With the Davidson algorithm as implemented in VASP, is there formally a variational principle operating? Does it apply before the algorithm reaches self consistency?
I asked this in a previous thread but I think got no reply.
Variational principle
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tjf
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Variational principle
Last edited by tjf on Thu Jan 18, 2007 12:29 pm, edited 1 time in total.
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admin
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Variational principle
The Davidson algorithm is for diagonalizing a matrix. For a fixed
Hamiltonian, it is thus an algorithm that determines the Minimum of
∑i elm occupied --> min
Of course if the algorithm (which one) has not reached selfconsistency, the variational principle is not fullfilled.
Hamiltonian, it is thus an algorithm that determines the Minimum of
∑i elm occupied --> min
Of course if the algorithm (which one) has not reached selfconsistency, the variational principle is not fullfilled.
Last edited by admin on Tue Jan 30, 2007 4:19 pm, edited 1 time in total.
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stephan
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Variational principle
Hi,
I'm new to the VASP world (and plane wave world in general) and the question might be stupid, but all along when using localized basis sets, the variational principle was always satisfied when applying DFT functionals, i.e., the energy for a given density, a given functional and basis set was always higher than the energy for the self-consistent density.
Now, in VASP this does not seem to be the case: at least when using
ICHARG=12
i.e. the superposition of atomic densities, the energy is, with -180.47033186 eV,well below the SCF energy (-152.72266618 eV).
Is the energy of VASP not guaranteed to be the energy of a Slater determinant, i.e., the wave function is not necessarily antisymmetric?
Would it be possible to recover the behavior I expect by setting some "special options"?
Thanks a lot in advance
I'm new to the VASP world (and plane wave world in general) and the question might be stupid, but all along when using localized basis sets, the variational principle was always satisfied when applying DFT functionals, i.e., the energy for a given density, a given functional and basis set was always higher than the energy for the self-consistent density.
Now, in VASP this does not seem to be the case: at least when using
ICHARG=12
i.e. the superposition of atomic densities, the energy is, with -180.47033186 eV,well below the SCF energy (-152.72266618 eV).
Is the energy of VASP not guaranteed to be the energy of a Slater determinant, i.e., the wave function is not necessarily antisymmetric?
Would it be possible to recover the behavior I expect by setting some "special options"?
Thanks a lot in advance
Last edited by stephan on Tue Dec 10, 2013 7:57 pm, edited 1 time in total.
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stephan
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Variational principle
Hi,
I'd just like to relaunch the question, as I have still not found the answer; from what I can see, the main difference between SCF and the Harris-functional energy (or rather the energy of the superposition of atomic densities), comes form EBANDS.
Speaking of which: indeed, the Harris-Foulkes functional is, as far as I understand, not variational. However
Non-selfconsistent calculations for a superposition of atomic charge densities. This is in the spirit of the non-selfconsistent Harris-Foulkes functional
Is not very clear: how are the non-self-consistent energies computed exactly in VASP? - Are they supposed to be variational or not? Are there some special switches that let the user choose between one or the other approach?
Thanks a lot in advance!
I'd just like to relaunch the question, as I have still not found the answer; from what I can see, the main difference between SCF and the Harris-functional energy (or rather the energy of the superposition of atomic densities), comes form EBANDS.
Speaking of which: indeed, the Harris-Foulkes functional is, as far as I understand, not variational. However
Non-selfconsistent calculations for a superposition of atomic charge densities. This is in the spirit of the non-selfconsistent Harris-Foulkes functional
Is not very clear: how are the non-self-consistent energies computed exactly in VASP? - Are they supposed to be variational or not? Are there some special switches that let the user choose between one or the other approach?
Thanks a lot in advance!
Last edited by stephan on Tue Jan 07, 2014 3:06 pm, edited 1 time in total.