As the waves travel with velocity $$v= f \lambda,$$ where $v$ is velocity, $f$ is frequency and $\lambda$ is wavelength. Here we can see that velocity of wave is directly proportional to wavelength. But in case of debroglie wavelength $ \lambda = \frac{h}{p}= \frac{h}{mv}$. Here velocity is inversely related to wavelength. Why we do not use $ v=f \lambda$ for the waves associated with matterphase particles (bosons and fermions)? Are these two relations related to each other in some way?

  • $\begingroup$ About de Broglie relations, what exactly is E ? Its energy of what? $\endgroup$
    – Frobenius
    Nov 27, 2018 at 11:25
  • $\begingroup$ According to de Broglie formulation : a "subluminal" particle with velocity $\;\mathbf{u}\;$ is accompanied by a "superluminal" plane phase wave of velocity $\;\mathbf{w}$. This picture is Lorentz invariant and the product of the speeds is a Lorentz scalar invariant : $\;u\cdot w=c^2$. In the rest frame of the particle ($\mathbf{u}_{0}=\boldsymbol{0}$) wave having $\;w_{0}=\infty, \lambda_{0}=\infty, \nu_{0}=mc^2/h\;$ represents a uniform in space periodic in time phase change. $\endgroup$
    – Frobenius
    Nov 27, 2018 at 11:51

2 Answers 2


The main thing you need to bring in here is the distinction between phase velocity and group velocity. Phase velocity is the velocity at which wavefronts move, group velocity is the velocity at which a change in the disturbance moves, such as a change in amplitude. Mathematically they are

$v_p = \frac{\omega}{k} = f \lambda \\ v_g = \frac{d\omega}{d k} = - \lambda^2 \frac{d f}{d \lambda}$

For de Broglie waves in free space we have

$ E = \hbar \omega = h f \\ p = \hbar k = h / \lambda \\ E^2 - p^2 c^2 = m^2 c^4 $


$v_p = \frac{E}{p} = \frac{\gamma m c^2}{\gamma mv} = \frac{c^2}{v}\\ v_g = \frac{d\omega}{dE} \frac{dE}{dp} \frac{dp}{dk} = \frac{dE}{dp} = v$

where $v = p c^2/E$ is the velocity associated with a classical particle of energy $E$ and momentum $p$. The information contained in a wave propagates at the group velocity, and so does the peak in a wave-packet. All told, it is a wavepacket which can model a particle-like behaviour.

The phase velocity does not have such a direct physical interpretation, and indeed its value depends on where we place the zero of energy. If we place it as I have done, by including the rest energy in $E$, then the phase of a de Broglie wave can be interpreted as an indication of which set of events are simultaneous.


The velocity $v$ is the velocity of the mass as a whole which is not the same as the velocity with which the waves of length $\lambda$ propagate. That velocity is called the phase velocity. So it's an issue of unclearly marked variables.


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