Difference between revisions of "Electric QuadrupoleMoment Forest NuclPhys I"

From New IAC Wiki
Jump to navigation Jump to search
 
(22 intermediate revisions by the same user not shown)
Line 18: Line 18:
 
You expand the electric potential in terms of spherical harmonics.
 
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>  
+
<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
 
because
Line 24: Line 24:
 
<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} = \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>\vec{E} = -\vec{\nabla} \Psi (r)</math>
  
 
<math>\Psi (r) =  \int \frac{\rho (r^')}{|\vec{r} - \vec{r^'}|}</math>
 
<math>\Psi (r) =  \int \frac{\rho (r^')}{|\vec{r} - \vec{r^'}|}</math>
Line 44: Line 44:
  
  
−
<math>\Psi (r) = \int \frac{\rho (r^') d^3 r^'}{|\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>
 +
 
 +
<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} <r^2> \int (Y_l^l)^* Y_0^2 Y_l^l d\Omega</math>
 +
 
 +
<math>= \sqrt{16 \pi}{5} <r^2> \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}<r^2>(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 <r^2> \frac{l}{(2l+3)}</math>
 +
 
 +
The Quadrupole moment of a single particle
 +
 
 +
<math><r^2> = \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<r^2> \times \frac{j - 1/2}{2(j + 1)}</math>  when l=j-1/2
 +
 
 +
<math>Q_{SingleParticle} = - 2<r^2> \times \frac{j + 1/2}{2(j + 1)}</math>  when l=j+1/2
 +
 
 +
If Q due to unpaired proton
 +
 
 +
then l=1 s=1/2
 +
 
 +
<math>Q = -2 <r^2> \frac{l}{()2l+3} = -2 (\frac{3}{5}R_0^2 A^{2/3}) \frac{l}{2l+3}</math>
 +
 
 +
<math>R_0^2 = (1.23fm)^2</math>  A=7
 +
 
 +
<math>= -2 \frac{3}{5}(1.23fm)^2 7^{2/3} \frac{1}{5} = -1.33 fm^2</math>
 +
 
 +
<math>1 barn = 100 fm^2</math>
 +
 
 +
-> <math> Q = (-1.33fm^2)(\frac{1 barn}{100fm^2}) = -0.013 barn</math>
 +
 
 +
exp: -0.04 barns
 +
 
 +
??????????????
 +
?????
 +
?????
 +
?????
 +
 
 +
 
 +
All of particles in a subshell could contribute to the quadrupole moment.
 +
 
 +
Max number in subshell =
 +
 
 +
<math>1\leq Number- of-Nucleons-in-Unfilled-Subshell \equiv N \leq 2j</math>
 +
 
 +
<math>Q_{tot} = Q_{SP} [1- \frac{2(N-1)}{2j-1}]</math>
 +
 
 +
B.) j=3/2 and N=3
 +
 
 +
<math>Q_{tot} = +0.0136</math> barns
 +
 
  
 
[[Forest_NucPhys_I]]
 
[[Forest_NucPhys_I]]

Latest revision as of 06:23, 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 [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

If Q due to unpaired proton

then l=1 s=1/2

[math]Q = -2 \lt r^2\gt \frac{l}{()2l+3} = -2 (\frac{3}{5}R_0^2 A^{2/3}) \frac{l}{2l+3}[/math]

[math]R_0^2 = (1.23fm)^2[/math] A=7

[math]= -2 \frac{3}{5}(1.23fm)^2 7^{2/3} \frac{1}{5} = -1.33 fm^2[/math]

[math]1 barn = 100 fm^2[/math]

-> [math] Q = (-1.33fm^2)(\frac{1 barn}{100fm^2}) = -0.013 barn[/math]

exp: -0.04 barns

?????????????? ????? ????? ?????


All of particles in a subshell could contribute to the quadrupole moment.

Max number in subshell =

[math]1\leq Number- of-Nucleons-in-Unfilled-Subshell \equiv N \leq 2j[/math]

[math]Q_{tot} = Q_{SP} [1- \frac{2(N-1)}{2j-1}][/math]

B.) j=3/2 and N=3

[math]Q_{tot} = +0.0136[/math] barns


Forest_NucPhys_I

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Electric_QuadrupoleMoment_Forest_NuclPhys_I&oldid=42133"