Electric Quadrupole Moment of a Nucleus
As in the dipole calculation we assume that the object is in a state such that its maximum total angular momentum is along the z-axis.
or [math]\Psi_{jm} = \Psi_{jj}[/math]
then
[math]Q = \lt \Psi_{jj} |3z^2 - r^2|\Psi_{jj}\gt [/math]
From definition of quadrupole moment for a single charged object/particle.
The origin of this comes from electron-statics.
You expand the electric potential in terms of spherical harmonics.
[math]\Phi(\vec{r}) = {\Sigma_{l=0}}^{\infty} {\Sigma_{m=-l}}^{l} \frac{4\pi}{2l + 1} q_{lm} \frac{Y_{lm}(\theta \psi)}{r^{l+1}}[/math]
because
[math]\vec{E} = \int \rho (\vec{r^'}) \frac{(\vec{r} - \vec{r^'})}{|r - r^'|^3} d^3r^' = - \vec{\nabla} \int \frac{\rho (r^')}{|\vec{r} - \vec{r^'}|}[/math]
\vec{E} = -\vec{\nabla} \Psi (r)
[math]\Psi (r) = \int \frac{\rho (r^')}{|\vec{r} - \vec{r^'}|}[/math]
Since
[math]\frac{1}{|\vec{r} - \vec{r^'}|} = 4\pi {\Sigma_{l=0}}^{\infty} {\Sigma_{m=-l}}^{l} \frac{1}{2l + 1} \frac{{r{\lt }^l}{{r{\gt }^{lm}} {Y_{lm}}^* ({\theta}^' {\psi}^'){Y_{l}} (\theta \psi)[/math]
[math]\frac{{r{\lt }^l}{{r{\gt }^{lm}}[/math]
[math]r_\lt = |\vec{r}|[/math] if [math]|\vec{r}|\lt |\vec{r^'}|[/math]
[math]r_\lt = |\vec{r^'}|[/math] if [math]|\vec{r^'}|\lt |\vec{r}|[/math]
[math]r_\gt = |\vec{r}|[/math] if [math]|\vec{r}|\gt |\vec{r^'}|[/math]
[math]r_\gt = |\vec{r^'}|[/math] if [math]|\vec{r^'}|\gt |\vec{r}|[/math]