De Broglie wavelength, frequency and velocity - interpretation

Two fundamental equations regarding wave-particle duality are: $$\lambda = \frac{h}{p}, \\ \nu = E/h .$$

We talk about de Broglie wavelength, is it meaningful to talk about de Broglie frequency ($\nu$ above) and de Broglie velocity ($\nu \lambda$)? Are these two equations independent or can one derive one from the other? Or mid-way, does one impose constraint on other? In case of light or photons we can relate frequency and wave, is there similar interpretation in case frequency and wavelengths in above equations? Comparing with that of light, if we multiple $\lambda$ and $\nu$ we get velocity, what does this velocity mean here?

If we do the above calculation for an average human, what would be the meaning of $\nu$ and $\nu \lambda$? Are we jiggling with that $\nu$?

-

Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with $$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$ which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.

This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity $$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$ where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.

Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been $v_{g}\leq c$, with $c$ the speed of light.

For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...

For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !

Yes, the formulae $p=h/\lambda$ and $E=h\nu$ (the same equations as yours, reverted a bit) are universal – they hold not only for photons but for any particle.
Also, these two equations aren't quite independent. Assuming special relativity, they both follow from the de Broglie form of the wave function which is pure phase: $$\psi(x,t) = C\cdot \exp(2\pi i (x/\lambda - t\nu))$$ and similarly, in any spacetime dimension, $$\psi\sim \exp(ix^\mu p_\mu / \hbar)$$ where $x^\mu$ has components $(t,x,y,z)$ with some conventional powers of $c$ and $p_\mu/\hbar$ has components $2\pi\nu=\omega$ and $\vec k$ where $|\vec k|=2\pi/\lambda$ and $\vec k$ is the wave number vector with the direction of the wave. Note that $\hbar = h/2\pi$ is the naturally reduced Planck constant.
In non-relativistic quantum mechanics, we often use the non-relativistic notion of energy which is $E - m_0 c^2$: we subtract the huge latent energy. This overall shift of energy by a constant makes no difference for physics and it's equivalent to redefining the phase of $\psi$ by $\psi\to \psi \exp(imc^2 t / \hbar)$. In this non-relativistic convention for the phase, we lose the manifest Lorentz symmetry between the momentum and energy but the symmetry is still there and may be restored if we return $mc^2$ to $E$.