FAQ¶

Where can I find the total energy in the BAND output?¶

BAND does not calculate the total energy. Instead, it calculates the formation energy with respect to the spherically symmetric spin-restricted atoms. The reason is that this number is easier to determine accurately using numerical integration then the total energy.

What to do with this PEDA ERROR? ERROR DETECTED: Fragments cannot be assigned by a simple translation!¶

The problem is that BAND is (internally) centralizing the atom positions with respect to the geometrical center of the structure. During this centralization an atom can be positioned on one side of the unit cell or the other. (the result for this individual calculation will be the same in both cases, but for a PEDA calculation that is not true) In your case, the centralization of the fragments and of the PEDA calculation gave results which cannot be transformed into one another by a simple translation.

The solution is to pre-centralize the whole system with respect to the geometrical center of the “bigger” fragment. (since this fragment usually gives this problem) E.g. if you study a surface-adsorbate interaction, you centralize the whole structure w.r.t. the geometrical center of the surface. In some cases the adsorbate fragment might be the reason of the problem. If so, we recommend redefining the unit cell so that the adsorbate is closer to the center of the unit cell than to any of the borders of the unit cell.

If this approach does not solve your problem please send us (support@scm.com) your input and output, so we can look into it.

What are the units in the Density Of States (DOS) plot?¶

The dimension of the “density of states per unit cell” (DOS)

\[D(E) = \int_k dk \sum_n \delta(E-\epsilon_n(k))\]

is [1/(energy)].

As the number of states at precisely one energy can be a very wild function an average is plotted (not affecting the unit)

\[D^{\rm smoothed}(E) = \frac{1}{\Delta_E} \int_E^{E+\Delta_E} D(e) \; de\]

The bin-width \(\Delta_E\) is controlled by the input option DOS%DeltaE. To obtain the exact non-smoothed dos, set the Dos%IntegrateDeltaE option to “false”. As stated before: the non-smoothed DOS will be a very wild function, and for instance bands with little dispersion may be completely missed.

Why do I get negative partial Density Of States (partial DOS)?¶

The negative values in the partial DOS are artifacts due to the fact that in a non-orthogonal basis set the definition of a “partial DOS” is somewhat arbitrary. The partial DOS can be a useful tool for understanding the physics/chemistry of your system, but it’s strictly speaking not a “physical quantity”.

Why is the DOS zero in an interval with bands?¶

Quite often there is missing DOS: in an energy interval where there are clearly bands there is no DOS. This is an artifact of using insufficient k-sampling. The simple solution is to redo the whole calculation with a finer k-grid. However, it turns out that the finer grid is only really needed for the DOS, and restarting the DOS with a finer k-grid is usually sufficient to solve such a problem.

Is the absolute value of the Fermi energy physically meaningful?¶

Absolute values of orbital energies are physically meaningful for non-periodic systems, 1D periodic systems and 2D periodic systems. For these cases, E=0 corresponds to the energy of a free electron and the absolute value of the Fermi energy is equal (to a first approximation) to the work function.

In 3D periodic systems the orbital energies (and Fermi energy) are defined up to a constant, ergo the absolute value of energy bands does not have a clear physical meaning (we nonetheless use the convention of setting V_{k=0} = 0, like most other programs).

Why does the band structure of Graphene look wrong?¶

There can be some issues. The first problem can be that a super cell is used. The smallest possible (primitive) cell has only two atoms in the unit cell. Only when using the primitive cell, the crossing of two bands at the fermi energy in the point with symmetry label “K” can be seen.

Even when using the primitive cell, the fermi energy may be off: it is not exactly where the two bands cross at “K”. This can happen when the grid used during the SCF (or during the SCC of DFTB) does not contain the point “K”. The simplest way to solve this is to switch to the symmetric grid and use for instance kInteg=5.

KSpace
    Type Symmetric
    Symmetric
        KInteg 5   # Should be an odd number bigger than 1. This value is good for Graphene
    End
end

You can also achieve this via amsinput. Go to Details -> K-Space integration. Set K-space grid type to “Symmetric” and enter 5 in “Accuracy”.