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]
Clebsch Gordon notation:
[math]|LM l_{1} l_{2}\gt = \Sigma_{m_1} \Sigma_{m_2} C_{m_1 m_2} | l_1 l_2 m_1 m_2 \gt [/math]
My notation + example
[math]{[Y^{l_2} Y^{l_1}]_M}^L = \Sigma_{m_1 m_2} {C_{m_1 m_2 M}}^{l_1 l_2 L} {Y_M}^L[/math]
[math]{[Y^{l_2} Y^{l_1}]_0}^L = \ sqrt{\frac{(2l_1 + 1)(2l_2 +1)}{4\pi (2L + 1)}} \Sigma_{m_1 m_2} {C_{0 0 0}}^{l_1 l_2 L} {Y_0}^L[/math]
You can also write in terms of [math]{Y_M}^L[/math] states
ie
[math]{Y_{m_2}}^{l_2}{Y_{m_1}}^{l_1} = \Sigma_{LM} {C_{m_1 m_2 M}}^{l_1 l_2 L} {[Y^{l_2}Y^{l_1}]_M}^L[/math]
[math]|{C_{m_1 m_2 M}}^{l_1 l_2 L} |^2[/math] = probability of finding states[math] {Y_{m_2}}^{l_2}[/math] and [math]{Y_{m_2}}^{l_2}[/math] with combined total angular momentum L and "z" component M.
Note: [math]({Y_m}^l)^* = (-1)^m {Y_{-m}}^l[/math] : result of taking complex conjugate.
[math]\lt Y_l^l | Y_0^2 | Y_l^l\gt = \int (Y_l^l)^* Y_0^2 (Y_l^l) d\Omega[/math]
[math]= \int (-1)^l (Y_{-l}^l) Y_0^2 (Y_l^l) d\Omega[/math], note that [math](Y_0^2) = (Y_0^2)^*[/math]
[math]= \int (-1)^l (Y_{-l}^l) (Y_l^l) (Y_0^2)^* d\Omega[/math]
[math]=\int (-1)^l \sum_{LM} C_{-l l M}^{l l L} [Y^l Y^l]_M^L (Y_0^2)^* d\Omega[/math]
[math][Y^l Y^l]_M^L = \sqrt{\frac{(2l_1 + 1)(2l_2 +1)}{4\pi (2L + 1)}} C_{0 0 0}^{l l L} Y_M^L[/math]
[math]= \int \sum_{LM} \frac{2l+1}{\sqrt{4\pi(2L+1)}} C_{-l l 0}^{l l L} (-1)^l C_{0 0 0}^{l l L} Y_M^L (Y_0^2)^* d\Omega[/math]
M=0 and L=2: Orthogonality of [math]Y_m^l[/math]'s
[math]= \frac{2l+1}{\sqrt{4\pi(2L+1)}} C_{0 0 0}^{l l 2} (-1)^l C_{-l l 0}^{l l 2}[/math]
[math]Q = \sqrt{16 \pi}{5} \lt r^2\gt \int (Y_l^l)^* Y_0^2 Y_l^l d\Omega[/math]
[math]= \sqrt{16 \pi}{5} \lt r^2\gt \frac{2l+1}{\sqrt{4\pi (5)}} C_{0 0 0}^{l l 2} (-1)^l C_{-l l 0}^{l l 2}[/math]
[math]C_{l_1 l_2 L}^{m_1 m_2 M} = (-1)^{l_1 - m_1} (\frac{2L+1}{2l_2+1})^{1/2} C_{l_1 L l_2}^{m_1 -M -m_2}[/math]
[math]C_{l l 2}^{0 0 0} = (-1)^l (\frac{5}{2l+1})^{1/2} C_{0 0 0}^{l 2 l}[/math]
[math]= (-1)^l \sqrt{\frac{5}{2l+1}} {\frac{3(l)^2 - l(l+1)}{\sqrt(2l-1)(l)(l+1)(2l+3)}}[/math]
[math]C_{l l 2}^{-l l 0} = (-1)^{l+l} (\frac{5}{2l+1})^{1/2} C_{-l 0 -l}^{l 2 l}[/math]
[math]=(-1)^{2l} \sqrt{\frac{5}{2l+1}} {\frac{3(l)^2 - l(l+1)}{\sqrt(2l-1)(l)(l+1)(2l+3)}}[/math]
So
[math]Q = \frac{2}{5}\lt r^2\gt (2l+1) ((-1)^l \sqrt{\frac{5}{2l+1}} \frac{[-l(l+1)]}{\sqrt{(2l-1)l(l+1)(2l+3)}}) \times \sqrt{\frac{5}{2l+1}} {\frac{3(l)^2 - l(l+1)}{\sqrt(2l-1)(l)(l+1)(2l+3)}} (-1)^l[/math]
After simplifying we get the following for Q:
[math]Q = -2 \lt r^2\gt \frac{l}{(2l+3)}[/math]
The Quadrupole moment of a single particle
[math]\lt r^2\gt = \frac{3}{5}R^2 = \frac{3}{5} R_0^2 A^{2/3}[/math]
[math]\vec{j} = \vec{l} + \vec{s}[/math] -> [math]\vec{l}=\vec{j} - \vec{s} = \vec{j} \pm \frac{1}{2}[/math]
[math]Q_{SingleParticle} = - 2\lt r^2\gt \times \frac{j - 1/2}{2(j + 1)}[/math] when l=j-1/2
[math]Q_{SingleParticle} = - 2\lt r^2\gt \times \frac{j + 1/2}{2(j + 1)}[/math] when l=j+1/2
Forest_NucPhys_I