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The computed normal \ modes are visualized through Mathematica\[LongDash]generated animations.", FontSlant->"Italic"]], "Abstract", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell["by Gustavo Demarco", "Author", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "This paper describes a purely classical discussion of the vibrations of \ molecules (regarded as systems of several interacting nuclei). Such a \ description is necessary prior to any quantum mechanical investigation of \ this problem.\n\nApplying the methods of group theory enables us to carry out \ a classification of the normal modes of vibration from the knowledge of the \ symmetry group of the molecule with the nuclei in its equilibrium positions. \ Then, the utilization of projection operators [1] allow us to solve the more \ complex problem of determining the normal coordinates or, what is the same, \ the displacements of the nuclei in each normal mode. This allows a beautiful \ visualization of the vibrating molecule with ", StyleBox["Mathematica", FontSlant->"Italic"], " generated animations. The functions necessary for doing these \ computations are bundled in the package ", StyleBox["Molecules.m", FontWeight->"Bold"], ". \n\nIn section 1, I summarize the group theoretic considerations \ necessary for the problem of molecular vibrations. The needed information \ about several point groups consists of a set of data files (one for each \ group) generated with previous packages ", StyleBox["charactr.m", FontWeight->"Bold"], " [2] and ", StyleBox["Projectors.m", FontWeight->"Bold"], " [3]. In section 2, I explain the use of these data files. Section 3 shows \ in more detail the utilization of the package ", StyleBox["Molecules.m", FontWeight->"Bold"], " solving the problem of vibrations in the ", Cell[BoxData[ FormBox[ StyleBox[\(NH\_3\), FontSlant->"Italic"], TraditionalForm]]], " molecule. Section 4 contains the strictly necessary steps to solve any \ molecule vibration problem using ", Cell[BoxData[ FormBox[ StyleBox[\(UF\_6\), FontSlant->"Italic"], TraditionalForm]]], " as example. Section 5 contains a discussion for the case when there are \ involved complex irreducible representations and section 6 gives a brief \ discussion about the force matrix. Section 7 contains the conclusions.\n\nThe \ largest molecule we considered was the fullerene ", Cell[BoxData[ \(TraditionalForm\`C\_60\)]], ". In this case, although the computations can be performed in symbolic \ form, we modified some of the functions of the package to do the calculations \ with floating-point arithmetic to avoid time consuming operations. The \ symbolic results are such large expressions that no insight is gained doing \ the computations in symbolic form. Also, the visualization of the vibrating \ ", Cell[BoxData[ \(TraditionalForm\`C\_60\)]], " molecule is done as a polyhedra with faces and not as a collection of \ points joined by bonds (as we have done in the other examples) so we have \ also adapted the visualization function for this case.\n\nIn the electronic \ distribution I included the three files (the notebook ", StyleBox["vib_C60num.nb", FontWeight->"Bold"], " and the data files ", StyleBox["C60.dat.num", FontWeight->"Bold"], " and ", StyleBox["bases.dat.num", FontWeight->"Bold"], ") necessary to perform the animation of the ", Cell[BoxData[ \(TraditionalForm\`C\_60\)]], " fullerene (buckyball). Also included in the electronic distribution are \ the data files for the 32 point groups and a notebook ", StyleBox["examples.nb", FontWeight->"Bold"], " with several examples: ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(C\_2\) H\_4\)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(C\_2\) H\_6\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(C\_6\) H\_6\)]], ", ", StyleBox["CH", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`Cl\_3\)], FontSlant->"Italic"], ", ", Cell[BoxData[ \(TraditionalForm\`\(H\_2\) O\)]], " and ", Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["Os", FontSlant->"Italic"], "F"}]], StyleBox["8", FontSlant->"Italic"]], FontSlant->"Italic"], TraditionalForm]]], ".\n\nAs a final remark, I should mention that throughout the paper, I \ start a fresh ", StyleBox["Mathematica", FontSlant->"Italic"], " session in each section." }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell["2. Group Theoretic Considerations.", "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "In general, a system of ", StyleBox["N", FontSlant->"Italic"], " nuclei have 3", StyleBox["N", FontSlant->"Italic"], " degrees of freedom. Of these, three correspond to a translational motion \ and three (two for a linear molecule) to a rotational motion of the system as \ a whole. This leaves 3", StyleBox["N", FontSlant->"Italic"], " \[Dash] 6 pure vibrational degrees of freedom (3", StyleBox["N", FontSlant->"Italic"], "-5 for a linear molecule).\n\nThe energy of the system executing small \ vibrations is a quadratic form in the velocities and in the cartesian \ coordinates of the particles. We can always find a set of coordinates ", Cell[BoxData[ \(TraditionalForm\`Q\_\(\[Alpha]\_i\)\)]], " (normal coordinates) such that the energy takes a diagonal form," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(E\ = \ 1\/2\ \(\[Sum]\+\(i, \[Alpha]\)\ Q\& . \_\(\[Alpha]\_i\)\%2\)\ + \ 1\/2\ \ \(\[Sum]\+\[Alpha]\ \[Omega]\_\[Alpha]\%2\ \ \(\[Sum]\+i\ \ Q\ \_\(\[Alpha]\_i\)\%2\)\)\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\_\[Alpha]\)]], " are the frequencies of vibration. The sum over ", StyleBox["i", FontSlant->"Italic"], " =1,2,...,", Cell[BoxData[ \(TraditionalForm\`f\_\[Alpha]\)]], " means that ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\_\[Alpha]\)]], " has associated ", Cell[BoxData[ \(TraditionalForm\`f\_\[Alpha]\)]], " normal coordinates.\n\nFrom symmetry considerations [4], it can be shown \ that the ", Cell[BoxData[ \(TraditionalForm\`f\_\[Alpha]\)]], " normal coordinates belonging to ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\_\[Alpha]\)]], " give some irreducible representation (IR) of the symmetry group of the \ molecule. To find the IR's present in ", "Equation", " 1, we need the reducible representation (RR) given by all the coordinates \ together (we call this RR the total representation TR). Decomposing it into \ irreducible parts, we determine the symmetry properties of the vibrations. We \ are interested only in the purely vibrational degrees of freedom. Thus we \ must exclude from the TR the six degrees of freedom corresponding to pure \ translations and rotations of the molecule as a whole. These six modes have \ null frequencies. We call the RR obtained by excluding these six modes the \ total vibrational representation (TVR).\n\nTo find the IR's contained in the \ TVR we must compute the character of this representation. Then we project it \ onto the characters of the IR's of the symmetry group G of the molecule using \ the orthogonality relation between these [5].\n \nThe character of the TVR \ for a symmetry operation T can be written in the following form," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(\[Chi]\^TVR\) \((T)\)\ = \ \(\(U\_M\) \((T)\)\(\ \)\(\[Times]\)\(\ \ \)\(U\_G\) \((T)\)\(\ \)\)\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(\(U\_M\) \((T)\)\)]], " depends on T and on the distribution of the atoms in the molecule and ", Cell[BoxData[ \(\(\(\ \)\(\(U\_G\) \((T)\)\(\ \)\)\)\)]], "depends only on the operation T.\n The symmetry transformation T can only \ be one of the following: pure rotation R, rotorreflexion S and inversion I. \ It can be shown [4] that ", Cell[BoxData[ \(\(\(\ \)\(\(U\_M\) \((T)\)\(\ \)\)\)\)]], " can take the following values," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(\(\(\(\(\(a\)\()\)\)\ \ \ \(U\_M\) \((R)\)\ = N\_R - 2\n b\)\()\)\)\ \ \ \(U\_M\) \((S)\)\ = \ N\_S\n c\)\()\)\)\ \ \ \(U\_M\) \((I)\)\ = \ N\_I\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`N\_i\)]], " is the number of atoms that do not change their position under the \ symmetry operation ", StyleBox["i. ", FontSlant->"Italic"], "It can also be shown [4] that," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(\(\(\(\(\(a\)\()\)\)\ \ \ \(U\_G\) \((R)\)\ = \ 1 + 2\ Cos[\[CapitalPhi]]\n b\)\()\)\)\ \ \ \(U\_G\) \((S)\)\ = \ \(-1\) + 2\ Cos[\[CapitalPhi]]\n c\)\()\)\)\ \ \ \(U\_G\) \((I)\)\ = \ \(-3\)\)], "NumberedEquation",\ TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where \[CapitalPhi] is the angle of rotation in the operation R or S \ (remember that any rotorreflexion S can be expressed as S = I\[CenterDot]R(\ \[CapitalPhi]) for a suitable chossen pure rotation R(\[CapitalPhi])).\n The \ numbers ", Cell[BoxData[ \(\(\(\ \)\(\(U\_G\) \((T)\)\)\)\)]], " depend only on G and are included in the data files for the point groups \ with the constant name ", StyleBox["vibrchar", FontWeight->"Bold"], ".\n Now we consider the problem of determining the displacements of the \ nuclei in each normal mode (a detailed discussion of this point can be viewed \ from reference [1]). First we should obtain the TR. This is a 3N dimensional \ representation acting on vectors of the form," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(v = {x\_1, y\_1, z\_1, x\_2, y\_2, z\_2, ... , x\_N, y\_N, z\_N}\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(x\_i, y\_i, z\_i\)]], " are the cartesian coordinates of the i nuclei. This TR of G is simply the \ direct product representation obtained from the three dimensional \ representation of G (that is, the representation of rotation matrices in ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], ") and the representation obtained by permuting the nuclei for each group \ operation, that is, an N dimensional permutation of G. For a given symmetry \ operation T we denote by ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TR\)]], "(T) the matrix that represents this operation acting on the vectors of ", "Equation", " (5), that is, ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TR\)]], "(T) are the matrices in TR. Now suppose that a given IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " of G is contained ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " times in TR. Thus, in the space ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^\(3 N\)\)]], " on which TR acts, there are ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " linearly independent vectors ", Cell[BoxData[ \(v\_j\)]], " (j=1,...,", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], ") that transform as the first row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". To obtain these ", Cell[BoxData[ \(TraditionalForm\`a\_\[Nu]\)]], " vectors we construct the following projector matrices [1]," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ RowBox[{\(\[WeierstrassP]\_11\%\((\[Nu])\)\), "=", " ", RowBox[{\(n\_\[Nu]\/g\), RowBox[{\(\[Sum]\+\(T \[Element] G\)\), RowBox[{\(\([\(\[CapitalGamma]\^\((\[Nu])\)\) \((T)\)]\)\_11\%*\), " ", FormBox[\(\[CapitalGamma]\^TR\), "TraditionalForm"], \(\((T)\)\(\ \ \)\(.\)\)}]}]}]}]], \ "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(n\_\[Nu]\)]], "is the dimension of the matrices ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], "(T) in the IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". g is the number of elements in G.\n Thus, ", Cell[BoxData[ \(\[WeierstrassP]\_11\%\((\[Nu])\)\)]], " is a 3N\[Times]3N matrix. This matrix has ", Cell[BoxData[ \(a\_\[Nu]\)]], " linearly independent rows which can be taken as the ", Cell[BoxData[ \(a\_\[Nu]\)]], " linearly independent vectors ", Cell[BoxData[ \(v\_j\)]], ". Each of these vectors transforms as the first row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ", and each of them allow us to generate the ", Cell[BoxData[ \(a\_\[Nu]\)]], " copies of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " contained in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TR\)]], ". However, since we are interested in the pure vibrational modes, we have \ to eliminate from the set {", Cell[BoxData[ \(v\_j\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(j = 1, ... , a\_\[Nu]\)\)]], " those vectors that correspond to the zero frequency modes (pure rotations \ and translations). We start eliminating the translations. For each vector ", Cell[BoxData[ \(v\_j\)]], " we extract the components ", Cell[BoxData[ \(u\_a\)]], "={", Cell[BoxData[ \(x\_a\)]], ",", Cell[BoxData[ \(y\_a\)]], ",", Cell[BoxData[ \(z\_a\)]], "} for each atom ", StyleBox["a", FontWeight->"Bold"], " (see Equation 5). In ", StyleBox["Mathematica", FontSlant->"Italic"], " notation," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\({u\_1, ... , u\_N} = Partition[v\_j, 3];\)\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "These ", Cell[BoxData[ \(u\_a\)]], ", when multiplied globally by an infinitesimal constant, are the \ infinitesimal displacements that transform as the first row of one of the \ modes \[Vee].\n The displacement of the centre of mass of the molecule \ corresponding to the displacement vectors ", Cell[BoxData[ \(v\_j\)]], " is ," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(u\_CM = \(1\/M\) \(\[Sum]\+a m\_a\ u\_a\)\ , \ \ \ \ \ \ \ \ M = \[Sum]\ \+a m\_a\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " Now we substract ", Cell[BoxData[ \(u\_CM\)]], " to each ", Cell[BoxData[ \(u\_a\)]], " ,and reconstruct from them the modified vectors ", Cell[BoxData[ \(v\_j\)]], " that do not contain translations," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(u\_a\%'\) = u\_a - u\_CM\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(\(v\_j\%'\) = Flatten[{\(u\_1\%'\), ... , \(u\_N\%'\)}];\)\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Some of these new vectors {", Cell[BoxData[ \(\(v\_j\%'\)\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(j = 1, ... , a\_\[Nu]\)\)]], " will be null vectors and correspond to pure translations. We discard \ them. \n Now we proceed with the global rotations. For simplicity, let us \ suppose that the centre of mass coincides with the origin of coordinates, \ that is, ", Cell[BoxData[ \(R\_CM\)]], "= 0. The \"angular momentum\" (strictly speaking, an infinitesimal \ rotation due to the fact that the ", Cell[BoxData[ \(u\_a\)]], " are infinitesimal displacements, not velocities) of the molecule to first \ order in ", Cell[BoxData[ \(u\_a\)]], " is," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(L = \[Sum]\+a m\_a\ r\_a\%0\[Times]\ \(u\_a\%'\)\)], "NumberedEquation",\ TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(r\_a\%0\)]], " are the equilibrium positions of the nuclei.\n If we perform a global \ infinitesimal rotation \[CapitalOmega] (an infinitesimal rotation can be \ caracterized by vector with direction given by the axis of rotation and \ modulus equal to the angle of rotation), L changes to ", Cell[BoxData[ \(\(L\^'\)\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(L\^'\) = \[Sum]\+a m\_a\ r\_a\%0\[Times]\ \(u\_a\%''\)\)], \ "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(\(\(\ \)\(u\_a\%''\)\)\)]], "= ", Cell[BoxData[ \(\(u\_a\%'\)\)]], "-", Cell[BoxData[ \(\(\(\ \)\(h\_a\)\)\)]], " and ", Cell[BoxData[ \(h\_a\)]], " = ", Cell[BoxData[ \(\(\(r\_a\%0\)\(\[Times]\)\)\)]], " \[CapitalOmega] .\n We have to find \[CapitalOmega] such that," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(L\^'\) = 0\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell["The answer is, in component notation,", "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\[CapitalOmega]\_j = \(\((U\^\(-1\))\)\_ji\) L\_i\)], "NumberedEquation",\ TextAlignment->Left, TextJustification->0], Cell["where", "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(U\_im = \[Sum]\+a m\_a[\(\((\ r\_a\%0\ )\)\_m\) \((\ r\_a\%0\ )\)\_i - \ \(\[Delta]\_\(\(im\)\(\ \)\)\) \((\ r\_a\%0\ \[CenterDot]r\_a\%0)\)]\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Now, from the vectors ", Cell[BoxData[ \(\(\(\ \)\(u\_a\%''\)\)\)]], " we obtain the vectors ", Cell[BoxData[ \(\(\(\ \)\(v\_j\%''\)\)\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(\(v\_j\%''\) = Flatten[{\(u\_1\%''\), ... , \(u\_N\%''\)}];\)\)], "NumberedEquation",\ TextAlignment->Left, TextJustification->0], Cell[TextData[{ "From this set, the non-null vectors are the pure vibrational modes that \ transform as the first row of the IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". Let ", Cell[BoxData[ \(b\_\[Nu]\)]], " be the number of non-null vectors and rename them as ", Cell[BoxData[ \(v\_j\)]], ". Now we construct the following ", Cell[BoxData[ \(n\_\[Nu]\)]], "-1 projection matrices [1]," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ RowBox[{\(\[WeierstrassP]\_\(1 m\)\%\((\[Nu])\)\), "=", " ", RowBox[{\(n\_\[Nu]\/g\), RowBox[{\(\[Sum]\+\(T \[Element] G\)\), RowBox[{\(\([\(\[CapitalGamma]\^\((\[Nu])\)\) \((T)\)]\)\_\(1 m\)\%*\), " ", FormBox[\(\[CapitalGamma]\^TR\), "TraditionalForm"], \(\((T)\)\(\ \ \)\(.\)\)}]}]}]}]], \ "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "From each vector ", Cell[BoxData[ \(v\_j\)]], " we obtain the following ", Cell[BoxData[ \(n\_\[Nu]\)]], "-1 vectors," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(v\_j\%\[Alpha] = \(v\_j\[CenterDot]\[WeierstrassP]\_\(1 \ \[Alpha]\)\%\((\[Nu])\)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[Alpha] \ = 2\), ... , n\_\[Nu]\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", Cell[BoxData[ \(v\_j\)]], "\[Congruent] ", Cell[BoxData[ \(v\_j\%1\)]], " is taken as a row matrix. \n The ", Cell[BoxData[ \(n\_\[Nu]\)]], " vectors {", Cell[BoxData[ \(v\_j\%\[Alpha]\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(\[Alpha] = 1, ... , n\_\[Nu]\)\)]], " transform as the IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". Thus, from the ", Cell[BoxData[ \(b\_\[Nu]\)]], " vectors {", Cell[BoxData[ \(v\_j\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(j = 1, ... , b\_\[Nu]\)\)]], " we obtain the ", Cell[BoxData[ \(b\_\[Nu]\)]], " copies of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " contained in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ", that is, we have ", Cell[BoxData[ \(b\_\[Nu]\)]], " linearly independent basis {", Cell[BoxData[ \(v\_j\%1\)]], ",...,", Cell[BoxData[ \(v\_j\%\(n\_\[Nu]\)\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(j = 1, ... , b\_\[Nu]\)\)]], " for the IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". However, in general, these do not correspond to the pure vibrational \ normal modes {", Cell[BoxData[ \(TraditionalForm\`Q\_\(\[Alpha]\_1\)\)]], ",...,", Cell[BoxData[ \(TraditionalForm\`Q\_\(\[Alpha]\_\(n\_\[Nu]\)\)\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(\[Alpha] = 1, ... , b\_\[Nu]\)\)]], " (see ", "Equation", " 1) but to a linear combination of them. This problem cannot be solved \ from purely group theoretic considerations. Instead, we have to take into \ account the force matrix, that is, an explicit knowledge from the dynamics is \ necessary. A few considerations about these points are relegated to section \ 7." }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["3. The Point Groups Database.", "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "With the help of our previous packages ", StyleBox["charactr.m", FontWeight->"Bold"], " [2] and ", StyleBox["Projectors.m", FontWeight->"Bold"], " [3], it is easy to obtain all the necessary information needed to treat \ molecular vibration problems. For several point groups, we generated a file \ ", StyleBox[".dat", FontSlant->"Italic"], ", where ", StyleBox["", FontSlant->"Italic"], " stands for a standard name of the group. When one of these files is \ loaded in ", StyleBox["Mathematica", FontSlant->"Italic"], ", it defines the following constants,\n\n ", StyleBox["comments \n \t\t", FontWeight->"Bold"], "this is a string that tells what rotation operations have been used to \ generate the group. This is necessary \n \t\tto give the location for the \ atoms that generate the molecule from the group operations.\n ", StyleBox["group", FontWeight->"Bold"], "\t\t\n \t\tthe representation of the group by rotation matrices in three \ diemensions.\n ", StyleBox["pos\n ", FontWeight->"Bold"], "\t\ta list whose elements are the classes of the group given by the \ position of their elements within the \n \t\tgroup.\n ", StyleBox["g", FontWeight->"Bold"], "\n \t\tthe number of elements within each class.\n ", StyleBox["ch", FontWeight->"Bold"], "\n \t\tthe character table.\n ", StyleBox["vibrchar", FontWeight->"Bold"], "\n \t\twe have already defined this quantity in the previous section (see \ ", "Equation", " 4 and the comments \n \t\tthereafter).\n ", StyleBox["symbols", FontWeight->"Bold"], "\n \t\tthe names for the IR's following Landau notation [4].\n ", StyleBox["reality", FontWeight->"Bold"], "\n \t\teach element from this list gives the reality properties of each \ IR. Its elements can take the following \n \t\tvalues:\n \t\t", StyleBox["1\t", FontWeight->"Bold"], " the IR is equivalent to as real representation (real representation).\n \ \t\t", StyleBox["0", FontWeight->"Bold"], "\tthe IR is complex and not equivalent to the complex conjugate \ representation (complex \n \t\t\trepresentation).\n \t\t", StyleBox["-1", FontWeight->"Bold"], "\tthe IR is equivalent to the complex conjugate representation but it is \ not equivalent to a real one \n \t\t\t(pseudo real representation).\n ", StyleBox["irreps", FontWeight->"Bold"], "\n \t\tthe matrices for the IR's. All the representations but the five \ dimensional representations for the \n \t\ticosahedral groups are unitary \ representations. The five dimensional representations for the \n \t\t\ icosahedral groups are equivalent to unitary representations, but the \ computation is very time \n \t\tconsuming (at least in our old 486)." }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "4. Vibrations of the ammonium molecule (", Cell[BoxData[ \(TraditionalForm\`NH\_3\)]], ")." }], "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " This molecule is a regular tetrahedron, with the N atom at the vertex \ and the H atoms at the corners of the base. The symmetry group of this \ molecule is ", Cell[BoxData[ \(TraditionalForm\`C\_\(3 v\)\)]], ". For any molecule with symmetry group ", StyleBox["G", FontSlant->"Italic"], ", the atoms can be grouped in disjoints sets (orbits). All the atoms \ within an orbit have positions that can be generated from the position of any \ one of them by applying to it all the symmetry operations of the group ", StyleBox["G", FontSlant->"Italic"], ". Thus, each orbit can be generated from one atom and the complete \ molecule can be generated from a minimal set of atoms (generators), one from \ each orbit. In the present case, the complete molecule can be obtained from \ the N atom and one of the H atoms. The orbit of the N atom contains only this \ atom and the orbit of the H atom contains the three H atoms. The ", StyleBox["comments", FontWeight->"Bold"], " in the data file ", StyleBox["C3v.dat", FontWeight->"Bold"], " helps us to choose an appropiate location for these generating atoms." }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(<< "\";\)\), "\n", \(\(<< "\";\)\), "\n", \(comments\)}], "Input", CellLabel->"In[1]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \("generators: group[[2]] (-2Pi/3 rotation through the z-axis) and \ group[[3]] (reflexion through the plane x=0)."\)], "Output", CellLabel->"Out[1]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["\<\ We put the N atom (atomic mass = 14) at {0,0,1} and an H generating \ atom (atomic mass = 1) at {0,1,0},\ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[{ \(\(locations = {{0, 0, 1}, {0, 1, 0}};\)\), "\n", \(\(masses = {14, 1};\)\), "\n", \(\(cls = {GrayLevel[0], RGBColor[0, 1, 0]};\)\)}], "Input", CellLabel->"In[2]:=", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where ", StyleBox["cls", FontWeight->"Bold"], " are the colors we assign to the generating atoms in the animations.\n \ Now we generate the coordinates, masses and colors of the remaining atoms. \ All the atoms in each orbit have the same mass and we assign the same color \ to all the atoms within each orbit," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(mol = GenerateMolecule[group, locations]\)], "Input", CellLabel->"In[3]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 0, 1}, {0, 1, 0}, {\(-\(\@3\/2\)\), \(-\(1\/2\)\), 0}, {\@3\/2, \(-\(1\/2\)\), 0}}\)], "Output", CellLabel->"Out[3]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(masses = GenerateMasses[group, mol, locations, masses]\)], "Input", CellLabel->"In[4]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({14, 1, 1, 1}\)], "Output", CellLabel->"Out[4]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(cls = GenerateMasses[group, mol, locations, cls]\)], "Input", CellLabel->"In[5]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({GrayLevel[0], RGBColor[0, 1, 0], RGBColor[0, 1, 0], RGBColor[0, 1, 0]}\)], "Output", CellLabel->"Out[5]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["\<\ Before classifying the nomal modes and obtaining the displacement \ vector for each normal mode, we will verify that we have correctly generated \ the molecule with a 3D plot. First we generate bonds joining the atoms from \ two generating bonds given by the atoms at the end of the bonds,\ \>", "Text",\ TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(bg = {{1, 2}, {2, 3}};\)\)], "Input", CellLabel->"In[6]:=", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(bonds = GenerateBonds[group, bg, mol]\)], "Input", CellLabel->"In[7]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}\)], "Output", CellLabel->"Out[7]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[BoxData[ \(\(bnscoor = Map[mol[\([#]\)] &, bonds];\)\)], "Input", CellLabel->"In[8]:=", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(\(Show[Graphics3D[Map[Line, bnscoor]], Graphics3D[{PointSize[0.05], \n\t\t\tMap[Point, mol]}]];\)\)], "Input",\ CellLabel->"In[9]:=", TextAlignment->Left, TextJustification->0], Cell[GraphicsData["PostScript", "\<\ %! 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TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(modes = NormalVibrations[group, pos, ch, vibrchar, mol]\)], "Input", CellLabel->"In[10]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({2, 0, 2}\)], "Output", CellLabel->"Out[10]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["We can see a standard notation for the modes,", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(symbols . modes\)], "Input", CellLabel->"In[11]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(2\ "\!\(A\_1\)" + 2\ "E"\)], "Output", CellLabel->"Out[11]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "The ", Cell[BoxData[ \("\!\(A\_1\)"\)]], " are symmetrical modes (non-degenerate) and the E modes are \ double-degenerate modes.\n Now we should obtain ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\(\(TR\)\(\ \)\)\)]], "(", StyleBox["totalrep", FontWeight->"Bold"], "). As we have said in section 2, this representation is the direct product \ representation of the representation obtained by permuting the atoms with the \ symmetry operations (", StyleBox["permrep", FontWeight->"Bold"], ") and the representation given by the rotation matrices in ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^3\)]], " (", StyleBox["group", FontWeight->"Bold"], "). In fact, ", StyleBox["permrep", FontWeight->"Bold"], " is a direct product of other permutation representations, one for each \ orbit," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(Map[ MatrixForm, \((permrep = Map[AtomsPermutations[mol, #] &, group])\)]\)], "Input", CellLabel->"In[12]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ RowBox[{"{", RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "1", "0", "0"}, {"0", "0", "1", "0"}, {"0", "0", "0", "1"} }], ")"}], (MatrixForm[ #]&)], ",", TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "0", "1", "0"}, {"0", "0", "0", "1"}, {"0", "1", "0", "0"} }], ")"}], (MatrixForm[ #]&)], ",", TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "1", "0", "0"}, {"0", "0", "0", "1"}, {"0", "0", "1", "0"} }], ")"}], (MatrixForm[ #]&)], ",", TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "0", "0", "1"}, {"0", "1", "0", "0"}, {"0", "0", "1", "0"} }], ")"}], (MatrixForm[ #]&)], ",", TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "0", "1", "0"}, {"0", "1", "0", "0"}, {"0", "0", "0", "1"} }], ")"}], (MatrixForm[ #]&)], ",", TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0", "0"}, {"0", "0", "0", "1"}, {"0", "0", "1", "0"}, {"0", "1", "0", "0"} }], ")"}], (MatrixForm[ #]&)]}], "}"}]], "Output", CellLabel->"Out[12]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[BoxData[ \(\(totalrep = MapThread[DirectProduct, {permrep, group}];\)\)], "Input", CellLabel->"In[13]:=", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "This is a 12\[Times]12 representation. \n Now we proceed to compute \ the set of ", Cell[BoxData[ \(a\_\[Nu]\)]], " linearly indepent rows of the projector ", Cell[BoxData[ \(\[WeierstrassP]\_11\%\((\[Nu])\)\)]], ", or what is the same, the vectors that transform as the first row (see \ Equation 6) for each of the IR's present in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\(\(TVR\)\(\ \)\)\)]], "(see the variable ", StyleBox["modes ", FontWeight->"Bold"], "before). We start with ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((1)\)\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vecs1 = FirstRowVectors[irreps[\([1]\)], totalrep]\)], "Input", CellLabel->"In[14]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}, {0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1}}\)], "Output", CellLabel->"Out[14]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "We can see that a linear combination of these three vectors should \ correspond to a pure translation or rotation since there are only two pure \ vibrational modes for ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((1)\)\)]], " (", StyleBox["modes[[1]]=2", FontWeight->"Bold"], "). To eliminate this zero frequency mode we start fixing the centre of \ mass of the molecule, that is, we compute the vectors ", Cell[BoxData[ \(\(v\_j\%'\)\)]], " (see Equation 10)," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vecs1 = FixTranslations[vecs1, masses]\)], "Input", CellLabel->"In[15]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 0, 1, 0, 0, \(-\(14\/3\)\), 0, 0, \(-\(14\/3\)\), 0, 0, \(-\(14\/3\)\)}, {0, 0, 0, 0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}}\)], "Output", CellLabel->"Out[15]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "There is no rotational mode remaining in ", StyleBox["vecs1", FontWeight->"Bold"], ". Since the ", Cell[BoxData[ \(A\_1\)]], " modes are one-dimensional, there are no additional vectors in these modes \ (the matrices in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((1)\)\)]], " have only one row, that is they are numbers).\n Now we proceed with ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((3)\)\)]], ", the other pure vibrational modes," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vecs3 = FirstRowVectors[irreps[\([3]\)], totalrep]\)], "Input", CellLabel->"In[16]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, \(-\(1\/\@3\)\), 0, 0, 1\/\@3, 0}, {0, 0, 0, 0, 0, 0, 1, 1\/\@3, 0, 1, \(-\(1\/\@3\)\), 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, \(-1\)}}\)], "Output", CellLabel->"Out[16]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(vecs3 = FixTranslations[vecs3, masses]\)], "Input", CellLabel->"In[17]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 0, 0, 0, 0, 0, \(-7\), \(-\(7\/\@3\)\), 0, \(-7\), 7\/\@3, 0}, {0, 0, 0, 1, 0, 0, \(-\(1\/2\)\), \(-\(\@3\/2\)\), 0, \(-\(1\/2\)\), \@3\/2, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, \(-1\)}}\)], "Output", CellLabel->"Out[17]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["\<\ Thus, a linear combination of these vectors corresponds to a pure \ rotation, and we should eliminate it,\ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vecs3 = FixRotations[vecs3, masses, mol]\)], "Input", CellLabel->"In[18]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 0, 0, 0, 0, 0, \(-7\), \(-\(7\/\@3\)\), \(-\(14\/\@3\)\), \(-7\), 7\/\@3, 14\/\@3}, {0, 0, 0, 1, 0, 0, \(-\(1\/2\)\), \(-\(\@3\/2\)\), 0, \(-\(1\/2\)\), \@3\/2, 0}}\)], "Output", CellLabel->"Out[18]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "These are the vectors that transform as the first row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((3)\)\)]], " in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ", but ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((3)\)\)]], "is two dimensional, thus we should compute the correspondig vectors that \ transform as the second row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((3)\)\)]], " (each E vibrational mode has a two-fold degenaration), that is, we should \ compute the vectors ", Cell[BoxData[ \(v\_j\%E\)]], ", j=2 (see Equation 18)." }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(base31 = GenerateBasisVectors[vecs3[\([1]\)], irreps[\([3]\)], totalrep]\)], "Input", CellLabel->"In[19]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 0, 0, 0, 0, 0, \(-7\), \(-\(7\/\@3\)\), \(-\(14\/\@3\)\), \(-7\), 7\/\@3, 14\/\@3}, {0, 1, 0, 0, \(-\(28\/3\)\), 28\/3, \(-\(7\/\@3\)\), \(-\(7\/3\)\), \(-\(14\/3\)\), 7\/\@3, \(-\(7\/3\)\), \(-\(14\/3\)\)}}\)], "Output", CellLabel->"Out[19]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(base32 = GenerateBasisVectors[vecs3[\([2]\)], irreps[\([3]\)], totalrep]\)], "Input", CellLabel->"In[20]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 0, 0, 1, 0, 0, \(-\(1\/2\)\), \(-\(\@3\/2\)\), 0, \(-\(1\/2\)\), \@3\/2, 0}, {0, 0, 0, 0, \(-1\), 0, \(-\(\@3\/2\)\), 1\/2, 0, \@3\/2, 1\/2, 0}}\)], "Output", CellLabel->"Out[20]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "Now from each of the vectors in ", StyleBox["vecs1", FontWeight->"Bold"], ", ", StyleBox["base31", FontWeight->"Bold"], " and ", StyleBox["base32", FontWeight->"Bold"], " we can make an animation for the vibrating molecule that corresponds to \ each of the vibrational modes. Remember however that these are not the pure \ vibrational modes but a linear combination of them. 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DGoo0018Ool00`00Oomoo`0NOol00`00Oomoo`1KOol4001EOol004Uoo`8001ioo`03001oogoo05Qo o`<005Uoo`00Bgoo00<007ooOol06goo00<007ooOol0E7oo1000G7oo001"], ImageRangeCache->{{{0, 287}, {242.375, 0}} -> {-0.000380689, 0.049297, \ 0.00336531, 0.00336531}}] }, Open ]] }, Open ]], Cell[TextData[{ "The option ", StyleBox["NumPic", FontWeight->"Bold"], " is the number of distincts pictures generated. The animation contains a \ series of ", StyleBox["NumPic", FontWeight->"Bold"], " pictures and the reversed sequence to give continuity to the movement. \ You can also control the amplitude of the movement and the viewpoint." }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["5. Uranium Hexafluoride.", "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " In the ", Cell[BoxData[ \(TraditionalForm\`UF\_6\)]], " molecule, the U atom is at the centre of an octahedron with the F atoms \ at the vertices. The symmetry group of the molecule is ", Cell[BoxData[ \(TraditionalForm\`O\_h\)]], ". We will solve the problem of molecular vibrations with the minimal set \ of necessary functions from the package ", StyleBox["Molecules.m", FontWeight->"Bold"], ". For simplicity we will assign atomic mass 2 to the U atom and 1 to the F \ atoms. (The U atom is much heavier than the F atoms thus the displacement for \ the U atom in each normal mode is much smaller than the displacements for the \ F atoms. Using the real atomic masses complicates the visualization of the \ animations.) \n The generators for the molecule are the U atom and one of \ the F atoms. We put the U atom at the origin and an F atom at {0,0,1}." }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(<< "\";\)\), "\n", \(\(<< "\";\)\), "\n", \(comments\)}], "Input", CellLabel->"In[1]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \("generators: group[[2]] (Pi/2 rotation about the z-axis) and \ group[[3]] (Pi/2 rotorreflexion about the x-axis)."\)], "Output", CellLabel->"Out[1]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(locations = {{0, 0, 0}, {0, 0, 1}};\)\), "\n", \(\(masses = {2, 1};\)\), "\n", \(\(cls = {GrayLevel[0], RGBColor[1, 0, 0]};\)\), "\n", \(mol = GenerateMolecule[group, locations]\)}], "Input", CellLabel->"In[2]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{\(-1\), 0, 0}, {0, \(-1\), 0}, {0, 0, \(-1\)}, {0, 0, 0}, {0, 0, 1}, {0, 1, 0}, {1, 0, 0}}\)], "Output", CellLabel->"Out[2]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(masses = GenerateMasses[group, mol, locations, masses]\)], "Input", CellLabel->"In[3]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({1, 1, 1, 2, 1, 1, 1}\)], "Output", CellLabel->"Out[3]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[BoxData[ \(\(cls = GenerateMasses[group, mol, locations, cls];\)\)], "Input", CellLabel->"In[4]:=", TextAlignment->Left, TextJustification->0], Cell[" The normal modes are,", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(modes = NormalVibrations[group, pos, ch, vibrchar, mol]\)], "Input", CellLabel->"In[5]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({1, 0, 1, 1, 0, 0, 0, 0, 1, 2}\)], "Output", CellLabel->"Out[5]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(modes . symbols\)], "Input", CellLabel->"In[6]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \("\!\(\(A\_1\)\_g\)" + "\!\(E\_g\)" + 2\ "\!\(\(F\_1\)\_u\)" + "\!\(\(F\_2\)\_g\)" + "\!\(\(F\_2\)\_u\)"\)], "Output", CellLabel->"Out[6]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["\<\ For visualization purposes we put bonds only between the F atoms,\ \ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(bg = {{5, 6}};\)\), "\n", \(bonds = GenerateBonds[group, bg, mol]\)}], "Input", CellLabel->"In[7]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 2}, {1, 3}, {1, 5}, {1, 6}, {2, 3}, {2, 5}, {2, 7}, {3, 6}, {3, 7}, {5, 6}, {5, 7}, {6, 7}}\)], "Output", CellLabel->"Out[7]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "Now we compute ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TR\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(totalrep = VibrationalRepresentation[group, mol];\)\)], "Input", CellLabel->"In[8]:=", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Now we obtain all the basis {", Cell[BoxData[ \(v\_j\%1\)]], ",...,", Cell[BoxData[ \(v\_j\%\(n\_\[Nu]\)\)]], Cell[BoxData[ \(TraditionalForm\`}\_\(j = 1, ... , b\_\[Nu]\)\)]], " (see Equation 18) for the pure vibrational modes for each ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(bases = GenerateVibrationalBasis[irreps, modes, totalrep, mol, masses];\)\)], "Input", CellLabel->"In[9]:=", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " ", StyleBox["bases", FontWeight->"Bold"], " is a list with length equal to the number of IR's of the symmetry group \ G. The element ", StyleBox["bases[[\[Nu]]]", FontWeight->"Bold"], " is a list with ", Cell[BoxData[ \(TraditionalForm\`b\_\[Nu]\)]], " elements, where ", Cell[BoxData[ \(TraditionalForm\`b\_\[Nu]\)]], " is the number of independent pure vibrational modes that transorm as ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". Finally, ", StyleBox["bases[[\[Nu], j]] ", FontWeight->"Bold"], "is the list of the ", Cell[BoxData[ \(n\_\[Nu]\)]], " vectors within each independent vibrational mode in correspondence with \ the rows of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " and with the same energy. For example, there are two modes ", Cell[BoxData[ \(\(F\_1\)\_u\)]], " (", StyleBox["symbols[[10]]=", FontWeight->"Bold"], Cell[BoxData[ \(\(F\_1\)\_u\)]], "), each of them containing three vibrations with the same energy (", Cell[BoxData[ \(n\_\(\(F\_1\)\_u\)\)]], "= 3). Thus, the animation for the vector that transforms as the second row \ in ", StyleBox["bases[[10, 2]] ", FontWeight->"Bold"], "should be obtained by executing the following comand," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(ShowVibratingMolecule[mol, bases[\([10, 2, 2]\)], bonds, Colors -> cls];\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " The modes ", Cell[BoxData[ \(\(A\_1\)\_g\)]], ", ", Cell[BoxData[ \(E\_g\)]], ", ", Cell[BoxData[ \(\(F\_2\)\_g\)]], " and ", Cell[BoxData[ \(\(F\_2\)\_u\)]], " appear only one time in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ". Thus, the basis vectors we have obtained for these modes are the pure \ vibrational modes; that is, we do not need to consider the force matrix to \ obtain them. This is not the case for the mode ", Cell[BoxData[ \(\(F\_1\)\_u\)]], " that appears two times in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], "." }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["6. Complex Representations.", "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " The symmetry groups of the two previous examples have only real \ representations. However, there are several point groups that have complex \ irreducible representations. We will not treat the case of pseudo-real \ representations, because none of the 32 point groups have this type of \ representation, although the treatment would be very similar to the case of \ complex representations. In this case, there are some displacement vectors \ obtained by the projector method that are complex. Obviously displacement \ vectors should be real. One must take into account another symmetry present \ in vibrating molecules, that is, the time reversal symmetry. Here, group \ theory tells us that there are extra degeneracies when the symmetry group has \ complex or pseudo-real representations. In the case of a normal mode \ transforming as a complex IR of dimension ", Cell[BoxData[ \(n\_\[Nu]\)]], ", there is another normal mode transforming as the non-equivalent \ conjugate IR with the same energy. Thus, there are 2", Cell[BoxData[ \(n\_\[Nu]\)]], " real, linearly independent combinations of these two normal modes that \ must be regarded as forming a single \"physically irreducible\" (real) [4] \ representation of twice the dimension ", Cell[BoxData[ \(n\_\[Nu]\)]], ". In the case of pseudo real representation, the IR and the complex \ conjugate one are equivalent but they are not equivalent to a real one. In \ this case, if there is a normal mode with the symmetry of this IR, there \ should be another normal mode transforming with the same IR and with the same \ energy such that the two normal modes form a physically irreducible \ representation.\n As an example we will consider a ficticious ", Cell[BoxData[ \(TraditionalForm\`H\_3\)]], " molecule with the ", StyleBox["H", FontSlant->"Italic"], " atoms at the vertex of an equilateral triangle. The symmetry group of \ this molecule is ", Cell[BoxData[ \(TraditionalForm\`D\_\(3 h\)\)]], " and has two normal modes, one one-dimensional with symmetry ", Cell[BoxData[ \(\(A\_1\%'\)\)]], " and the other, bidimensional, with symmetry ", Cell[BoxData[ \("\!\(\(E\^'\)\)"\)]], ". ", Cell[BoxData[ \(TraditionalForm\`D\_\(3 h\)\)]], " has only real representations. So we will forget some of the symmetries \ of the molecule and consider ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], " as the symmetry group. This group has two one-dimensional complex IR's. \ As we will see, time-reversal symmetry forces the same degeneracy as in the \ case of ", Cell[BoxData[ \(TraditionalForm\`D\_\(3 h\)\)]], ". It is possible to consider molecules that have only the ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], " symmetry, but they have at least six atoms and so the displacement \ vectors are eighteen dimensional. For pedagogical purpuses, it is sufficient \ to consider the ", Cell[BoxData[ \(TraditionalForm\`H\_3\)]], " molecule with ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], " symmetry," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[{ \(\(<< "\";\)\), "\n", \(\(<< "\";\)\)}], "Input", CellLabel->"In[1]:=", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(points = {{0, 1, 0}};\)\), "\n", \(mol = GenerateMolecule[group, points]\)}], "Input", CellLabel->"In[2]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 1, 0}, {\(-\(\@3\/2\)\), \(-\(1\/2\)\), 0}, {\@3\/2, \(-\(1\/2\)\), 0}}\)], "Output", CellLabel->"Out[2]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(masses = GenerateMasses[group, mol, points, {1}];\)\), "\n", \(\(cls = GenerateMasses[group, mol, points, {GrayLevel[0]}];\)\), "\n", \(bonds = GenerateBonds[group, {{1, 2}}, mol]\)}], "Input", CellLabel->"In[3]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{1, 2}, {1, 3}, {2, 3}}\)], "Output", CellLabel->"Out[3]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(modes = NormalVibrations[group, pos, ch, vibrchar, mol]\)], "Input", CellLabel->"In[4]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({1, 1, 1}\)], "Output", CellLabel->"Out[4]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(symbols\)], "Input", CellLabel->"In[5]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({"A", "\!\(\(E\_+\)\)", "\!\(\(E\_-\)\)"}\)], "Output", CellLabel->"Out[5]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ " The types of the IR's are given by the quantity ", StyleBox["reality", FontWeight->"Bold"], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(reality\)], "Input", CellLabel->"In[6]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({1, 0, 0}\)], "Output", CellLabel->"Out[6]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "Thus ", Cell[BoxData[ \(\(E\_+\)\)]], " and ", Cell[BoxData[ \("\!\(\(E\_-\)\)"\)]], " are complex conjugate representations and the modes transforming as these \ IR's have the same energy, according to time-reversal symmetry. Now we \ compute the displacement vectors for each normal mode," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(vibrep = VibrationalRepresentation[group, mol];\)\), "\n", \(\(basis = GenerateVibrationalBasis[irreps, modes, vibrep, mol, masses];\)\), "\n", \(basis = ComplexExpand[basis]\)}], "Input", CellLabel->"In[7]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{{{0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}}}, {{{1, I, 0, \(-\(1\/2\)\) + \(I\ \@3\)\/2, \(-\(I\/2\)\) - \@3\/2, 0, \(-\(1\/2\)\) - \(I\ \@3\)\/2, \(-\(I\/2\)\) + \@3\/2, 0}}}, {{{1, \(-I\), 0, \(-\(1\/2\)\) - \(I\ \@3\)\/2, I\/2 - \@3\/2, 0, \(-\(1\/2\)\) + \(I\ \@3\)\/2, I\/2 + \@3\/2, 0}}}}\)], "Output", CellLabel->"Out[7]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "As we can see, the basis vectors for ", Cell[BoxData[ \(\(E\_+\)\)]], " and ", Cell[BoxData[ \("\!\(\(E\_-\)\)"\)]], " are complex and conjugate to each other and together form a real, two \ dimensional physically irreducible representation.\n From a complex \ normalized basis vector ", Cell[BoxData[ \(v\_j\%\[Alpha]\)]], " (where \[Alpha] indicates an IR, j the row and we do not indicate the \ index that tells to which of the ", Cell[BoxData[ \(TraditionalForm\`b\_\[Alpha]\)]], " copies of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Alpha])\)\)]], " present in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], "belongs this vector) like ", StyleBox["basis[[2,1,1]]", FontWeight->"Bold"], " and the complex conjugate (", Cell[BoxData[ \(v\_j\%\[Alpha]\)]], Cell[BoxData[ \(TraditionalForm\`\()\^*\)\)]], " (", StyleBox["basis[[3,1,1]]", FontWeight->"Bold"], ") we can construct two real normalized independent vectors [1]," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[{ RowBox[{\(a\_j\%\[Alpha]\), "=", RowBox[{\(1\/\@2\), RowBox[{"{", RowBox[{\(v\_j\%\[Alpha]\), "+", RowBox[{"(", \(v\_j\%\[Alpha]\), FormBox[\()\^*\), "TraditionalForm"]}]}], "}"}]}]}], "\n", RowBox[{ RowBox[{\(d\_j\%\[Alpha]\), "=", RowBox[{ FractionBox["1", RowBox[{ StyleBox["i", FontSlant->"Italic"], \(\@2\)}]], RowBox[{"{", RowBox[{\(v\_j\%\[Alpha]\), "-", RowBox[{"(", \(v\_j\%\[Alpha]\), FormBox[\()\^*\), "TraditionalForm"]}]}], "}"}]}]}], "\n", "\t"}]}], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Thus ", Cell[BoxData[ \(a\_j\%\[Alpha]\)]], " and ", Cell[BoxData[ \(d\_j\%\[Alpha]\)]], " do not transform as the j row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Alpha])\)\)]], ", but rather they are basis vectors of the physically irreducible \ representation," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ RowBox[{ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { RowBox[{\(2\^\(\(-1\)/2\)\), StyleBox["1", FontWeight->"Bold"]}], RowBox[{ RowBox[{"-", SuperscriptBox[ StyleBox[ RowBox[{ StyleBox["i", FontSlant->"Italic"], "2"}]], \(\(-1\)/2\)]}], StyleBox["1", FontWeight->"Bold"]}]}, { RowBox[{\(2\^\(\(-1\)/2\)\), StyleBox["1", FontWeight->"Bold"]}], RowBox[{ SuperscriptBox[ StyleBox[ RowBox[{ StyleBox["i", FontSlant->"Italic"], "2"}]], \(\(-1\)/2\)], StyleBox["1", FontWeight->"Bold"]}]} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { FormBox[\(\[CapitalGamma]\^\((\[Alpha])\)\), "TraditionalForm"], "0"}, {"0", FormBox[\(\(\[CapitalGamma]\^\((\[Alpha])\)\)\^*\), "TraditionalForm"]} }], "\[NegativeThinSpace]", ")"}], RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ { RowBox[{\(2\^\(\(-1\)/2\)\), StyleBox["1", FontWeight->"Bold"]}], RowBox[{ RowBox[{"-", SuperscriptBox[ StyleBox[ RowBox[{ StyleBox["i", FontSlant->"Italic"], "2"}]], \(\(-1\)/2\)]}], StyleBox["1", FontWeight->"Bold"]}]}, { RowBox[{\(2\^\(\(-1\)/2\)\), StyleBox["1", FontWeight->"Bold"]}], RowBox[{ SuperscriptBox[ StyleBox[ RowBox[{ StyleBox["i", FontSlant->"Italic"], "2"}]], \(\(-1\)/2\)], StyleBox["1", FontWeight->"Bold"]}]} }], "\[NegativeThinSpace]", ")"}]}]], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "which is a representation contained in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], " and where ", StyleBox["1", FontWeight->"Bold"], " is the identity matrix of dimension ", Cell[BoxData[ \(n\_\[Nu]\)]], ".\n The function ", StyleBox["RealVibrationalBasis", FontWeight->"Bold"], " allows us to compute the real basis vectors. The second argument is a \ list whose elements are lists of two integers that give the positions in the \ output from ", StyleBox["GenerateBasisVectors", FontWeight->"Bold"], " of the complex representations and their conjugate representations. In \ the present case, the second and thirth IR's are conjugate to each other, so \ the second argument should be {{2,3}}. The output gives the ", Cell[BoxData[ \(a\_j\%\[Alpha]\)]], " vectors in the position formerly ocupied by the second IR, and the ", Cell[BoxData[ \(d\_j\%\[Alpha]\)]], " in the position occupied by the third IR," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(realbasis = RealVibrationalBasis[basis, {{2, 3}}]\)], "Input", CellLabel->"In[8]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{{{0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}}}, {{{\@2, 0, 0, \(-\(1\/\@2\)\), \(-\@\(3\/2\)\), 0, \(-\(1\/\@2\)\), \@\(3\/2\), 0}}}, {{{0, \(-\@2\), 0, \(-\@\(3\/2\)\), 1\/\@2, 0, \@\(3\/2\), 1\/\@2, 0}}}}\)], "Output", CellLabel->"Out[8]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["\<\ Now we could perform the animations with the following \ commands,\ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(ShowVibratingMolecule[mol, realbasis[\([1, 1, 1]\)], bonds, Colors -> cls];\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(ShowVibratingMolecule[mol, realbasis[\([2, 1, 1]\)], bonds, Colors -> cls];\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(ShowVibratingMolecule[mol, realbasis[\([3, 1, 1]\)], bonds, Colors -> cls];\)\)], "Input", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["7. The Force Matrix.", "Section", TextAlignment->Left, TextJustification->0], Cell["\<\ The potential energy of a molecule executing small vibrations in \ terms of the cartesian coordinates of the individual atoms is of the \ form,\ \>", "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(V = \(v\^T\) F\ v\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "where v is the vector of ", "Equation", " 5 considered as a column matrix and F is the force matrix. F is a real, \ symmetrical matrix with positive (pure vibrations) and zero (translations and \ rotations) eigenvalues. The number of independent parameters in F is \ determined by the symmetry properties of the molecule. F acquires a \ particularly (although in general not real if there are complex IR's) simple \ form when considered in a basis of displacement vectors that transform as the \ rows of the IR's that we have calculated in the previous sections. (Remember \ that these may not be the pure normal modes but a linear combination of \ them). If a given IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " of dimension ", Cell[BoxData[ \(n\_\[Nu]\)]], " appears ", Cell[BoxData[ \(b\_\[Nu]\)]], " times in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ", we will denote the basis vectors for the m copy (m=1,...,", Cell[BoxData[ \(TraditionalForm\`b\_\[Nu]\)]], ") with ", Cell[BoxData[ \(v\_mj\%\[Nu]\)]], " where j indicates the row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " according to which ", Cell[BoxData[ \(v\_mj\%\[Nu]\)]], " transforms.\n Group theory tells us that the matrix elements of F in \ this basis are of the form," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\[LeftAngleBracket]v\_mj\%\[Nu] | F | v\_nk\%\[Mu]\[RightAngleBracket] = \[Delta]\_\[Nu]\[Mu]\ \ \[Delta]\_jk\ F\_mn\%\((\[Nu])\)\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "that is, F has non-null matrix elements only between vectors that \ transform with the same row in the same IR, and the value does not depend on \ the row.\n To obtain the energy of vibrations (eigenvalues of F) and the \ normal modes (eigenvectors of F) we have to diagonalize the force matrix. \ Thus, when the given IR ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " appears in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ", we have to consider only the submatrix of F whose matrix elements are," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(f\_mn\%\((\[Nu])\) = \(\[LeftAngleBracket]v\_m1\%\[Nu] | F | v\_n1\%\[Nu]\[RightAngleBracket] = \ F\_mn\%\((\[Nu])\)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m\), n = 1, ... , b\_\[Nu]\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "This ", Cell[BoxData[ \(b\_\[Nu]\)]], "\[Times] ", Cell[BoxData[ \(b\_\[Nu]\)]], " matrix appears ", Cell[BoxData[ \(n\_\[Nu]\)]], "times in the diagonal of F, due to ", "Equation", " 22. Thus the eigenvalues of ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], "have multiplicity ", Cell[BoxData[ \(n\_\[Nu]\)]], " in F. The eigenvectors determine the coefficients for the linear \ combinations of vectors ", Cell[BoxData[ \(v\_m1\%\[Nu]\)]], " that are the pure normal modes of the molecule that transform as the \ first row of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " (if they are complex we should make the unitary transformation of the \ previous section). These coefficients are the same for the combinations of \ the other rows ", Cell[BoxData[ \(v\_mj\%\[Nu]\)]], ". The number of independent coefficients in ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], " depends on the reality of ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], ". If ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " is real, ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], " is a real symmetrical matrix. If ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^\((\[Nu])\)\)]], " is complex, ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], " is hermition and the IR ", Cell[BoxData[ \(TraditionalForm\`\(\(\[CapitalGamma]\^\(\((\[Nu])\)\(\ \ \)\)\)\^*\)\)]], ", due to time reversal symmetry, has ", Cell[BoxData[ \(\(\(f\^\((\[Nu])\)\)\^*\)\)]], " (the conjugate of ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], ") as the force submatrix.\n As a simple example, consider a fictitious \ molecule like the one of the previous section but with an additional H atom \ at the centre of the triangle. Again we consider the symmetry of the molecule \ as being ", Cell[BoxData[ \(TraditionalForm\`C\_3\)]], "," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[{ \(\(<< "\";\)\), "\n", \(\(<< "\";\)\)}], "Input", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(points = {{0, 1, 0}, {0, 0, 0}};\)\), "\n", \(\(masses = {1, 1};\)\), "\n", \(mol = GenerateMolecule[group, points]\)}], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{0, 0, 0}, {0, 1, 0}, {\(-\(\@3\/2\)\), \(-\(1\/2\)\), 0}, {\@3\/2, \(-\(1\/2\)\), 0}}\)], "Output", CellLabel->"Out[2]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ \(\(masses = GenerateMasses[group, mol, points, masses];\)\), "\n", \(\(modes = NormalVibrations[group, pos, ch, vibrchar, mol];\)\), "\n", \(modes . symbols\)}], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(2\ "A" + 2\ "\!\(\(E\_-\)\)" + 2\ "\!\(\(E\_+\)\)"\)], "Output", CellLabel->"Out[3]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "Thus, the F matrix has three 2\[Times]2 ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], " submatrices, one for each IR (in this case, since all the IR's are one \ dimensional, the submatrices ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], "appears only one time on the diagonal, that is, there are only ", Cell[BoxData[ \(v\_m1\%\[Nu]\)]], " vectors because ", Cell[BoxData[ \(n\_\[Nu]\)]], "=1 for all the IR's). We call these three submatrices as ", Cell[BoxData[ \(f\^A\)]], ", ", Cell[BoxData[ \(f\^\(E\_-\)\)]], "and ", Cell[BoxData[ \(f\^\(E\_+\)\)]], ". Due to time reversal, ", Cell[BoxData[ \(f\^\(E\_+\)\)]], " is the conjugate of ", Cell[BoxData[ \(f\^\(E\_-\)\)]], " and in general they are complex. ", Cell[BoxData[ \(f\^A\)]], " is real because A is a real IR. \n We will find the pure normal modes \ given the coefficients of the ", Cell[BoxData[ \(f\^\((\[Nu])\)\)]], " matrices. We should first compute the displacement vectors as in the \ previous sections," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[{ \(\(vibrep = VibrationalRepresentation[group, mol];\)\), "\n", \(\(basis = GenerateVibrationalBasis[irreps, modes, vibrep, mol, masses];\)\), "\n", \(basis = ComplexExpand[basis]\)}], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{{{0, 0, 1, 0, 0, \(-\(1\/3\)\), 0, 0, \(-\(1\/3\)\), 0, 0, \(-\(1\/3\)\)}}, {{0, 0, 0, 0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}}}, {{{1, \(-I\), 0, 0, \(2\ I\)\/3, 0, \(-\(1\/2\)\) + I\/\(2\ \@3\), I\/6 - 1\/\(2\ \@3\), 0, \(-\(1\/2\)\) - I\/\(2\ \@3\), I\/6 + 1\/\(2\ \@3\), 0}}, {{0, 0, 0, 1, I, 0, \(-\(1\/2\)\) + \(I\ \@3\)\/2, \(-\(I\/2\)\) - \@3\/2, 0, \(-\(1\/2\)\) - \(I\ \@3\)\/2, \(-\(I\/2\)\) + \@3\/2, 0}}}, {{{1, I, 0, 0, \(-\(\(2\ I\)\/3\)\), 0, \(-\(1\/2\)\) - I\/\(2\ \@3\), \(-\(I\/6\)\) - 1\/\(2\ \@3\), 0, \(-\(1\/2\)\) + I\/\(2\ \@3\), \(-\(I\/6\)\) + 1\/\(2\ \@3\), 0}}, {{0, 0, 0, 1, \(-I\), 0, \(-\(1\/2\)\) - \(I\ \@3\)\/2, I\/2 - \@3\/2, 0, \(-\(1\/2\)\) + \(I\ \@3\)\/2, I\/2 + \@3\/2, 0}}}}\)], "Output", CellLabel->"Out[4]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ " We start with the A representation. The vectors ", Cell[BoxData[ \(v\_11\%A\)]], " and ", Cell[BoxData[ \(v\_21\%A\)]], " are," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vA11 = basis[\([1, 1, 1]\)]\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({0, 0, 1, 0, 0, \(-\(1\/3\)\), 0, 0, \(-\(1\/3\)\), 0, 0, \(-\(1\/3\)\)}\)], "Output", CellLabel->"Out[5]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(vA21 = basis[\([1, 2, 1]\)]\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({0, 0, 0, 0, 1, 0, \(-\(\@3\/2\)\), \(-\(1\/2\)\), 0, \@3\/2, \(-\(1\/2\)\), 0}\)], "Output", CellLabel->"Out[6]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "The most general form for the ", Cell[BoxData[ \(f\^A\)]], "matrix is," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\((FA = {{V1, c1}, {c1, V2}})\)]\)], "Input", CellLabel->"In[7]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {"V1", "c1"}, {"c1", "V2"} }], ")"}], (MatrixForm[ #]&)]], "Output", CellLabel->"Out[7]//MatrixForm=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "The eigenvalues of ", StyleBox["FA", FontWeight->"Bold"], " are the coefficients for the linear combinations of the vectors ", StyleBox["vA11 ", FontWeight->"Bold"], " and ", StyleBox["vA21", FontWeight->"Bold"], " that give the true normal A modes," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \({coeff1, coeff2} = Eigenvectors[FA]\)], "Input", CellLabel->"In[8]:=", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({{\(-\(\(\(-V1\) + V2 + \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(2\ \ c1\)\)\), 1}, {\(-\(\(\(-V1\) + V2 - \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(2\ \ c1\)\)\), 1}}\)], "Output", CellLabel->"Out[8]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell["Thus, the pure normal A modes are,", "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(purevA11 = coeff1 . {vA11, vA21}\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({0, 0, \(-\(\(\(-V1\) + V2 + \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(2\ \ c1\)\)\), 0, 1, \(\(-V1\) + V2 + \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(6\ \ c1\), \(-\(\@3\/2\)\), \(-\(1\/2\)\), \(\(-V1\) + V2 + \@\(4\ c1\^2 + V1\^2 - \ 2\ V1\ V2 + V2\^2\)\)\/\(6\ c1\), \@3\/2, \(-\(1\/2\)\), \(\(-V1\) + V2 + \ \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(6\ c1\)}\)], "Output", CellLabel->"Out[9]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(purevA21 = coeff2 . {vA11, vA21}\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({0, 0, \(-\(\(\(-V1\) + V2 - \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(2\ \ c1\)\)\), 0, 1, \(\(-V1\) + V2 - \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(6\ \ c1\), \(-\(\@3\/2\)\), \(-\(1\/2\)\), \(\(-V1\) + V2 - \@\(4\ c1\^2 + V1\^2 - \ 2\ V1\ V2 + V2\^2\)\)\/\(6\ c1\), \@3\/2, \(-\(1\/2\)\), \(\(-V1\) + V2 - \ \@\(4\ c1\^2 + V1\^2 - 2\ V1\ V2 + V2\^2\)\)\/\(6\ c1\)}\)], "Output", CellLabel->"Out[10]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "Giving values to the coefficients ", StyleBox["V1", FontWeight->"Bold"], ", ", StyleBox["V2 ", FontWeight->"Bold"], "and ", StyleBox["c1", FontWeight->"Bold"], ",", " we could compute the numerical values for these vectors and perform the \ animations. We will not do this here.\n Now we can start with the complex \ representations ", Cell[BoxData[ \(\(E\_+\)\)]], " and ", Cell[BoxData[ \(\(E\_-\)\)]], ". Since they are conjugate, we only need to obtain the pure complex \ displacement vectors for ", Cell[BoxData[ \(\(E\_+\)\)], "NumberedEquation"], " and then we can apply the procedure of the previous section to obtain the \ pure real displacement vectors.\n The vectors ", Cell[BoxData[ \(v\_11\%\(E\_+\)\)]], " and ", Cell[BoxData[ \(v\_21\%\(E\_+\)\)]], " are," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(vEplus11 = basis[\([2, 1, 1]\)]\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({1, \(-I\), 0, 0, \(2\ I\)\/3, 0, \(-\(1\/2\)\) + I\/\(2\ \@3\), I\/6 - 1\/\(2\ \@3\), 0, \(-\(1\/2\)\) - I\/\(2\ \@3\), I\/6 + 1\/\(2\ \@3\), 0}\)], "Output", CellLabel->"Out[11]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(vEplus21 = basis[\([2, 2, 1]\)]\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \({0, 0, 0, 1, I, 0, \(-\(1\/2\)\) + \(I\ \@3\)\/2, \(-\(I\/2\)\) - \@3\/2, 0, \(-\(1\/2\)\) - \(I\ \@3\)\/2, \(-\(I\/2\)\) + \@3\/2, 0}\)], "Output", CellLabel->"Out[12]=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "The most general form for the ", Cell[BoxData[ \(f\^\(E\_+\)\)]], "matrix is," }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[\((FEplus = {{W1, b1 + I\ d1}, {b1 - I\ d1, W2}})\)]\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {"W1", \(b1 + I\ d1\)}, {\(b1 - I\ d1\), "W2"} }], ")"}], (MatrixForm[ #]&)]], "Output", CellLabel->"Out[13]//MatrixForm=", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[TextData[{ "The eigenvalues of ", StyleBox["FEplus", FontWeight->"Bold"], " are the coefficients for the linear combinations of the vectors ", StyleBox["vEplus11 ", FontWeight->"Bold"], " and ", StyleBox["vEplus21", FontWeight->"Bold"], " that give the pure normal ", Cell[BoxData[ \(\(E\_+\)\)]], " modes," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\({coeff1, coeff2} = Eigenvectors[FEplus];\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Finally, the pure normal ", Cell[BoxData[ \(\(E\_+\)\)]], " modes are," }], "Text", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(purevEplus11 = coeff1 . {vEplus11, vEplus21};\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(\(purevEplus21 = coeff2 . {vEplus11, vEplus21};\)\)], "Input", TextAlignment->Left, TextJustification->0], Cell[TextData[{ " These are complex vectors. Now we can perform the unitary transformation \ of ", "Equation", "s 19 and 20 to obtain the pure real displacement vectors that correspond \ to the two copies of the physically irreducible representation ", StyleBox["E", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\(E\_+\)\)]], " \[CirclePlus] ", Cell[BoxData[ \(TraditionalForm\`\(E\_-\)\)]], " that appear in ", Cell[BoxData[ \(TraditionalForm\`\[CapitalGamma]\^TVR\)]], ".\n Finally, we will consider the same molecule but with the full \ symmetry group, that is ", Cell[BoxData[ \(TraditionalForm\`D\_\(3 h\)\)]], ". 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This is due to the fact \ that to solve small molecules, there are usually some tricks that one can \ exploit to avoid the use of projectors. For example, in reference [6], in the \ computation of the displacement vectors for the ", Cell[BoxData[ \(TraditionalForm\`NH\_3\)]], " molecule, three methods are explained, each with more algebra than the \ previous one, that can be used only in certain limited problems. The \ projector method is usually deferred as the last recourse if the tricks do \ not work. Although it is certainly true that the work needed to perform a \ calculation with the projector method by hand is tedious, it is the only with \ general applicability and it is conceptually simple. The use of ", StyleBox["Mathematica", FontSlant->"Italic"], " allows one to concentrate on the general aspects of group theory that \ have general applicability to this kind of problem. Also these principles \ appear in other physical problems. For example, the projector method is \ applicable to the molecular orbitals (MO) method using the linear \ combination of atomic orbitals (LCAO) method for solving electronic structure \ problems in molecules. Basically, one has only to replace the first argument \ in the function ", StyleBox["VibrationalRepresentation", "Input", FontWeight->"Plain"], StyleBox[" ", FontWeight->"Bold"], " by the atomic orbital representation under study, that is the restriction \ to the symmetry group of the molecule of one of the irreducible \ representations ", Cell[BoxData[ \(TraditionalForm\`D\^\((j)\)\)]], " of the rotation group. In fact, the package ", StyleBox["Molecules.m", "Input", FontWeight->"Plain"], " is a subset of a more general package we developed that includes tools to \ treat MO problems by the LCAO method." }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["REFERENCES", "Section", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "[1] Cornwell, J. F., ", StyleBox["Group Theory in Physics, Academic Press", FontSlant->"Italic"], ", 1984.\n\n[2] Sosnovsky, A. and Demarco, G., ", StyleBox["Math in Ed. and Res. ", FontSlant->"Italic"], StyleBox["6", FontWeight->"Bold"], " (1997) 5.\n\n[3] Sosnovsky, A. and Demarco, G., submmited to ", StyleBox["Math. in Ed. and Res.", FontSlant->"Italic"], " (1997) .\n\n[4] Landau, L. D. and Lifshitz E.M., ", StyleBox["Quantum Mechanics", FontSlant->"Italic"], ", Pergamon, 1989\n\n[5] Hamermesh M., ", StyleBox["Group Theory and Its Application to Physical Problems,", FontSlant->"Italic"], " Addison-Wesley, 1962.\n\n[6] Lax M., ", StyleBox["Symmetry Principles in Solid State and Molecular Physics,", FontSlant->"Italic"], " Wiley, 1974." }], "Reference", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell["ABOUT THE AUTHOR", "Subsection", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "Gustavo Demarco obtained a Ph.D. in Physics from Instituto Balseiro in \ 1992. Since that time he has been at Centro Atomico Bariloche working in the \ Isotope Separation Group and now in the Advanced Designs and Economic \ Evaluation Division. He is also an Assistant Researcher of CONICET and an \ Assistant at Instituto Balseiro. 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Argentina demarco@cab.cnea.edu.ar \ \>", "Text", TextAlignment->Left, TextJustification->0] }, Open ]], Cell[CellGroupData[{ Cell["ELECTRONIC SUBSCRIPTIONS", "Subsection", TextAlignment->Left, TextJustification->0], Cell[TextData[{ StyleBox["Included in the distribution for each electronic subscription is \ the file", FontFamily->"Times New Roman", FontSize->12], StyleBox[" ", "Input", FontFamily->"Times New Roman", FontSize->12], StyleBox["molvib.nb", "Input", FontWeight->"Plain"], StyleBox[", containing ", FontFamily->"Times New Roman", FontSize->12], StyleBox["Mathematica", FontFamily->"Times New Roman", FontSize->12, FontSlant->"Italic"], StyleBox[" code for the material described in this article.", FontFamily->"Times New Roman", FontSize->12] }], "Text", TextAlignment->Left, TextJustification->0] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"4.0 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 695}}, WindowToolbars->"EditBar", WindowSize->{812, 629}, WindowMargins->{{63, Automatic}, {Automatic, 13}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "Magnification"->1}, Magnification->1.25, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of \ all cells in a given style. 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