# Group velocity in solid state physics force equation

When reading about electron holes here https://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics), the group velocity is introduced as the reciprocal space gradient of the dispersion relation.

The group velocity is then used in the expression $$F=mdv/dt$$. This makes no sense to me, shouldn't the group velocity be in position space for this to work? The left hand side of the equation seems to be in position space since it is quoted as being one of Newton's laws. Is the article I am reading hiding away some isomorphism or Fourier transform to save some digital ink? Could someone please write it in position space for me? Perhaps the time derivative commutes with the Fourier transform?

The group velocity of a wave packet with dispersion relation $$\omega=\omega(k)$$ is $$v_g = \frac{d\omega}{dk}$$. For a quasi-free particle moving through a solid, $$\omega = E/\hbar$$ and $$k=p/\hbar$$, so $$v_g = \frac{dE}{dp}$$.

It's not really clear what you mean about the group velocity being in position space. In the classical physics of a free particle, $$E= \frac{p^2}{2m}$$, and $$\frac{dE}{dp} = \frac{p}{m}=v$$. This is just a semi-classical extension of that idea to a quantum mechanical system.

• I reworked the question, to give the context which confuses me. – Emil Aug 28 '20 at 20:19
• @Emil There is no position space vs momentum space here. $E$ is not a wavefunction. The dispersion relation gives $E$ as a function of $p$, and $\frac{dE}{dp}$ tells you how quickly a wave packet with momentum $p$ will move through space. – J. Murray Aug 28 '20 at 20:26
• This does not clarify to me how they are using the group velocity expanded in the reciprocal basis in a vector field equation for position space. – Emil Aug 28 '20 at 20:33
• @Emil What do you mean by the group velocity expanded in the reciprocal basis? $E$ is just a function which depends on $p$, and you obtain $v_g$ by differentiating it. Using the result in the equation $F=m\frac{dv}{dt}$ gives a semiclassical equation of motion which can provide some intuition for how electrons in crystals respond to external forces. – J. Murray Aug 28 '20 at 21:32
• The article posits as I read it $F(r,t) = [F_i dx^i](r,t)= d_t \partial_{k_i}\omega(k) dk^i$, there is a mismatch in the vector basis, and I don't even know what manifold $dk^i$ lives in, but I doubt its points are $(r,t)$. (My attempt att expanding the gradient in the reciprocal space) – Emil Aug 28 '20 at 22:26