The following are 3 different representations for the \(\textbf{same}\) state on a quantum ring for \(r_0=1\) \begin{equation} \left|{\Phi_a}\right\rangle = i\sqrt{\frac{ 2}{12}}\left|{3}\right\rangle - \sqrt{\frac{ 1}{12}}\left|{1}\right\rangle +\sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i\sqrt{\frac{ 2}{ 12}}\left|{-1}\right\rangle +\sqrt{\frac{ 4}{12}}\left|{-3}\right\rangle \end{equation} \begin{equation} \left| \Phi_b\right\rangle \doteq \left( \begin{matrix} \vdots \\ i\sqrt{\frac{ 2}{12}}\\ 0 \\ -\sqrt{\frac{ 1}{12}} \\ \sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}} \\ -i\sqrt{\frac{ 2}{12}}\\ 0 \\ \sqrt{\frac{4}{12} }\\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0 \end{matrix} \end{equation} \begin{equation} \Phi_c(\phi) \doteq \sqrt{\frac{1}{24 \pi}} \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) \end{equation}
If you measured the energy of the state to be \(\frac{9}{2}\frac{\hbar^2}{I}\), what would the state of the particle be immediately after the measurement is made?