0
$\begingroup$

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)

$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.