If I understand you rigth, you want to simulate a NMR spectrum with an
external field (and in addition there is a internal field transfered
from the Co moments.
I think what you should do is using x lapwdm with a proper input for
the dipolar contribution.
This will integrate the spin densitywithin the Li sphere.
Dipan only gives you contriutions from further away, which is usually
very small and I've never used it.
However, the main contribution should be the orbital contribution and
you should use the NMR package (x_nmr).
Am 10.09.2020 um 15:46 schrieb 林敏:
Dear Wien2K experts,
I am calculating the hyperfine field of lithium in battery materials,
which usually is transition metal oxide, HFF of Li comes from transition
metal ions.
I clear that HFFXXX(in case.scf) corresponds to isotropic paramagnetic
shift, while I also have to simulate NMR spectrum (the shape of spin
sideband) with hyperfine field anisotropic.
After careful check the mail list, I believe DIPAN is what I have to
use. Am I right?
If so, I have several questions:
1. DIPAN use a lattice summation over the magnetic moments of all sites,
which attribute all spin density to certain nucleus as point magnetons.
Other codes, such as CP2K, CRYSTAL and VASP integral the spin density
directly. What is the pros and cons for this two schemes?
2. A example hyperfine field anisotropic output of CP2k is:
-1.5614239048 5.8480253506
-4.2472955850
A_ani [Mhz] 5.8480253506 4.0184112292
0.8464507946
-4.2472955850 0.8464507946
-2.4569873244
Which is a symmetric 3x3 traceless matrix.
While in DIPAN, I get :
Matrix of dipolar fields
0.0000 -0.0001 0.0000 0.0000 0.0448
0.0448 -0.0028 -0.0028
-0.0005 0.0041 -0.0005 0.0041
-0.0003 0.0000 0.0000 0.0000 0.0244
0.0244 -0.0024 -0.0024
0.0041 -0.0013 0.0041 -0.0013
0.0000 0.0000 0.0000 0.0000 -0.1359
0.0260 0.0030 -0.0017
-0.0024 0.0045 -0.0019 -0.0028
0.0000 0.0000 0.0000 0.0000 0.0260
-0.1359 -0.0017 0.0030
-0.0019 -0.0028 -0.0024 0.0045
0.0001 0.0000 -0.0001 0.0000 0.0022
0.0413 -0.0015 0.0057
0.0045 -0.0020 -0.0039 -0.0027
0.0001 0.0000 0.0000 -0.0001 0.0413
0.0022 0.0057 -0.0015
-0.0039 -0.0027 0.0045 -0.0020
-0.0004 -0.0001 0.0001 -0.0001 -0.0968
0.3749 0.0000 -0.0022
0.0009 0.0012 -0.0002 0.0015
-0.0004 -0.0001 -0.0001 0.0001 0.3749
-0.0968 -0.0022 0.0000
-0.0002 0.0015 0.0009 0.0012
-0.0001 0.0001 -0.0001 -0.0001 0.2796
-0.2378 0.0008 -0.0002
0.0000 -0.0022 0.0015 -0.0003
0.0005 -0.0000 0.0001 -0.0001 -0.1254
-0.1691 0.0011 0.0014
-0.0022 0.0000 -0.0003 -0.0010
-0.0001 0.0001 -0.0001 -0.0001 -0.2378
0.2796 -0.0002 0.0008
0.0015 -0.0003 0.0000 -0.0022
0.0005 -0.0000 -0.0001 0.0001 -0.1691
-0.1254 0.0014 0.0011
-0.0003 -0.0010 -0.0022 0.0000
Using the case.indipan:
56. 1 Rmax (a.u.); iprint
0.00539 magnetic moment of 1 atom (:MMI001)
0.00146 magnetic moment of 2 atom (:MMI002)
0.00143 magnetic moment of 3 atom (:MMI003)
2.73152 magnetic moment of 4 atom (:MMI004)
0.04182 magnetic moment of 5 atom (:MMI005)
0.04424 magnetic moment of 6 atom (:MMI006)
682.51653 Volume of unit cell (:VOL)
3
1. 1. 1. 1st direction of magnetic moment
1. 1. 1. 2nd direction of magnetic moment
1. 1. 1. 2nd direction of magnetic moment
So, how to make those results in same convention and comparable?
Thanks.
———————————————
Min Lin
2018 Ph. D student
Physical Chemistry
Chemistry Department Chemistry & College of Chemistry and
Chemical Engineering
Xiamen University
China
e-mail: lin...@stu.xmu.edu.cn <mailto:lin...@stu.xmu.edu.cn>
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