- #1
Spylock
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- 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.
$$\frac{\partial\Psi(x-ut,t)}{\partial t}=-u\frac{\partial\Psi(x-ut,t)}{\partial(x-ut)}+\frac{\partial\Psi(x-ut,t)}{\partial t}$$
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.
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