This is usually not true, except when EMAX is set to 100 Ry or so.
We use for lapwso as basisset the lapw1 eigenstates, thus the dimension
of the lapwso eigenvalue problem is only 2 * NE, where NE is typically
(depending on EMAX) 20-30% of NMAT of lapw1.
Am 11.06.2023 um 16:33 schrieb
Let me make a different comment. If you have 2hr for lapw1, 9 hr for
lapwso, it will be almost impossible to get results in less than weeks.
Therefore I suggest looking more carefully:
1) Do you need all the kpoints?
2) Are your RMTs too small?
3) Is your RKMAX too large?
Think.
Then add -so
Dear Lukasz,
The difference in computation time between lapw1 and lapwso
calculations is expected in band calculations. The lapwso step
involves the calculation of spin-orbit coupling, which can be
computationally more demanding compared to the lapw1 step that
calculates the bands without
The matrix that lapw1 -up solves is the spin up part of the Hamiltonian and
it should be much smaller than the matrix that lapwso solves.
On Sunday, June 11, 2023, Peter Blaha wrote:
> The speed of lapwso depends on EMAX in case.in1, which limits the number
> of eigenvalues calculated in lapw1
The speed of lapwso depends on EMAX in case.in1, which limits the number
of eigenvalues calculated in lapw1 and used as basis for lapwso.
With EMAX=5.0 the speed of lapw1 and lapwso is usually similar.
With larger emax lapwso may take much more time.
Am 11.06.2023 um 12:36 schrieb pluto via
Dear All,
When calculating bands for a large slab I have following sequence:
Sun May 14 12:33:03 PM CEST 2023> (x) lapw1 -band -up -p
Sun May 14 02:25:26 PM CEST 2023> (x) lapw1 -band -dn -p
Sun May 14 04:17:22 PM CEST 2023> (x) lapwso -up -p
Mon May 15 01:30:05 AM CEST 2023> (x) qtl -up -p
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