Electric QuadrupoleMoment Forest NuclPhys I - Revision history

Oborn: /* Pages 104-111 */

2009-04-07T06:23:41Z

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2009-04-07T06:21:21Z

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2009-04-07T05:46:37Z

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@@ Line 174: / Line 174: @@
 Max number in subshell =
-<math>1\leq Number- Nucleons-in-unfilled subshell \equiv N \leq 2j</math>
+<math>1\leq Number- of-Nucleons-in-Unfilled-Subshell \equiv N \leq 2j</math>
 [[Forest_NucPhys_I]]

@@ Line 174: / Line 174: @@
 Max number in subshell =
-<math>1\leq #</math>  nucleons in unfilled subshell <math>\equiv N \leq 2j</math>
+<math>1\leq Number- Nucleons-in-unfilled subshell \equiv N \leq 2j</math>
 [[Forest_NucPhys_I]]

@@ Line 174: / Line 174: @@
 Max number in subshell =
-<math>1 \leq #</math> nucleons in unfilled subshell <math>\equiv N \leq 2j</math>
+<math>1\leq #</math>  nucleons in unfilled subshell <math>\equiv N \leq 2j</math>
 [[Forest_NucPhys_I]]

← Older revision		Revision as of 06:20, 7 April 2009
Line 147:		Line 147:

	<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 #</math> nucleons in unfilled subshell <math>\equiv N \leq 2j</math>
		+
		+

	[[Forest_NucPhys_I]]		[[Forest_NucPhys_I]]

← Older revision		Revision as of 06:11, 7 April 2009
Line 134:		Line 134:
	<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>		<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

	[[Forest_NucPhys_I]]		[[Forest_NucPhys_I]]

@@ Line 124: / Line 124: @@
 <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(?)^2 - l(l+1)}{\sqrt(2l-1)(l)(l+1)(2l+3)}}</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>

← Older revision	Revision as of 06:01, 7 April 2009
Line 114:	Line 114:


	+	<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(?)^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>

← Older revision		Revision as of 05:51, 7 April 2009
Line 106:		Line 106:

	<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>[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>

← Older revision		Revision as of 05:46, 7 April 2009
Line 105:		Line 105:
	<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>=\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>[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>

← Older revision		Revision as of 05:46, 7 April 2009
Line 97:		Line 97:
	Note: <math>({Y_m}^l)^* = (-1)^m {Y_{-m}}^l</math> : result of taking complex conjugate.		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></math>	+	<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>
		+
		+