(a) Check that the eigenvalues of the hermitian operator in Example 3.1 are real. Show that the eigenfunctions (for distinct eigenvalues) are orthogonal.

(b) Do the same for the operator in Problem 3.6.

The eigenfunctions are given by:
$\mathit{f}\mathbf{\left(}\mathit{\varphi }\mathbf{\right)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{iq\varphi }}$

a)

The eigenfunctions are given by:

$f\left(\varphi \right)=A{e}^{-iq\varphi }$

And the eigenvalues of the operator are:

$q=0,\pm 1,\pm 2,\pm 3,\dots$

Therefore, the spectra of the given operator is real. It is important to notice that they are real due to restrictions that we required on the functions.

The inner product between eigenfunction $f\left(\varphi \right)=A_f{e}^{-in\varphi }$corresponding to the eigenvalue $n$and $g\left(\varphi \right)=A_g{e}^{-im\varphi }$corresponding to the eigenvalue $\mathrm{m}$.

$$\begin{gathered} ⟨f\mid g⟩={\int }_0^{2\pi }{f}^{*}\left(\varphi \right)g\left(\varphi \right)d\varphi \\ =A_f^{*}{A}_g{\int }_0^{2\pi }{e}^{in\varphi }{e}^{-im\varphi }d\varphi \\ =A_f^{*}{A}_g{\int }_0^{2\pi }{e}^{i\varphi (n-m)}d\varphi \\ ={\left.A_f^{*}{A}_g\frac{e^{i\varphi (n-m)}}{i(n-m)}\right|}_0^{2\pi } \\ =0 \end{gathered}$$

The latter vanishes because the lower limit gives 1 and so does the upper one since $n$and $m$are integers, therefore their difference $(n-m)\equiv k$is an integer and $e^{2\pi ik}$is equal to 1 if $k\in \mathrm{\mathbb{Z}}$. We have proved that the eigenfunctions corresponding to different eigenvalues are orthogonal.

b)

The eigenfunctions of the given operator,

We start by assuming that the function $z\left(\varphi \right)$is an eigenfunction of the given operator with a eigenvalue $-{\lambda }^2$.

$\begin{array}{rcl}\frac{d^2}{d{\varphi }^2}z\left(\varphi \right)& =& -{\lambda }^{2}z\left(\varphi \right)\\ & \Rightarrow & z^{\text{'}\text{'}}\left(\varphi \right)+{\lambda }^{2}z\left(\varphi \right)=0\\ & & \end{array}$

This is a simple linear differential equation and the solution to the problem is given by:

$f\left(\varphi \right)~e^{\pm i\lambda \varphi }$

The plus-minus sign corresponding to the two different linear independent solutions. Equivalent choice for solutions would be independent functions $\sin \left(\lambda \varphi \right)$and $\cos \left(\lambda \varphi \right)$. We have chosen our eigenvalue to be explicitly negative, because if it were positive, the solutions would be $f\left(\varphi \right)~e^{\pm \lambda \varphi }$ and we would not be able to satisfy the condition that $f(\varphi +2\pi )=f\left(\varphi \right)$. Also, generally, the option $\lambda =0$ would be possible as well, but then the eigenfunction would be a linear function which as well cannot satisfy $f(\varphi +2\pi )=f\left(\varphi \right)$. Therefore, we have chosen our eigenvalue to be negative as it is the only option that can satisfy the latter. By requiring the latter condition, we obtain:

$$\begin{gathered} e^{\pm i\lambda (\varphi +2\pi )}=e^{\pm i\lambda \varphi } \\ \Rightarrow e^{2\pi \lambda }=1 \\ \Rightarrow \lambda \in \mathrm{\mathbb{N}} \end{gathered}$$

We have not allowed our values of$\lambda$to be negative because then it would simply overlap with the second linear independent solution. Since the eigenvalues are natural numbers, they are trivially real. To check whether they are mutually orthogonal for different values of $\lambda$, we do same as before, assuming some constants for two different eigenfunctions with two different eigenvalues:

$\begin{array}{rcl}⟨f& \mid & g⟩={\int }_0^{2\pi }{f}^{*}\left(\varphi \right)g\left(\varphi \right)d\varphi \\ & =& A_f^{*}{A}_g{\int }_0^{2\pi }{e}^{\mp i\lambda \varphi }{e}^{\pm i\overline{\lambda }\varphi }d\varphi \\ & =& 0\end{array}$

It should be noted that there is one key difference here. For the same value of $\lambda$there are two different eigenfunctions corresponding to the same eigenvalue. Therefore, we conclude that the eigenfunctions corresponding to different eigenvalues, ergo the $\pm$sign. They are not mutually orthogonal, but any two eigenfunctions corresponding to different eigenvalues are mutually orthogonal.