Why the unit of LDOS is different between ATK2008.10 and ATK2011.2?

[jtchzhy]

Dear everyone,
The unit of LDOS in ATK2008.10 is (Angstrom)^(-3)*(eV)^(-1),and in ATK2011.2 (Angstrom)^(-3). Why they are different? Which one is right? Or are they both right?
Thanks a lot.

[ziand]

This is a good question, so I repeat it. (I use ATK 11.2.3)

I calculated the LDDOS and when I plot is with VNL, the unit is given as 1/Ang**3.
When loading the same LDDOS via lddos=nlread('lddos.nc') and doing lddos.evaluate(0*Ang,0*Ang,0*Ang),
the unit comes out as 1/Bohr**1.5. This should not happen, no matter what the real unit of LDDOS is!

Finally I think the unit of LDDOS should be (1/LengthUnit**3)*(1/EnergyUnit).
(So that integrating it e.g. over a unit cell would give the DOS in 1/eV. This in turn could be integrated over all energies and weighted by the Fermi function to get the total number of electrons within that unit cell.)

Or is there a significant difference between LDDOS and the normal LDOS that may explain this?

[Anders Blom]

Something is fishy - we'll look into it. Good thing is the values are correct

Update: There is a clear bug in "nlread", it forgets the eV when reading back in the data. Also, it's odd that VNL shows another unit than what nlread actually gives. But in general, we're not sure we agree with your unit (or ours...), because there is also an intergration over kx and ky. So our current feeling is that the proper unit should be 1/(eV*Å), but we need to triple-check before making a change.

[ziand]

Well, okay, I will look at it too.

If you say, the values are correct, then surely one has to specify the coorect unit too.
I am used to think (and plot my data) in Angström or even nano meter and eV. If some value has a unit of Bohr**-1.5 or Bohr**-1 it does make quite a difference when going to Ang**-1.5 or Ang**-1.

About my unit considerations:
Geee, there is much uncertainty as how the DOS is defined. Some include a volume element, others not. My first statement did not!

Here is what I learned:
I did not look into how the LDOS is actually calculated but started with the formula for the normal DOS of the free electron gas:
DOS(E) = [ (2m)**3/2 / (2pi hbar**3) ] * E**1/2
What comes out is in units of (1/LengthUnit**3)*(1/EnergyUnit).
That means, the such defined DOS is the number of states at a given Energy within a unit energy interval and within a unit volume.
If one does multiply it by a volume (e.g. a crystal) and and integrate (Fermi weighted) over the energy, one should get the number of electrons within that volume.

Now, the other version of DOS, and as it seems the case in ATK too, is DOS(E) = [ V (2m)**3/2 / (2pi hbar**3) ] * E**1/2 .
A volume, e.g. unit cell volume, is already included and the unit is simply 1/eV.


Another thing:
Recently I made a feature request about having LDOS and not only LDDOS.
Prof. Stokbro mentioned that one can do that by a summing Block states.
I did not do it, but a definition of the LDOS is indeed LDOS(E,r) = Sum_i|psi_i(E,r)|**2. The summation is obviously over k points.
Now, which unit does come out of this???

[Anders Blom]

What I meant by the values being correct is that
a) They are correct given the internal atomic units used in ATK. The confusion is just about what that unit is for LDOS, but that's more a notation point than a numerical one.
b) You rarely care about the actual value of the LDOS - what matters is where is it large and where is it small within a structure, i.e. the relative behavior within the device, and in this case neither the unit or any overall normalization (constant factor) actually matters too much.

But of course we should and will get it correct.

The two different version of the DOS you have encountered are basically due to the fact that a crystal contains infinitely many electrons. So demanding that integrating over energy should sum up to the "number of electrons in the system" becomes ill-defined. What we can demand, however, is that it sums up to the number of electrons in the unit cell. This is still a number (no unit), but it can also be seen as a concentration (electrons/volume of unit cell).

About LDOS, there is still a k-dependence not accounted for by your formula...

[peter77]

In solid-state and condensed matter physics, the density of states (DOS) of a system describes the number of states per interval of energy at each energy level that are available to be occupied by electrons. Unlike isolated systems, like atoms or molecules in gas phase, the density distributions are not discrete like a spectral density but continuous. A high DOS at a specific energy level means that there are many states available for occupation. A DOS of zero means that no states can be occupied at that energy level. In general a DOS is an average over the space and time domains occupied by the system. Local variations, most often due to distortions of the original system, are often called local density of states (LDOS). If the DOS of an undisturbed system is zero, the LDOS can locally be non-zero due to the presence of a local potential.

[Anders Blom]

While the above comment is correct, it's copy/pasted from some other online resource, with the sole purpose of posting the signature which essentially constitutes spam. The user will be banned from this Forum, we have zero tolerance for spam.