©2001 by P. Blaha, K. Schwarz and J. Luitz

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Eigenvalues, core states, linearization energies, the Fermi energy,... are given in Rydberg units in wien2k, and are relative to a zero-point that is defined as the average of the potential in the interstitial region. This is an arbitrary choice (one half of this average, or 2.6128 times this average would have been good as well). Therefore, it might happen that for some compounds (solids) your Fermi energy (:FER) is positive, while for others (surfaces or molecules with lots of vacuum) it is negative. This has no physical meaning, and it is also not an indication of a possible error, it is just due to the choice of the zero-point.

Why is there a need to make such a choice? For a free atom, the choice of zero is arbitrary as well, but if one makes the usual convention that the potential at infinity is zero, then negative and positive energies indicate bound and unbound states, respectively. This is intuitively appealing. But for our infinite solids, there is nothing special about infinity: the solid is everywhere, also at infinity. Therefore, another convention has to be made, like the one explained above.

Please note, the TOTAL-Energy does NOT depend on this choice!!!). A constant shift of the potential shifts the all eigenvalues by this constant (gives a total energy contribution of NE * V-const (with NE=number of electrons), but this contribution is cancelled by the total-energy contribution " - integral (rho * V-const)".

For more info about total energy calculations check the references by Wimmer et al. and Weinert (see UG).

Peter Blaha,