Author Topic: Hückel convergence in 0 steps  (Read 3269 times)

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Offline ziand

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Hückel convergence in 0 steps
« on: August 31, 2011, 16:11 »
When calculating a perfect infinite CNT, using the Hückel method (defaults + cerda_graphite_basis) I get (here a (6,0) CNT as example):
Code
+------------------------------------------------------------------------------+
| Density Matrix Report                        V_H         DM        DD        |
+------------------------------------------------------------------------------+
|   0   C   [  8.400 ,  6.050 ,  2.842 ]    0.00000 Ha   4.00001   0.00001     |
|   1   C   [  8.085 ,  7.225 ,  3.552 ]    0.00000 Ha   4.00000  -0.00000     |
|   2   C   [  8.400 ,  6.050 ,  1.421 ]    0.00000 Ha   4.00001   0.00001     |
|   3   C   [  7.225 ,  8.085 ,  2.842 ]    0.00000 Ha   4.00000  -0.00000     |
|   4   C   [  6.050 ,  8.400 ,  3.552 ]    0.00000 Ha   4.00001   0.00001     |
|   5   C   [  8.085 ,  7.225 ,  0.710 ]    0.00000 Ha   4.00000  -0.00000     |
|   6   C   [  7.225 ,  8.085 ,  1.421 ]    0.00000 Ha   4.00000  -0.00000     |
|   7   C   [  4.875 ,  8.085 ,  2.842 ]    0.00000 Ha   4.00000  -0.00000     |
|   8   C   [  4.015 ,  7.225 ,  3.552 ]    0.00000 Ha   4.00000  -0.00000     |
|   9   C   [  6.050 ,  8.400 ,  0.710 ]    0.00000 Ha   4.00001   0.00001     |
|  10   C   [  4.875 ,  8.085 ,  1.421 ]    0.00000 Ha   4.00000  -0.00000     |
|  11   C   [  3.700 ,  6.050 ,  2.842 ]    0.00000 Ha   4.00001   0.00001     |
|  12   C   [  4.015 ,  4.875 ,  3.552 ]    0.00000 Ha   4.00000  -0.00000     |
|  13   C   [  4.015 ,  7.225 ,  0.710 ]    0.00000 Ha   4.00000  -0.00000     |
|  14   C   [  3.700 ,  6.050 ,  1.421 ]    0.00000 Ha   4.00001   0.00001     |
|  15   C   [  4.875 ,  4.015 ,  2.842 ]    0.00000 Ha   4.00000  -0.00000     |
|  16   C   [  6.050 ,  3.700 ,  3.552 ]    0.00000 Ha   4.00001   0.00001     |
|  17   C   [  4.015 ,  4.875 ,  0.710 ]    0.00000 Ha   4.00000  -0.00000     |
|  18   C   [  4.875 ,  4.015 ,  1.421 ]    0.00000 Ha   4.00000  -0.00000     |
|  19   C   [  7.225 ,  4.015 ,  2.842 ]    0.00000 Ha   4.00000  -0.00000     |
|  20   C   [  8.085 ,  4.875 ,  3.552 ]    0.00000 Ha   4.00000  -0.00000     |
|  21   C   [  6.050 ,  3.700 ,  0.710 ]    0.00000 Ha   4.00001   0.00001     |
|  22   C   [  7.225 ,  4.015 ,  1.421 ]    0.00000 Ha   4.00000  -0.00000     |
|  23   C   [  8.085 ,  4.875 ,  0.710 ]    0.00000 Ha   4.00000  -0.00000     |
+------------------------------------------------------------------------------+
|   0 E = -51.7466 dE =  8.096459e-01 dM =  5.335291e-06 dH =  0.000000e+00    |
+------------------------------------------------------------------------------+
| Calculation Converged in 0 steps                                             |
|                                                                              |
| Fermi Level  = -5.328540 eV                                                  |
+------------------------------------------------------------------------------+
+------------------------------------------------------------------------------+
|                                                                              |
| Huckel Calculation  [Finished Wed Aug 31 15:58:53 2011]                      |
|                                                                              |
+------------------------------------------------------------------------------+
Could you give me a hint, why it stops at this point? If we look at
Code
|   0 E = -51.7466 dE =  8.096459e-01 dM =  5.335291e-06 dH =  0.000000e+00    |
the energy difference (and also dH) are well above default tolerance. When I change the CNT structure a bit (shifting some atoms around), then there is really some convergence loop visible and dE, dM and dH go down.

Offline Anders Blom

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Re: Hückel convergence in 0 steps
« Reply #1 on: August 31, 2011, 21:40 »
The Huckel method is inherently not really self-consistent, but we have added a step that makes it so. However, if all atoms in the calculation are the same (all carbon, for instance) and sit in the same environment (all bond lengths very similar, for instance), then there is no charge transfer between the atoms and so it converges in one step. This is normal behavior.

Offline ziand

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Re: Hückel convergence in 0 steps
« Reply #2 on: September 1, 2011, 13:36 »
Yes, I know that there should not be much charge transfer and convergence should be fast. I am just asking about the criterion that is used to determine convergence. Is dM the difference in Mulliken populations? Why is only dM below the given tolerance and not dE and dH? Could you please comment on the strategy used to determine convergence? In a past version there was a keyword "criterion". Within the latest manual (version 11.2) the information on tolerance does not give a clear statement about the criterion used.

Another small question: Is there serious reason why the Huckel method does not support to calculate the electrostatic potential or the electron density? Or is it just not implemented yet?

Offline Anders Blom

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Re: Hückel convergence in 0 steps
« Reply #3 on: September 1, 2011, 14:07 »
dM is the change in the density matrix, but convergence is checked on the Hamiltonian, which is dH, and as you see this is 0, thus well below any convergence criterion ;)

The only convergence criterion currently available in ATK is the Hamiltonian (and therefore there is no option or parameter for it), and the definition of the tolerance in this respect is described in the "Note" at the bottom of http://quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.iterationcontrolparameters.html.

I'll revert later on the other things. Some of it can be done (later, as new features), although it's not necessarily meaningful :)

Offline ziand

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Re: Hückel convergence in 0 steps
« Reply #4 on: September 2, 2011, 11:40 »
Oh sorry, that was a mistake from my side. When I inspected dH I saw ...e+00 and overlooked that there is actually nothing else than zero in front. It was the first time this happened as I used DFT in the past, where this is really uncommon...

For the second question, I pretty much expected your answer but wanted to be really sure.

Thank you very much.