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Abragam (p. 232) writes the quadrupolar interaction for $I\ge 1$ as

${\cal H} = -h \nu_L I_z + \frac {e^2 qQ} {4I(2I-1)}\left[ 3I_z^2 -I(I+1) +\frac \eta 2 (I_+^2 + I_-^2) \right]$

and then derives a quadrupole frequency

$\nu_Q= \frac {3e^2qQ}{2hI(2I-1)}$

so that

$(1) \qquad\qquad {\cal H} = -h \nu_L I_z + \frac {h\nu_Q} {6}\left[ 3I_z^2 -I(I+1) +\frac \eta 2 (I_+^2 + I_-^2) \right]$

We show here how this is derived from

$(2) \qquad\qquad {\cal H} = -h \nu_L I_z + \frac {h\nu_Q} {3V_{zz}}\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I}$

remembering that the electric field gradient ($eq=V_{zz}$) is a traceless tensor, $V_{\alpha\beta}=\frac {\partial E_\alpha} {\partial x_\beta}$, i.e.

$V_{xx}+V_{yy}+V_{zz}=0,$

hence with the rhombicity parameter $\eta=\frac{V_{xx}-V_{yy}}{V_{zz}}$ we can write its principal component rations as

$\frac {V_{xx}}{V_{zz}}=\frac {\eta -1} 2 \qquad \qquad \qquad \frac {V_{yy}}{V_{zz}}=-\frac {\eta +1} 2$

and finally that

$I_x^2=\frac 1 4 (I_+^2+I_+I_-+I_-I_++I_-^2)$

$I_x^2=-\frac 1 4 (I_+^2-I_+I_--I_-I_++I_-^2)$

With the above relations one may write

$\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I} = V_{zz}\left[ I_z^2 + \eta \frac{I_+^"+I_-^"} 4 -\frac {I_+I_-+I_-I_+} 4 \right]$

In the last term on the right hand side the anticommutator of $I_+$ with $I_-$ is equal to $2(I_x^"+I_y^2)$, and we can rewrite it as $2(I(I+1)-I_z^2)$, considering that

$I(I+1)=I_x^2+I_y^2+I_z^2$

The result is

$\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I} = \frac {V_{zz}} 2 \left[ 3I_z^2 -I(I+1)+ \frac\eta 2 ({I_+^2+I_-^2}) \right]$

and it is easy to check that substitution into Eq. (2) yields Eq. (1)

Page last modified on July 15, 2010, at 01:33 PM