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How do we calculate the effective mass of an electron in an Aluminium lattice? Is there any simple analytical way to work it out?

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I think the answer is "no", almost without qualification... –  genneth Mar 17 '11 at 23:18
@genneth please correct me if I'm wrong, but isn't the inverse effective mass given by an expression of the form $m^{-1} = \partial^2 E(k)/\partial k_a \partial k_b $, where $E(k)$ is the dispersion relation, $k_a, k_b$ are directions in momentum space, and for simplicity we assume that $m^{-1}_{ab} := m^{-1}$? So given the energy spectrum of Aluminum one can calculate the effective mass for any excitation of momentum $k$ in its brillouin zone? –  user346 Mar 18 '11 at 0:33
The last sentence is meant as a statement, not a question. –  user346 Mar 18 '11 at 1:05
That gives the band mass, but actual electron masses (the things one measures via heat capacity, for example) also get renormalisations from the Coulomb interaction. The 2nd bit is where the calculation breaks down. However, I don't remember which dominates in Aluminium --- it's a bit of an odd material electronically speaking. In any case, even a band-structure calculation is not "simple analytic" --- I have a whole group of people around me who spend their lives working out better/faster ways to do it with massive supercomputers :-) –  genneth Mar 18 '11 at 11:49

1 Answer 1

The effective mass can be determined from a exactly known band structure. As genneth already mentioned there is no simple analytical way to get a from first principles to the Aluminium band structure.

In practice there are a number of very successful numerical approaches to calculate band structures, e.g. Hartree-Fock, LDA+U, DMFT, Tight binding and the list goes on depending on the exact material and properties you are interested in.

If you use a number of material parameters as the electron density, Fermi energy or others you can express the effective mass in those terms but that is more a reformulation than a way to calculate it.

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