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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
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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.

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