I'm trying to solve the following:
Suppose $\psi(x,t)$ is a solution to the time-dependent Schrodinger Equation with zero potential; show that $$ f(x,t)=\psi(x-ut,t)e^{ikx}e^{-i \omega t } $$ is also a solution, for $k$ and $\omega$ chosen in terms of $u$. The problem arises when I try to calculate $\frac{\partial}{\partial t} \psi(x-ut,t)$ in terms of $\psi_{t}(x-ut,t)$ and $\psi_{x}(x-ut,t)$. I've tried using the multivariate chain rule but keep going in circles. What's the right approach here?
A suggestion:
Now first consider time independent schrödinger equation.(Omitting all physical constants) $ - \frac{\partial^2} {\partial x^2} \psi = E \psi $ Now set $E=k^2$ now solutions are of the form $e^{-ikx} $ and multiplyied with $e^{iE(k) t} $ it is a solution to time dependendent equation can be varified by pluging in.
A general solution $ \psi(x, t) =\int_{-\infty} ^{\infty} \tilde{\psi} (k) e^{-ikx}e^{iE(k) t} dk $. Assume $ \psi $ is of this form then show $f$ is of this form.
I assume hear all solution of equation must be of "the form" and I am not sure if it is true.