[Pw_forum] virtual states in TDDFT (turbo_lanczos.x) calculations
O. Baris Malcioglu
baris.malcioglu at gmail.com
Mon May 28 18:33:00 CEST 2012
Understanding the negative values is quite simple actually.
Imagine two compatible Hilbert spaces. Let's say Cartesian coordinates
and spherical coordinates. Let's be in 2D to make things simple.
Transforming a coordinate like d=135 degrees and r=1 cm (which are
both positive numbers) will give you 0.707107y, -0.707107x
so it is possible to have a transformation that gives you negative
values, although you start with positive values.
Mathematically speaking, as long as you are dealing with spaces that
obey closure/completeness you can choose whatever to represent your
results. The KS states constitute a complete set (assuming you have
infinite/enough virtual orbitals calculated), so you can choose it to
describe your results, but it is very important to remember that the
interpretation should be adapted to new situation.
O.k. This part needs a risky allegory.
Imagine a particle that can move only in a line. The Hilbert space
spanned by the corresponding Hamiltonian will then be on a line as
well, say it consists of +X and -X
When you add a perturbation to the system, say a slight perturbation
in the Y direction, the new Hilbert space is +X,-X,+Y and -Y
Since this is a perturbation, the magnitude of the Y component is
negligible in comparison to magnitude of X component. The result is
almost exactly equivalent to re-adjusting the length of the X in the
previous calculation. So let's assume your Hilbert space remains the
same in both cases.
When you want to represent the perturbed result using +X and -X
Hilbert space, irrespective if you have a constraint in the system
that restricts the particle to move only in +X direction, when you
"project" the +Y or -Y component in your perturbed system to the
vectors spanning the unperturbed Hilbert space, you will see a -X.
So what can be the values that will serve our best interest in our case?
After a lengthy discussion, and I totally agree with it, the most
physically useful value we come up with are the oscillator strengths.
So, those values are to be interpreted as corrections to the
oscillator strengths due to transitions that "look like" mentioned
elements. (In a vaguely similar manner how a movement in Y direction
looks like -X, but since we are talking about a Hilbert space that
obeys closure and can span both Hamiltonians, this time the
resemblance is more "exact")
and I would advise you to use more virtual states, otherwise you will
have too much spilling (i.e. error due to not satisfying closure)
On Fri, May 25, 2012 at 11:19 AM, Giuseppe Mattioli
<giuseppe.mattioli at ism.cnr.it> wrote:
> Dear all
> I've performed a TDDFT calculation of the absorption spectrum of a molecule, and I am playing around
> with the new (QE 5.0) "stage 2" calculations, which should indicate the contribution of virtual
> states to the absorption coefficients, as explained in
> The "stage 2" calculation has been pointed on the first absorption band found by turbo_spectrum.x; my
> molecule has 98 occupied states, and 10 unoccupied have been calculated by pw.x (nband=108) and I
> suppose to be quite satisfied (for all I know of this molecule) with these lines
> 98 1 0.28374685E+04 -0.82400471E+04 2.33588
> 98 2 0.27984762E+04 -0.82731097E+04 2.34525
> 98 3 -0.52066456E-05 0.13245374E-04 0.00000
> 98 4 -0.93634066E-05 0.23337916E-04 0.00000
> 98 5 0.52748267E-05 -0.14618050E-04 0.00000
> 98 6 -0.78431969E+01 0.20124113E+02 -0.00570
> 98 7 -0.78379694E+01 0.20102365E+02 -0.00570
> 98 8 -0.10283141E-04 0.18715947E-04 0.00000
> 98 9 -0.29200741E-04 0.36618310E-04 0.00000
> 98 10 0.38666959E-05 -0.12074233E-04 0.00000
> which seem to indicate that the onset of the absorption spectum can be attributed to single-particle
> transitions involving the highest occupied state and the first and second virtual states (I daresay
> HOMO connected with LUMO and LUMO+1...).
> But what about these lines?
> 92 1 -0.67396365E+03 0.17328962E+04 -0.49124
> 92 2 -0.68404352E+03 0.17306015E+04 -0.49059
> 92 3 0.94419011E-05 -0.60010033E-04 0.00000
> 92 4 -0.18274042E-04 0.26892556E-04 0.00000
> 92 5 -0.51970803E-04 0.11972278E-03 0.00000
> 92 6 -0.38956134E+01 0.48423966E+01 -0.00137
> 92 7 -0.38946693E+01 0.48428589E+01 -0.00137
> 92 8 -0.20625070E-01 -0.59183925E-04 0.00000
> 92 9 -0.17290402E-02 -0.34538054E-04 0.00000
> 92 10 0.59116529E-05 -0.18839515E-04 0.00000
> That is, what do negative (and not so small) coefficients means?
> Thank you in advance
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