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Due to its rotational symmetry a nucleus with I > 1/2 cannot be perfectly spherical. This means that the wavefunction of its charges component (quarks) must be also non-spherically symmetric. The nucleus must posses an electric quadrupole moment {$e\mathbf Q$} and it must reorient in an electric field gradient, {$\mathbf V = \boldsymbol{\nabla}\mathbf{E}$}, such as that due to surrounding charges in a crystal lattice. Both are rank-2 tensors and thanks to Wigner-Eckart theorem it is possible to rewrite the quadrupole interaction energy {$- e\mathbf Q\cdot \mathbf V$} in terms of the angular momentum operators {$\mathbf I, I_z$}.
Abragam (p. 232) writes the Zeeman plus the quadrupolar interaction for {$ I\ge 1$} as
{$$ \tag{1} {\cal H} = -h \nu_L I_z + \frac {e^2 qQ} {4I(2I-1)}\left[ 3I_z^2 -I(I+1) +\frac \eta 2 (I_+^2 + I_-^2) \right] $$}
and then derives a quadrupole frequency
{$$ \nu_Q= \frac {3e^2qQ}{2hI(2I-1)} = \frac 3 2 \frac {eQV_{zz}} {2hI(2I-1)} $$}
so that
{$$ \tag{2}
{\cal H} = -h \nu_L I_z + \frac {h\nu_Q} {6}\left[ 3I_z^2 -I(I+1) +\frac \eta 2 (I_+^2 + I_-^2) \right]$$}
Notice that this assumes that the field is along the principal axis, {$\hat z$}, of the electric field gradient, {$\mathbf V$}, with {$V_{zz}=\hbar e^2qQ$}. We show here how this is derived from
{$$ \tag{3} {\cal H} = -h \nu_L I_z + \frac {h\nu_Q} {3V_{zz}}\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I}$$}
Indeed, remembering that the electric field gradient is a traceless tensor, {$ V_{\alpha\beta}=\frac {\partial E_\alpha} {\partial x_\beta}$}, i.e.
{$$ V_{xx}+V_{yy}+V_{zz}=0,$$}
We can define the rhombicity parameter {$\eta=\frac{V_{xx}-V_{yy}}{V_{zz}}$} and write its principal component ratios 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
{$$ \tag{4}
\begin{align}
I_x^2&=\frac 1 4 (I_+^2+I_+I_-+I_-I_++I_-^2)\\
I_y^2&=-\frac 1 4 (I_+^2-I_+I_--I_-I_++I_-^2) \end{align}$$}
With the above relations one may write
{$$\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I} = V_{zz}\left[ I_z^2 + \eta \frac{I_+^2+I_-^2} 4 -\frac {I_+I_-+I_-I_+} 4 \right] $$}
In the last term on the right hand side the anticommutator of {$I_+$} with {$I_-$} is equal to {$2(I_x^2+I_y^2)$}, 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
{$$ \tag{5}\mathbf{I}\cdot\mathbf{V}\cdot\mathbf{I} = V_{zz} \mathbf{I}\cdot \begin{bmatrix}
\frac {\eta-1} 2 & 0 & 0 \\
0 & -\frac {\eta+1} 2 & 0\\
0 & 0 & 1
\end{bmatrix} \cdot\mathbf{I} = \frac {V_{zz}} 2 \left[ 3I_z^2 -I(I+1)+ \frac\eta 2 ({I_+^2+I_-^2}) \right] $$}
and it is easy to check that substitution into Eq. 3 yields Eq. 2.
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