View source for 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 = < \Psi_{jj} |3z^2 - r^2|\Psi_{jj}></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<}^l}{{r>}^{lm}} {Y_{lm}}^* ({\theta}^' {\psi}^'){Y_{l}} (\theta \psi)</math>

<math>\frac{{r<}^l}{{r>}^{lm}}</math>

<math>r_< = |\vec{r}|</math> if <math>|\vec{r}|<|\vec{r^'}|</math>

<math>r_< = |\vec{r^'}|</math> if <math>|\vec{r^'}|<|\vec{r}|</math>


<math>r_> = |\vec{r}|</math> if <math>|\vec{r}|>|\vec{r^'}|</math>

<math>r_> = |\vec{r^'}|</math> if <math>|\vec{r^'}|>|\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_< = |\vec{r^'}|</math>  <math>r_> = |\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 = <\Phi_{jj} | 4 \sqrt{\frac{\pi}{5}} r^2 Y_{20}| \Phi_{jj} ></math>

let 

<math>\Phi_{jj} = R(r)Y_{ll}</math> = general wave function (l=m for maximum projection)

then

<math>Q = <R(r)Y_{ll} | \sqrt{\frac{16 \pi}{5}} r^2 Y_{20} | R(r) Y_{ll}></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><r^2> = \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}> = \Sigma_{m_1} \Sigma_{m_2} C_{m_1 m_2} | l_1 l_2 m_1 m_2 ></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><Y_l^l | Y_0^2 | Y_l^l> = \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>










[[Forest_NucPhys_I]]