Eigenfunction in momentum space

Question

For the wavefunction $\Psi(x, 0)=A e^{-a x^{2}}$, its momentum representation is as follow: \begin{equation} \begin{aligned} \tilde{\psi}(p, 0) &=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} e^{-i(p / \hbar) x} \psi(x, 0) d x \\ &=\left(\frac{1}{2a\pi \hbar^{2}}\right)^{1/4} \cdot e^{\frac{-p^{2}}{4a\hbar^{2}}} \end{aligned} \end{equation} To find the time development of the wavefunction I use the TDSE to yield: \begin{equation} i \hbar \frac{\partial}{\partial t} \tilde{\psi}(p, t)=\frac{\hat{p}^{2}}{2 m} \tilde{\psi}(p, t) \end{equation} \begin{equation} \tilde{\psi}(p , t)=e^{-(i / \hbar) E(p)t} \tilde{\psi}(p , 0)\end{equation} When I plot the real part of $\tilde{\psi}(p , t)$, the magnitude is ofc a sinusoidal function of time. However, in this video (https://www.youtube.com/watch?v=F2Tt80NhmyQ) $\tilde{\psi}(p , t)$ in momentum space should be independent of time. Im not sure why. A mathematical as well as intuitive explanation would be very much appreciated.(The calculated $\left\langle p^{2}\right\rangle=a\hbar^{2}$ also supports the statement that $\tilde{\psi}(p , t)$ is independent of time)

Answer 1

They are plotting the mod of the wavefunction. So the time dependence which is just a phase cancels out.