INTERATOMIC SPIN-ORBIT COUPLING

Download Report

Transcript INTERATOMIC SPIN-ORBIT COUPLING

INTERATOMIC SPIN-ORBIT
COUPLING: A MECHANISM FOR SPINSPIRAL-CAUSED FERROELECTRICITY
T. A. Kaplan and S. D. Mahanti
Michigan State University
APS March, 2007
Introduction
We had considered a simple model that gave the effect.
The model involves inter-atomic spin-orbit coupling, and
considers just ‘direct hopping’ between magnetic atoms a
and b. O indicates the mid-point.
a
O
b
An initial crude estimate suggested realism, perhaps.
Very recently: discovered an additional, physically real
contribution that appreciably reduces the size; now
probably too small.
But precise model physically correct; still gives qualitative
insight; further, interatomic s.o. coupling might be important
in superexchange. For our simplified model we show:
• For O a crystal center of inversion, and the direction of the
spiral wave vector along the line including a and b, the
ferroelectric polarization,
P  eˆab  (sa  sb ),
in agreement with Katsura, Nagaosa, Balatzky (KNB). This
unchanged for a Cr-Cr pair in CoCr2O4, where O not a ctr.
of inv., but there are mirror planes through O. However,
direction of P does not satisfy this in general.
• P does not change sign when one of a n.n. pair of magnetic
ions changes from <1/2-filled to >1/2-filled shell. (Interest via
Co-Cr vs Cr-Cr pairs in CoCr2O4 .)
The Model
Start with pair of atoms, e.g. hydrogens, or Cr 3+ ions. For
Cr’s, each ion: (3d) 3 in t 2g orbitals. The atoms have
different spin directions.
y
O
a
x
b
z
Spin states  a , b , (  ,  quantized along z):
i a
 a   cos( a / 2)  sin( a / 2)e  , and a
i
 a    sin( a / 2)  cos( a / 2)e 
a
b
Cobalt chromite
Co Cr 2O4
From Menyuk review, 1970
THE CALCULATION
Spin orbit coupling: Vso  c0V  p  s
Interatomic matrix elements:

a a


a

 p  | Vso | tb b    ic0 ( p | V  | tb )  (  a | s | b )
p
t
s
a
b
Cr-Cr
a
b
H-H
The wave function to 1st order (in Vso ) gives the
electron density to 1st order
n( r )  n0 ( r )
in
fin

'
2  f ( r ) f ' ( r ) m '  Im (    |   '     ' | s |   ) /   '
bond charge
m '  c0  f | V   | f ' 
y
x
For spiral, contribution to dipole moment d from spin zcomponent same for each bond,  ferroelectricity
(similar to KNB).
So only the z-component of m enters:
m '  c0  f |  xV  y   yV  x | f ' 
We considered V having the same symmetry as
that of a n.n. pair of Cr’s in the spinel, namely,
V(x,y,z)=V(-x,y, z)=V(x,-y, z). Origin at O (O not a
center of inversion). This symmetry leads to the
vanishing of certain of these quantities. E.g. for the
H-H model, one set of such quantities is
g  pa |  xV  y   yV  x | sb ,   x, y, z.
This vanishes for  = x and z (by y-integral),
leaving only g y ;  +, s  p y overlap bond
charge  ferroelectric polarization in y direction.
This direction given by
P  eab  ( sa  sb )
found by KNB who assumed O is a center of inversion. We
find conclusion unchanged for O not a coi., but with symmetry
of Cr-Cr pair in CoCr2O4 .
3
Cr
Same conclusion for pair of
ions.
However, for general V ( x, y, z ) , direction of P not that of
above. (It will also have a component along Sa  Sb ).
Sign change of P under
< ½-  > ½-filled shell?
Imagine one atom having d-levels in a crystal field
such that there are just two low-lying orbitals d1 and d2 .
Case I: 1 electron at b; Case II: 3 electrons at b; 1
electron at a in both cases. Put
p
p
“down”
d2
“up”
d2
s
d1
d1
a
b
1-electron energy-level scheme
 '  Im(  a | b '   b ' | sz |  a )
Can show that
           
Contribution to the charge density from electron
hopping from a to b:
Case I, 1 electron on b:
 n( r )
a b

1
1
  M ( r ) 



   J p d1   J p d1




(Sign difference from     .)
Case II, 3 electrons on b:
 n( r )
a b

1
1
  M ( r ) 

   Jp d   Jp d

 2
 2





Intra-atomic exchange J < , 1-electron energy
difference; required for stability of model ground state.
Then sign of d same in cases I and II.
Argument shows how  n  0 for filled shell.