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pp hybridization

asked by Riccardo Piombo (2019/12/03 12:37)

Dear all,

I would like to simulate the energy diagram of a system formed by a Cu ion with 10 3d states surrounded by 4 planar Oxygens that hybridizing their 2p states give birth to a band of 5 Ligand-P states with the same symmetry of the 3d levels. It is clear to me how to insert the hybridization between d and P the states but how about the pp hybridization?

In particular I would like to simulate the case in which, when the P-d hybridization is zero, omitting the crystalline field on the d states then the pp hybridization separates the p states into four levels with energies:

Ea1u = Delta_pd + Tpp
Eb1u = Delta_pd - Tpp
Eb2u = Ea1u
Eeu = Delta_pd

Schematically, at Tpd == 0, the energy diagram should be

d states

------- 5 3d states -------



P states

------- Ea1u, Eb2u -------

------- Eeu -------

------- Eb1 -------




I think that this is not right:

Delta_pd = -2.75
Tpp = -1.0

Ea1u = Delta_pd + Tpp
Eb1u = Delta_pd - Tpp
Eb2u = Ea1u
Eeu = Delta_pd

Akm_pp = PotentialExpandedOnClm(“Oh”,1,{Ea1u,Eb1u,Eb2u,Eeu})
hybrid_pp = NewOperator(“CF”, NF, IndexUp_Ld, IndexDn_Ld, Akm_pp)
hybrid_pp.Name = “p-p hybridization”

Thanks in advance to all

Answers

, 2019/12/03 23:22, 2019/12/04 20:13

Dear Ricardo

The code you need is listed below. The basic idea is that the symmetry of the Hamiltonian acting on the d-shell is the same as for the Ligand d shell or for the hopping between them. If you expand this Hamiltonian on spherical harmonics you thus get the same expansion coefficients. Which coefficients are allowed you can find in the point-group tables.

Best wishes, Maurits

NF=20
NB=0
IndexDn_3d={ 0, 2, 4, 6, 8}
IndexUp_3d={ 1, 3, 5, 7, 9}
IndexDn_Ld={10,12,14,16,18}
IndexUp_Ld={11,13,15,17,19}

-- number of electrons (formal valence)
nd = 9

-- parameters from experiment (core level PES)
-- J. Ghijsen, L. H. Tjeng, J. van Elp, H. Eskes, J. Westerink, G. A. Sawatzky, and M. T. Czyzyk, Phys. Rev. B 38, 11322 (1988).
-- H. Eskes, L. H. Tjeng, and G. A. Sawatzky, Phys. Rev. B 41, 288 (1990).

U       =  6.0
Delta   =  2.2
-- Slater integrals for multipole part of Coulomb interaction
F2dd    = 10.00
F4dd    =  6.25

-- Onsite splitting of the 3d shell (numbers from the references above)
Eda1g   =  0
Edb1g   =  0
Edb2g   =  0
Edeg    =  0

-- Onsite splitting of the Ligand d shell (numbers from the references above)
ELa1g   =   1.25
ELb1g   =  -1.25
ELb2g   =   1.25
ELeg    =   0

-- Hopping between the Ligand and 3d shell (numbers from the references above)
TdLa1g   =  2.3 / sqrt(3.0)
TdLb1g   =  2.3
TdLb2g   =  2.3 / 2
TdLeg    =  2.3 / (2*sqrt(2))
 
-- turning U and Delta to onsite energies (Including the transformation from U to F0)
 
ed      = (10*Delta-nd*(19+nd)*U/2)/(10+nd)
eL      = nd*((1+nd)*U/2-Delta)/(10+nd)
F0dd    = U+(F2dd+F4dd)*2/63

-- expansion of a potential on shperical Harmonics
-- Format for D4h symmetry copied from
-- http://quanty.org/physics_chemistry/point_groups/d4h/orientation_zxy
-- we set A(k=0,m=0) to ed or eL for the d and ligand shell respectively
-- The energy difference between d^n L^10 and d^n+1 L^9 the becomes delta


Akmd  = {{0, 0, ed},
         {2, 0, Eda1g + (-1)*(Edb1g) + (-1)*(Edb2g) + Edeg} ,
         {4, 0, (3/10)*((6)*(Eda1g) + Edb1g + Edb2g + (-8)*(Edeg))} ,
         {4,-4, (3/2)*((sqrt(7/10))*(Edb1g + (-1)*(Edb2g)))} ,
         {4, 4, (3/2)*((sqrt(7/10))*(Edb1g + (-1)*(Edb2g)))} }

AkmL  = {{0, 0, eL},
         {2, 0, ELa1g + (-1)*(ELb1g) + (-1)*(ELb2g) + ELeg} ,
         {4, 0, (3/10)*((6)*(ELa1g) + ELb1g + ELb2g + (-8)*(ELeg))} ,
         {4,-4, (3/2)*((sqrt(7/10))*(ELb1g + (-1)*(ELb2g)))} ,
         {4, 4, (3/2)*((sqrt(7/10))*(ELb1g + (-1)*(ELb2g)))} }

AkmLd = {{0, 0, (1/5)*(TdLa1g + TdLb1g + TdLb2g + (2)*(TdLeg))} ,
         {2, 0, TdLa1g + (-1)*(TdLb1g) + (-1)*(TdLb2g) + TdLeg} ,
         {4, 0, (3/10)*((6)*(TdLa1g) + TdLb1g + TdLb2g + (-8)*(TdLeg))} ,
         {4,-4, (3/2)*((sqrt(7/10))*(TdLb1g + (-1)*(TdLb2g)))} ,
         {4, 4, (3/2)*((sqrt(7/10))*(TdLb1g + (-1)*(TdLb2g)))} }



OppOnsite3d = NewOperator("CF", NF, IndexUp_3d, IndexDn_3d, Akmd)
OppOnsiteLd = NewOperator("CF", NF, IndexUp_Ld, IndexDn_Ld, AkmL)
OppHopLd3d  = NewOperator("CF", NF, IndexUp_3d,IndexDn_3d, IndexUp_Ld,IndexDn_Ld,AkmLd) +  NewOperator("CF", NF, IndexUp_Ld,IndexDn_Ld, IndexUp_3d,IndexDn_3d,AkmLd)


OppU =NewOperator("U", NF, IndexUp_3d, IndexDn_3d, {F0dd,F2dd,F4dd})


Hamiltonian =  OppU + OppOnsite3d + OppOnsiteLd + OppHopLd3d
             
 
-- we now can create the lowest Npsi eigenstates:
Npsi=20
-- in order to make sure we have a filling of nd electrons we need to define some restrictions
StartRestrictions = {NF, NB, {"1111111111 0000000000",nd,nd}, {"0000000000 1111111111",10,10}}
 
psiList = Eigensystem(Hamiltonian, StartRestrictions, Npsi)
oppList={Hamiltonian, OppOnsite3d, OppOnsiteLd, OppHopLd3d}

-- print of some expectation values
 
print("  #    <E>      <E3d>    <ELd>    <Ehyb> ");
for i = 1,#psiList do
  io.write(string.format("%3i ",i))
  for j = 1,#oppList do
    expectationvalue = Chop(psiList[i]*oppList[j]*psiList[i])
    io.write(string.format("%8.3f ",expectationvalue))
  end
  io.write("\n")
end
, 2019/12/04 17:18

Thanks a lot professor, You've been very kind to me

Riccardo

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