Electric Quadrupole Moment of a Nucleus
Pages 104-111
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}) = \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell} \frac{4\pi}{2l + 1} q_{lm} \frac{Y_{lm}(\theta ,\phi)}{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]
[math]\vec{E} = -\vec{\nabla} \Psi (r)[/math]
[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]
[math]\Psi (r) = \int \frac{\rho (r^') d^3 r^'}{|\vec{r} - \vec{r^'}|} = 4\pi {\Sigma_{l=0}}^{\infty} {\Sigma_{m=-l}}^{l} \frac{1}{2l + 1} [\int \frac{{Y_{lm}^8}(\theta^' \psi^') (r^')^l \rho (r^') Y_{lm}}{r^{l+1}} d^3r^' ][/math]
potential ar [math]r^{''}[/math] due to charge distribution at [math]\vec{r^'}[/math]
[math]r_\lt = |\vec{r^'}|[/math] [math]r_\gt = |\vec{r}|[/math] for outside of charged sphere.
[math]\vec{r^'}[/math] is fixed.
[math][\int {Y_{lm}^*} (r^')^l \rho(r^')d^3r^'] \equiv q_{lm}[/math] = multiple moments
[math]q_{20} = \frac{1}{2} \sqrt{\frac{5}{4\pi}} [(3z^')^2 - (r^')^2] \rho (r^') d^3r^' = \frac{1}{2} \sqrt{\frac{5}{4\pi}} Q_{33}[/math] quadrupole moment
[math]Y_{20} = \frac{1}{4} \sqrt{\frac{5}{\pi}} (3 cos^2 \theta -1) = \frac{1}{4} \sqrt{\frac{5}{\pi}} \frac{3z^2 - r^2}{r^2}[/math]
[math]Q = \lt \Phi_{jj} | 4 \sqrt{\frac{\pi}{5}} r^2 Y_{20}| \Phi_{jj} \gt [/math]
let
[math]\Phi_{jj} = R(r)Y_{ll}[/math] = general wave function (l=m for maximum projection)
then
[math]Q = \lt R(r)Y_{ll} | \sqrt{\frac{16 \pi}{5}} r^2 Y_{20} | R(r) Y_{ll}\gt [/math]
[math]= \int R^*(r){Y_{ll}}^* { \sqrt{\frac{16 \pi}{5}} r^2 Y_{20} } R(r) Y_{ll} r^2 dr d\Omega[/math]
[math]= \sqrt{\frac{16 \pi}{5}} \int r^2 R^*(r) R(r) dr \int {Y_{ll}}^* Y_{20} Y_{ll} d\Omega[/math]
[math]\lt r^2\gt = \int r^2 R^*(r) R(r) dr[/math] mean square radius.
[math]\int {Y_{ll}}^* Y_{20} Y_{ll} d\Omega = ?[/math]
Forest_NucPhys_I