What does $m^*>m_e$ imply? (the effective mass of electron is larger than its rest mass) From what I understand, the concept of effective mass is just something people come up with to make electrons and holes obey the equation of motion
$$
\vec{F}=m^* \vec{a}
$$
without dealing with the charge carrier and the crystal at the same time. But how could $m^*$ be compared with $m_e$? They do not seem to be related at first sight. What does $m^*>m_e$ imply?  Could anyone shed some light on this? Thanks!
 A: Second derivative of kinetic energy with respect to momentum equals inverse mass of a particle. In a metal, you have a band structure defined through the dispersion relation of the form E(k) where k is wave vector of electron. Second derivative of this expression can be also taken to be some sort of inertia of a particle, as you can see by analogy with a classical particle whose energy is described by the simple formula for kinetic energy. So, you can think of electron as moving in a cristal potential or as moving with effective mass as a free particle...Why is this mass larger then real mass? Well, I dont see why it has to be that way, derivative can diverge for some value of k, but also can become smaller why not? Simplest form of this inertia tensor is one for parabolic band which becomes constant..
A: Consider two models:


*

*A wave packet of a free electron with $m_e$ with negligible mean energy relative to rest state

*A wave packet with same parameters of an electron in crystal with $m^*$ with negligible mean energy relative to band edge


Assuming that wave packet is large enough for the effective mass approximation to hold (i.e. its uncertainty of position is much larger than Bravais lattice constant, and energy is small enough to consider effective mass constant), we can study how the wave packet will be affected by external potential.
Applying linear potential, we'll get a usual Schroedinger equation for electron in linear potential in both cases - but in the second one it would have effective mass instead of free electron mass. What does it imply? It implies that the wave packet of the crystal electron will accelerate faster than that of free electron, because the equations are the same and masses differ. The group velocities of the electrons are different for the same (quasi)momentum.
So, if you make a race between vacuum electron and electron in crystal, starting with the above described wave packets and equal external potentials, the electron in crystal will arrive at the destination first. Caveat: effective mass must remain constant for this statement to be true, so you're limited in the energy range you can use for such a race.
A: It implies that the band in question would have a narrower bandwidth than would be expected from an electron with free electron mass. In turn, this also means that the electron finds it harder to hop from site to site meaning that the electron is more localized that would be an electron with free electron mass. 
