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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 57500, 2229]*) (*NotebookOutlinePosition[ 59026, 2277]*) (* CellTagsIndexPosition[ 58982, 2273]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "Calculating the Matrix Elements for ", Cell[BoxData[ \(TraditionalForm\`\(\[InvisiblePrefixScriptBase]\^39\)K\)]] }], "Title"], Cell[CellGroupData[{ Cell[TextData[{ "Determining the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(3 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket]\)]], " matrix element from the measured lifetime" }], "Section"], Cell[TextData[{ "The lifetime of the ", Cell[BoxData[ \(TraditionalForm\`L = 1\)]], " excited states in ", Cell[BoxData[ \(TraditionalForm\`\(\[InvisiblePrefixScriptBase]\^39\)K\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Tau]\)]], ", has been measured to be ", Cell[BoxData[ \(TraditionalForm\`25.8 ns\)]], ". All but one of the excited states can spontaneously decay to several \ ground states. Because ", Cell[BoxData[ \(TraditionalForm \`\(\[VerticalSeparator] F' = 3, \ m' = 3\[RightAngleBracket]\)\)]], " decays only to ", Cell[BoxData[ \(TraditionalForm \`\(\(\[VerticalSeparator] F\) = 2, \ m = 2\[RightAngleBracket]\)\)]], ", the spontaneous lifetime is related to ", Cell[BoxData[ \(TraditionalForm \`\[Mu] = \[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(3 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket]\)]], " by ", Cell[BoxData[ FormBox[ StyleBox[\(1\/\[Tau]\_sp = \(\(4 k\^3\)\/\(3 \[HBar]\)\) \[Mu]\^2\), FontColor->GrayLevel[0]], TraditionalForm]], FontColor->RGBColor[1, 0, 0]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\[HBar] = 1.054\[Times]10\^\(-27\)\ (*\ erg\[CenterDot]s\ *) ; \ k = \(2 \[Pi]\)\/\(766.7\[Times]10\^\(-7\)\)\ (*\ cm\^\(-1\)\ *) ; \ \[Tau] = 25.8\[Times]10\^\(-9\)\ (*\ s\ *) ; \n \[Mu] = \@\(\(3 \[HBar]\)\/\(4 \( k\^3\) \[Tau]\)\)\ (*\ esu\[CenterDot]cm\ *) \)\)], "Input"], Cell[BoxData[ \(7.46121365717074169`*^-18\)], "Output"] }, Open ]], Cell[TextData[{ "Generally, \[Mu] is put into Debye (D). ", Cell[BoxData[ \(TraditionalForm\`1 D = 10\^\(-18\)\ esu\[CenterDot]cm\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Mu] = \[Mu]\/10\^\(-18\)\)], "Input"], Cell[BoxData[ \(7.46121365717074258`\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the reduced matrix element ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(3 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket]\)]] }], "Subsection"], Cell[TextData[{ "Using JET 92.1:\n", Cell[BoxData[ FormBox[ RowBox[{ \(\[LeftAngleBracket]n' \((l' s)\) j' m' | \[Mu]\_q\%1 | \(n(l\ s)\) j\ m\[RightAngleBracket]\), "=", RowBox[{\(\((\(-1\))\)\^\(j' - m'\)\), RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(j'\), "1", "j"}, {\(-m'\), "q", "m"} }], "\[NegativeThinSpace]", ")"}], \(\[LeftAngleBracket]n' \((l' s)\) j' || \[Mu]\^1 || \ \(n(l\ s)\) j\[RightAngleBracket]\)}]}], TraditionalForm]]], "\nwe can calculate the matrix elements for other transitions between these \ two F-levels. Knowing the left hand side of the equation for one transition \ allows us to calculate the reduced matrix element in the ", Cell[BoxData[ FormBox[ StyleBox[\(F = I + J\), FontWeight->"Bold"], TraditionalForm]]], " basis. Rewriting 92.1: \n", Cell[BoxData[ FormBox[ RowBox[{ \(\[LeftAngleBracket]n' \((J' I)\) F' m' | \[Mu]\_q\%1 | \(n(J\ I)\) F\ m\[RightAngleBracket]\), "=", RowBox[{\(\((\(-1\))\)\^\(F' - m'\)\), RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(F'\), "1", "F"}, {\(-m'\), "q", "m"} }], "\[NegativeThinSpace]", ")"}], \(\[LeftAngleBracket]n' \((J' I)\) F' || \[Mu]\^1 || \ \(n(J\ I)\) F\[RightAngleBracket]\)}]}], TraditionalForm]]], "\nThe following function calculates the entire coefficient on the right \ side of the equation:" }], "Text"], Cell[BoxData[ \(threeJcoeff[fp_, mp_, q_, f_, m_] := \((\(-1\))\)\^\(fp - mp\)\[Times]\n\t\t ThreeJSymbol[{fp, \(-mp\)}, {1, q}, {f, m}]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(threeJcoeff[3, 3, 1, 2, 2]\)], "Input"], Cell[BoxData[ \(1\/\@7\)], "Output"] }, Open ]], Cell[TextData[{ "This gives ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(3 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \(\@7\) \[Mu]\)]], ". Using this result and 92.1 repetitively, we an determine all of the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2, \ m\[RightAngleBracket]\)]], " matrix elements." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[TextData[{ "We will express all matrix elements in terms of \[Mu] as calculated above. \ The function f232 (2 for the ", Cell[BoxData[ \(TraditionalForm\`D\_2\)]], " line, 3 for ", Cell[BoxData[ \(TraditionalForm\`F' = 3\)]], " and 2 for ", Cell[BoxData[ \(TraditionalForm\`F = 2\)]], ") performs the calculation: " }], "Text"], Cell[BoxData[ \(f232[mp_, q_, m_] := threeJcoeff[3, mp, q, 2, m] \@7\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f232[3, 1, 2]\)], "Input"], Cell[BoxData[ \(1\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f232[2, 0, 2]\)], "Input"], Cell[BoxData[ \(1\/\@3\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f232[1, \(-1\), 2]\)], "Input"], Cell[BoxData[ \(1\/\@15\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f232[2, 1, 1]\)], "Input"], Cell[BoxData[ \(\@\(2\/3\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f232[1, 0, 1]\)], "Input"], Cell[BoxData[ \(2\ \@\(2\/15\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f232[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(1\/\@5\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(f232[1, 1, 0]\n\)\)], "Input"], Cell[BoxData[ \(\@\(2\/5\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\nf232[0, 0, 0]\)\)], "Input"], Cell[BoxData[ \(\@\(3\/5\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(3 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = 1\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(2 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = 1\/\@3\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(1 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = 1\/\@15\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(2 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = \@\(2\/3\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(1 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = 2 \@\( 2\/15\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = 1\/\@5\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(1 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = \@\(2\/5\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 3, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = \@\(3\/5\)\)]], "\nThe remaining elements can be determined using the following rule:\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ \(\[LeftAngleBracket]F', \ m' | \[Mu]\_q\%1 \[VerticalSeparator] F, \ m\[RightAngleBracket]\), "=", " ", RowBox[{ RowBox[{\(\((\(-1\))\)\^\(F' - m'\)\), RowBox[{"(", GridBox[{ {\(F'\), "1", "F"}, {\(-m'\), "q", "m"} }], ")"}], \(\@7\), "\[Mu]"}], "=", RowBox[{ RowBox[{\(\((\(-1\))\)\^\(\(-m'\) - F - 1\)\), RowBox[{"(", GridBox[{ {\(F'\), "1", "F"}, {\(m'\), \(-q\), \(-m\)} }], ")"}], \(\@7\), "\[Mu]"}], "=", \(\(\((\(-1\))\)\^\(\(-F'\) - F - 2 m' - 1\)\) \[LeftAngleBracket]F', \ \(-m'\) | \[Mu]\_\(-q\)\%1 \[VerticalSeparator] F, \ \(-m\)\[RightAngleBracket]\)}]}]}], "\n"}], TraditionalForm]]], " Since for ", Cell[BoxData[ \(TraditionalForm\`\(\[InvisiblePrefixScriptBase]\^39\)K\)]], " the 2\[Times]m will be even, we can write:\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]F', \ m' | \[Mu]\_q\%1 \[VerticalSeparator] F, \ m \[RightAngleBracket]\)]], "=", Cell[BoxData[ \(TraditionalForm \`\(\((\(-1\))\)\^\(\(-F'\) - F - 1\)\) \[LeftAngleBracket]F', \ \(-m'\) | \[Mu]\_\(-q\)\%1 \[VerticalSeparator] F, \ \(-m\)\[RightAngleBracket]\)]], "\nFor transitions between these two levels,the quantity ", Cell[BoxData[ \(TraditionalForm \`\[Phi] = \(\((\(-1\))\)\^\(\(-F'\) - F - 1\) = \(\((\(-1\))\)\^\(\(-3\) - 2 - 1\) = 1\)\)\)]], ".\nThe matrix elements can be checked by using the sum-rule:\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(m', m \)\((\(|\[LeftAngleBracket]F', \ m' | \[Mu]\_q\%1 \[VerticalSeparator] F, \ m \[RightAngleBracket] | \))\)\^2 = \(\((\(|\[LeftAngleBracket]F' || \[Mu]\^1 || F\[RightAngleBracket] | \))\)\^2\/3 = \(7 \[Mu]\)\/3\)\)]], " \nFor ", Cell[BoxData[ \(TraditionalForm\`q = \(\[PlusMinus]1\)\)]], ":" }], "Text", TextAlignment->Left], Cell[CellGroupData[{ Cell[BoxData[ \(1 + 1\/15 + 2\/3 + 1\/5 + 2\/5 == 7\/3\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`q = 0\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2\[Times]\((1\/3 + 8\/15)\) + 3\/5 == 7\/3\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the other ", Cell[BoxData[ \(TraditionalForm\`D\_2\)]], " matrix elements" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "The reduced matrix elements in the ", Cell[BoxData[ \(TraditionalForm\`D\_2\)]], " line" }], "Subsection"], Cell[TextData[{ "To calculate the matrix elements between different F-levels of the ", Cell[BoxData[ \(TraditionalForm\`D\_2\)]], "line, we can use JET 104.1:\n", Cell[BoxData[ FormBox[ RowBox[{ \(\[LeftAngleBracket]\((J' I)\) F' || \[Mu]\^1 || \((J\ I)\) F\[RightAngleBracket]\), " ", "=", " ", RowBox[{ \(\((\(-1\))\)\^\(J' + I + F + 1\)\), \(\@\(\((2 F' + 1)\) \((2 F + 1)\)\)\), RowBox[{"{", GridBox[{ {\(J'\), "I", \(F'\)}, {"F", "1", "J"} }], "}"}], \(\[LeftAngleBracket]J' || \[Mu]\^1 || J \[RightAngleBracket]\)}]}], TraditionalForm]]], "\nThe function sixJcoeff evaluates the coefficient on the right side of \ the equation:" }], "Text"], Cell[BoxData[ \(sixJcoeff[jp_, i_, fp_, j_, f_] := \(\((\(-1\))\)\^\(jp + i + f + 1\)\) \(\@\(\((2\ fp + 1)\) \((2 f + 1)\)\)\) SixJSymbol[{jp, i, fp}, {f, 1, j}]\)], "Input"], Cell[TextData[{ "We know ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(3 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \(\@7\) \[Mu]\)]], ", so we can calculate ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]J' = \(3\/2 || \[Mu]\^1 || J = 1\/2\)\[RightAngleBracket]\ by\ \(evaluating : \)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sixJcoeff[3/2, 3/2, 3, 1/2, 2]\)], "Input"], Cell[BoxData[ \(\@7\/2\)], "Output"] }, Open ]], Cell[TextData[{ "This gives ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]J' = \(3\/2 || \[Mu]\^1 || J = 1\/2\)\[RightAngleBracket]\ = 2 \( \[Mu].\)\)]], "We can use this result and 104.1 again to calculate other reduced matrix \ elements in the ", Cell[BoxData[ \(TraditionalForm\`D\_2\)]], " line." }], "Text"], Cell[BoxData[ \(g2[fp_, f_] := sixJcoeff[3/2, 3/2, fp, 1/2, f]\[Times]2\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(g2[3, 2]\)], "Input"], Cell[BoxData[ \(\@7\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g2[2, 2]\)], "Input"], Cell[BoxData[ \(\(-\@\(5\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g2[1, 2]\)], "Input"], Cell[BoxData[ \(1\/\@2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g2[2, 1]\)], "Input"], Cell[BoxData[ \(\@\(5\/2\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g2[1, 1]\)], "Input"], Cell[BoxData[ \(\(-\@\(5\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g2[0, 1]\)], "Input"], Cell[BoxData[ \(1\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[""], "Input"], Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(3 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \@7\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(2 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \(-\@\(5\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(1 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = 1\/\@2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(2 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1 \)\[RightAngleBracket] = \@\(5\/2\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(1 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1 \)\[RightAngleBracket] = \(-\@\(5\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(0 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1 \)\[RightAngleBracket] = 1\)]], "\nTo check these results, we can use the following sum-rule:\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(F'\)\(( \(|\[LeftAngleBracket]\((J' I)\) F' || \[Mu]\^1 || \((J\ I)\) F\[RightAngleBracket] | \))\)\^2 = \(\((\(|\[LeftAngleBracket]J' || \[Mu]\^1 || J\[RightAngleBracket] | \))\)\^2\) 2 F + 1\/\(2 J + 1\)\)]], "\nFor ", Cell[BoxData[ \(TraditionalForm\`F = 2, \ J = 1\/2\)]], ", we have: " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ \(\((\@7)\)\^2\), "+", \(\((\(-\@\(5\/2\)\))\)\^2\), " ", "+", " ", SuperscriptBox[ RowBox[{"(", FormBox[\(1\/\@2\), "TraditionalForm"], ")"}], "2"]}], "==", " ", \(\(2\^2\) \(2 \((2)\)\ + 1\)\/\(2 \((1/2)\) + 1\)\)}]], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`F = 1, J = 1\/2\)]], ", we have: " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ FormBox[\(\((\@\(5\/2\))\)\^2\), "TraditionalForm"], " ", "+", SuperscriptBox[ RowBox[{"(", FormBox[\(-\@\(5\/2\)\), "TraditionalForm"], ")"}], "2"], "+", \(\((1)\)\^2\)}], "==", " ", \(\(2\^2\) \(2 \((1)\)\ + 1\)\/\(2 \((1/2)\) + 1\)\)}]], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[TextData[{ "With ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = \(2 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \(-\@\(5\/2\)\)\)]], " and JET 92.1 we can calculate the matrix elements between ", Cell[BoxData[ \(TraditionalForm\`F' = 2\)]], " and ", Cell[BoxData[ \(TraditionalForm\`F = 2\)]], ". Defining f222:" }], "Text"], Cell[BoxData[ \(f222[mp_, q_, m_] := threeJcoeff[2, mp, q, 2, m]\ \[Times]\ \(-\@\(5\/2\)\)\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f222[2, 0, 2]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\@3\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[1, \(-1\), 2]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\@6\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[2, 1, 1]\)], "Input"], Cell[BoxData[ \(1\/\@6\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[1, 0, 1]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\(2\ \@3\)\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(\(-\(1\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[1, 1, 0]\)], "Input"], Cell[BoxData[ \(1\/2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f222[0, 0, 0]\)], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(2 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = \(-\(1\/\@3\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(1 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = \(-\(1\/\@6\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(2 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = 1\/\@6\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(1 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = \(-\(1\/\(2\ \@3\)\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\(\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = \)\)]], Cell[BoxData[ \(TraditionalForm\`\(-\(1\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(1 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = 1\/2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = 0\)]], "\n\nThe \[Phi] for these two levels is ", Cell[BoxData[ \(TraditionalForm \`\[Phi] = \(\((\(-1\))\)\^\(\(-2\) - 2 - 1\) = \(-1.\)\)\)]] }], "Text"], Cell[TextData[{ "Checking these results using the sum-rule ", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(m', m \)\(( | \[LeftAngleBracket]F', \ m' | \[Mu]\_q\%1 \[VerticalSeparator] F, \ m \[RightAngleBracket] | )\)\^2 = \(( | \[LeftAngleBracket]F' || \[Mu]\^1 || F\[RightAngleBracket] | ) \)\^2\/3\)]], ":\nFor ", Cell[BoxData[ \(TraditionalForm\`q = \(\[PlusMinus]1\)\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\(-1\)\/\@6)\)\^2 + \((1\/\@6)\)\^2 + \((\(-1\)\/2)\)\^2 + \((1\/2)\)\^2 == \(1\/3\) \((\(-\@\(5\/2\)\))\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell["For q=0:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"2", RowBox[{"(", RowBox[{ FormBox[\(\((\(-1\)\/\@3)\)\^2\), "TraditionalForm"], "+", " ", \(\((\(-1\)\/\(2 \@ 3\))\)\^2\)}], ")"}]}], "+", "0"}], "==", \(\(1\/3\) \((\(-\@\(5\/2\)\))\)\^2\)}]], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[BoxData[ \(f212[mp_, q_, m_] := threeJcoeff[1, mp, q, 2, m]\ \[Times]1\/\@2\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f212[1, \(-1\), 2]\)], "Input"], Cell[BoxData[ \(1\/\@10\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f212[1, 0, 1]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\(2\ \@5\)\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f212[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(1\/\(2\ \@5\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f212[1, 1, 0]\)], "Input"], Cell[BoxData[ \(1\/\(2\ \@15\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f212[0, 0, 0]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\@15\)\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(1 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 2\[RightAngleBracket] = 1\/\@10\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(1 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = \(-\(1\/\(2\ \@5\)\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 1\[RightAngleBracket] = 1\/\(2\ \@5\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(1 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = 1\/\(2\ \@15\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 2\), \ m = 0\[RightAngleBracket] = \(-\(1\/\@15\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[Phi] = \ \(\((\(-1\))\)\^\(\(-1\) - 2 - 1\) = 1\)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((1\/\@10)\)\^2 + \((1\/\(2 \@ 5\))\)\^2 + \((1\/\(2 \@ 15\))\)\^2 == \(1\/3\) \((1\/\@2)\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(2 \((\(-\(1\/\(2\ \@5\)\)\))\)\^2 + \((\(-1\)\/\@15)\)\^2 == \(1\/3\) \((1\/\@2)\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[BoxData[ \(f221[mp_, q_, m_] := threeJcoeff[2, mp, q, 1, m]\ \[Times]\@\(5\/2\)\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f221[2, 1, 1]\)], "Input"], Cell[BoxData[ \(1\/\@2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f221[1, 0, 1]\)], "Input"], Cell[BoxData[ \(1\/2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f221[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(1\/\(2\ \@3\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f221[1, 1, 0]\)], "Input"], Cell[BoxData[ \(1\/2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f221[0, 0, 0]\)], "Input"], Cell[BoxData[ \(1\/\@3\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in terms of \[Mu]"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(2 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = 1\/\@2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(1 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = 1\/2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = 1\/\(2 \@ 3\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(1 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 0\[RightAngleBracket] = 1\/2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 2, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 0\[RightAngleBracket] = 1\/\@3\)]], "\n", Cell[BoxData[ \(TraditionalForm\`\[Phi] = \(\((\(-1\))\)\^\(\(-2\) - 1 - 1\) = 1\)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((1\/\@2)\)\^2 + \((1\/\(2 \@ 3\))\)\^2 + \((1\/2)\)\^2 == \(1\/3\) \((\@\(5\/2\))\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(2 \((1\/2)\)\^2 + \((1\/\@3)\)\^2 == \(1\/3\) \((\@\(5\/2\))\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[BoxData[ \(f211[mp_, q_, m_] := threeJcoeff[1, mp, q, 1, m]\ \[Times]\ \(-\@\(5\/2\)\)\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f211[1, 0, 1]\)], "Input"], Cell[BoxData[ \(\(-\(\@\(5\/3\)\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f211[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(\(-\(\@\(5\/3\)\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f211[1, 1, 0]\)], "Input"], Cell[BoxData[ \(\@\(5\/3\)\/2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f211[0, 0, 0]\)], "Input"], Cell[BoxData[ \(0\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(1 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = \(-\(\@\(5\/3\)\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = \(-\(\@\(5\/3\)\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(1 | \[Mu]\_1\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 0\[RightAngleBracket] = \@\(5\/3\)\/2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 1, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 0\[RightAngleBracket] = 0\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[Phi] = \(\((\(-1\))\)\^\(\(-1\) - 1 - 1\) = \(-1\)\)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\(\(-1\)\/2\) \@\(5\/3\))\)\^2 + \((\(1\/2\) \@\(5\/3\))\)\^2 == \(1\/3\) \((\(-\@\(5\/2\)\))\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(2 \((\(\(-1\)\/2\) \@\(5\/3\))\)\^2 + 0 == \(1\/3\) \((\(-\@\(5\/2\)\))\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 0, \ m' | \[Mu]\_q\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1, \ m\[RightAngleBracket]\)]], " matrix elements" }], "Subsection"], Cell[BoxData[ \(f201[mp_, q_, m_] := threeJcoeff[0, mp, q, 1, m]\ \[Times]\ 1\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(f201[0, \(-1\), 1]\)], "Input"], Cell[BoxData[ \(1\/\@3\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(f201[0, 0, 0]\)], "Input"], Cell[BoxData[ \(\(-\(1\/\@3\)\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData["Results in \[Mu] units"], "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 0, \ m' = \(0 | \[Mu]\_\(-1\)\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 1\[RightAngleBracket] = 1\/\@3\)]], "\n\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 3\/2, I = 3\/2)\) F' = 0, \ m' = \(0 | \[Mu]\_0\%1 \[VerticalSeparator] \((J = 1\/2, I = 3\/2)\) F = 1\), \ m = 0\[RightAngleBracket] = \(-\(1\/\@3\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[Phi] = \(\((\(-1\))\)\^\(\(-1\) - 1 - 1\) = \(-1\)\)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((1\/\@3)\)\^2 == \(1\/3\) \((1)\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\((\(-1\)\/\@3)\)\^2 == \(1\/3\) \((1)\)\^2\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Calculating the ", Cell[BoxData[ \(TraditionalForm\`D\_1\)]], " matrix elements" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ "Determining the ", Cell[BoxData[ \(TraditionalForm \`\(\[InvisibleComma] \[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(1\/2 || \[Mu]\^1 || \((l = 0, s = 1\/2)\) J = 1\/2\)\[RightAngleBracket]\)\)]], " matrix element" }], "Subsection"], Cell[TextData[{ "To get ", Cell[BoxData[ \(TraditionalForm \`\(\[InvisibleComma] \[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(1\/2 || \[Mu]\^1 || \((l = 0, s = 1\/2)\) J = 1\/2\)\[RightAngleBracket]\)\)]], " and the reduced matrix elements in the ", Cell[BoxData[ \(TraditionalForm\`D\_1\)]], " line, we must reduce the ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(3\/2 || \[Mu]\^1 || \((l = 0, s = 1\/2)\) J = 1\/2\)\[RightAngleBracket]\)]], " to the ", Cell[BoxData[ \(TraditionalForm\`l\)]], " basis usingJET 104.1 again: ", Cell[BoxData[ FormBox[ RowBox[{ \(\[LeftAngleBracket]\((l' s)\) J' || \[Mu]\^1 || \((l\ s)\) J\[RightAngleBracket]\), " ", "=", " ", RowBox[{ \(\((\(-1\))\)\^\(\(l'\)' + s + J + 1\)\), \(\@\(\((2 J' + 1)\) \((2 J + 1)\)\)\), RowBox[{"{", GridBox[{ {\(l'\), "s", \(J'\)}, {"J", "1", "l"} }], "}"}], \(\[LeftAngleBracket]l' = \(1 || \[Mu]\^1 || l = 0\)\[RightAngleBracket]\)}]}], TraditionalForm]]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sixJcoeff[1, 1/2, 3/2, 0, 1/2]\)], "Input"], Cell[BoxData[ \(2\/\@3\)], "Output"] }, Open ]], Cell[TextData[{ "With ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(3\/2 || \[Mu]\^1 || \((l = 0, \ s = 1\/2)\) J = 1\/2\)\[RightAngleBracket] = 2 \[Mu]\)]], ":\n\n ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]l' = \(1 || \[Mu]\^1 || l = 0\)\[RightAngleBracket] = \(\@3\) \[Mu]\)]], ". \n \n Then, in \[Mu] units, ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(1\/2 || \[Mu]\^1 || \((l = 0, \ s = 1\/2)\) J = 1\/2\)\[RightAngleBracket]\)]], " is:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(sixJcoeff[1, 1/2, 1/2, 0, 1/2]\[Times]\@3\)], "Input"], Cell[BoxData[ \(\(-\@2\)\)], "Output"] }, Open ]], Cell[TextData[Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(1\/2 || \[Mu]\^1 || \((l = 0, \ s = 1\/2)\) J = 1\/2\)\[RightAngleBracket] = \(-\@2\) \[Mu]\)]]], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "The reduced matrix elements in the ", Cell[BoxData[ \(TraditionalForm\`D\_1\)]], " line" }], "Subsection"], Cell[TextData[{ "With ", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((l' = 1, s = 1\/2)\) J' = \(1\/2 || \[Mu]\^1 || \((l = 0, \ s = 1\/2)\) J = 1\/2\)\[RightAngleBracket]\)]], "=", Cell[BoxData[ \(\(-\@2\)\)]], "\[Mu], the ", Cell[BoxData[ \(TraditionalForm\`D\_1\)]], " reduced matrix elements are given by ", Cell[BoxData[ FormBox[ RowBox[{ \(\[LeftAngleBracket]\((J' I)\) F' || \[Mu]\^1 || \((J\ I)\) F\[RightAngleBracket]\), " ", "=", " ", RowBox[{ \(\((\(-1\))\)\^\(J' + I + F + 1\)\), \(\@\(\((2 F' + 1)\) \((2 F + 1)\)\)\), RowBox[{"{", GridBox[{ {\(J'\), "I", \(F'\)}, {"F", "1", "J"} }], "}"}], \(\[LeftAngleBracket]J' || \[Mu]\^1 || J \[RightAngleBracket]\)}]}], TraditionalForm]]], " using the function g1:" }], "Text"], Cell[BoxData[ \(g1[fp_, f_] := \ sixJcoeff[1/2, 3/2, fp, 1/2, f]\[Times]\((\(-\@2\))\)\)], "Input"], Cell[CellGroupData[{ Cell["Calculations", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(g1[2, 2]\)], "Input"], Cell[BoxData[ \(\(-\@\(5\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g1[1, 2]\)], "Input"], Cell[BoxData[ \(\@\(5\/2\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g1[2, 1]\)], "Input"], Cell[BoxData[ \(\(-\@\(5\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(g1[1, 1]\)], "Input"], Cell[BoxData[ \(1\/\@2\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Results", "Subsubsection"], Cell[TextData[{ Cell[BoxData[""], "Input"], Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 1\/2, I = 3\/2)\) F' = \(2 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \(-\@\(5\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 1\/2, I = 3\/2)\) F' = \(1 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 2 \)\[RightAngleBracket] = \@\(5\/2\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 1\/2, I = 3\/2)\) F' = \(2 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1 \)\[RightAngleBracket] = \(-\@\(5\/2\)\)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\[LeftAngleBracket]\((J' = 1\/2, I = 3\/2)\) F' = \(1 || \[Mu]\^1 || \ \((J = 1\/2, \ I = 3\/2)\) F = 1 \)\[RightAngleBracket] = 1\/\@2\)]], "\nChecking with the sum-rule:\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(F'\)\(( \(|\[LeftAngleBracket]\((J' I)\) F' || \[Mu]\^1 || \((J\ I)\) F\[RightAngleBracket] | \))\)\^2 = \(\((\(|\[LeftAngleBracket]J' || \[Mu]\^1 || J\[RightAngleBracket] | \))\)\^2\) \(2 F + 1\)\/\(2 J + 1\)\)]], "\nFor ", Cell[BoxData[ \(TraditionalForm\`F = 2, \ J = 1\/2\)]], ", we have: " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\((\(-\@\(5\/2\)\))\)\^2 + \((\@\(5\/2\))\)\^2 == \(\((\(-\@2\))\)\^2\) \(2\ \((2)\) + 1\)\/\(2 \((1\/2)\) + 1\)\)], "Input"], Cell[BoxData[ \(True\)], "Output"] }, Open ]], Cell[TextData[{ Cell[BoxData[ \(\ \)], "Input"], " For ", 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