notes sr2iro4

7/28/2019 Notes Sr2IrO4

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2nd NN: Hoppings for the second nearest neigbors:

lmn=110

Txy =xyyz

zx

xy yz xz

3V 2V + V 0 00 V + V V V0 V V V + V

H(2) (k) =

dNN(2)

eikd0|H|d = ei(kx+ky)aTxy + ei(kx+ky)aTxy + ei(kxky)aTxy + ei(kxky)aTxy

= 2eikya cos(kxa)Txy + 2eikya cos(kxa)Txy = 4cos(kxa)

p

cos(kya) q

Txy = 4pqTxy

H(2)(k) = 4pq

3V(2) 2V(2) + V(2) 0 0

0 V(2) + V(2) V

(2)

V(2)

0 V(2) V(2) V(2) + V(2)

Parameters from Harrisonsbook [1]:

Vdd = 45

2r3dmd5

109.14846 r3d

d5= 139.41

d5eVA = 0.195 eV

Vdd = +30

2r3d

md5+72.76564

r3dd5

= +92.94

d5eVA = +0.106 eV

Vdd = 15

2r3dmd5

18.19141r3d

d5= 23.24

d5eVA = 0.026 eV

We have used that: rd=1.085 A (r3d=1.277289125 A3), and also2

m =7.62 eVA2 and e2=14.4 eVA. The

distance between nearest Ir atoms is: d(Ir-Ir)=

2

2alat.=

2

25.4846 A=3.8782 A. According to the

scaling which is for d-electrons 1d5

, we notice that the values for hopping parameters for the second

nearest neigbors (NN) is125=0.177, and for the third NN it is

1

25=0.031. Therefore, we may neglect

the third NN. For the second NN we have: V(2)dd=-0.035 eV, V(2)dd=0.019 eV and V

(2)dd=-0.005 eV. We

may also neglect V(2)dd . Then the hamiltonian is:

H = 2 pV + qV 0 00 pV + qV 00 0 pV + qV

+ 2 6pqV(2)

4pqV

(2) 0 0

0 2pqV(2) 2pqV(2)

0 2pqV(2) 2pqV

(2)

(1)We have one solution immediately: 1(k) = 2(p(k) + q(k))V + 4p(k)q(k)(3V(2) 2V(2) )For the other two we have to solve eigenvalue problem for the next 2 2 matrix:

H(k) = 2

pV + qV + 2pqV

(2) 2pqV

(2)

2pqV(2) pV + qV + 2pqV

(2)

2

A + C D

D B + C

, (2)

where A = pV + qV, B = pV + qV, C = 2pqV(2) + V

(2)dd and D = 2pqV

(2)

V

(2)dd . If we neglect

V(2)dd , then C = D. The solutions are: 2/3(k) =

A + CB + C

=

pV + qV + 2pqV(2)

pV + qV + 2pqV(2)

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1.2 t2g and eg states

1th NN: Hoppings for the nearest neigbors:

lmn=100

Tx =

xyyzzxx2

z2

xy yz xz x2

z2

V 0 0 0 00 V 0 0 00 0 Vpi 0 00 0 0 0 00 0 0 0 0

and similar for lmn = 200 (though, hoppings are different). For planar case Tz is dropped.

2 Crystal structure

Translation vectors for Body-centered tetragonal (bct) cell in real space:(vq 4/mmm(D4h))

12

(a,a,c); 12

(a, a, c); 12

(a,a, c))and in reciprocal space:(vq 4/mmm(D4h))

2ca

(0, c , a); 2ca

(c, 0, a); 2ca

(a,a, 0))

3 Magnetic properties

The LDA+SO+U predicts the ground state with weak ferromagnetism with weak ferromagnetismresulting from a canted AFM order with 11 canting angle (net 22) in the plane. The predictedlocal moment is 0.36 B/Ir with 0.10 B spin and 0.26 B orbital contribution. This value is onlyone-third of the ionic value 1 B (0.33 B spin and 0.67 B orbital ones) for Jeff=1/2 but still retainsthe respective ratio close to 1:2. [2]

4 Group theory

For inversion symmerty:1) Bethes notation: j , where + stands for even (gerade) and - for odd (ungerade) states.2) Mulliken-Herzberg notation uses superscripts g and u.

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x

y

X

XY

Y

Figure 1: Atomic positions for oneIrO2layer. Large circles filled andopen are the Ir atoms on two sublat-tices, while small open circles repre-

sent oxygen.

eg

t2g

SO

L =1eff

J5/2

J3/2

Jeff

=3/2

J =1/2eff

SO

coupling

Crystal

field

SO

coupling

Crystal

field

5d 5d10Dq

Figure 2: 5d levels splitting by the crystal field and SO coupling.

5 SO interaction

It is relatively easy to show (see Ref.[3] and Appendix A) that time reversal symmetry leads to therestriction:.

E(k, ) = E(k, ). (3)

If the lattice has inversion symmetry (i.e. the operatorr r does not change the crystal lattice),one finds:

E(k, ) = E(k, ) (4)

References

[1] W. A. Harrison, Electronic Structure and the Properties of Solids (Dover, New York, 1989).

[2] B. J. Kim et al., Phys. Rev. Lett. 101, 076402 (2008).

[3] C. Kittel, Quantum Theory of Solids, 2nd Revised Edition (John Wiley and Sons, New York,1964).

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