2.6.6 So­lu­tion herm-f

Ques­tion:

Show that the op­er­a­tor ${\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ is not a Her­mit­ian op­er­a­tor, but ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ is, as­sum­ing that the func­tions on which they act van­ish at the ends of the in­ter­val $a$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $b$ on which they are de­fined. (Less re­stric­tively, it is only re­quired that the func­tions are pe­ri­odic; they must re­turn to the same value at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$ that they had at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$.)

An­swer:

You first need to show that

$$\bigg\langle f\bigg\vert\frac{{\rm d}}{{\rm d}x} g\bigg\rangle$$

is not the same as

$$\bigg\langle\frac{{\rm d}}{{\rm d}x} f\bigg\vert g\bigg\rangle$$

in or­der for ${\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ not to be a Her­mit­ian op­er­a­tor.

By de­f­i­n­i­tion,

$$\bigg\langle f\bigg\vert\frac{{\rm d}}{{\rm d}x} g\bigg\rangle = \int_a^b f^*\frac{{\rm d}g}{{\rm d}x}{ \rm d}x.$$

You can use in­te­gra­tion by parts, [1, p. 64], to move the de­riv­a­tive from $g$ to $f$:

$$\bigg\langle f\bigg\vert\frac{{\rm d}}{{\rm d}x} g\bigg\rang... ... \int_a^b \left(\frac{{\rm d}f}{{\rm d}x}\right)^* g{ \rm d}x$$

(the dif­fer­en­ti­a­tion can be moved in­side the com­plex con­ju­gate since it is a real op­er­a­tion.) Since the func­tions $f$ and $g$ are the same at the end points $a$ and $b$ you have

$$\bigg\langle f\bigg\vert\frac{{\rm d}}{{\rm d}x} g\bigg\rang... ...\bigg\langle\frac{{\rm d}}{{\rm d}x} f\bigg\vert g\bigg\rangle$$

This makes ${\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ a skew-Her­mit­ian op­er­a­tor, rather than a Her­mit­ian one: flip­ping over the op­er­a­tor to the other side changes the sign of the in­ner prod­uct.

To get rid of the change of sign, you can add a fac­tor ${\rm i}$ to the op­er­a­tor, since the ${\rm i}$ adds a com­pen­sat­ing mi­nus sign when you bring it in­side the com­plex con­ju­gate:

$$\bigg\langle f\bigg\vert{\rm i}\frac{{\rm d}}{{\rm d}x} g\bi... ...angle{\rm i}\frac{{\rm d}}{{\rm d}x} f\bigg\vert g\bigg\rangle$$

This makes ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ a Her­mit­ian op­er­a­tor.