91Ó°ÊÓ

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

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

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

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

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

$⟨gâˆ\pounds \hat{Q}fâŸ\copyright =⟨\hat{Q}gâˆ\pounds fâŸ\copyright$

Consider the operator,

$\hat{Q}=\frac{d^2}{d{\mathrm{Ï•}}^2}$

Where, $\mathrm{Ï•}$is the azimuthal angle in polar coordinates. We must demonstrate that this operator is Hermitian, as follows:

$$\begin{gathered} ⟨fâˆ\pounds \hat{Q}gâŸ\copyright ={∫}_0^{2\mathrm{Ï€}}{f}^{*}\frac{d^{2}g}{d{\mathrm{Ï•}}^2}d\mathrm{Ï•} \\ ={\left.f^{*}\frac{dg}{d\mathrm{Ï•}}\right|}_0^{2\mathrm{Ï€}}-{∫}_0^{2\mathrm{Ï€}}\frac{d{f}^{*}}{d\mathrm{Ï•}}\frac{dg}{d\mathrm{Ï•}}d\mathrm{Ï•} \\ ={\left.f^{*}\frac{dg}{d\mathrm{Ï•}}\right|}_0^{2\mathrm{Ï€}}-{\left.\frac{d{f}^{*}}{d\mathrm{Ï•}}g\right|}_0^{2\mathrm{Ï€}}+{∫}_0^{2\mathrm{Ï€}}\frac{d^2{f}^{*}}{d{\mathrm{Ï•}}^2}gd\mathrm{Ï•} \end{gathered}$$

Due to the function's periodicity, the values of $f\left(\mathrm{Ï•}\right)$and $f\left(\mathrm{Ï•}\right)$are the same at 0 and $2\mathrm{Ï€}$, so:

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

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

$\frac{d^{2}f}{d{\mathrm{Ï•}}^2}=q^{2}f$

This problem has two linearly independent solutions:

$f_1=e^{q\mathrm{Ï•}}$

$f_2=e^{-q\mathrm{Ï•}}$

The periodicity condition requires that:

$f_1\left(0\right)=f_1\left(2\mathrm{Ï€}\right)$

$1=e^{2\mathrm{Ï€}q}$

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

$q=ni$

So, the eigenvalues are:

$q=-n^2,(n=0,1,2,…)$

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.