Difference between revisions of "Electric QuadrupoleMoment Forest NuclPhys I"
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<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>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> | ||
Revision as of 05:21, 7 April 2009
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
then
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.
because
Since
if
if
if
if
potential ar
due to charge distribution atfor outside of charged sphere.
is fixed.
= multiple moments
quadrupole moment
let
= general wave function (l=m for maximum projection)
then
mean square radius.