(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.1' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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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[ 38455, 1452]*) (*NotebookOutlinePosition[ 39417, 1482]*) (* CellTagsIndexPosition[ 39373, 1478]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Answers to Exam II (Nov. 4-7 2005)", "Subtitle"], Cell[CellGroupData[{ Cell["Prob 1", "Subsection"], Cell[CellGroupData[{ Cell["LS coupling", "Subsubsection"], Cell[TextData[{ "For LS coupling of identical particles, we have the restriction L+S is \ even. Therefore,\n(3s3s) \[RightArrow] ", Cell[BoxData[ \(TraditionalForm\`\(\^1\) S\)]], " 1 substate\n", Cell[BoxData[ \(TraditionalForm\`\(\((3 p3p)\)\( \[RightArrow] \^1\)\)\)]], "S, ", Cell[BoxData[ \(TraditionalForm\`\^3\)]], "P, ", Cell[BoxData[ \(TraditionalForm\`\^1\)]], "D 1+3\[Times]3+5 = 15 substates\n", Cell[BoxData[ \(TraditionalForm\`\(\((3 d3d)\)\( \[RightArrow] \^1\)\)\)]], "S, ", Cell[BoxData[ \(TraditionalForm\`\^3\)]], "P, ", Cell[BoxData[ \(TraditionalForm\`\^1\)]], "D, ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\(\ \)\(\^3\)\), "TraditionalForm"], \(F \(\(\(,\)\(\ \)\)\^1\) G\)}], TraditionalForm]]], " 1+3\[Times]3+5+3\[Times]7+9 = 45 substates\n", Cell[BoxData[ \(TraditionalForm\`\(\((3 s3d)\)\( \[RightArrow] \^1\)\)\)]], "D, ", Cell[BoxData[ \(TraditionalForm\`\^3\)]], "D 5+3\[Times]5 = 20 substates\nTotal number \ of substates = 1+15+45+20=81" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["jj coupling", "Subsubsection"], Cell[TextData[{ "For jj coupling of identical particles J is even. Therefore,\n", Cell[BoxData[ \(TraditionalForm\`\((3 s\_\(1/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 s\_\(1/2\)\)]], ") \[RightArrow] [0] 1 substate\n", Cell[BoxData[ \(TraditionalForm\`\((3 p\_\(1/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(1/2\)\)]], ") \[RightArrow] [0] 1 substate\n", Cell[BoxData[ \(TraditionalForm\`\((3 p\_\(1/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(3/2\)\)]], ") \[RightArrow] [1], [2] 3+5 = 8 substates\n", Cell[BoxData[ \(TraditionalForm\`\((3 p\_\(3/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(3/2\)\)]], ") \[RightArrow] [0], [2] 1+5 = 6 substates \n", Cell[BoxData[ \(TraditionalForm\`\((3 d\_\(3/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 d\_\(3/2\)\)]], ") \[RightArrow] [0], [2] 1+5 = 6 substates\n", Cell[BoxData[ \(TraditionalForm\`\((3 d\_\(3/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 d\_\(5/2\)\)]], ") \[RightArrow] [1], [2], [3], [4] 3+5+7+9 = 24 substates\n", Cell[BoxData[ \(TraditionalForm\`\((3 d\_\(5/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 d\_\(5/2\)\)]], ") \[RightArrow] [0], [2], [4] 1+5+9 = 15 substates\n", Cell[BoxData[ \(TraditionalForm\`\((3 s\_\(1/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 d\_\(3/2\)\)]], ") \[RightArrow] [1], [2] 3+5 = 8 substates\n", Cell[BoxData[ \(TraditionalForm\`\((3 s\_\(1/2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 d\_\(5/2\)\)]], ") \[RightArrow] [2], [3] 5+7 = 12 substates\nTotal \ number of substates = 1+(1+8+6)+(6+24+15)+(8+12)=1+15+45+20=81" }], "Text"], Cell["\<\ A mathematica version of the abpbe tables is constructed as \ follows:\ \>", "Text"], Cell[BoxData[ \(tLS\ := TableForm[{{"\", "\<\>", "\<\>", "\<\>", "\<\>", "\<\>", \ "\"}, \[IndentingNewLine]{"\<(3s3s)\>", \*"\"\<\!\(\(\(\\\ \ \)\(\^1\)\)\)S\>\"", "\<\>", "\<\>", "\<\>", "\<\>", 1}, {"\<(3p3p)\>", \*"\"\<\!\(\(\(\\\ \)\(\^1\)\)\)S\>\"", \ \ \*"\"\<\!\(\(\(\\\ \)\(\^3\)\)\)P\>\"", \*"\"\<\!\(\(\(\\\ \ \)\(\^1\)\)\)D\>\"", "\<\>", "\<\>", 1 + \ 3\ 3 + 5}, \ {"\<(3d3d)\>", \*"\"\<\!\(\(\(\\\ \)\(\^1\)\)\)S\>\"", \ \ \*"\"\<\!\(\(\(\\\ \)\(\^3\)\)\)P\>\"", \*"\"\<\!\(\(\(\\\ \ \)\(\^1\)\)\)D\>\"", \ \*"\"\<\!\(\(\(\\\ \)\(\^3\)\)\)F\>\"", \ \*"\"\<\!\(\(\(\\\ \)\(\^1\)\)\)G\>\"", 1 + 3\ 3\ + 5 + \ 3\ 7\ + \ 9}, {"\<(3s3d)\>", \*"\"\<\!\(\(\(\\\ \)\(\^1\)\)\)D\>\"", \ \ \*"\"\<\!\(\(\(\\\ \)\(\^3\)\)\)D\>\"", "\<\>", "\<\>", "\<\>", 5 + 3\ 5}}]\)], "Input", CellLabel->"In[1]:="], Cell[BoxData[ \(tjj\ := TableForm[{{"\", "\<\>", "\<\>", "\<\>", "\<\>", "\"}, \ \[IndentingNewLine]{\*"\"\<(3\!\(s\_\(1/2\)\)3\!\(s\_\(1/2\)\))\>\"", \ "\<[0]\>", "\<\>", "\<\>", "\<\>", 1}, {\*"\"\<(3\!\(p\_\(1/2\)\)3\!\(p\_\(1/2\)\))\>\"", "\<[0]\>", \ \ "\<\>", "\<\>", "\<\>", 1}, {\*"\"\<(3\!\(p\_\(1/2\)\)3\!\(p\_\(3/2\)\))\>\"", "\<[1]\>", \ \ "\<[2]\>", "\<\>", "\<\>", 3 + 5}, \ \ \[IndentingNewLine]{\*"\"\<(3\!\(p\_\(3/2\)\)3\!\(p\_\(3/2\)\))\>\"", \ "\<[0]\>", \ "\<[2]\>", "\<\>", "\<\>", 1 + \ 5}, \ \ {\*"\"\<(3\!\(d\_\(3/2\)\)3\!\(d\_\(3/2\)\))\>\"", \ "\<[0]\>", \ "\<[2]\>", "\<\>", \ "\<\>", 1 + 5}, {\*"\"\<(3\!\(d\_\(3/2\)\)3\!\(d\_\(5/2\)\))\>\"", "\<[1]\ \>", \ "\<[2]\>", "\<[3]\>", "\<[4]\>", 3 + 5 + 7 + 9}, \[IndentingNewLine]{\*"\"\<(3\!\(d\_\(5/2\)\)3\!\(d\_\(5/2\)\ \))\>\"", "\<[0]\>", \ "\<[2]\>", "\<[4]\>", \ "\<\>", 1 + 5 + 9}, \[IndentingNewLine]{\*"\"\<(3\!\(s\_\(1/2\)\)3\!\(d\_\ \(3/2\)\))\>\"", "\<[1]\>", \ "\<[2]\>", "\<\>", "\<\>", 3 + 5}, {\*"\"\<(3\!\(s\_\(1/2\)\)3\!\(d\_\(5/2\)\))\>\"", "\<[2]\ \>", \ "\<[3]\>", "\<\>", "\<\>", 5 + 7}}]\)], "Input", CellLabel->"In[2]:="], Cell[BoxData[ \(LStot\ := \ Sum[tLS[\([1, k, Length[tLS[\([1, 2]\)]]]\)], {k, 2, Length[tLS[\([1]\)]]}]\)], "Input", CellLabel->"In[3]:="], Cell[BoxData[ \(jjtot\ := \ Sum[tjj[\([1, k, Length[tjj[\([1, 2]\)]]]\)], {k, 2, Length[tjj[\([1]\)]]}]\)], "Input", CellLabel->"In[4]:="], Cell[CellGroupData[{ Cell[BoxData[ \(tLS\)], "Input", CellLabel->"In[5]:="], Cell[BoxData[ TagBox[GridBox[{ {"\<\"LS\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", \ "\<\"\"\>", "\<\"Sum\"\>"}, {"\<\"(3s3s)\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^1\\)\\)\\)S\"\>", \ "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "1"}, {"\<\"(3p3p)\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^1\\)\\)\\)S\"\>", \ "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^3\\)\\)\\)P\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\ \\^1\\)\\)\\)D\"\>", "\<\"\"\>", "\<\"\"\>", "15"}, {"\<\"(3d3d)\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^1\\)\\)\\)S\"\>", \ "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^3\\)\\)\\)P\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\ \\^1\\)\\)\\)D\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^3\\)\\)\\)F\"\>", \ "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^1\\)\\)\\)G\"\>", "45"}, {"\<\"(3s3d)\"\>", "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^1\\)\\)\\)D\"\>", \ "\<\"\\!\\(\\(\\(\\\\ \\)\\(\\^3\\)\\)\\)D\"\>", "\<\"\"\>", "\<\"\"\>", \ "\<\"\"\>", "20"} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], Function[ BoxForm`e$, TableForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[5]//TableForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(tjj\)], "Input", CellLabel->"In[6]:="], Cell[BoxData[ TagBox[GridBox[{ {"\<\"jj\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", \ "\<\"Sum\"\>"}, {"\<\"(3\\!\\(s\\_\\(1/2\\)\\)3\\!\\(s\\_\\(1/2\\)\\))\"\>", \ "\<\"[0]\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "1"}, {"\<\"(3\\!\\(p\\_\\(1/2\\)\\)3\\!\\(p\\_\\(1/2\\)\\))\"\>", \ "\<\"[0]\"\>", "\<\"\"\>", "\<\"\"\>", "\<\"\"\>", "1"}, {"\<\"(3\\!\\(p\\_\\(1/2\\)\\)3\\!\\(p\\_\\(3/2\\)\\))\"\>", \ "\<\"[1]\"\>", "\<\"[2]\"\>", "\<\"\"\>", "\<\"\"\>", "8"}, {"\<\"(3\\!\\(p\\_\\(3/2\\)\\)3\\!\\(p\\_\\(3/2\\)\\))\"\>", \ "\<\"[0]\"\>", "\<\"[2]\"\>", "\<\"\"\>", "\<\"\"\>", "6"}, {"\<\"(3\\!\\(d\\_\\(3/2\\)\\)3\\!\\(d\\_\\(3/2\\)\\))\"\>", \ "\<\"[0]\"\>", "\<\"[2]\"\>", "\<\"\"\>", "\<\"\"\>", "6"}, {"\<\"(3\\!\\(d\\_\\(3/2\\)\\)3\\!\\(d\\_\\(5/2\\)\\))\"\>", \ "\<\"[1]\"\>", "\<\"[2]\"\>", "\<\"[3]\"\>", "\<\"[4]\"\>", "24"}, {"\<\"(3\\!\\(d\\_\\(5/2\\)\\)3\\!\\(d\\_\\(5/2\\)\\))\"\>", \ "\<\"[0]\"\>", "\<\"[2]\"\>", "\<\"[4]\"\>", "\<\"\"\>", "15"}, {"\<\"(3\\!\\(s\\_\\(1/2\\)\\)3\\!\\(d\\_\\(3/2\\)\\))\"\>", \ "\<\"[1]\"\>", "\<\"[2]\"\>", "\<\"\"\>", "\<\"\"\>", "8"}, {"\<\"(3\\!\\(s\\_\\(1/2\\)\\)3\\!\\(d\\_\\(5/2\\)\\))\"\>", \ "\<\"[2]\"\>", "\<\"[3]\"\>", "\<\"\"\>", "\<\"\"\>", "12"} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], Function[ BoxForm`e$, TableForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[6]//TableForm="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(LStot\)], "Input", CellLabel->"In[7]:="], Cell[BoxData[ \(81\)], "Output", CellLabel->"Out[7]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(jjtot \[Equal] LStot\)], "Input", CellLabel->"In[8]:="], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[8]="] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Prob 2", "Subsection"], Cell[TextData[{ "The ground-state energy is E[ ", Cell[BoxData[ \(TraditionalForm\`\((1 s)\)\^2\)]], " ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\^1\), "TraditionalForm"], \(S\_0\)}], TraditionalForm]]], "] = 2", Cell[BoxData[ \(TraditionalForm\`\[Epsilon]\_\(1 s\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`R\_0\)]], "(1s1s1s1s) - ", Cell[BoxData[ \(TraditionalForm\`2 U\_\(1 s1s\)\)]], ", where U = ", Cell[BoxData[ \(TraditionalForm\`5\/16\)]], Cell[BoxData[ \(TraditionalForm\`1\/r\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Epsilon]1s\ = \ \(-Zs^2\)/2\)], "Input", CellLabel->"In[9]:="], Cell[BoxData[ \(\(-\(Zs\^2\/2\)\)\)], "Output", CellLabel->"Out[9]="] }, Open ]], Cell[BoxData[ \(P1s[r_]\ := \ 2\ Zs^\((3/2)\)\ r\ Exp[\(-Zs\)\ r]\)], "Input", CellLabel->"In[10]:="], Cell[CellGroupData[{ Cell[BoxData[ \(U1s1s\ = \ \(\((5/16)\) \(Integrate[P1s[r]^2\ /r\ , {r, 0, Infinity}, Assumptions \[Rule] \ Zs > 0]\)\(\ \)\)\)], "Input", CellLabel->"In[11]:="], Cell[BoxData[ \(\(5\ Zs\)\/16\)], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[BoxData[ \(Y0s[r_]\ := Integrate[P1s[x]^2, {x, 0, r}, Assumptions \[Rule] \ Zs > 0]/r\ + Integrate[P1s[x]^2/x, {x, r, Infinity}, Assumptions \[Rule] \ Zs > 0]\)], "Input", CellLabel->"In[12]:="], Cell[CellGroupData[{ Cell[BoxData[ \(R0s\ = \ Integrate[\ P1s[r]^2\ Y0s[r], {r, 0, Infinity}, Assumptions \[Rule] \ Zs > 0]\)], "Input", CellLabel->"In[13]:="], Cell[BoxData[ \(\(5\ Zs\)\/8\)], "Output", CellLabel->"Out[13]="] }, Open ]], Cell[TextData[{ "The final result for the ground-state energy is ", Cell[BoxData[ \(TraditionalForm\`\(-Zs\^2\)\)]], " (as should be expected from first section of Chap 4)" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e1S0\ = \ 2\ \[Epsilon]1s\ + R0s\ - 2 U1s1s\)], "Input", CellLabel->"In[14]:="], Cell[BoxData[ \(\(-Zs\^2\)\)], "Output", CellLabel->"Out[14]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(e1S0\ /. \ Zs \[Rule] 5\ - \ 5/16 // N\)], "Input", CellLabel->"In[15]:="], Cell[BoxData[ \(\(-21.97265625`\)\)], "Output", CellLabel->"Out[15]="] }, Open ]], Cell[TextData[{ "Compare to NIST : IP = 2091960 ", Cell[BoxData[ \(TraditionalForm\`cm\^\(-1\)\)]], " [ground state energy = ", Cell[BoxData[ \(TraditionalForm\`E\_ion\)]], " - IP = -", Cell[BoxData[ \(TraditionalForm\`Z\^2\/2\)]], "-IP(a.u.)] " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eNist\ = \ \(-12.5\)\ - \ 2091960/219474.624\)], "Input", CellLabel->"In[16]:="], Cell[BoxData[ \(\(-22.03167141546168`\)\)], "Output", CellLabel->"Out[16]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(del\ = \ 100 \((\((e1S0 /. \ Zs \[Rule] 5 - 5/16)\) - eNist)\)/eNist\)], "Input",\ CellLabel->"In[17]:="], Cell[BoxData[ \(\(-0.2678651308328062`\)\)], "Output", CellLabel->"Out[17]="] }, Open ]], Cell["The ground state energy differs from measured value by 0.27%", "Text"], Cell["", "Text"], Cell[TextData[{ "Now let's look at the (1s2p) singlet and triplet P states. The energy of \ the 2P level is \n", Cell[BoxData[ \(TraditionalForm\`\(\(E\ \ [\(\(\((1 s2p)\)\(\ \ \)\)\^\(2 S + 1\)\) P]\)\(\ \)\)\)]], " = ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ FormBox[\(\[Epsilon]\_\(1 s\)\), "TraditionalForm"], "+", \(\[Epsilon]\_\(2 p\)\)}], "TraditionalForm"], "+", \(\(R\_0\)(1 s2p1s2p)\), " ", "+"}], TraditionalForm]]], " ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(\((\(-1\))\)\^S\/3\), "TraditionalForm"], \(R\_1\)}], TraditionalForm]]], "(1s,2p,2p,1s) -U(1s1s)-U(2p2p). \nNote that the counter potential U for \ the 2p state is ", Cell[BoxData[ \(TraditionalForm\`1\/r\)]], "\nAs a first step , we calculate the integrals ", Cell[BoxData[ \(TraditionalForm\`R\_0\)]], "(1s2p1s2p) ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(&\)\(\ \)\(R\_1\)\)\)\)]], "(1s,2p,2p,1s)" }], "Text"], Cell[BoxData[ \(P2p[r_]\ := \ Zp^\((5/2)\)/\((2 Sqrt[6])\)\ r^2 Exp[\(-Zp\)\ r/2]\)], "Input", CellLabel->"In[18]:="], Cell[CellGroupData[{ Cell[BoxData[ \(R0 = Integrate[\ P2p[r]^2\ Y0s[r], {r, 0, Infinity}, Assumptions \[Rule] \ {Zp > 0, Zs > 0}]\)], "Input", CellLabel->"In[19]:="], Cell[BoxData[ \(1\/4\ Zp\^5\ \((1\/Zp\^4 - \(Zp + 6\ Zs\)\/\((Zp + 2\ Zs)\)\^5)\)\)], \ "Output", CellLabel->"Out[19]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(R0\ /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[20]:="], Cell[BoxData[ \(0.9807861609592465`\)], "Output", CellLabel->"Out[20]="] }, Open ]], Cell[BoxData[ \(Y1[r_]\ := Integrate[P2p[x] P1s[x]\ x, {x, 0, r}, Assumptions \[Rule] \ {Zp > 0, Zs > 0}]/r^2\ + r\ Integrate[P2p[x] P1s[x]/x^2, {x, r, Infinity}, Assumptions \[Rule] \ {Zp > 0, Zs > 0}]\)], "Input", CellLabel->"In[21]:="], Cell[CellGroupData[{ Cell[BoxData[ \(R1\ = \ Integrate[\ P2p[r]\ P1s[r]\ Y1[r], {r, 0, Infinity}, Assumptions \[Rule] \ {Zp > 0, Zs > 0}]\)], "Input", CellLabel->"In[22]:="], Cell[BoxData[ \(\(112\ Zp\^5\ Zs\^3\)\/\((Zp + 2\ Zs)\)\^7\)], "Output", CellLabel->"Out[22]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(R1\ /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[23]:="], Cell[BoxData[ \(0.1542713523780816`\)], "Output", CellLabel->"Out[23]="] }, Open ]], Cell["Now, we evaluate the counter term U(2p2p)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(U2p2p\ = \ Integrate[P2p[r]^2\ /r\ , {r, 0, Infinity}, Assumptions \[Rule] \ Zp > 0]\)], "Input", CellLabel->"In[24]:="], Cell[BoxData[ \(Zp\/4\)], "Output", CellLabel->"Out[24]="] }, Open ]], Cell[TextData[{ "The lowest order energy ", Cell[BoxData[ \(TraditionalForm\`\[Epsilon]\_\(\(2\) \(p\)\(\ \)\)\)]], "of the 2p orbital is " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Epsilon]2p\ = \ \(-Zp^2\)/8\)], "Input", CellLabel->"In[25]:="], Cell[BoxData[ \(\(-\(Zp\^2\/8\)\)\)], "Output", CellLabel->"Out[25]="] }, Open ]], Cell["Put together everything for singlets then triplets", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e2pS\ = \ \(\(\[Epsilon]1s\)\(+\)\(\[Epsilon]2p\)\(\ \)\(+\)\(\ \)\(R0\ \)\(\ \)\(+\)\(\ \)\(\((1/3)\) R1\)\(\ \)\(-\)\(U1s1s\)\(\ \)\(-\)\(U2p2p\)\(\ \)\)\)], "Input", CellLabel->"In[26]:="], Cell[BoxData[ \(\(-\(Zp\/4\)\) - Zp\^2\/8 - \(5\ Zs\)\/16 - Zs\^2\/2 + \(112\ Zp\^5\ Zs\^3\)\/\(3\ \((Zp + 2\ Zs)\)\^7\) + 1\/4\ Zp\^5\ \((1\/Zp\^4 - \(Zp + 6\ Zs\)\/\((Zp + 2\ Zs)\)\^5)\)\)], \ "Output", CellLabel->"Out[26]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(e2pT\ = \ \[Epsilon]1s + \[Epsilon]2p\ + \ R0\ - \ \((1/3)\) R1\ - U1s1s\ - U2p2p\)], "Input", CellLabel->"In[27]:="], Cell[BoxData[ \(\(-\(Zp\/4\)\) - Zp\^2\/8 - \(5\ Zs\)\/16 - Zs\^2\/2 - \(112\ Zp\^5\ Zs\^3\)\/\(3\ \((Zp + 2\ Zs)\)\^7\) + 1\/4\ Zp\^5\ \((1\/Zp\^4 - \(Zp + 6\ Zs\)\/\((Zp + 2\ Zs)\)\^5)\)\)], \ "Output", CellLabel->"Out[27]="] }, Open ]], Cell["\<\ The multiplet ine separation is e2pS-e2pT. The NIST value is \ 0.095935 (Theory gives a value about 5% too large)\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e2pS - e2pT\ /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[28]:="], Cell[BoxData[ \(0.10284756825205439`\)], "Output", CellLabel->"Out[28]="] }, Open ]], Cell["\<\ Next, we set up the singlet and triplet energies relative to the \ ground state (These are the excitation energies!)\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e2pS\ - \ e1S0 /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[29]:="], Cell[BoxData[ \(7.553694320085275`\)], "Output", CellLabel->"Out[29]="] }, Open ]], Cell["The NIST value is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(nistS\ = \ 1658020/219474.624\)], "Input", CellLabel->"In[30]:="], Cell[BoxData[ \(7.554495229480379`\)], "Output", CellLabel->"Out[30]="] }, Open ]], Cell["For the triplet state", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(e2pT\ - \ e1S0 /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[31]:="], Cell[BoxData[ \(7.45084675183322`\)], "Output", CellLabel->"Out[31]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(nistT\ = \ 1636882/219474.624\)], "Input", CellLabel->"In[32]:="], Cell[BoxData[ \(7.458183411673141`\)], "Output", CellLabel->"Out[32]="] }, Open ]], Cell["The percentage differences with NIST are", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]S\ = \ 100 \((e2pS - e1S0 - nistS)\)/nistS\ /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[33]:="], Cell[BoxData[ \(\(-0.010601759227799799`\)\)], "Output", CellLabel->"Out[33]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]T\ = \ 100 \((e2pT - e1S0 - nistT)\)/nistT\ /. \ {Zp \[Rule] 4, \ Zs \[Rule] 5 - 5/16}\ // N\)], "Input", CellLabel->"In[34]:="], Cell[BoxData[ \(\(-0.0983706009219051`\)\)], "Output", CellLabel->"Out[34]="] }, Open ]], Cell["\<\ The predicted excitation energies within 0.1% of NIST, while the \ predicted splitting is about 5% too large.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Prob 3", "Subsection"], Cell[TextData[{ "The reduced matrix element ", Cell[BoxData[ \(TraditionalForm\`R\_E1\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(<\)\(\((1 s)\)\^2\)\)\)]], Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(1\^1\)\)\)]], "S || r || (2p1s) ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(\(2\)\(\ \)\)\^1\) P\)\(>\)\)\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\@2\)\)\)]], " <0 ||C1||1> ", Cell[BoxData[ \(TraditionalForm\`R\_\(1 s\ 2 p\)\)]], " , where \n ", Cell[BoxData[ \(TraditionalForm\`R\_\(1 s\ 2 p\)\)]], " = \[Integral] ", Cell[BoxData[ \(TraditionalForm\`P\_\(2 p\)\)]], "(r) r ", Cell[BoxData[ \(TraditionalForm\`P\_\(1 s\)\)]], "(r) dr is the radial dipole integral." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Rps\ = \ Integrate[P2p[r] r\ P1s[r], {r, 0, Infinity}, Assumptions \[Rule] \ {Zp > 0, Zs > 0}]\)], "Input", CellLabel->"In[35]:="], Cell[BoxData[ \(\(128\ \@6\ Zp\^\(5/2\)\ Zs\^\(3/2\)\)\/\((Zp + 2\ Zs)\)\^5\)], "Output",\ CellLabel->"Out[35]="] }, Open ]], Cell["Let cps = <0 ||C1||1>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(cps\ = \ Sqrt[3]\ ThreeJSymbol[{0, 0}, {1, 0}, {1, 0}]\)], "Input", CellLabel->"In[36]:="], Cell[BoxData[ \(\(-1\)\)], "Output", CellLabel->"Out[36]="] }, Open ]], Cell["Then, we find that the reduced matrix element is:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(RE1\ = \ Sqrt[2]\ cps\ Rps\)], "Input", CellLabel->"In[37]:="], Cell[BoxData[ \(\(-\(\(256\ \@3\ Zp\^\(5/2\)\ Zs\^\(3/2\)\)\/\((Zp + 2\ \ Zs)\)\^5\)\)\)], "Output", CellLabel->"Out[37]="] }, Open ]], Cell["The associated line strength is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(SE1\ = \ RE1^2\)], "Input", CellLabel->"In[38]:="], Cell[BoxData[ \(\(196608\ Zp\^5\ Zs\^3\)\/\((Zp + 2\ Zs)\)\^10\)], "Output", CellLabel->"Out[38]="] }, Open ]], Cell["The transition energy from the NIST web site in 1/cm is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Eexp\ = \ 1657980\)], "Input", CellLabel->"In[39]:="], Cell[BoxData[ \(1657980\)], "Output", CellLabel->"Out[39]="] }, Open ]], Cell["The corresponding wavelength (Angstrom) is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda]\ = \ N[10^8/Eexp]\)], "Input", CellLabel->"In[40]:="], Cell[BoxData[ \(60.314358436169314`\)], "Output", CellLabel->"Out[40]="] }, Open ]], Cell[TextData[{ "The transition rate is A = ", Cell[BoxData[ \(TraditionalForm\`\(2.02613\ 10\^18\)\/\[Lambda]\^3\)]], " ", Cell[BoxData[ \(TraditionalForm\`SE1\/g\_a\)]], " in 1/s, where ", Cell[BoxData[ \(TraditionalForm\`g\_a\)]], " is the initial state degeneracy (3 for a singlet P state)" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"A", " ", "=", " ", RowBox[{ RowBox[{ FormBox[\(\(2.02613\ 10\^18\)\/\[Lambda]\^3\), "TraditionalForm"], " ", FormBox[\(SE1\/g\_a\), "TraditionalForm"]}], " ", "/.", " ", \({Zs \[Rule] 5 - 5/16, \ Zp \[Rule] \ 4, \ g\_a \[Rule] \ 3}\)}]}]], "Input", CellLabel->"In[41]:="], Cell[BoxData[ \(3.483942605007109`*^11\)], "Output", CellLabel->"Out[41]="] }, Open ]], Cell["The lifetime is \[Tau]= 1/A (s)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Tau]\ = \ 1/A\)], "Input", CellLabel->"In[42]:="], Cell[BoxData[ \(2.8703113494545056`*^-12\)], "Output", CellLabel->"Out[42]="] }, Open ]], Cell[TextData[{ "This compares favorably with the \"exact\" result 2.69 ", Cell[BoxData[ \(TraditionalForm\`10\^\(-12\)\)]], " s." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Prob 4", "Subsection"], Cell[CellGroupData[{ Cell["H 3p state", "Subsubsection"], Cell[TextData[{ "The 3p level in H decays by E1 emission to both 2s and 1s levels. The \ A-coefficient is ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(2.02613\ 10\^18\)\/\[Lambda]\^3\)\)\)]], Cell[BoxData[ \(TraditionalForm\`SE1\/3\)]], " for each transition. The total rate is the sum of the partial rates to \ each final state. The measured wavelengths in Angstrom of the transitions \ to the 2s and 1s states are ", Cell[BoxData[ \(6564.5\)]], " and ", Cell[BoxData[ \(1025.7\)]], ", respectively." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Details of a H 3p rate calculation.", "Subsubsection"], Cell[BoxData[ \(P3p[r_]\ := \ 8 Z^\((5/2)\)/\((27\ Sqrt[6])\)\ r^2\ \((1 - Z\ r/6)\)\ Exp[\(-Z\)\ r/3]\)], "Input", CellLabel->"In[43]:="], Cell[BoxData[ \(P2s[r_]\ := \ Z^\((3/2)\)/Sqrt[2]\ r\ \((1 - Z\ r/2)\)\ Exp[\(-Z\)\ r\ /2]\)], "Input",\ CellLabel->"In[44]:="], Cell["The radial matrix elements are ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(R32\ = \ Integrate[\ P2s[r]\ r\ P3p[r], {r, 0, Infinity}, Assumptions \[Rule] \ Z > 0]\)], "Input", CellLabel->"In[45]:="], Cell[BoxData[ \(\(27648\ \@3\)\/\(15625\ Z\)\)], "Output", CellLabel->"Out[45]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(R31\ = \ Integrate[\((\ P1s[r] /. Zs \[Rule] Z)\)\ \ r\ P3p[r], {r, 0, Infinity}, Assumptions \[Rule] \ \ Z > 0]\)], "Input", CellLabel->"In[46]:="], Cell[BoxData[ \(\(27\ \@\(3\/2\)\)\/\(64\ Z\)\)], "Output", CellLabel->"Out[46]="] }, Open ]], Cell["The angular factor is cps = ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(cps\ = \ Sqrt[3]\ ThreeJSymbol[{0, 0}, {1, 0}, {1, 0}]\)], "Input", CellLabel->"In[47]:="], Cell[BoxData[ \(\(-1\)\)], "Output", CellLabel->"Out[47]="] }, Open ]], Cell[TextData[{ "The line strengths for the 3p \[RightArrow] 2s transitions is S32 = \^2\)\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`R32\^2\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(S32\ = \ cps^2\ R32^2\)], "Input", CellLabel->"In[48]:="], Cell[BoxData[ \(2293235712\/\(244140625\ Z\^2\)\)], "Output", CellLabel->"Out[48]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(S31\ = \ cps^2\ R31^2\)], "Input", CellLabel->"In[49]:="], Cell[BoxData[ \(2187\/\(8192\ Z\^2\)\)], "Output", CellLabel->"Out[49]="] }, Open ]], Cell[TextData[{ "The transition energy (au) for the 3p \[RightArrow] 2s transitions is \ \[CapitalDelta]E32 = ", Cell[BoxData[ \(TraditionalForm\`Z\^2\)]], "/(2 ", Cell[BoxData[ \(TraditionalForm\`2\^2\)]], ") - ", Cell[BoxData[ \(TraditionalForm\`Z\^2\)]], "/(2 ", Cell[BoxData[ \(TraditionalForm\`3\^2\)]], ")" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]E32\ = \ \(Z^2/2\)/2^2\ - \ \(Z^2/2\)/3^2\)], "Input", CellLabel->"In[50]:="], Cell[BoxData[ \(\(5\ Z\^2\)\/72\)], "Output", CellLabel->"Out[50]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]E31\ = \ \(Z^2/2\)/1^2\ - \ \(Z^2/2\)/3^2\)], "Input", CellLabel->"In[51]:="], Cell[BoxData[ \(\(4\ Z\^2\)\/9\)], "Output", CellLabel->"Out[51]="] }, Open ]], Cell["The conversion from (a.u.) to 1/cm is 219474.62 ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(con\ = \ 219474.62\)], "Input", CellLabel->"In[52]:="], Cell[BoxData[ \(219474.62`\)], "Output", CellLabel->"Out[52]="] }, Open ]], Cell["The (32) and (31) transition energies in 1/cm are", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]cm32\ = \ con\ \[CapitalDelta]E32\)], "Input", CellLabel->"In[53]:="], Cell[BoxData[ \(15241.293055555556`\ Z\^2\)], "Output", CellLabel->"Out[53]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[CapitalDelta]cm31\ = \ con\ \[CapitalDelta]E31\)], "Input", CellLabel->"In[54]:="], Cell[BoxData[ \(97544.27555555555`\ Z\^2\)], "Output", CellLabel->"Out[54]="] }, Open ]], Cell["The transition wavelengths (angstrom) are", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda]32\ = \ 10^8/\[CapitalDelta]cm32\)], "Input", CellLabel->"In[55]:="], Cell[BoxData[ \(6561.123103892377`\/Z\^2\)], "Output", CellLabel->"Out[55]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda]31\ = \ 10^8/\[CapitalDelta]cm31\)], "Input", CellLabel->"In[56]:="], Cell[BoxData[ \(1025.175484983184`\/Z\^2\)], "Output", CellLabel->"Out[56]="] }, Open ]], Cell[TextData[{ "The Einstein A-coefficients are ", Cell[BoxData[ RowBox[{\(A\_31\), "=", RowBox[{ FormBox[\(\(2.02613\ 10\^18\)\/\[Lambda]\_31\^3\), "TraditionalForm"], FormBox[\(S31\/3\), "TraditionalForm"], " ", "and", " "}]}]]], " ", Cell[BoxData[ RowBox[{\(A\_32\), "=", RowBox[{ FormBox[\(\(2.02613\ 10\^18\)\/\[Lambda]\_32\^3\), "TraditionalForm"], FormBox[\(S32\/3\), "TraditionalForm"]}]}]]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(A31\ = \ 2.02631\ 10^\((18)\)/\[Lambda]31^3\ S31/3\)], "Input", CellLabel->"In[57]:="], Cell[BoxData[ \(1.6735890871944028`*^8\ Z\^4\)], "Output", CellLabel->"Out[57]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(A32\ = \ 2.02631\ 10^\((18)\)/\[Lambda]32^3\ S32/3\)], "Input", CellLabel->"In[58]:="], Cell[BoxData[ \(2.2462532488882676`*^7\ Z\^4\)], "Output", CellLabel->"Out[58]="] }, Open ]], Cell["The total rate is the sum of partial rates", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(A3\ = \ A31 + A32\)], "Input", CellLabel->"In[59]:="], Cell[BoxData[ \(1.8982144120832294`*^8\ Z\^4\)], "Output", CellLabel->"Out[59]="] }, Open ]], Cell[TextData[{ "The lifetime (s) of the 3p state in H is ", Cell[BoxData[ \(TraditionalForm\`A\_3\%\(-1\)\)]], " evaluated with Z=1" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Tau]\ = \ 1/A3\ /. \ Z \[Rule] 1\)], "Input", CellLabel->"In[60]:="], Cell[BoxData[ \(5.2681087744062174`*^-9\)], "Output", CellLabel->"Out[60]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`Al\^\(\(2\)\(+\)\)\)]], " (Na-like) 3p state" }], "Subsubsection"], Cell[TextData[{ "The 3p state in ", Cell[BoxData[ \(TraditionalForm\`Al\^\(\(2\)\(+\)\)\)]], " decays to the 3s state by E1 emission. The A-coefficient is ", Cell[BoxData[ \(TraditionalForm\`\(2.02613\ 10\^18\)\/\[Lambda]\^3\)]], Cell[BoxData[ \(TraditionalForm\`SE1\/3\)]], "\nThe average transition energy is (53684.1+2\[Times]53916.6)/3 ", Cell[BoxData[ \(TraditionalForm\`cm\^\(-1\)\)]], "and the corresponding average wavelength in Angstrom is" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda] = \ 10\^8\ 3/\((53684.1 + 2\ 53916.6)\)\)], "Input", CellLabel->"In[61]:="], Cell[BoxData[ \(1857.3861747317474`\)], "Output", CellLabel->"Out[61]="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`Al\)]], " ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(3/2\)\)]], " state" }], "Subsubsection"], Cell[TextData[{ "The ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(3/2\)\)]], " state in Al decays to the ", Cell[BoxData[ \(TraditionalForm\`3 p\_\(1/2\)\)]], "ground state by M1 emission. The A-coefficient is ", Cell[BoxData[ \(TraditionalForm\`\(2.69735\ 10\^13\)\/\[Lambda]\^3\)]], Cell[BoxData[ \(TraditionalForm\`S\_M1\/4\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`S\_M1\)]], "= | ", Cell[BoxData[ \(TraditionalForm\`\(\(<\)\(p\_\(1/2\)\)\)\)]], "|| ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(L + 2 S || p\_\(\(3/2\)\(\ \)\)\), "TraditionalForm"], ">"}], " ", \( | \^2\)}], TraditionalForm]]], " The transition energy is 112.04 ", Cell[BoxData[ \(TraditionalForm\`cm\^\(-1\)\)]], " and the corresponding wavelength in Angstrom is" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda] = \ 10\^8/112.04\)], "Input", CellLabel->"In[62]:="], Cell[BoxData[ \(892538.3791503033`\)], "Output", CellLabel->"Out[62]="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(Ba\^+\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`5 d\_\(3/2\)\)]], "state" }], "Subsubsection"], Cell[TextData[{ "The ", Cell[BoxData[ \(TraditionalForm\`5 d\_\(3/2\)\)]], " state in ", Cell[BoxData[ \(TraditionalForm\`\(Ba\^+\)\)]], " decays to the ", Cell[BoxData[ \(TraditionalForm\`6 s\_\(1/2\)\)]], " ground state by E2 emission. The A-coefficient is ", Cell[BoxData[ \(TraditionalForm\`\(1.1198\ 10\^18\)\/\[Lambda]\^5\)]], Cell[BoxData[ \(TraditionalForm\`S\_E2\/4\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`S\_E2\)]], "= | ", Cell[BoxData[ \(TraditionalForm\`\(\(<\)\(5 d\_\(3/2\)\)\)\)]], "|| ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(Q\^\((2)\) || 6 s\_\(\(1/2\)\(\ \)\)\), "TraditionalForm"], ">"}], " ", \( | \^2\)}], TraditionalForm]]], " The transition energy is 4873.85 ", Cell[BoxData[ \(TraditionalForm\`cm\^\(-1\)\)]], " and the corresponding wavelength in Angstrom is" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Lambda] = \ 10\^8/4873.85\)], "Input", CellLabel->"In[63]:="], Cell[BoxData[ \(20517.660576341084`\)], "Output", CellLabel->"Out[63]="] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Prob 5", "Subsection"], Cell[CellGroupData[{ Cell["H 2s decay", "Subsubsection"], Cell[TextData[{ "The only state in H below the ", Cell[BoxData[ \(TraditionalForm\`2 s\_\(1/2\)\)]], " state is the ", Cell[BoxData[ \(TraditionalForm\`1 s\_\(1/2\)\)]], " state. The multipolarity J of the transition is limited by the angular \ momentum selections to 0= |1/2-1/2|\[LessSlantEqual] J \ \[LessSlantEqual]1/2+1/2 = 1. Therefore, since the parity of the two states \ is relatively even, only an M1 transition is possible. Since the \ nonrelativistic M1 amplitude is nonvanishing only for ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[\(n\_a\), "TraditionalForm"], "=", \(n\_b\)}], ","}], TraditionalForm]]], " the rate ", Cell[BoxData[ \(TraditionalForm\`A\_\(1 s\ 2 s\)\%M1\)]], " vanishes and the lifetime to single-photon decay is infinite. (This state \ actually decays to the ground state by a two-photon transition and has a \ lifetime of about 1/7 s.)" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "He (1s2s) ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \^1\)\(S\_0\)\)\)]] }], "Subsubsection"], Cell[TextData[{ "This state cannot decay by single-photon emission to the ", Cell[BoxData[ \(TraditionalForm\`\((1 s)\)\^2\)]], " ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\^1\)\) S\_0\)]], " ground state, since the triangle relation would require multipolarity \ J=0! 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