# Shouldn't the velocity of the wave associated with a particle be equal to the velocity of the particle?

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!

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 $$v_p=\lambda \nu$$
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 ($$v_p= v_g$$)