $e^{{{\rm i}}kx}$ into $({{\rm i}}k)^2e^{{{\rm i}}kx}$; the eigen­value is there­fore $({{\rm i}}k)^2$ which equals $\vphantom{0}\raisebox{1.5pt}{$-$}$$k^2$.