[Wien] hyperfine field anisotropic of Li in battery materials

[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: linmin at stu.xmu.edu.cn





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