Electric QuadrupoleMoment Forest NuclPhys I

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}) = {\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]

[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

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