how to find the effective mass of a hole

how can we find out the effective mass of a hole,since a hole in the valence band is just an absence of electron?

-
Conceptually, a hole is a virtual object. That virtual object behaves like a real particle, occupying energy levels, "moving" in the lattice, etc. Mathematically, then, since mass affects these properties, we can assign an effective mass that explains the observed numerical quantities associated with those properties. –  Mitchell Apr 28 '11 at 16:29
Are you interested in how we measure the mass experimentally or calculate it theoretically? Experimentally, you can measure the effective mass by, e.g., seeing how the energy and momentum of a system containing a hole are related to each other, and applying $E=p^2/(2m)$. –  Ted Bunn Apr 28 '11 at 18:54

Actually the concept of hole is an idealization because is easier to describe how electrons behave in a band describing the behavior of the holes rather than that of all the electrons that compose it. So the hole is our own definition and is defined as the absence of an electron and all its properties are obtained according to this definition. in particular we find that

$( \frac{1}{m})_{ij}^{(h)}=-( \frac{1}{m})_{ij}^{(e)}$

where $m_{ij}$ is the effective mass tensor.

-
This is correct, but it should be emphasized that for a filled band the electronic effective mass is negative, so that it is the hole mass which is positive. –  Ron Maimon Oct 14 '11 at 0:55

To understand hole dynamics, consider an idealized model of a band-gap, where the electronic energy at wavenumber k has two branches, one with energy rising quadratically from a minimum value,

$$E_1(k) = A + B k^2$$

and one with energy falling quadratically from a maximum value

$$E_2(k) = - A - Bk^2$$

the energy gap for the insulator is 2A, while the effective mass of a small number of added electrons in the upper band is given by the usual Schrodinger formula:

$$B = {1\over 2m}$$

The Schrodinger equation describes any excitation quadratically rising energy/wavenumber dependence.

The electrons are Fermions, and you assume that the lower band is completely filled. When you add a few electrons (doping with donor atoms), the new electrons make a free Schrodinger Fermi gas, which has a usual Fermi-surface and conductivity. The effective mass is as above.

When you remove electrons, however, the electron excitations near the Fermi surface have a negative value for B, so their dynamics is opposite intuition. If you apply a force to them, they will accelerate away from the force. This property is best treated by reversing the notion of creation and annihilation operator in the Schrodinger field and calling the positively charged annihilation operators creation operators for holes. This reverses the sign of the kinetic term, and gives the holes a positive mass. The description of the semiconductor in the acceptor atom regime is of a Fermi sea of holes, with the usual Fermi sea conductivity, except for positively charged positive mass carriers.

You can equivalently think of this as a reverse-Fermi surface for negative mass electrons. The physics is identical (it only differs by what operator you call annihilation and what you call creation). The negative carriers have a Hall voltage which is opposite the usual case because they have a negative mass.

The aversion to negative mass Schrodinger equation means that the entire literature works exclusively in terms of positive mass effective holes. I think this is a bit unfortunate, because there are situations where a negative mass hole description is the best one available, such as the description of Moseley's law for heavy atoms.

-