## magnetic moments dependent on reducebf document.SUBSCRIPTION_OPTIONS = { "thing": "thread", "subscribed": false, "url": "subscribe", "icon": { "css": "fa fa-envelope-o" } };

Elk Users
2012-12-19
2013-06-11
• Andrea Floris - 2012-12-19

Dear users, dear Kay,
I have a probably naive question: I am calculating the band structure of a cuprate system that should not be magnetic in GGA, but is AFM adding some U value.  In GGA,  I  set some initial local B fields (everything collinear) on two different Cu atoms (let's call them Cu1 and Cu2).  Let's say +2 on Cu1 and -2 on Cu2. I set reducebf = 1. (so no action on the B field in each scf step, right?). I obtain a converged solution to a finite magnetic moment locally on the two Cu atoms (actually with a reversed sign, i.e.  - 0.13 on Cu1 and +0.13 on Cu2).

The point is that when I set reducebf = 0.5 , the system converges to a non-magnetic solution (zero local moments on Cu1 and Cu2), with clearly  strong modification of the band structure close to Fermi, which is exactly what I want to check with an all electron calculation.

So I have a reducebf-dependent solution. Am I doing something wrong? The total energies differ on the third decimal.
Is there any "proper" choice of reducebf parameter to resolve this ambiguity?
Thanks!

Andrea

• Sangeeta Sharma - 2012-12-20

The stating b-field should be quite large if you want to keep reducebf to 0.5. Can you send me the input file and  can test it.

• Lars Nordström - 2012-12-22

Dear Andreas,

You get what one expects.

The effect of reducebf is that  the initial b-field is reduced with the given factor in each iteration. When reducebf<1 the magnetic field is vanishingly small at convergence. However, with reducebf=1 you keep the given applied field throughout. Hence you should expect different solutions for the two cases, reducebf <1 or =1, as you also see. From the small Cu moment it consistent with that it is only induced by the applied field in the reducebf=1 case.

Best,
Lars

• Sangeeta Sharma - 2012-12-22

Exactly what Lars said. But if the starting bfield (bfeildmt or bfieldc) are small, the difference in moment in two calculations cannot be this large. I would say increase the starting bfield by an order in magnitude (something like 0 0 1.5 ) and then take reducebf to 0.5 and you should get the same answer as for the case of reducebf 1.

• Andrea Floris - 2013-01-14

Dear Sangeeta and Lars, thank you for your answers,  that clarified my problem
Best,
Andrea