I would not do it with scf calculations, but using the "force theorem".

Choose a symmetry, in which both magnetization directions can be described 
(maybe even
P1).
runsp -ec 0.000001     with VERY good k-mesh (SO anisotropy is VERY sensitive 
to k-mesh)

M in (001) in case.inso
x lapwso -up
x lapw2 -up -so
x lapw2 -dn -so
add the two "sum of eigenvalues" energies at the bottom of scf2up/dn

M in (100) direction; and repeat the above steps.
The difference of these energies gives you the MAE.

Papers: I don't know without checking, but there should be several papers by 
Igor Mazin


Am 24.09.2013 10:53, schrieb Madhav Ghimire:
Dear Prof. Blaha and wien users,

I tried to calculate the magnetic anisotropy energy (MAE) of double-perovskites 
A2BB'O6 with magnetic ions in A (lanthanides) and B' (transition metals) sites.
Following the definition and methods given in Phys. Rev. B 65 134422, i obtain 
the MAE using
E=E(001)-E(100)
which is un-expectedly large (~ 1.0 eV)
I would be glad to get your opinion:
(i) In Wien2k, Can we calculate MAE as the total energy-difference between 
E(easy axis magnetization) and E(hard axis) or need additional term to include
(ii) Any relevant paper discussing about MAE based on Wien2k calculations

Thanks

Madhav Ghimire

--
MANA, Nano-System Theoretical Physics Unit
NIMS, Tsukuba, Japan


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Peter Blaha
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email: pbl...@theochem.tuwien.ac.at
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