Traveling Wavefunction: Differentiating & Relating $\Psi$

[Spylock]

Homework Statement

Hi, I want to check how to show $\Phi(x,t)$ is a solution to TDSE given that $\Psi(x,t)$ is a solution to TDSE. Note that there is no potential.

Relevant Equations

$$\Phi(x,t)=\Psi(x-ut,t)e^{i(k(x-\omega t)}$$

I am not sure whether I have differentiated correctly
$$\frac{\partial\Psi(x-ut,t)}{\partial t}=-u\frac{\partial\Psi(x-ut,t)}{\partial(x-ut)}+\frac{\partial\Psi(x-ut,t)}{\partial t}$$

and
$$\frac{\partial^2\Psi(x-ut,t)}{\partial x^2}=\frac{\partial^2\Psi(x-ut,t)}{\partial(x-ut)^2}$$

So, we have
$$\frac{\partial\Phi(x,t)}{\partial t}=\frac{\partial\Psi(x-ut,t)}{\partial t}e^{i(kx-\omega t)}-i\omega\Psi(x-ut,t)e^{i(kx-\omega t)}$$
$$\frac{\partial^2\Phi(x,t)}{\partial x^2}=\frac{\partial^2\Psi(x-ut,t)}{\partial x^2}e^{i(kx-\omega t)}-k^2\Psi(x-ut,t)e^{i(kx-\omega t)}+2ik\Psi(x-ut,t)e^{i(kx-\omega t}$$

Also, how do I relate \Psi(x,t) with \Psi(x-ut,t) given that
$$i\hbar\frac{\partial\Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}$$

Thank you for advice.

 

[Charles Link]

Your equation needs to read: $\frac{d \Psi(y,t)}{dt}=(\frac{\partial{\Psi(y,t)}}{\partial{y}})_t(\frac{\partial{y}}{\partial{t}})_x+(\frac{\partial{\Psi(y,t)}}{\partial{t}})_y$, where $y=x-ut$.
Also, in the first line, your exponent needs to read $kx-\omega t$. (You have a parenthesis in the wrong place). $u=\frac{\omega}{k}$.
With the corrections I gave you above, you should be able to finish it up.

 

[Charles Link]

Also, be careful when taking $(\frac{\partial{\Phi(x,t)}}{\partial{t}})_x$. It basically means taking $\frac{d}{dt}$ of $\Psi(x-ut)e^{i(kx-\omega t)}$, but treating $x$ as a constant.