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\(\(\(=\)\(\(\(e\^2\)\(\ \)\)\/\(2 \[Pi]\^2\)\ \(\[Integral]\+\(-\ \[Infinity]\)\%\[Infinity]\ \ \(1\/q\) ln \((\(q/2 + \ k\ - i0\)\/\(q/2 - \ k\ - i0\))\) \((k\^2 + m\^2 - q\^2/ 4)\) \(k \(\(\[DifferentialD]k\)\(\ \)\)\)\/\@\(k\^2 + m\^2\ \)\)\)\)\), "\[IndentingNewLine]", \(\)}], "NumberedEquation"], Cell[BoxData[ \(\(-P\) \((q\^2)\) = \(\(2\) \( e\^2\)\(\ \)\)\/\[Pi]\ \ \(\[Integral]\+\(-\[Infinity]\)\%\[Infinity]\ \ \(1\/q\) ln \((\(q/2 + \ k\ - i0\)\/\(q/2 - \ k\ - i0\))\) \((k\^2 + m\^2 - q\^2/ 4)\) \(k \(\(\[DifferentialD]k\)\(\ \)\)\)\/\@\(k\^2 + m\^2\ \)\)\)], "NumberedEquation"], Cell[BoxData[{ \(\(-P\) \((q\^2)\) = \(\(\(2\) \( e\^2\)\(\ \)\)\/\[Pi]\ \(\[Integral]\+\ \(-\[Infinity]\)\%\[Infinity]\ \ \(1\/q\) ln \((\(q/2 + \ k\ - i0\)\/\(q/2 - \ k\ - i0\))\) \ \[DifferentialD]\([\((\(k\^2 + m\^2\)\/3 - q\^2\/4)\) \@\(k\^2 + m\^2\)\ ]\)\) = \ \[IndentingNewLine]\(-\(\(\(2\) \( e\^2\)\(\ \)\)\/\[Pi]\)\)\ \ \(\[Integral]\+\(-\[Infinity]\)\%\[Infinity]\ \ \(1\/q\) \((1\/\(q/2 + \ k\ \ - i0\) + 1\/\(q/2 - \ k\ - i0\))\)\)\)\), "\[IndentingNewLine]", \(\t\t\t\([\((\(k\^2 + m\^2\)\/3 - q\^2\/4)\) \@\(k\^2 + m\^2\)\ ]\) \[DifferentialD]k = \ \[IndentingNewLine]\(-\(\(\(2\) \( e\^2\)\(\ \)\)\/\[Pi]\)\) \(\[Integral]\+\(-\[Infinity]\)\ \%\[Infinity]\ \ \(1\/\(q\^2/4 - \ k\^2\ - i0\)\) \((\(k\^2 + m\^2\)\/3 - q\^2\/4)\) \@\(k\^2 + m\^2\)\ \[DifferentialD]k\)\)}], \ "NumberedEquation"], Cell["\<\ Again throwing away a const (having in mind the renormalization, see below)\ \>", "Text", CellFrame->True], Cell[BoxData[ \(P \((q\^2)\) = \(\(\(2\) \( e\^2\)\(\ \)\)\/\[Pi]\ \(\[Integral]\+\(-\ \[Infinity]\)\%\[Infinity]\ \ \(1\/\(q\^2/4 - \ k\^2\ - i0\)\) \((\(m\^2 - 2 k\^2\)\/3\ )\) \@\(k\^2 + m\^2\)\ \ \[DifferentialD]k\) = \[IndentingNewLine]\(\(2\)\(\ \)\(e\^2\)\(\ \)\)\/\(3 \ \[Pi]\)\ \(\[Integral]\+\(-\[Infinity]\)\%\[Infinity]\ \ \(1\/\(q\^2/4 - \ k\^2\ - i0\)\) \((m\^2 - 2 k\^2\ )\) \@\(k\^2 + m\^2\)\ \[DifferentialD]k\)\)\)], \ "NumberedEquation"], 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the upper semiplane \ of the complex plane k. The point k = i m on the imaginary axes is a branch \ point. Make the integration counter C to begin at k=i\[Infinity], going down \ to k=i m, then circle clockwise and return back to k=i\[Infinity]. Making a \ substitution k\[Rule]i k one finds\ \>", "Text", CellFrame->True], Cell[BoxData[ \(P \((q\^2)\) = \(i\^2\) \(\(4\)\(\ \)\)\/\(3 \[Pi]\)\ \(e\^2\) \(\ \[Integral]\+m\%\[Infinity]\ \ \(1\/\(\(q\^2/ 4\)\(+\)\(\ \)\(k\^2\)\(\ \)\)\) \((m\^2 + 2 k\^2\ )\) \@\(k\^2 - m\^2\)\ \ \[DifferentialD]k\)\)], \ "NumberedEquation"], Cell["\<\ Here an additional coefficient 2 comes from the two branches of the \ counter.\ \>", "Text", CellFrame->True], Cell[BoxData[ \(P \((q\^2)\) = \(-\(\(16\ e\^2\)\/\(3 \[Pi]\)\)\)\ \(\[Integral]\+m\%\ \[Infinity]\ \ \(1\/\(q\^2 + \(\(4\)\(\ \)\(k\^2\)\(\ \)\)\)\) \((m\^2 + 2 k\^2\ )\) \@\(k\^2 - m\^2\)\ \ \[DifferentialD]k\)\)], \ "NumberedEquation"], Cell[BoxData[{ \(\(\(4 k\^2 = t'\)\(\[IndentingNewLine]\) \)\), "\[IndentingNewLine]", \(k = \(1\/2\) \@t'\), "\[IndentingNewLine]", \(dk = \(1\/4\) dt'\/\@t'\)}], "NumberedEquation"], Cell[BoxData[ \(\(\(\[IndentingNewLine]\)\(P \((q\^2)\) \[Rule] \(-\(\(\(16\)\(\ \ \)\)\ \/\(3 \[Pi]\)\)\)\ \(e\^2\) \(1\/4\) \(1\/2\) \(1\/2\) \(\[Integral]\+\(4 m\ \^2\)\%\[Infinity]\ \ \(1\/\(\(q\^2\)\(+\)\(t'\)\(\ \)\)\) \((\ 2 m\^2 + t')\) \@\(t' - 4\ m\^2\)\ \ \[DifferentialD]t'\/\@t'\) = \ \[IndentingNewLine]\[IndentingNewLine]\(\(-\(\(\(\ \)\(e\^2\)\)\/\(3 \[Pi]\)\ \)\)\ \(\[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \(1\/\(\(q\^2\)\(+\)\(t'\)\(\ \ \)\)\) \((2 m\^2 + t'\ )\) \@\(t - 4\ m\^2\)\ \ \[DifferentialD]t'\/\@t'\) = \ \[IndentingNewLine]\(P \((q\^2)\) = \(-\(\(\(e\^2\)\(\ \ \)\)\/\(3\ \[Pi]\)\)\ \)\ \(\[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \(1\/\(\(q\^2\)\(+\)\(t'\)\(\ \ \)\)\) \((t' + 2 m\^2\ )\) \@\(\(t' - 4\ m\^2\)\/t'\)\ \ \ \[DifferentialD]t'\)\)\)\)\)\)], "NumberedEquation"], Cell["Renormalization", "Subtitle", Background->None], Cell[BoxData[ \(\(P\_Reg\) \((q\^2)\) = P \((q\^2)\) - P \((0)\) - \(q\^2\) P' \((0)\)\)], "NumberedEquation"], Cell[BoxData[{ \(The\ constant\ was\ thrown\ away\ a\ couple\ of\ times\ above . \ This\ makes\ no\ \), "\[IndentingNewLine]", \(difference, \ since\ the\ renormalization\ takes\ the\ constant\ away\ \ \(\(anyway\)\(.\)\)\)}], "Text", CellFrame->True], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[ 1\/\(q\^2 + t'\) - Normal[Series[ 1\/\(\(q\^2\)\(+\)\(t'\)\(\ \)\), {q, 0, 2}]]]\)], "Input", Background->None], Cell[BoxData[ FractionBox[\(q\^4\), RowBox[{ SuperscriptBox[ RowBox[{"(", SuperscriptBox["t", "\[Prime]", MultilineFunction->None], ")"}], "2"], " ", RowBox[{"(", RowBox[{\(q\^2\), "+", SuperscriptBox["t", "\[Prime]", MultilineFunction->None]}], ")"}]}]]], "Output", Background->None] }, Open ]], Cell[BoxData[{\(Thus, \ regularization\ means\), "\[IndentingNewLine]", RowBox[{\(1\/\(q\^2 + t'\)\), "\[Rule]", FractionBox[\(q\^4\), RowBox[{ SuperscriptBox[ RowBox[{"(", SuperscriptBox["t", "\[Prime]", MultilineFunction->None], ")"}], "2"], " ", RowBox[{"(", RowBox[{\(q\^2\), "+", SuperscriptBox["t", "\[Prime]", MultilineFunction->None]}], ")"}]}]]}]}], "NumberedEquation"], Cell[BoxData[ RowBox[{\(\(P\_Reg\) \((q\^2)\)\), "=", RowBox[{\(-\(\(\(\ \)\(e\^2\)\)\/\(3 \[Pi]\)\)\), " ", RowBox[{\(\[Integral]\+\(4 m\^2\)\%\[Infinity]\), " ", RowBox[{ FractionBox[\(q\^4\), RowBox[{ SuperscriptBox[ RowBox[{"(", SuperscriptBox["t", "\[Prime]", MultilineFunction->None], ")"}], "2"], " ", RowBox[{"(", RowBox[{\(q\^2\), "+", SuperscriptBox["t", "\[Prime]", MultilineFunction->None]}], ")"}]}]], \((t' + 2 m\^2\ )\), \(\@\(\(t' - 4\ m\^2\)\/t'\)\), " ", \(\[DifferentialD]t'\)}]}]}]}]], "NumberedEquation"], Cell[BoxData[ RowBox[{\(P \((q\^2)\)\), "=", RowBox[{\(-\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \[Pi]\)\)\), \(q\^4\), StyleBox[\(\[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \ \(1\/\(\(t'\)\(+\)\(q\^2\)\(-\)\(i0\)\(\ \)\)\) \(t' + 2 m\^2\)\/t'\^2\ \@\(\ \(t' - 4\ m\^2\)\/t'\)\ \ \[DifferentialD]t'\), FontSize->18]}]}]], "NumberedEquation", CellFrame->True], Cell[BoxData[{ RowBox[{"t", "=", RowBox[{"-", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"]}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalP]", FontWeight->"Bold"], \((t)\)}], "=", " ", RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalP]", FontWeight->"Bold"], RowBox[{"(", RowBox[{"-", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"]}], ")"}]}], "=", RowBox[{"P", RowBox[{"(", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"], ")"}]}]}]}]}], "NumberedEquation"], Cell[BoxData[ RowBox[{ RowBox[{"t", "=", RowBox[{"-", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"]}]}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ StyleBox["\[ScriptCapitalP]", FontWeight->"Bold"], \((t)\)}], "=", \(\(-\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \[Pi]\)\)\) \(t\^2\) \(\ \[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \(1\/\(\(t'\)\(-\)\(t\)\(-\)\(i0\)\(\ \ \)\)\) \(t' + 2 m\^2\)\/t'\^2\ \@\(\(t' - 4\ m\^2\)\/t'\)\ \ \ \[DifferentialD]t'\)\)}]}]], "NumberedEquation"], Cell[BoxData[{ \(t' = 4 \( m\^2\) \[Zeta]\^2\), "\[IndentingNewLine]", \(P \((q\^2)\) = \(\(-\(\(\(\ \)\(e\^2\)\)\/\(3 \[Pi]\)\)\) \(q\^4\) \(\ \[Integral]\+1\%\[Infinity]\ \ \(1\/\(\(4 \( m\^2\) \[Zeta]\^2\)\(+\)\(q\^2\)\(\ \)\)\) \ \(\[Zeta]\^2 + 1/2\)\/\[Zeta]\^4\ \@\(\[Zeta]\^2 - 1\)\/\[Zeta]\ \ 2 \[Zeta] \ \[DifferentialD]\[Zeta]\) = \(-\(\(2\ e\^2\)\/\(3 \[Pi]\)\)\) \(q\^4\) \(\ \[Integral]\+1\%\[Infinity]\ \ \(1\/\(\(4 \( m\^2\) \[Zeta]\^2\)\(+\)\(q\^2\)\(\ \)\)\) \ \(\[Zeta]\^2 + 1/2\)\/\[Zeta]\^4\ \@\(\[Zeta]\^2 - 1\)\ \ \[DifferentialD]\ \[Zeta]\)\)\)}], "NumberedEquation"], Cell[BoxData[ \(P \((q\^2)\) = \(-\(\(2\ \[Alpha]\)\/\(3 \[Pi]\)\)\) \(q\^4\) \(\ \[Integral]\+1\%\[Infinity]\ \ \(1\/\(\(4 \( m\^2\) \[Zeta]\^2\)\(+\)\(q\^2\)\(\ \)\)\) \((1 + 1\/\(2 \[Zeta]\^2\))\) \(\@\(\[Zeta]\^2 - 1\)\ \ \ \[DifferentialD]\[Zeta]\)\/\[Zeta]\^2\)\)], "NumberedEquation"], Cell[BoxData[ \(P \((q\^2)\)\ \[TildeEqual] \(-\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \ \[Pi]\)\)\) \(q\^2\) ln \((q\^2\/m\^2)\)\ , \ \ q\^2 \[GreaterGreater] m\^2\)], "NumberedEquation"], Cell["Scalar particles", "Subtitle", Background->None], Cell["For spinor particles it was found that", "Text", CellFrame->True], Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["P", FontSize->16], RowBox[{ StyleBox["(", FontSize->16], SuperscriptBox[ StyleBox["q", FontSize->16], "2"], ")"}]}], "=", RowBox[{ StyleBox[\(\(-\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \[Pi]\)\)\) \(q\^4\) \(\ \[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \ \(1\/\(\(t'\)\(+\)\(q\^2\)\(-\)\(i0\)\(\ \)\)\) \(t' + 2 m\^2\)\/t'\^2\ \@\(\ \(t' - 4\ m\^2\)\/t'\)\ \ \[DifferentialD]t'\)\), FontSize->16], StyleBox["=", FontSize->16], StyleBox["\[IndentingNewLine]", FontSize->24], StyleBox[\(\(-\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \[Pi]\)\)\) \(q\^4\) \(\ \[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \ \(1\/\(\(t'\)\(+\)\(q\^2\)\(-\)\(i0\)\(\ \)\)\) \(2 m\^2 - t'/2 + 3 \ t'/2\)\/t'\^2\ \@\(\(t' - 4\ m\^2\)\/t'\)\ \ \[DifferentialD]t'\)\), FontSize->16]}]}]], "NumberedEquation"], Cell[TextData[{ "Here ", Cell[BoxData[ \(TraditionalForm\`m\^2 - t'/2\)]], " comes from the charge contribution, while ", Cell[BoxData[ \(TraditionalForm\`3 t'/2\)]], " originates from the spin. These two terms have different signs. The \ charge tends to make P(q) positive, while the spin results in a opposite, \ negative contribution, which dominates." }], "Text", CellFrame->True], Cell[TextData[{ "Calculating the polarization for scalar particles one needs to make three \ amendments:\n1. To change the total sign, which is related to statistics \ (i.e. to eliminate the negative energy)\n2. To eliminate the spin \ contribution, i.e. the term ", Cell[BoxData[ \(TraditionalForm\`3 t'/2\)]], ". \n3.To reduce the result by a factor of 2, which exists for fermions due \ to their spin 1/2, but is absent for scalars.\nThen the polarization ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["P", FontSlant->"Plain"], "0"], TraditionalForm]]], " created by scalar particles reads" }], "Text", CellFrame->True], Cell[BoxData[ RowBox[{\(\(P\_0\) \((q\^2)\)\), "=", RowBox[{ RowBox[{ RowBox[{"-", RowBox[{"(", StyleBox[\(-\(1\/2\)\), FontSize->16], StyleBox[")", FontSize->16]}]}], StyleBox[\(\(\(\ \)\(\[Alpha]\)\)\/\(3 \[Pi]\)\), FontSize->16], StyleBox[\(q\^4\), FontSize->16], StyleBox[\(\[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \ \(1\/\(\(t'\)\(+\)\(q\^2\)\(-\)\(i0\)\(\ \)\)\) \(2 m\^2 - t'/2\)\/t'\^2\ \@\ \(\(t' - 4\ m\^2\)\/t'\)\ \ \[DifferentialD]t'\), FontSize->16]}], StyleBox["=", FontSize->16], "\[IndentingNewLine]", RowBox[{\(-\(1\/2\)\), StyleBox[\(\(\(\ \)\(\[Alpha]\)\)\/\(6 \[Pi]\)\), FontSize->16], StyleBox[\(q\^4\), FontSize->16], StyleBox[\(\[Integral]\+\(4 m\^2\)\%\[Infinity]\ \ \ \(1\/\(\(t'\)\(+\)\(q\^2\)\(-\)\(i0\)\(\ \)\)\) \(t' - 4 m\^2\)\/t'\^2\ \@\(\ \(t' - 4\ m\^2\)\/t'\)\ \ \[DifferentialD]t'\), FontSize->16]}]}]}]], "NumberedEquation"], Cell[BoxData[ RowBox[{ RowBox[{\(\(P\_0\) \((q\^2)\)\), "\[TildeEqual]", RowBox[{ RowBox[{"-", StyleBox[\(\(\(\ \)\(\[Alpha]\)\)\/\(12 \[Pi]\)\), FontSize->16]}], StyleBox[\(q\^2\), FontSize->16], StyleBox["ln", FontSize->16], RowBox[{ StyleBox["(", FontSize->16], FractionBox[ SuperscriptBox[ StyleBox["q", FontSize->16], "2"], \(m\^2\)], ")"}]}]}], ",", " ", RowBox[{\(q\^2\), "=", RowBox[{ SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"], "\[GreaterGreater]", \(m\^2\)}]}]}]], "NumberedEquation"], Cell["\<\ The sign of the vacuum polarization is same for fermions ( Landau, \ Abrikosov, Khalatnikov, 1954) ) and scalars ( Fradkin, 1955? ).\ \>", "Text", CellFrame->True], Cell["Uehling potential", "Subtitle", Background->None], Cell[BoxData[{ RowBox[{\(\[Delta]U \((r)\)\), "=", RowBox[{\(-4\), "\[Pi]", " ", "Z", " ", \(e\^2\), " ", "Re", RowBox[{"\[Integral]", RowBox[{ FractionBox[\(P \((q\^2)\)\), SuperscriptBox[ RowBox[{"(", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"], ")"}], "2"]], SuperscriptBox["e", RowBox[{ StyleBox[ RowBox[{"i", StyleBox["q", FontWeight->"Bold"]}]], StyleBox["\[CenterDot]", FontWeight->"Bold"], StyleBox["r", FontWeight->"Bold"]}]], " ", FractionBox[ RowBox[{\(\[DifferentialD]\^3\), StyleBox["q", FontWeight-> "Bold"]}], \(\((2 \[Pi])\)\^3\)]}]}]}]}], "\ \[IndentingNewLine]", \(P \((q\^2)\) = \(-\(\(2\ \[Alpha]\)\/\(3 \[Pi]\)\)\) \ \(q\^4\) \(\[Integral]\+1\%\[Infinity]\ \ \(1\/\(\(4 \( m\^2\) \[Zeta]\^2\)\(+\)\(q\^2\)\(\ \)\)\) \((1 + 1\/\(2 \[Zeta]\^2\))\) \(\@\(\[Zeta]\^2 - 1\)\/\[Zeta]\^2\) \ \[DifferentialD]\[Zeta]\)\)}], "NumberedEquation"], Cell[BoxData[ RowBox[{ RowBox[{"\[Integral]", RowBox[{\(1\/\(q\^2 + \[Mu]\^2\)\), SuperscriptBox["e", RowBox[{ StyleBox[ RowBox[{"i", StyleBox["q", FontWeight->"Bold"]}]], StyleBox["\[CenterDot]", FontWeight->"Bold"], StyleBox["r", FontWeight->"Bold"]}]], FractionBox[ RowBox[{\(\[DifferentialD]\^3\), StyleBox["q", FontWeight->"Bold"]}], \(\((2 \[Pi])\)\^3\)]}]}], "=", \(\(exp \((\(-\[Mu]r\))\)\)\/\(4 \[Pi]\ r\)\)}]], \ "NumberedEquation"], Cell[BoxData[ RowBox[{"\[Delta]U", "=", RowBox[{"4", "\[Pi]", " ", "Z", " ", \(e\^2\), " ", "Re", RowBox[{"\[Integral]", RowBox[{ FractionBox[\(P \((q\^2)\)\), SuperscriptBox[ RowBox[{"(", SuperscriptBox[ StyleBox["q", FontWeight->"Bold"], "2"], ")"}], "2"]], SuperscriptBox["e", RowBox[{ StyleBox[ RowBox[{"i", StyleBox["q", FontWeight->"Bold"]}]], StyleBox["\[CenterDot]", FontWeight->"Bold"], StyleBox["r", FontWeight->"Bold"]}]], " ", FractionBox[ RowBox[{\(\[DifferentialD]\^3\), StyleBox["q", FontWeight-> "Bold"]}], \(\((2 \[Pi])\)\^3\)]}]}]}]}]], \ "NumberedEquation"], Cell[BoxData[ RowBox[{\(\[Delta]U \((r)\)\), "=", RowBox[{ RowBox[{ "4", "\[Pi]", " ", "Z", " ", \(e\^2\), \((\(-\(\(2\ \[Alpha]\)\/\(3 \[Pi]\)\)\))\), RowBox[{\(\[Integral]\+1\%\[Infinity]\), RowBox[{\(\[DifferentialD]\[Zeta]\), " ", RowBox[{"\[Integral]", " ", RowBox[{ FractionBox[ RowBox[{\(\[DifferentialD]\^3\), StyleBox["q", FontWeight->"Bold"]}], \(\((2 \[Pi])\)\^3\)], FractionBox[ SuperscriptBox["e", RowBox[{ StyleBox[ RowBox[{"i", StyleBox["q", FontWeight->"Bold"]}]], StyleBox["\[CenterDot]", FontWeight->"Bold"], StyleBox["r", FontWeight->"Bold"]}]], \(\(4 \( m\^2\) \[Zeta]\^2\)\(+\)\(q\^2\)\(\ \)\)], \((1 + 1\/\(2 \[Zeta]\^2\))\), \(\(\(\@\(\[Zeta]\^2 - 1\)\)\(\ \ \)\)\/\[Zeta]\^2\)}]}]}]}]}], "=", "\[IndentingNewLine]", "\[IndentingNewLine]", \(\(-\(\(Z\ e\^2\)\/r\)\) \(\(2\ \ \[Alpha]\)\/\(3 \[Pi]\)\) \(\[Integral]\+1\%\[Infinity]\[DifferentialD]\ \[Zeta]\ \ \((1 + 1\/\(2 \[Zeta]\^2\))\) \(\(\(\@\(\[Zeta]\^2 - 1\)\)\(\ \ \)\)\/\[Zeta]\^2\) e\^\(\(-2\) mr\[Zeta]\)\)\)}]}]], "NumberedEquation"], Cell[BoxData[{ \(U \((r)\) = \(-\(\(Z\ e\^2\)\/r\)\)\), "\[IndentingNewLine]", \(U \((r)\) + \[Delta]U = \(-\(Ze\^2\/r\)\) \((1 + \(\(2\ \[Alpha]\)\/\(3 \ \[Pi]\)\) \(\[Integral]\+1\%\[Infinity]\ \((1 + 1\/\(2 \[Zeta]\^2\))\) \(\(\(\@\(\[Zeta]\^2 - 1\)\)\(\ \ \)\)\/\[Zeta]\^2\) e\^\(\(-2\) mr\[Zeta]\)\ \ \[DifferentialD]\[Zeta]\)\ \ )\)\)}], "NumberedEquation"], Cell[BoxData[{ \(Consider\ \ r \[GreaterGreater] 1\/m\), "\[IndentingNewLine]", \(\[Delta]U \[TildeEqual] \(-\(Ze\^2\/r\)\) \(\(2\ \[Alpha]\)\/\(3 \[Pi]\ \)\) \(\[Integral]\+1\%\[Infinity]\[DifferentialD]\[Zeta]\ \ \((1 + 1\/2)\) \@\(2 \((\[Zeta] - 1)\)\)\ \ e\^\(\(-2\) \ mr\[Zeta]\)\) = \[IndentingNewLine]\(\(-\(Ze\^2\/r\)\) \(\(\@2\ \[Alpha]\)\/\ \[Pi]\) \(\[Integral]\+1\%\[Infinity]\[DifferentialD]\[Zeta] \@\( \[Zeta] - 1\ \)\ \ e\^\(\(-2\) mr\[Zeta]\)\)\( = \+\(\[Zeta] \[Rule] \[Zeta] + 1\)\)\[IndentingNewLine]\(\(-\(Ze\^2\/r\)\) \(\(\@2\ \[Alpha]\ \)\/\[Pi]\) \(\[Integral]\+0\%\[Infinity]\[DifferentialD]\[Zeta] \@ \ \[Zeta]\ \ e\^\(\(-2\) mr \((\[Zeta] + 1)\)\)\)\( = \+\(\[Zeta] = x\^2\)\)\[IndentingNewLine]\(\(-\(Ze\^2\/r\)\) \(\@2\ \ \[Alpha]\)\/\[Pi]\ \(e\^\(\(-2\) mr\)\) 2 \(\[Integral]\+0\%\[Infinity]\[DifferentialD]x\ x\^2\ \ e\^\(\ \(-2\) mrx\^2\)\) = \[IndentingNewLine]\(\(\(\(-\(Ze\^2\/r\)\) \(\@2\ \ \[Alpha]\)\/\[Pi]\ \(e\^\(\(-2\) mr\)\) \((\(-\(d\/d\[Lambda]\)\))\) \@\(\[Pi]\/\ \[Lambda]\)\)\( | \+\(\[Lambda] = 2 mr\)\)\) = \[IndentingNewLine]\(-\(Ze\^2\/r\)\) \(\(\ \ \)\(\[Alpha]\)\)\/\@\(2 \[Pi]\)\ e\^\(\(-2\) mr\)\/\((2 \ mr)\)\^\(3/2\)\)\)\)\)\)}], "NumberedEquation"], Cell[BoxData[ \(\[Delta]U \[TildeEqual] \(-\(Ze\^2\/r\)\) \(\(\ \)\(\[Alpha]\)\)\/\@\(2 \ \[Pi]\)\ e\^\(\(-2\) mr\)\/\((2 mr)\)\^\(3/2\), \ r \[GreaterGreater] 1\/m\)], "NumberedEquation"], Cell[BoxData[ \(\[Delta]U \[TildeEqual] \(-\(Ze\^2\/r\)\) \(\(2\ \[Alpha]\)\/\(3 \[Pi]\ \)\) ln \((1\/mr)\), \ \ \ \ \ \ \ r \[LessLess] 1\/m\)], "NumberedEquation"], Cell["Landau (Moscow) pole", "Subtitle", CellFrame->True, Background->None], Cell[BoxData[ \(\(e\^2\) \((r)\) \[TildeEqual] \(e\^2\) \((\ 1 + \(\(2\ \[Alpha]\)\/\(3 \[Pi]\)\) ln \((1 + 1\/mr)\)\ \ \ )\)\)], "NumberedEquation"], Cell["\<\ More accurate account of polarization can be obtained by iterating the vacuum \ polarization, which gives\ \>", "Text", CellFrame->True], Cell[BoxData[ StyleBox[\(\(e\^2\) \((r)\) \[TildeEqual] e\^2\/\(1 - 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