Consider the operator $\hat{\mathbf{Q}}\mathbf{=}{\mathit{d}}^{\mathbf{2}}\mathbf{/}\mathit{d}{\mathit{\varphi }}^{\mathbf{2}}$, where (as in Example 3.1)$\mathit{\varphi }$ is the azimuthal angle in polar coordinates, and the functions are subject to Equation 3.26. Is $\hat{\mathbf{Q}}$Hermitian? Find its eigenfunctions and eigenvalues. What is the spectrum of $\hat{\mathbf{Q}}$? Is the spectrum degenerate?

For a Hermitian operator (say $Q$), the following condition must be satisfied:

${\int }_a^b{g}^{*}Qfdx={\int }_a^{b}f(Qg{)}^{*}dx$

For a Hermitian operator (say $Q$), the following condition must be satisfied:

${\int }_a^b{g}^{*}Qfdx={\int }_a^{b}f(Qg{)}^{*}dx$

This condition can be written as follows using the more compact bracket notation:

$⟨g\mid \hat{Q}f⟩=⟨\hat{Q}g\mid f⟩$

Consider the operator,

$\hat{Q}=\frac{d^2}{d{\varphi }^2}$

Where, $\varphi$is the azimuthal angle in polar coordinates. We must demonstrate that this operator is Hermitian, as follows:

$$\begin{gathered} ⟨f\mid \hat{Q}g⟩={\int }_0^{2\pi }{f}^{*}\frac{d^{2}g}{d{\varphi }^2}d\varphi \\ ={\left.f^{*}\frac{dg}{d\varphi }\right|}_0^{2\pi }-{\int }_0^{2\pi }\frac{d{f}^{*}}{d\varphi }\frac{dg}{d\varphi }d\varphi \\ ={\left.f^{*}\frac{dg}{d\varphi }\right|}_0^{2\pi }-{\left.\frac{d{f}^{*}}{d\varphi }g\right|}_0^{2\pi }+{\int }_0^{2\pi }\frac{d^2{f}^{*}}{d{\varphi }^2}gd\varphi \end{gathered}$$

Due to the function's periodicity, the values of $f\left(\varphi \right)$and $f\left(\varphi \right)$are the same at 0 and $2\pi$, so:

So, yes $\hat{Q}$is Hermitian.

Now we'll look for the operator's eigenvalues:

$\frac{d^{2}f}{d{\varphi }^2}=q^{2}f$

This problem has two linearly independent solutions:

$f_1=e^{q\varphi }$

$f_2=e^{-q\varphi }$

The periodicity condition requires that:

$f_1\left(0\right)=f_1\left(2\pi \right)$

$1=e^{2\pi q}$

We can deduct from this that q must be imaginary and is limited to the value:

$q=ni$

So, the eigenvalues are:

$q=-n^2,(n=0,1,2,\dots )$

For a given $n$there are two eigenfunctions which are the plus sign or the minus sign, in the exponent Therefore, the spectrum is doubly degenerate. A special case for $n=0$, which is not degenerate.