Quadrupole interaction in high field.
Contributor: Y. Millot

Home and Applets > Quadrupole Interaction > Definition > Hamiltonian in High Field

## Quadrupole interaction in high magnetic field

The first two terms in the Magnus expansion of the quadrupole interaction are:
the first-order quadrupole interaction (一階四極矩作用力),

$H Q ( 1 ) = eQ 4 I ( 2 I - 1 ) ħ 6 3 [ 3 I z 2 - I ( I + 1 ) ] V ( 2,0 )$

and the second-order quadrupole interaction (二階四極矩作用力),

$H Q ( 2 ) = - 1 ω 0 ( eQ 2 I ( 2 I - 1 ) ħ ) 2 { 1 2 V ( 2 , - 1 ) V ( 2,1 ) [ 4 I ( I + 1 ) - 8 I z 2 - 1 ]$ $+ 1 2 V ( 2 , - 2 ) V ( 2,2 ) [ 2 I ( I + 1 ) - 2 I z 2 - 1 ] } I z$

with $ω 0 = γ B 0$

where $V ( 2,i )$ are the spherical components of the second-rank EFG tensor. In its principal-axis system, the components of this tensor are:

$V ( 2 , 0 ) PAS = 3 2 eq , V ( 2 , ± 1 ) PAS = 0 , V ( 2 , ± 2 ) PAS = 1 2 eq η$

The direct products of the EFG second-rank irreducible tensors can be expressed as higher rank spherical tensors W using the Clebsch-Gordan coefficients. Since $V ( 2 , - 2 ) V ( 2,2 ) = V ( 2,2 ) V ( 2 , - 2 )$ and $V ( 2 , - 1 ) V ( 2,1 ) = V ( 2,1 ) V ( 2 , - 1 )$, we simply have

$V ( 2 , - 1 ) V ( 2,1 ) = V ( 2,1 ) V ( 2 , - 1 ) = 8 35 W ( 4,0 ) + 1 14 W ( 2,0 ) - 1 5 W ( 0,0 )$ $V ( 2 , - 2 ) V ( 2,2 ) = V ( 2,2 ) V ( 2 , - 2 ) = 1 70 W ( 4,0 ) + 2 7 W ( 2,0 ) + 1 5 W ( 0,0 )$

where $W ( 3,0 ) = W ( 1,0 ) = 0$.

The second-order quadrupole interaction becomes:

$H Q ( 2 ) = - 1 ω 0 ( eQ 2 I ( 2 I - 1 ) ħ ) 2 { 70 140 W ( 4,0 ) [ 18 I ( I + 1 ) - 34 I z 2 - 5 ]$ $+ 14 28 W ( 2,0 ) [ 8 I ( I + 1 ) - 12 I z 2 - 3 ]$ $- 1 5 W ( 0,0 ) [ I ( I + 1 ) - 3 I z 2 ] } I z$

The eigen values of the fourth-rank EFG tensor W in the spherical tensor representation are:

$W ( 0,0 ) PAS = 1 5 [ 2 ( V ( 2,2 ) PAS ) 2 + ( V ( 2,0 ) PAS ) 2 ] = 5 10 ( eq ) 2 ( η 2 + 3 )$

$W ( 2,0 ) PAS = 14 7 [ 2 ( V ( 2,2 ) PAS ) 2 - ( V ( 2,0 ) PAS ) 2 ] = 1 14 ( eq ) 2 ( η 2 - 3 )$

$W ( 2 , ± 2 ) PAS = 4 14 V ( 2,2 ) PAS V ( 2,0 ) PAS = 3 7 ( eq ) 2 η$

$W ( 4,0 ) PAS = 2 70 [ ( V ( 2,2 ) PAS ) 2 + 3 ( V ( 2,0 ) PAS ) 2 ] = 1 70 ( eq ) 2 ( 1 2 η 2 + 9 )$

$W ( 4 , ± 2 ) PAS = 6 42 V ( 2,2 ) PAS V ( 2,0 ) PAS = 3 14 7 ( eq ) 2 η$

$W ( 4 , ± 4 ) PAS = ( V ( 2,2 ) PAS ) 2 = 1 4 ( eq ) 2 η 2$

## References

### Solid-state NMR bibliography for:

[Contact me] - Last updated August 30, 2020