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".


Which vacuum gap should I use? How to deal with polarized surfaces and solvation?
=================================================================================

BAND as a linear combination of atomic-orbital (LCAO) based code. As
such it can treat any periodicity. Slabs can be treated as pure 2D
systems. Unlike plane-wave periodic DFT codes, which always use 3D
periodic boundary conditions, you do not need use a slab-gap
approximation. So with BAND you do not need to converge with respect
to vacuum size between repeating slabs. Additional advantages of the
proper 2D representation in BAND are that you can have polarized
surface without the need for a dipole correction, and you can apply
proper electric fields, as opposed to the sawtooth potential needed in
plane-wave codes. Furthermore the polarizable continuum model (COSMO)
works very naturally for surfaces in BAND.
