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

Physics Asked by Teethandfur YT on November 26, 2020

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 $t$?

One Answer

The following may be a useful approach.

A simple traveling wave can be written as:

$$ y=sinleft(kx-omega t right) $$

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.

Answered by ad2004 on November 26, 2020

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