Definition

• EFG tensor
• EFG tensor
  (mathML)
• EFG tensor
  (mathML + SVG)
• Hamiltonian in high field
• Hamiltonian in high field
  (mathML)
• Static crystal
• Rotating crystal
• V20 static
• W20 static
• W40 static
• V20 VAS (MAS)
• W20 VAS (MAS)
• W40 VAS (MAS)

One-pulse MAS

MQ-MAS

Reference

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Quadrupole interaction in high magnetic field

The first two terms in the Magnus expansion of the quadrupole interaction are:

(1) the first-order quadrupole interaction (一階四極矩作用力),

$${\displaystyle H_Q^{\left(1\right)}=\frac{\mathrm{eQ}}{4I(2I-1)ħ}\frac{\sqrt{6}}{3}[3I_z^2-I(I+1\left)\right]V_{\left(\mathrm{2,0}\right)}}$$
and (2) the second-order quadrupole interaction (二階四極矩作用力),

$${\displaystyle H_Q^{\left(2\right)}=-\frac{1}{{\omega }_0}{\left[\frac{\mathrm{eQ}}{2I(2I-1)ħ}\right]}^2\left\{\frac{1}{2}{V}_{(2,-1)}{V}_{\left(\mathrm{2,1}\right)}\right[4I(I+1)-8I_z^2-1]}$$ $${\displaystyle +\frac{1}{2}{V}_{(2,-2)}{V}_{\left(\mathrm{2,2}\right)}\left[2I\right(I+1)-2I_z^2-1]\}{I}_z}$$
with $${\displaystyle {\omega }_0=\gamma B_0}$$
where ${\displaystyle V_{\left(\mathrm{2,i}\right)}}$ are the spherical components of the second-rank EFG tensor. In its principal-axis system, the components of this tensor are:

$${\displaystyle V_{(2,0)}^{\mathrm{PAS}}=\sqrt{\frac{3}{2}}\mathrm{eq},V_{(2,\pm 1)}^{\mathrm{PAS}}=0,V_{(2,\pm 2)}^{\mathrm{PAS}}=\frac{1}{2}\mathrm{eq}\eta }$$

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 ${\displaystyle V_{(2,-2)}{V}_{\left(\mathrm{2,2}\right)}=V_{\left(\mathrm{2,2}\right)}{V}_{(2,-2)}}$ and ${\displaystyle V_{(2,-1)}{V}_{\left(\mathrm{2,1}\right)}=V_{\left(\mathrm{2,1}\right)}{V}_{(2,-1)}}$, we simply have
$${\displaystyle V_{(2,-1)}{V}_{\left(\mathrm{2,1}\right)}=V_{\left(\mathrm{2,1}\right)}{V}_{(2,-1)}=\sqrt{\frac{8}{35}}{W}_{\left(\mathrm{4,0}\right)}+\frac{1}{\sqrt{14}}{W}_{\left(\mathrm{2,0}\right)}-\frac{1}{\sqrt{5}}{W}_{\left(\mathrm{0,0}\right)}}$$ $${\displaystyle V_{(2,-2)}{V}_{\left(\mathrm{2,2}\right)}=V_{\left(\mathrm{2,2}\right)}{V}_{(2,-2)}=\frac{1}{\sqrt{70}}{W}_{\left(\mathrm{4,0}\right)}+\sqrt{\frac{2}{7}}{W}_{\left(\mathrm{2,0}\right)}+\frac{1}{\sqrt{5}}{W}_{\left(\mathrm{0,0}\right)}}$$
where ${\displaystyle W_{\left(\mathrm{3,0}\right)}=W_{\left(\mathrm{1,0}\right)}=0}$.

The second-order quadrupole interaction becomes:

$${\displaystyle H_Q^{\left(2\right)}=-\frac{1}{{\omega }_0}{\left[\frac{\mathrm{eQ}}{2I(2I-1)ħ}\right]}^2\left\{\frac{\sqrt{70}}{140}{W}_{\left(\mathrm{4,0}\right)}\right[18I(I+1)-34I_z^2-5]}$$ $${\displaystyle +\frac{\sqrt{14}}{28}{W}_{\left(\mathrm{2,0}\right)}\left[8I\right(I+1)-12I_z^2-3]}$$ $${\displaystyle -\frac{1}{\sqrt{5}}{W}_{\left(\mathrm{0,0}\right)}\left[I\right(I+1)-3I_z^2]\}{I}_z}$$

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

• ${\displaystyle W_{\left(\mathrm{0,0}\right)}^{\mathrm{PAS}}=\frac{1}{\sqrt{5}}[2{\left(V_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}\right)}^2+{\left(V_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}\right)}^2]=\frac{\sqrt{5}}{10}{\left(\mathrm{eq}\right)}^2({\eta }^2+3)}$
• ${\displaystyle W_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}=\frac{\sqrt{14}}{7}[2{\left(V_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}\right)}^2-{\left(V_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}\right)}^2]=\frac{1}{\sqrt{14}}{\left(\mathrm{eq}\right)}^2({\eta }^2-3)}$
• ${\displaystyle W_{\left(2\mathrm{,}\mathrm{\pm }2\right)}^{\mathrm{PAS}}=\frac{4}{\sqrt{14}}{V}_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}{V}_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}=\sqrt{\frac{3}{7}}{\left(\mathrm{eq}\right)}^2\eta }$
• ${\displaystyle W_{\left(\mathrm{4,0}\right)}^{\mathrm{PAS}}=\frac{2}{\sqrt{70}}[{\left(V_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}\right)}^2+3{\left(V_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}\right)}^2]=\frac{1}{\sqrt{70}}{\left(\mathrm{eq}\right)}^2(\frac{1}{2}{\eta }^2+9)}$
• ${\displaystyle W_{\left(4\mathrm{,}\mathrm{\pm }2\right)}^{\mathrm{PAS}}=\frac{6}{\sqrt{42}}{V}_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}{V}_{\left(\mathrm{2,0}\right)}^{\mathrm{PAS}}=\frac{3}{14}\sqrt{7}{\left(\mathrm{eq}\right)}^2\eta }$
• ${\displaystyle W_{\left(4\mathrm{,}\mathrm{\pm }4\right)}^{\mathrm{PAS}}={\left(V_{\left(\mathrm{2,2}\right)}^{\mathrm{PAS}}\right)}^2=\frac{1}{4}{\left(\mathrm{eq}\right)}^2{\eta }^2}$