# Why is the speed of a quantum particle defined as coefficient of $t$ over coefficient of $x$?

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 $$t$$?

• Did you square away their dimensions? – 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.