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Cell[CellGroupData[{
Cell["Problem 6.20 (Engel)", "Title"],

Cell[TextData[{
  "We did not discuss entangled photons in class, but the book describes the \
situation and the problem is quite straightforward even if one knows nothing \
about the subject.\n\nFirst, we assume the two-photon system is described by \
a wave function composed of eigenfunctions of a \"polarization\" operator, ",
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  ", where the operator works on photon ",
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  " only.",
  " The eigenfunctions of this operator are ",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_i\)]],
  " where ",
  StyleBox["i",
    FontSlant->"Italic"],
  " also identifies the  photon. The eigenvalues of these functions depend on \
the photon's polarization,",
  " ",
  StyleBox["H",
    FontSlant->"Italic"],
  " or ",
  StyleBox["V",
    FontSlant->"Italic"],
  ", such that a horizontally polarized photon produces an eigenvalue of -1 \
and a vertically polarized photon produces an eigenvalue of +1.\n\nSuppose we \
know that one photon is horizontally polarized and the other is vertically \
polarized, but we don't know which photon is which. We might write the \
two-photon wave function as:",
  "\n\n",
  Cell[BoxData[
      \(TraditionalForm\`\[Psi]\_12\  = \ \(\(1\/\@2\) \((\(\(\[Psi]\_1\)(
                  H)\) \(\(\[Psi]\_2\)(V)\)\  + \(\(\[Psi]\_1\)(
                  V)\) \(\(\[Psi]\_2\)(H)\))\)\(\ \)\)\)]],
  "\n",
  StyleBox["\nPart A.",
    FontWeight->"Bold"],
  " Show that ",
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  " is not an eigenfunction of ",
  Cell[BoxData[
      \(TraditionalForm\`\(P\&^\)\_1\)]],
  " or ",
  Cell[BoxData[
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  "."
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Cell["Solution to Part A", "Subtitle"],

Cell[TextData[{
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      \(TraditionalForm\`\(\(P\&^\)\_1\) \[Psi]\_12\  = \ \(\(1\/\@2\) \
\((\(\(P\&^\)\_1\) \(\(\[Psi]\_1\)(H)\) \(\(\[Psi]\_2\)(
                  V)\)\  + \(\(P\&^\)\_1\) \(\(\[Psi]\_1\)(
                  V)\) \(\(\[Psi]\_2\)(H)\))\)\(\ \)\)\)],
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                      V)\)\  + \((\(+\(\(\[Psi]\_1\)(
                          V)\)\))\) \(\(\[Psi]\_2\)(H)\))\)\(\ \)\)\),
          FontWeight->"Plain"], TraditionalForm]]],
  "\n\t\n\t",
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      FormBox[
        StyleBox[\(\(=\)\(\(1\/\@2\) \((\(-\(\(\[Psi]\_1\)(
                        H)\)\) \(\(\[Psi]\_2\)(V)\)\  + \(\(\[Psi]\_1\)(
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  "\n\n",
  StyleBox["The operator creates a function that is not a multiple of the \
original function, so the original function is not an eigenfunction. The same \
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Cell[TextData[{
  "Part B.",
  StyleBox[" Show that each term ",
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  StyleBox["in ",
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    FontWeight->"Plain"],
  StyleBox[".",
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}], "Text",
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Cell["Solution to Part B", "Subtitle"],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\(\(P\&^\)\_1\) \(\(\[Psi]\_1\)(H)\) \(\(\[Psi]\_2\)(
            V)\)\ \  = \ \(\(-\ \(\(\[Psi]\_1\)(H)\)\) \(\(\[Psi]\_2\)(
            V)\)\(\ \)\)\)],
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  "\n\n",
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      \(TraditionalForm\`\(\(P\&^\)\_2\) \(\(\[Psi]\_1\)(H)\) \(\(\[Psi]\_2\)(
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                H)\)\),
          FontWeight->"Plain"], TraditionalForm]]],
  "\n"
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Cell[TextData[{
  "Part C.",
  StyleBox[" What is the average value of the polarization ",
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          FontWeight->"Plain"], TraditionalForm]]],
  " ",
  StyleBox["measured on identically prepared systems? No calculation is \
needed.",
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}], "Text",
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Cell["Solution to Part C", "Subtitle"],

Cell[TextData[{
  StyleBox["The average value is the expectation value <",
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  StyleBox[" with eigenvalues of +1 and -1. The two eigenfunctions contribute \
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  Cell[BoxData[
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  StyleBox[" so the eigenvalues contribute equally to the weighted average of \
eigenvalues. In other words, <",
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  StyleBox["> = 0.",
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