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If a free particle of mass $m$ is moving with a velocity $v$, then it's kinetic energy is $\frac{mv^2}{2}$, therefore its frequency is $\nu = \frac{E}{h} = \frac{mv^2}{2h}$ where $h$ is Planck's constant, and it's wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$, then the velocity of the wave associated with this particle is $V = \nu \lambda = \frac{v}{2}$, hence we see that the Velocity of the wave $\neq$ Velocity of the particle, but shouldn't they be equal?

Where is the mistake here? We've just started learning Quantum Mechanics in this semester so please help me understand this. Thanks a lot!

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The velocity you're talking about is the phase velocity $v_p$

$v_p$ is defined as $v_p = \frac{\omega}{k}$ or more commonly as $v_p=\lambda \nu$

The problem with the phase velocity is that it is only defined for purely sinusoïdal waves.

When we deal with wavepackets (made of a continuous sum of purely sinusoïdal waves) we use instead the group velocity $v_g$ defined as $v_g = \frac{\partial \omega}{\partial k}_{|k_0}$ (the derivation can be found here, you will probably see it in class with your teacher).

There is a special case however, where the phase velocity can be pertinent for a wavepacket: it is when the medium is non-dispersive. In that case, the dispersion relation $\omega = f(k)$ is linear (note that for $k=0$, $\omega$ should always be $0$, so it can't be affine) and the phase velocity is equal to the group velocity ($v_p= v_g$)

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    $\begingroup$ This is routine in macroscopic physics. See youtube.com/watch?v=lWi_KpBy8kU. Watch the waves move faster than the group. The group corresponds to the "particle". $\endgroup$
    – John Doty
    Sep 15, 2022 at 16:43

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