(************** Content-type: application/mathematica **************
                     CreatedBy='Mathematica 5.1'

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Notebook[{

Cell[CellGroupData[{
Cell["Lamb shift", "Title",
  FontSize->16],

Cell[CellGroupData[{

Cell["Introduction", "Section",
  FontSize->16],

Cell["\<\
\tSuppose there is  a one-electron state 0. An interation  of electrons with \
the electromaelegnetic field  mixes this state with a set of states N, in  \
which the electron occupies  some state n, and additionally there is a \
photon, which can be described by  its momentum and polarization. As a \
result, in the second order of the perturbation theory there appears an \
energy shift of the level |0\[RightAngleBracket]. This shift, as well as \
physical reasons producing it,  is called the Lab shift. 
\tA potential initiating this effect can be written in terms of the velocity \
of the electron and the vector potential \
\>", "Text",
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  CellMargins->{{Inherited, 0}, {Inherited, Inherited}},
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  TextAlignment->Left,
  TextJustification->0,
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Cell[BoxData[
    \(V = \(-e\)\ \(v\& \[Rule] \)\[CenterDot]\(A\& \[Rule] \)\)], \
"NumberedEquation",
  FontSize->16],

Cell["\<\
Here the vector potential describes a photon with the given momentum and \
polarization \
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\(\(\(A\& \[Rule] \) = \(\@\(\(4  \[Pi]\)\/\(2  \[Omega]\)\)\) \(\
\[Epsilon]\& \[Rule] \)\ Exp[\[ImaginaryI] \((\(k\& \[Rule] \)\[CenterDot]\(r\
\& \[Rule] \) - \[Omega]\ t)\)]\)\(,\)\)\)], "NumberedEquation",
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Cell[BoxData[{
    RowBox[{
      RowBox[{\(k\& \[Rule] \), " ", ",", 
        RowBox[{\(\[Epsilon]\& \[Rule] \), " ", 
          StyleBox["are",
            FontFamily->"Times New Roman"], 
          StyleBox[" ",
            FontFamily->"Times New Roman"], 
          StyleBox["the",
            FontFamily->"Times New Roman"], 
          StyleBox[" ",
            FontFamily->"Times New Roman"], 
          StyleBox["momentum",
            FontFamily->"Times New Roman"], 
          StyleBox[" ",
            FontFamily->"Times New Roman"], 
          StyleBox["and",
            FontFamily->"Times New Roman"], 
          StyleBox[" ",
            FontFamily->"Times New Roman"], 
          StyleBox["polarization",
            FontFamily->"Times New Roman"]}], 
        StyleBox[",",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox[\(\[Omega] = 
            k\ is\ the\ photon\ frequency . \ \ The\ normalization\),
          FontFamily->"Times New Roman"]}], 
      StyleBox[" ",
        FontFamily->"Times New Roman"]}], "\[IndentingNewLine]", 
    RowBox[{
      RowBox[{
        StyleBox["factor",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox[\(\@\(2  \[Pi]/\[Omega]\)\),
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox["is",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox["justified",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        RowBox[{
          StyleBox["below",
            FontFamily->"Times New Roman"], 
          StyleBox[".",
            FontFamily->"Times New Roman"], "\[IndentingNewLine]", 
          StyleBox[
            RowBox[{"\t", 
              StyleBox[" ",
                FontFamily->"Times New Roman"]}]], 
          StyleBox["In",
            FontFamily->"Times New Roman"]}], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox["the",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox["dipole",
          FontFamily->"Times New Roman"], 
        StyleBox[" ",
          FontFamily->"Times New Roman"], 
        StyleBox["approximation",
          FontFamily->"Times New Roman"]}], 
      StyleBox[",",
        FontFamily->"Times New Roman"], 
      StyleBox[\(which\ we\ will\ use\ below\),
        FontFamily->"Times New Roman"], 
      StyleBox[",",
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      StyleBox[" ",
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      StyleBox[\(the\ vector - potential\ simplifies\),
        FontFamily->"Times New Roman"]}]}], "Text",
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Cell[BoxData[
    \(\(A\& \[Rule] \) \[TildeEqual] \@\(\(4  \[Pi]\)\/\(2  \
\[Omega]\)\)\ \ \(\[Epsilon]\& \[Rule] \)\ \ Exp[\(-\[ImaginaryI]\[Omega]\)\ \
t]\)], "NumberedEquation",
  FontSize->16],

Cell[BoxData[
    StyleBox[\(The\ matrix\ element, \ 
      which\ mixes\ the\ states\ 0\ and\ N\ reads\),
      FontFamily->"Times New Roman"]], "Text",
  FontSize->16],

Cell[BoxData[
    \(V\_\(N, 0\) = \(-e\)\ \@\(\(4  \[Pi]\)\/\(2  \[Omega]\)\)\ \(\(\(v\& \
\[Rule] \)\_\(n, 0\)\ \[CenterDot]\ \ \(\[Epsilon]\& \[Rule] \)\)\(\ \)\)\)], \
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Cell["For the energy shift one finds ", "Text",
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Cell[BoxData[{
    \(\[Delta]E\_0 = \[Sum]\+N\(\(|\)\(V\_\(N, 0\)\)\( | \^2\)\)\/\(E\_0 - \
E\_N\)\), "\n", 
    \(\(\(=\)\(4  \[Pi]\ \(e\^2\) \(\[Sum]\+\(k, \[Epsilon]\)\(\[Sum]\+n\( 
                    1\/\(2  \[Omega]\)\) \(\(|\)\(\(v\& \[Rule] \)\_\(n, 0\)\ \
\[CenterDot]\ \ \(\[Epsilon]\& \[Rule] \)\)\(\ \)\( | \^2\)\)\/\(E\_0 - E\_n \
- \[Omega]\)\)\) = \[IndentingNewLine]4  \[Pi]\ \(e\^2\) \(\[Sum]\+\(\
\[Epsilon] = 1, 2\)\[Integral]\(\[DifferentialD]\^3 
                      k\/\((2  \[Pi])\)\^3\) \(\[Sum]\+n\( 
                        1\/\(2  \[Omega]\)\) \(\(|\)\(\(v\& \[Rule] \)\_\(n, \
0\)\ \[CenterDot]\ \ \(\[Epsilon]\& \[Rule] \)\)\(\ \)\( | \^2\)\)\/\(E\_0 - \
E\_n - \[Omega]\)\)\)\)\)\)}], "NumberedEquation",
  FontSize->16],

Cell["\<\
Here it is taken into account that the summation over the photon intemediate \
states results in the integral over the photon momentum and summation over \
its polarixation.  \
\>", "Text",
  FontSize->16]
}, Open  ]],

Cell[CellGroupData[{

Cell["Normalization of the vector potential", "Section",
  FontSize->16],

Cell[CellGroupData[{

Cell["Preliminary remarks", "Subsection",
  FontSize->16],

Cell["\<\
Suppose one needs to fulfill summation over a set of a continuum states of \
some particle. One can  take some large volume, call it V, and assume that it \
is greater than the volume in which physical events considered take place. \
Then instead of summation over the continuum states one fulfilles summation \
over quasi-continuum set of states, for which periodical boundary conditions \
in the volume V are valid. After that one rewrites the summation as the \
integral using the following formulae  \
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[Sum]\+n\( \[Psi]\_\(\(p\& \[Rule] \)\_n\)\) \((\(r\& \[Rule] \))\) \(\
\[Psi]\_\(\(p\& \[Rule] \)\_n\)\%*\) \((\(r\& \[Rule] \)')\)\ \( \[Rule] \
\+\(V \[Rule] \[Infinity]\)\)\ 
      V \(\[Integral]\(\[DifferentialD]\^3 
                  p\/\((2  \[Pi])\)\^3\) \(\[Psi]\_\(p\& \[Rule] \
\)\%\((V)\)\) \((\(r\& \[Rule] \))\) \(\[Psi]\_\(p\& \[Rule] \)\%\(\((V)\)\(*\
\)\)\) \((\(r\& \[Rule] \)')\)\)\)], "NumberedEquation",
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Cell["\<\
Here the superscript (V) indicates that the wave functions on the right-hand \
side are normalized in the volume V. Instead of this normalization it is \
convenient to introduce normalization for unit volume, in which the following \
obvious identity holds \
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\(\[Psi]\_\(p\& \[Rule] \)\%\((V)\)\) \((\(r\& \[Rule] \))\) = \(1\/\@V\
\) \(\[Psi]\_\(p\& \[Rule] \)\%\((V = 
                1)\)\) \((\(r\& \[Rule] \))\)\)], "NumberedEquation",
  FontSize->16],

Cell["Then the necessary summation reads", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[Sum]\+n\( \[Psi]\_\(\(p\& \[Rule] \)\_n\)\) \((\(r\& \[Rule] \))\) \(\
\[Psi]\_\(\(p\& \[Rule] \)\_n\)\%*\) \((\(r\& \[Rule] \)')\)\  = \ \
\[Integral]\(\[DifferentialD]\^3 
                p\/\((2  \[Pi])\)\^3\) \(\[Psi]\_\(p\& \[Rule] \)\) \((\(r\& \
\[Rule] \))\) \(\[Psi]\_\(p\& \[Rule] \)\%*\) \((\(r\& \[Rule] \)')\)\)], \
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Cell[TextData[{
  "where the superscript (V=1) is suppressed.\n\tThus, to fulfil summation \
over the continuum set of states one can integrate  over the momentum ",
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]\^3 p/\((2  \[Pi])\)\^3\)]],
  ",  normalizing wave functions as \"one particle per unit volume\"."
}], "Text",
  FontSize->16]
}, Open  ]],

Cell[CellGroupData[{

Cell["Photons", "Subsection",
  FontSize->16],

Cell[BoxData[
    \(\[LeftAngleBracket]0 | \ \(\(A\& \[Rule] \)\_\(k\& \[Rule] \)\^*\) \
\(A\& \[Rule] \)\_\(k\& \[Rule] \) | 
          0\[RightAngleBracket] = \(\[Sum]\+m\[LeftAngleBracket]0 | \ \(\(A\& \
\[Rule] \)\_\(k\& \[Rule] \)\^*\) | 
                m\[RightAngleBracket] \[LeftAngleBracket]m | \(A\& \[Rule] \)\
\_\(k\& \[Rule] \) | 
                0\[RightAngleBracket] = \(\[LeftAngleBracket]0 | \ \(\(A\& \
\[Rule] \)\_\(k\& \[Rule] \)\^*\) | 
                1\[RightAngleBracket] \[LeftAngleBracket]1 | \(A\& \[Rule] \)\
\_\(k\& \[Rule] \) | 
                0\[RightAngleBracket] = \(\(|\)\(\[LeftAngleBracket]1 | \(A\& \
\[Rule] \)\_\(k\& \[Rule] \) | 
                0\[RightAngleBracket]\)\( | \^2\)\)\)\)\)], "NumberedEquation",\

  FontSize->16],

Cell["\<\
A photon represents a harmonic oscillator. Therefore the avaraged kinetic and \
potential energies equal half of the total energy. In the vacuum this total \
energy is half of the frequency. In this case\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\(T\&-\) = \(\(U\&-\) = \(\(1\/2\) 
            E = \(1\/4\) \[Omega]\)\)\)], "NumberedEquation",
  FontSize->16],

Cell["\<\
The kinetic energy of the oscillator can be expressed using the energy of the \
electromagnetic field in a unit volume, which reads\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(d\[CurlyEpsilon]\/dV = \(\(|\)\(\(E\& \[Rule] \)\_\(k\& \[Rule] \)\)\( \
| \^2\)\(+\(\(|\)\(\(H\& \[Rule] \)\_\(k\& \[Rule] \)\)\( | \^2\)\)\)\)\/\(8  \
\[Pi]\)\)], "NumberedEquation",
  FontSize->16],

Cell["\<\
Here the energy is calculated \"per unit volume\", precisely as is necessary \
for the purposes of integration over the continuum states of the photon, see \
above. Since\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\(E\& \[Rule] \)\_\(k\& \[Rule] \) = \(-\(\[PartialD]\/\[PartialD]t\)\) \
\(A\& \[Rule] \)\_\(k\& \[Rule] \)\)], "NumberedEquation",
  FontSize->16],

Cell["the first term in the energy represents the kinetic energy ", "Text",
  FontSize->16],

Cell[BoxData[{
    \(T = \(\(\(|\)\(\(E\& \[Rule] \)\_\(k\& \[Rule] \)\)\( | \^2\)\) = \(\
\[Omega]\^2\) \(\(A\& \[Rule] \)\_\(k\& \[Rule] \)\^*\) \(A\& \[Rule] \)\_\(k\
\& \[Rule] \)\)\), "\[IndentingNewLine]", 
    \(\(T\&-\) = \(\[LeftAngleBracket]0 | T | 
            0\[RightAngleBracket] = \(\[Omega]\^2\/\(8  \[Pi]\)\) \
\[LeftAngleBracket]0 | \(\(A\& \[Rule] \)\_\(k\& \[Rule] \)\^*\) \(A\& \
\[Rule] \)\_\(k\& \[Rule] \) | 0\[RightAngleBracket]\)\)}], "NumberedEquation",\

  FontSize->16],

Cell["Combining this with the previous disscussion one finds", "Text",
  FontSize->16],

Cell[BoxData[
    \(\(T\&-\)\  = \ \(\[Omega]\^2\/\(8  \[Pi]\) | \[LeftAngleBracket]1 | \(A\
\& \[Rule] \)\_\(k\& \[Rule] \) | 
              0\[RightAngleBracket]\( | \^2\) = \(1\/4\) \[Omega]\)\)], \
"NumberedEquation",
  FontSize->16],

Cell["\<\
Thus considering creation (or annialation) of one photn one should have \
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[LeftAngleBracket]1 | \(A\& \[Rule] \)\_\(k\& \[Rule] \) | 
          0\[RightAngleBracket] = \@\(\(4  \[Pi]\)\/\(2  \[Omega]\)\)\)], \
"NumberedEquation",
  FontSize->16]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Energy sfift", "Section",
  FontSize->16],

Cell[BoxData[
    \(\[Delta]E\_0 = \(4  \[Pi]\ \(e\^2\) \(\[Sum]\+\(\[Epsilon] = 1, 2\)\
\[Integral]\(\[DifferentialD]\^3 
                      k\/\((2  \[Pi])\)\^3\) \(\[Sum]\+n\( 
                        1\/\(2  \[Omega]\)\) \(\(|\)\(\(v\& \[Rule] \)\_\(n, \
0\)\ \[CenterDot]\ \ \(\[Epsilon]\& \[Rule] \)\)\(\ \)\( | \^2\)\)\/\(E\_0 - \
E\_n - \[Omega]\)\)\) = 
        4  \[Pi]\ \(e\^2\) \(\[Sum]\+n\[Integral]\+0\%\[Infinity]\(\(\( \
\[Omega]\^2\) \[DifferentialD]\[Omega]\)\/\((2  \[Pi])\)\^3\) \(1\/\(2  \
\[Omega]\)\) \(1\/\(E\_0 - 
                        E\_n - \[Omega]\)\) \(\[Integral]\[Sum]\+\(\[Epsilon] \
= 1, 2\)\)\) | \(v\& \[Rule] \)\_\(n, 0\)\ \[CenterDot]\ \ \(\[Epsilon]\& \
\[Rule] \)\ \( | \^2\)\[DifferentialD]\[CapitalOmega]\)\)], "NumberedEquation",\

  FontSize->16],

Cell["\<\
Here \[DifferentialD]\[CapitalOmega] represents integration over the \
spherical angle of the photon momentum\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[Integral]\[Sum]\+\(i = 1, 2\) | \(v\& \[Rule] \)\_\(n, 0\)\ \
\[CenterDot]\ \ \(\[Epsilon]\& \[Rule] \)\_i\ \( | \^2\)\[DifferentialD]\
\[CapitalOmega] = \(\[Integral]\((\(|\)\(\(v\& \[Rule] \)\_\(n, 
                    0\)\)\( | \^2\)\(-\(\(|\)\(\(v\& \[Rule] \)\_\(n, 0\)\
\[CenterDot]\(n\& \[Rule] \)\)\( | \^2\)\)\))\)\ \[DifferentialD]\
\[CapitalOmega] = \(\(\(|\)\(\(v\& \[Rule] \)\_\(n, 
                0\)\)\( | \^2\)\(\[Integral]\+\(4  \[Pi]\)\((1 - 
                    Cos[\[Theta]]\^2)\) \[DifferentialD]\[CapitalOmega]\)\) = \
\[IndentingNewLine]\(\(\(|\)\(\(v\& \[Rule] \)\_\(n, 
                  0\)\)\( | \^2\)\(2  \[Pi] \(\[Integral]\+0\%\[Pi]\((1 - 
                        Cos[\[Theta]]\^2)\) 
                    Sin[\[Theta]] \[DifferentialD]\[Theta]\)\)\) = \(2\/3\) 
                4  \[Pi] | \(v\& \[Rule] \)\_\(n, 0\)\( | \^2\)\)\)\), \
\[IndentingNewLine]\(n\& \[Rule] \) = \(\(\(k\& \[Rule] \)\/k\)\(.\)\)\)], \
"NumberedEquation",
  FontSize->16],

Cell["\<\
Thus, summation over the polarization and integration over the angles for a \
dipole transition gives a factor 2/3\
\>", "Text",
  FontSize->16],

Cell[BoxData[{
    \(\[Delta]E\_0 = \(\(4  \[Pi]\ e\^2\)\/\((2  \[Pi])\)\^3\) \(1\/2\) \
\(2\/3\) 4  \[Pi] \(\(\ \ \)\(\[Sum]\)\)\+n | \(v\& \[Rule] \)\_\(n, 0\)\( | \
\^2\)\[Integral]\+0\%\[Infinity]\[DifferentialD]\[Omega]\ \(\[Omega]\^2\) \(1\
\/\[Omega]\) 1\/\(E\_0 - E\_n - \[Omega]\)\), "\[IndentingNewLine]", 
    \(\(\(=\)\(\(2\/\(3  \[Pi]\)\) \(e\^2\) \(\(\ \)\(\[Sum]\)\)\+n | \(v\& \
\[Rule] \)\_\(n, 0\)\( | \^2\)\[Integral]\+0\%\[Infinity]\[DifferentialD]\
\[Omega]\ \[Omega]\/\(E\_0 - E\_n - \[Omega]\)\)\)\)}], "NumberedEquation",
  FontSize->16],

Cell[CellGroupData[{

Cell["Renormalization", "Subsection",
  FontSize->16],

Cell["Repeating the argument for the vacuum one can write", "Text",
  FontSize->16],

Cell[BoxData[
    \(\((\[Delta]E\_0)\)\_vac = \ \(\(2 
                 e\^2\)\/\(3\ \[Pi]\)\) \(\(\ \)\(\[Sum]\)\)\+n | \(v\& \
\[Rule] \)\_\(n, 0\)\( | \^2\)\[Integral]\+0\%\[Infinity]\[DifferentialD]\
\[Omega]\ \[Omega]\/\(-\[Omega]\)\)], "NumberedEquation",
  FontSize->16],

Cell["\<\
since in the vacuum the operator of velocity does not have nondiagonal matrix \
elements. For the energy difference one finds\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\((\[Delta]E\_0)\)\_ren = \(\[Delta]E\_0 - \((\[Delta]E\_0)\)\_vac = \
\[IndentingNewLine]\(\(=\)\(\(\(2 
                       e\^2\)\/\(3  \[Pi]\)\) \(\(\ \)\(\[Sum]\)\)\+n | \(v\& \
\[Rule] \)\_\(n, 0\)\( | \^2\)\[Integral]\+0\%\[Infinity]\[DifferentialD]\
\[Omega]\ \((\[Omega]\/\(E\_0 - E\_n - \[Omega]\) + 1)\) = \(\(2 
                       e\^2\)\/\(3  \[Pi]\)\) \(\(\ \)\(\[Sum]\)\)\+n | \(v\& \
\[Rule] \)\_\(n, 0\)\( | \^2\)\[Integral]\+0\%\[Infinity]\[DifferentialD]\
\[Omega]\ \(E\_0 - E\_n\)\/\(E\_0 - E\_n - \[Omega]\)\)\)\)\)], \
"NumberedEquation",
  FontSize->16],

Cell["\<\
In the following discussion the subscript ren (for renormalized) is \
suppressed.
\tThe  physical picture considered can be only correct in the nonrelativistic \
approximation. Therefore the integration over \[Omega] should run only up to \
m (not to infinity, where different formulas need to be worked out).\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[Delta]E\_0 \[Implies] \(\(2 
                   e\^2\)\/\(3  \[Pi]\)\) \(\(\ \)\(\[Sum]\)\)\+n | \(v\& \
\[Rule] \)\_\(n, 0\)\( | \^2\)\[Integral]\+0\%m\[DifferentialD]\[Omega]\ \
\(E\_0 - E\_n\)\/\(E\_0 - E\_n - \[Omega]\) = \(-\(\(2 
                   e\^2\)\/\(3  \[Pi]\)\)\) \(\(\(\ \)\(\[Sum]\)\)\+n\) \
\((E\_0 - E\_n)\) | \(v\& \[Rule] \)\_\(n, 0\)\( | \^2\)Log[
          m\/\(E\_0 - E\_n\)]\)], "NumberedEquation",
  FontSize->16],

Cell[BoxData[
    \(\(v\& \[Rule] \)\_\(n, 0\) = \(\(1\/m\) \(p\& \[Rule] \)\_\(n, 0\) \
\[Congruent] \(1\/m\) \[LeftAngleBracket]n | \(p\& \[Rule] \) | 
                0\[RightAngleBracket] = \(\(\[ImaginaryI]\/m\) \
\[LeftAngleBracket]n | \[ImaginaryI][H \( r\& \[Rule] \)] | 
                0\[RightAngleBracket] = \[ImaginaryI] \(\( E\_0 - 
                    E\_n\)\/m\) \[LeftAngleBracket]n | \(r\& \[Rule] \) | 
                  0\[RightAngleBracket] \[Congruent] \(\(E\_0 - 
                    E\_n\)\/m\) \(r\& \[Rule] \)\_\(n, 0\)\)\)\)], \
"NumberedEquation",
  FontSize->16],

Cell[BoxData[
    \(\[Delta]E\_0 = \(\(-\(\(2 
                     e\^2\)\/\(3  \[Pi]\)\)\) \(\(\(\ \)\(\[Sum]\)\)\+n\) \
\((E\_0 - E\_n)\)\^3 | \(r\& \[Rule] \)\_\(n, 0\)\( | \^2\)Log[
            m\/\(E\_0 - E\_n\)] = \(\(2 
                   e\^2\)\/\(3  \[Pi]\)\) \(\(\(\ \)\(\[Sum]\)\)\+n\) \((E\_n \
- E\_0)\)\^3 | \(r\& \[Rule] \)\_\(n, 0\)\( | \^2\)Log[
            m\/\(E\_0 - E\_n\)]\)\)], "NumberedEquation",
  FontSize->16],

Cell["\<\
This expression has a real and imaginary parts. The latter will be discussed \
when radiative widths are considered. For the real part, which represents a \
proper energy, one finds\
\>", "Text",
  FontSize->16],

Cell[BoxData[
    \(Re[\[Delta]E\_0] = \(\(2 
                 e\^2\)\/\(3  \[Pi]\)\) \(\(\(\ \)\(\[Sum]\)\)\+n\) \((E\_n - \
E\_0)\)\^3 | \(r\& \[Rule] \)\_\(n, 0\)\( | \^2\)Log[
          m\/\(\(|\)\(E\_n - E\_0\)\(|\)\)]\)], "NumberedEquation",
  FontSize->16],

Cell["In absolute units", "Text",
  FontSize->16],

Cell[BoxData[
    \(Re[\[Delta]E\_0] = \(2\/\(3  \[Pi]\ c\^3\)\) \(\(\(\ \
\)\(\[Sum]\)\)\+n\) \[Omega]\_\(n, 0\)\%3 | \(d\& \[Rule] \)\_\(n, 0\)\( | \
\^2\)Log[\(m\ c\^2\)\/\(\(|\)\(E\_n - E\_0\)\(|\)\)]\)], "NumberedEquation",
  FontSize->16],

Cell["where", "Text",
  FontSize->16],

Cell[BoxData[
    \(\[Omega]\_\(n, 0\) = \(E\_n - E\_0\)\/\:f7d8\)], "NumberedEquation",
  FontSize->16],

Cell[CellGroupData[{

Cell["Large Log approximation ", "Subsubsection",
  FontSize->16],

Cell["For atoms", "Text",
  FontSize->16],

Cell[BoxData[
    \(E\_n - E\_0 \[Tilde] \(Z\^2\) Ry\)], "NumberedEquation",
  FontSize->16],

Cell[BoxData[
    \(Log[\(m\ c\^2\)\/\(\(|\)\(E\_n - E\_0\)\(|\)\)] \[TildeEqual] 
      Log[1\/\((Z\[Alpha])\)\^2]\  \[GreaterGreater] \ 
      1\)], "NumberedEquation",
  FontSize->16],

Cell["assuming that (Z\[Alpha])\[LessLess]1.", "Text",
  FontSize->16],

Cell[BoxData[
    \(Re[\[Delta]E\_0] \[TildeEqual] \(\(2 
                 e\^2\)\/\(\(3\)  \(\[Pi]\)\(\ \)\)\) 
          Log[1\/\((Z\[Alpha])\)\^2] \(\(\(\ \)\(\[Sum]\)\)\+n\) \((E\_n - \
E\_0)\)\^3 | \(r\& \[Rule] \)\_\(n, 0\)\( | \^2\)\)], "NumberedEquation",
  FontSize->16]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
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In the log approximation only the s-states exibit the Lamb shift. Other \
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For a quasistationary state 0 the energy has its real and imaginary parts\
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Correspondingly, the probability of the radiative decay of the state 0 is\
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is the probability of the decay into some given state n, while W is the total \
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It follows from here that the probability of radiation of one quantum is\
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