Consider the system of Example 4.6, now with a time-dependent flux Φ(t) through the solenoid. Show that
\Psi(t)=\frac{1}{\sqrt{2 \pi}} e^{i n \phi} e^{-i f(t)} .
with
f(t)=\frac{1}{\hbar} \int_{0}^{t} \frac{\hbar^{2}}{2 m b^{2}}\left(n-\frac{q \Phi\left(t^{\prime}\right)}{2 \pi \hbar}\right)^{2} d t^{\prime}is a solution to the time-dependent Schrödinger equation.
We simply need to plug into the Schrödinger equation. Taking the time derivative
i \hbar \frac{\partial \Psi}{\partial t}=\hbar f^{\prime}(t) \Psi(t)=\frac{\hbar^{2}}{2 m b^{2}}\left(n-\frac{q \Phi(t)}{2 \pi \hbar}\right)^{2} \Psi .
and then evaluating the right side of Schrödinger’s equation
H \Psi=\frac{1}{2 m}\left[-\frac{\hbar^{2}}{b^{2}} \frac{\partial^{2}}{\partial \phi^{2}}+\left(\frac{q \Phi(t)}{2 \pi b}\right)^{2}+i \frac{\hbar q \Phi(t)}{\pi b^{2}} \frac{\partial}{\partial \phi}\right] \Psi .
=\frac{1}{2 m}\left[\frac{\hbar^{2}}{b^{2}} n^{2}+\left(\frac{q \Phi(t)}{2 \pi b}\right)^{2}-\frac{\hbar q \Phi(t)}{\pi b^{2}} n\right] \Psi=\frac{\hbar^{2}}{2 m b^{2}}\left(n-\frac{q \Phi(t)}{2 \pi \hbar}\right)^{2} \Psi .
we see that these are equal.