2.6.9 So­lu­tion herm-i

Ques­tion:

A com­plete set of or­tho­nor­mal eigen­func­tions of the op­er­a­tor ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ that are pe­ri­odic on the in­ter­val 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$ are the in­fi­nite set of func­tions

$$\ldots , \frac{e^{-3{\rm i}x}}{\sqrt{2\pi}}, \frac{e^{-2{\rm... ... i}x}}{\sqrt{2\pi}}, \frac{e^{3{\rm i}x}}{\sqrt{2\pi}}, \ldots$$

Check that these func­tions are in­deed pe­ri­odic, or­tho­nor­mal, and that they are eigen­func­tions of ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ with the real eigen­val­ues

$$\ldots , 3, 2, 1, 0 , -1, -2, -3, \ldots$$

Com­plete­ness is a much more dif­fi­cult thing to prove, but they are. The com­plete­ness proof in the notes cov­ers this case.

An­swer:

Any eigen­func­tion of the above list can be writ­ten in the generic form $e^{k{{\rm i}}x}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ where $k$ is a whole num­ber, in other words where $k$ is an in­te­ger, one of ..., $\vphantom{0}\raisebox{1.5pt}{$-$}$3, $\vphantom{0}\raisebox{1.5pt}{$-$}$2, $\vphantom{0}\raisebox{1.5pt}{$-$}$1, 0, 1, 2, 3, ... If you show that the stated prop­er­ties are true for this generic form, it means that they are true for every eigen­func­tion.

Now pe­ri­od­ic­ity re­quires that $e^{k{\rm i}2\pi}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^0$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$, and the Euler for­mula ver­i­fies this: sines and cosines are the same if the an­gle changes by a whole mul­ti­ple of $2\pi$. (For ex­am­ple, $2\pi$, $4\pi$, $\vphantom{0}\raisebox{1.5pt}{$-$}$$2\pi$, etcetera are phys­i­cally all equiv­a­lent to a zero an­gle.)

The de­riv­a­tive of $e^{k{{\rm i}}x}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ with re­spect to $x$ is $k{{\rm i}}e^{k{\rm i}{x}}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$, and mul­ti­ply­ing by ${\rm i}$ you get $\vphantom{0}\raisebox{1.5pt}{$-$}$$ke^{k{\rm i}{x}}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$, so $e^{k{{\rm i}}x}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ is an eigen­func­tion of ${\rm i}{\rm d}$$\raisebox{.5pt}{$/$}$​${\rm d}{x}$ with eigen­value $\vphantom{0}\raisebox{1.5pt}{$-$}$$k$.

To see that $e^{k{{\rm i}}x}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ is nor­mal­ized, check that its norm is unity:

$$\left\vert\left\vert\frac{e^{k{\rm i}x}}{\sqrt{2\pi}}\right\... ...\rm d}x} = \sqrt{\int_0^{2\pi} \frac{1}{2\pi} { \rm d}x} = 1.$$

To ver­ify that $e^{k{\rm i}{x}}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ is or­thog­o­nal to every other eigen­func­tion, take the generic other eigen­func­tion to be $e^{l{\rm i}{x}}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ with $l$ an in­te­ger dif­fer­ent from $k$. You must then show that the in­ner prod­uct of these two eigen­func­tions is zero. Since the nor­mal­iza­tion con­stants do not make any dif­fer­ence here, you can just show that $\big\langle{e}^{k{\rm i}{x}}\big\vert e^{l{\rm i}{x}}\big\rangle$ is zero. You get

$$\Big\langle e^{k{\rm i}x} \Big\vert e^{l{\rm i}x} \Big\rangl... ...rac{1}{(l-k){\rm i}} e^{(l-k){\rm i}x} \Big\vert _0^{2\pi} = 0$$

since $e^{(l-k){\rm i}2\pi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. So dif­fer­ent eigen­func­tions are or­thog­o­nal, their in­ner prod­uct is zero.