I’m currently studying quantum mechanics from Introduction to Quantum Mechanics by Griffiths. In his free particle section, he says that the speed of a particle is the coefficient of $t$ over the coefficient of $x$. Shouldn’t it be the coefficient of $x$ over the coefficient of $t$?
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$\begingroup$ Did you square away their dimensions? $\endgroup$ – Cosmas Zachos Oct 23 '20 at 0:02
The following may be a useful approach.
A simple traveling wave can be written as:
$$ y=\sin\left(kx-\omega t \right) $$
We want to follow the position of the wave at a constant phase, $\phi$. Let that phase=0 which leads to:
$$ \phi=kx - \omega t = 0 $$
$$ kx=\omega t $$
$$ x=\frac{\omega}{k} t $$
Then the velocity will be
$$ v=\dot{x}=\frac{\omega}{k} $$
which is the ratio of the coefficient of $t$ over the coefficient of $x$.
I hope this helps.