Question 3.8: (a) Check that the eigenvalues of the hermitian operator in ...

(a) The eigenvalues (Eq. 3.29) are 0 , ±1, ±2 ,…. which are obviously real. For any two eigenfunctions, f=A_{q} e^{-i q \phi} and g=A_{q^{\prime}} e^{-i q^{\prime} \phi} (Eq. 3.28), we have

e^{-i q 2 \pi}=1 \quad \Rightarrow \quad q=0, \pm 1, \pm 2, \ldots     (3.29).

f(\phi)=A e^{-i q \phi}       (3.28).

\langle f \mid g\rangle=A_{q}^{*} A_{q^{\prime}} \int_{0}^{2 \pi} e^{i q \phi} e^{-i q^{\prime} \phi} d \phi=\left.A_{q}^{*} A_{q^{\prime}} \frac{e^{i\left(q-q^{\prime}\right) \phi}}{i\left(q-q^{\prime}\right)}\right|_{0} ^{2 \pi}=\frac{A_{q}^{*} A_{q^{\prime}}}{i\left(q-q^{\prime}\right)}\left[e^{i\left(q-q^{\prime}\right) 2 \pi}-1\right] .

But q and q′ are integers, so e^{i\left(q-q^{\prime}\right) 2 \pi}=1 , and hence \langle f \mid g\rangle=0 (provided q \neq q^{\prime} , so the denominator is nonzero).

(b) In Problem 3.6 the eigenvalues are q=-n^{2} with n = 0, 1, 2, …., which are obviously real.

For any two eigenfunctions, f=A_{q} e^{\pm i n \phi} and g=A_{q^{\prime}} e^{\pm i n^{\prime} \phi} , we have

\langle f \mid g\rangle=A_{q}^{*} A_{q^{\prime}} \int_{0}^{2 \pi} e^{\mp i n \phi} e^{\pm i n^{\prime} \phi} d \phi=\left.A_{q}^{*} A_{q^{\prime}} \frac{e^{\pm i\left(n^{\prime}-n\right) \phi}}{\pm i\left(n^{\prime}-n\right)}\right|_{0} ^{2 \pi}=\frac{A_{q}^{*} A_{q^{\prime}}}{\pm i\left(n^{\prime}-n\right)}\left[e^{\pm i\left(n^{\prime}-n\right) 2 \pi}-1\right]=0 .

(provided n \neq n^{\prime} ).