Before you begin, recall that an arbitrary state \(\left|\Phi\right\rangle\) can be written in the \(L_z\) eigenbasis as \begin{equation} \left| \Phi\right\rangle \doteq \left( \begin{matrix} \vdots \\ \left\langle {2}\middle|{\Phi}\right\rangle \\ \left\langle {1}\middle|{\Phi}\right\rangle \\ \left\langle {0}\middle|{\Phi}\right\rangle \\ \left\langle {-1}\middle|{\Phi}\right\rangle \\ \left\langle {-2}\middle|{\Phi}\right\rangle \\ \vdots \end{matrix}\right) = \left( \begin{matrix} \vdots \\ a_{2} \\ a_{1} \\ a_{0} \\ a_{-1} \\ a_{-2} \\ \vdots \end{matrix} \right) \end{equation}
For this question, you will carry out calculations on each of the following normalized quantum states on a ring: \begin{equation} \left|{\Phi_a}\right\rangle = \sqrt\frac{ 4}{15}\left|{4}\right\rangle + \sqrt\frac{ 1}{15}\left|{2}\right\rangle +\sqrt\frac{ 4}{15}\left|{1}\right\rangle +\sqrt\frac{ 3}{ 15}\left|{0}\right\rangle +\sqrt\frac{ 1}{15}\left|{-3}\right\rangle +\sqrt\frac{ 2}{15}\left|{-4}\right\rangle \end{equation} \begin{equation} \left| \Phi_b\right\rangle \doteq \left( \begin{matrix} \vdots \\ \sqrt\frac{4}{15} \\ 0 \\ \sqrt\frac{1}{15} \\ \sqrt\frac{4}{15} \\ \sqrt\frac{3}{15} \\ 0 \\ 0 \\ \sqrt\frac{1}{15} \\ \sqrt\frac{2}{15} \\ \vdots \end{matrix}\right) \begin{matrix} \color{\red}{\leftarrow m=0} \end{matrix} \end{equation} \begin{equation} \Phi_c(\phi) = \sqrt{\frac{2}{15\pi}} ~ e^{i4\phi} + \sqrt{\frac{1}{30\pi}} ~ e^{i2\phi} +\sqrt{\frac{2}{15\pi}} ~ e^{i\phi} + \sqrt{\frac{1}{10\pi}} + \sqrt{\frac{1}{30\pi}} ~ e^{-i3\phi} + \sqrt{\frac{1}{15\pi}} ~ e^{-i4\phi} \end{equation}
For each of states, answer the following questions. Perform your calculations in the same representation as the given state and, if are giving any answers by inspection, throughly describe your reasoning.