this is probably a daring attempt considerung the difficulties of f-electron systems but I was curious if someone could give me hints on how to improve the magnetic moments for lanthanides. I attempted Holmium which should have a magnetic moment of ~10uB per atom but I am getting around 7.
Charges :
core : 92.00000000
valence : 42.00000000
interstitial : 3.564420513
muffin-tins (core leakage)
species : 1 (Ho)
atom 1 : 65.21778974 ( 0.3348166125E-05)
atom 2 : 65.21778974 ( 0.3348166125E-05)
total in muffin-tins : 130.4355795
total calculated charge : 133.9973517
total charge : 134.0000000
error : 0.2648312164E-02
Moments :
interstitial : -0.2460401870E-18 -0.1887594982E-18 -0.3229774976
muffin-tins
species : 1 (Ho)
atom 1 : 0.8321849799E-21 -0.1336464142E-19 -3.454048165
atom 2 : 0.8321849799E-21 -0.1336464142E-19 -3.454048165
total in muffin-tins : 0.1664369960E-20 -0.2672928283E-19 -6.908096330 total moment : -0.2443758170E-18 -0.2154887810E-18 -7.231073827
My input is attached below. Thank you in advance for your assistance!
my experience with magnetic moments so far would say to try one (or all of the following)
1. Use DFT+U and "fit" U to match the magenetic moment
2. try to relax the unit cell since (speaking as an experimental physicist) the DFT should result in a pressure that is different from the experimental value of the lattice constant. This in turn should affect the occupation of f-electrons.
3. Try a different functional (like xc=3 and xc=21). This is however more to test things, since one should not search for a functional that "by chance" fits your expected results.
4. Use the Fixed Spin Momentum calculation of elk to match the moment.
best,
Eike
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Yes the atomic moment for Ho is 10 Bohrs per atom, decomposed into a spin moment of 4 and orbital moment of 6. Hund's rules tell you that you have 7 spin up electrons and 3 spin down, hence spin moment 4. The spin up electrons do not give rise to any orbital moment, but the 3 spin down states add up with 3+2+1=6. As the 4f shell is more than half filled these two moments add upp to 10 bohr magnetons per atom.
Now, you have to realise that the numbers you have given says that the two Ho atoms have spin moments of 3.45 each, or 6.9 in total. (There is no information there about the orbital moments, but you can find them elsewhere.) This is then actually a fairly good result, but I have a feeling that within GGA this is fortuitous. I would recommend that you try DFT+U as suggested by Eike. However, I would think that the magnetic moments should converge in the magnitude of U when the splitting of the 4f shell is large enough (and hopefully correct). Please let me know your results if you try out.
Best regards,
Lars
Last edit: Lars Nordström 2018-08-21
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Dear all,
this is probably a daring attempt considerung the difficulties of f-electron systems but I was curious if someone could give me hints on how to improve the magnetic moments for lanthanides. I attempted Holmium which should have a magnetic moment of ~10uB per atom but I am getting around 7.
My input is attached below. Thank you in advance for your assistance!
Best,
Chris
Hi Chris,
my experience with magnetic moments so far would say to try one (or all of the following)
1. Use DFT+U and "fit" U to match the magenetic moment
2. try to relax the unit cell since (speaking as an experimental physicist) the DFT should result in a pressure that is different from the experimental value of the lattice constant. This in turn should affect the occupation of f-electrons.
3. Try a different functional (like xc=3 and xc=21). This is however more to test things, since one should not search for a functional that "by chance" fits your expected results.
4. Use the Fixed Spin Momentum calculation of elk to match the moment.
best,
Eike
Dear Chris,
Yes the atomic moment for Ho is 10 Bohrs per atom, decomposed into a spin moment of 4 and orbital moment of 6. Hund's rules tell you that you have 7 spin up electrons and 3 spin down, hence spin moment 4. The spin up electrons do not give rise to any orbital moment, but the 3 spin down states add up with 3+2+1=6. As the 4f shell is more than half filled these two moments add upp to 10 bohr magnetons per atom.
Now, you have to realise that the numbers you have given says that the two Ho atoms have spin moments of 3.45 each, or 6.9 in total. (There is no information there about the orbital moments, but you can find them elsewhere.) This is then actually a fairly good result, but I have a feeling that within GGA this is fortuitous. I would recommend that you try DFT+U as suggested by Eike. However, I would think that the magnetic moments should converge in the magnitude of U when the splitting of the 4f shell is large enough (and hopefully correct). Please let me know your results if you try out.
Best regards,
Lars
Last edit: Lars Nordström 2018-08-21