Symmetry part 2 2

1
Symmetry part 2 ??????? ??? 2
http//www.tau.ac.il/chemistry/undergraduate/unde
rgraduate-courses.html
Israel Schek ?????
???? ?????????? ?? ???? Tel Aviv University,
Israel
2
Molecular Motion

• A non-linear molecule with N atoms has (3N-6)
  internal vibrational degrees of freedom
• 3 translational 3 rotational 6 is the number
  of the center of mass degrees of freedom.
• A linear polyatomic molecule, (e.g. a diatomic),
  has only two rotational degrees of freedom, and
  thus 3N-5 vibrational modes.
• A change in any of these internal coordinates
  generates a distortion of the molecule without
  changing its center of mass, or creating an
  overall rotation.

3
Molecular Motion

• If a single vibrational mode (e.g. an
  inter-atomic bond) is excited, the two relevant
  atoms start to vibrate.
• This in turn causes other modes connected with
  these atoms to start vibrating, and thus
  vibrational coupling mechanisms practically shake
  the whole molecule.
• Denote the displacement of each nucleus from its
  equilibrium xa,ya,za the kinetic energy is

4
Molecular Motion- Potential Expansion

• The potential energy function of a polyatomic
  molecule, a generalization of the diatomic
  function, is expanded in power series
• where ?a is any one of the Cartesian coordinates,
  and the quadratic form ?a?ß is any of the
  combinations
• The mixed high-order derivatives
• are the coupling terms mixing the vibrations.

5
Molecular Motion - Potential Expansion

• To simplify the expressions, define the
  mass-weighted coordinates
• In terms of these 3N coordinates the kinetic
  energy is
• The potential energy is expanded as

6
Molecular Motion

• At equilibrium of a diatomic molecule, the
  potential curve exhibits a minimum
• For a polyatomic molecule the potential function
  is not a mere curve but a hyper-surface.
• Local minima are exhibited where first
  derivatives vanish
• The molecule is considered to have a local
  equilibrium with respect to the relevant
  coordinate.

7
Potential Surfaces

• To calculate the potential surface by solving the
  Schrödinger equation is a huge task even for a
  small molecule.
• The surface is practically obtained by empirical
  functions using spectroscopic data.

Atkins, Quanta
8
Molecular Motion Harmonic Approximation

• For small vibrations, the second-order expansion
  is sufficient (harmonic approximation)
• where the mixed Hook force constant is
• In matrix notation, the kinetic energy is
  diagonal
• is a column vector with elements
• The dot above the vector denotes a vector of
  derivatives.

9
Molecular Motion Harmonic Approximation

• The potential energy in the harmonic
  approximation is
• where the symmetric force constants matrix is
  Kij
• It is symmetric since
• Thus the potential energy is definitely
  non-diagonal, causing difficulties in handling
  the problem.
• We desire to find a representation Qi, which
  diagonalizes the total energy (without cross
  terms)

10
Molecular Motion - Normal Modes

• The set Qi is a linear combination of the
  original set qi
• It represents a set f uncoupled harmonic
  oscillators, each one vibrates by independently
  (uncoupled) by itself, called normal modes
• We look for a transformation L which diagonalize
  the matrix of force constants K
• where ? is the diagonal matrix of the eigenvalues
  ?m
• We need to solve the secular equation

11
Molecular Motion - Normal Modes

• Having the eigenvalues ?m, the linear
  simultaneous equations
• are solved, L(m) is the m-th column of the
  orthogonal matrix L.
• Explicitly, the equation to solve is
• A new set of coordinates Q is obtained by
  rotation of the original set q

12
Molecular Motion - Normal Modes

• The potential energy in terms of the rotated
  basis becomes
• Explicitly
• The kinetic energy is
• Explicitly
• Since the potential energy becomes a non-mixed
  sum of quadratic terms and the kinetic energy
  stays as such, the new diagonal set is called the
  set of normal coordinates.

13
Molecular Motion - Normal Modes

• So far the modes were treated in a way as if all
  the 3N modes were independent.
• There are six relations between the qi, which
  eliminate six non-vibrational degrees of freedom.
• As a result, six roots of the secular equation
  (five for a linear molecule) are identically
  zero, corresponding to the translational and
  rotational degrees of freedom.
• Hence the summation over the normal coordinates
  of the potential energy need to be taken only to
  3N-6, not 3N.

14
Molecular Motion - Normal Modes

• According to the orthogonal transformation, the
  second-order derivative of the normal coordinate
  is
• Due to the chain rule of differentiation
• These two expressions are equal

15
Normal Modes, Frequency

• Using the explicit quadratic form of the
  potential energy, the equation of motion in terms
  of a normal coordinate is
• This is the well-known equation for the harmonic
  oscillator
• Bk the amplitue, bk the phase, determined
  according to given initial conditions.
• This mode of vibration, which is independent of
  other ones, is called a normal mode of vibration,
  oscillating in an obvious frequency

16
Normal Modes and Atomic Stretches

• The mode as a whole vibrates harmonically. In
  general, all atoms take part in this mode in a
  well-orchestrated manner that is resulting in one
  common vibration of the molecule.
• Transforming back into the original mass-weighted
  Cartesian basis qi, The atomic coordinate
  develops in time according to
• The transformation coefficient is

17
Molecular Motion - Normal Modes

• Thus a normal mode is a superposition of the
  atomic oscillations, and inversely, the atomic
  motion may be considered as a linear
  superposition of the normal modes.
• Each atom moves in a way that generally
  incorporates all the normal modes, each one with
  its own amplitude.
• It is important to realize that when a single
  bond is excited, it induces a motion of all
  neighboring bonds, which consequently shakes the
  whole molecule, where each bond has vibrates in
  its own frequency.
• On the other hand, when a normal mode is excited,
  the whole molecule vibrates as a corporate
  entity, with a certain frequency, where each atom
  plays its specific role of this mode.

18
Molecular Motion - Normal Modes

• Each normal mode behaves as a single uncoupled
  harmonic oscillator, with a certain effective
  mass, force constant and frequency.
• The effective mass is a combination of the
  individual masses of the involved atoms,
  according to how much each atom moves in this
  mode.
• Likewise, each force constant is a combination of
  the force constants of the involved bonds.
• The quantum energy of the normal modes ?i with
  excitations ni, i1, s

19
Molecular Motion - Normal Modes

• In spectroscopy activation of these normal modes
  are treated to certain approximation as if they
  were uncoupled oscillators, with proper selective
  rules.
• Electric dipole excitation occurs if the normal
  mode exhibits a change of the electric dipole
  moment (e.g. CO2 symmetric stretch is NOT IR
  active, while the 3 other modes are IR active)

20
Bond Motion

• Thus, in principle it is hardly possible to
  isolate and excite a single bond.
• However, if a bond is loosely coupled to other
  atoms, and if the mass of the rest is high, it
  may not be a bad approximation to handle this
  mode separately.
• This is the basis of IR analysis of bonds, angles
  and chemical groups

21
Symmetry of Molecular Motion

• Consider a molecule which belongs to asymmetry
  group G.
• The molecular motion defines a symmetry group H
• Theorem H is always a sub-group of G
• Example the linear homonuclear diatomic H2 of
  symmetry D8h
• Attach an arrow to each atom in the direction
  of its motion

22
Nonlinear Symmetric Triatomic Molecules

• The following analysis are according to Atoms
  Molecules by Karplus and Porter (chapter 6)
• Non linear Y-X-Y molecule has 3 vibrational
  coordinates.
• Motions of atoms out of the molecular plane
  correspond to either translation or rotation.
• Restricted in plane we treat ?x1, ?y1, ?x2, ?y2,
  ?x3, ?y3
• Convenient to introduce combinations of these
  displacements, appropriate to the molecular
  symmetry (a plane of symmetry).
• Symmetric coordinates are either symmetric or
  antisymmetric with respect to reflection.

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Nonlinear Symmetric Triatomic Molecules

• Vibrational displacements (a) and (b) are
  unchanged by reflection in the yz plane
• ?x1 ? ?x1
• ?x2 ? ?x2
• ?x3 ? ?x3
• ?y1 ? ?y1
• ?y2 ? ?y2
• ?y3 ? ?y3

S1 symmetric
S2 anti-symmetric
24
Nonlinear Symmetric Triatomic Molecules

• The center of mass remains fixed (frozen
  translation)
• Hence
• In terms of the displacement coordinates S1 and
  S2, respectively

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Nonlinear Symmetric Triatomic Molecules

• The third vibrational displacement is
  antisymmetric with respect to reflection in the
  yz plane.
• ?x2 and ?x3 have the same magnitude and sign
• ?y2 and ?y3 have the same magnitude and opposite
  signs
• To keep a stationary center of mass

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Nonlinear Symmetric Triatomic Molecules

• To prevent rotation ? a zero angular momentum
  about the z axis
• For small displacements
• where x10, y10, are the equilibrium
  coordinate values
• Substitute
• 
  where a is the equilibrium bond angle
• Simply take

27
Nonlinear Symmetric Triatomic Molecules

• Now any arbitrary vibration of the symmetric
  triatomic can be expressed in terms of the 3
  symmetry displacements S1, S2, S3
• The corresponding Cartesian coordinate
  displacements

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Nonlinear Symmetric Triatomic Molecules

• The vibrational kinetic energy in Cartesian
  coordinates
• Using relations as
• the kinetic energy in terms of the displacement
  coordinates is diagonal since they are
  orthogonal
• where

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Nonlinear Symmetric Triatomic Molecules

• The vibrational potential energy is not diagonal
  in terms of the 3 symmetry displacements S1, S2,
  S3
• On reflection in the yz plane S3 is replaced by
  S3, while S1 and S2 are unchanged.
• The potential energy function, being a scalar, is
  invariant with respect to any symmetry operation
  (the two Y atoms are interchanged)

30
Nonlinear Symmetric Triatomic Molecules

• Adding these two expressions and dividing by 2
• The two terms in the potential function that
  change sign on reflection (products of
  coordinates with different symmetries) must have
  zero coefficients
• Now we write Newton equations of motion according
  to Tvib and Vvib

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Nonlinear Symmetric Triatomic Molecules

• Consequently the displacement coordinate S3Q3 is
  a normal mode, with frequency
• The other two coordinates are not normal modes !
• Try
• so that
• Substitute the mixed S1, S2 equations and
  rearrange

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Nonlinear Symmetric Triatomic Molecules

• Since the two displacements coordinates S1 and S2
  are independent, they have arbitrary values.
• Take S20 and S1?0 or S10 and S2?0 we get two
  linear homogeneous equations for g1 and g2
• We get the ratio
• depending on which ? (or which normal mode)
  we take

33
Nonlinear Symmetric Triatomic Molecules

• Pitaron ?
• ? Simultaneous solution only if the secular
  determinant of the coefficient vanishes
• depending through C11, C22, C12 upon the
  inter-atomic forces.

34
Nonlinear Symmetric Triatomic Molecules

• Assume that the potential function may be
  expressed in terms of the bonds stretching and
  angle changes, take the changes ?r1, ?r2, ?r3 of
  the bonds X1Y2, X1Y3, Y2Y3 respectively, and d
  the increase in the bond angle.
• For small displacements, r0 is the equilibrium
  X-Y bond

35
Nonlinear Symmetric Triatomic Molecules

• A simple model the central-force field
• Presently k2k1. Two parameters k1 and k3 should
  be fitted to the three normal-mode frequencies.
• A third parameter can be considered by additional
  terms to the central-force field, namely an
  interaction between the bonds

36
Nonlinear Symmetric Triatomic Molecules

• The potential function coefficients are then
• Unfortunately, when substituted into the secular
  equation imaginary values are obtained for k1 and
  k12, hence imaginary values for ?1,2
• If k1 and k12 are both real the observed
  frequencies must obey

37
Nonlinear Symmetric Triatomic Molecules

• A better model the modified valence
  central-force field
• Whence
• Often the value of the mixed parameter k12 is
  small (compared to k1). The pure valence-force
  field is when we set k120, yielding a
  satisfactory two-parameter potential function.

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Symmetry Adapted Linear Combinations
Maurits Cornelis Escher
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Symmetry Adapted Orbitals (a)(????????? ???????
???????)

• A common problem to in theoretical chemistry,
  solid state theory and molecular dynamics is
• How to create orthonormal linear combinations
  of atomic orbitals (AOs) or molecular internal
  coordinates.
• These combinations should form bases for the
  irreps of the symmetry group of the system.
• These functions are referred to as
• Symmetry adapted linear combinations (SALCs).

40
Symmetry Adapted Orbitals (b)

• Among these problems
• Construction of hybrid orbitals
• Construction of molecular orbitals (MOs)
• Adaptation of orbitals to ligand field
• Analysis of molecular vibrations

41
Symmetry Adapted Orbitals (c)

• Reasons for looking for SALCs
• Fundamental reason
• Only these functions form acceptable
  eigenfunctions of the Schrödinger equation.
• Practical reason
• When SALCs are used the symmetry properties
  of the functions are defined according to which
  irreps they span.
• Then matrix elements (or integrals) are treated
  without numerical calculations to find whether
  they vanish.

42
Wavefunctions Forming Bases for Irreps (a)

• Look for solutions of the stationary Schrödinger
  equation H?E? with reference to symmetry
  considerations.
• The Hamiltonian, the sum of the potential and the
  kinetic energies, has a fundamental property
• The Hamiltonian is unchanged (invariant) with
  respect to interchanging two or more like
  particles (either electrons or nuclei) by any
  symmetry operation of the system.

43
Wavefunctions Forming Bases for Irreps (b)

• The symmetry operation definitely creates a
  system which is physically indistinguishable from
  the original one.
• The energy, the eigenvalue of the Hamiltonian is
  obviously invariant.
• The Hamiltonian H commutes with each symmetry
  operator R of the system
• HRRH or
  H,R0.

44
Wavefunctions Forming Bases for Irreps (c)

• Look at a nondegenerate normalized eigenfunction
  ?i with the eigenvalue Ei, H?i Ei?i
• Multiplying from the left by the symmetry
  operator R, RH?iREi?i.
• Due to commutativity, RH?iHR?iREi?iEiR?i
• Hence
• R?i is an eigenfunction of the Hamiltonian,
  with the same eigenvalue as ?i.

45
Wavefunctions Forming Bases for Irreps (d)

• Since ?i is nondegenerate, it is required that
  R?ic?i where c is a constant.
• Since ?i is normalized c1.
• Applying each symmetry operator to a
  nondegenerate eigenfunction generates a
  one-dimensional irrep of the group (Gi(R)1).

46
Wavefunctions Forming Bases for Irreps (e)

• Look at a degenerate normalized set of
  eigenfunction ?ik k1,,n all with the common
  eigenvalue Ei, H?ik Ei?ik.
• Any linear combination of the degenerate set is
  also an eigenfunction of the Hamiltonian
• Application of a symmetry operator R on a
  degenerate eigenfunction may yield a linear
  combination of the degenerate set
• For brevity the index i, which characterizes the
  degenerate set, is omitted from the
  transformation matrix, i.e. rijk?rjk

47
Wavefunctions Forming Bases for Irreps (f)

• Likewise, for another symmetry element S
• Since R and S are members of a group, there is a
  third element T which is their product TSR.
• The operation of T transforms the set as
• But according to the rule of consecutive
  operations
• The product rule between the three related
  matrices means

48
Wavefunctions Forming Bases for Irreps (g)

• Hence this set of matrices created by applying
  the symmetry operators on the n-fold degenerate
  set ?ik k1,,n is an n-dimensional
  representation of the group.
• A representation built by wavefunctions basis is
  an irrep.
• If it were reducible, then the n-fold degenerate
  set could be divided into subsets, such that each
  subset would separately serve as a basis to
  represent the group operators.
• i.e. a member of each set would be transformed
  into a linear combination of members merely of
  its own subset.

49
Wavefunctions Forming Bases for Irreps (h)

• But then the eigenvalues for members of one
  subset could be different from the eigenvalues of
  other subsets.
• This contradicts the assumption that all members
  of the set belong to the same eigenvalue Ei.

50
Wavefunctions Bases for C3v (a)

• Example NH3 (C3v group).
• The real functions 2px and 2py are written as
  px?rsin?cosf py?rsin?sinf, where ?r is the
  angle-independent term, which is symmetry
  invariant.
• Apply the symmetry operations on a line directed
  at (?,f).
• The angle ? is not affected, so that
  sin?1sin?2.
• The operation C3 rotates by 2p/3 and f2f12p/3
• Thus

51
Wavefunctions Bases for C3v (b)

• Reflection in the xz plane yields f2-f1
• Hence cosf2cosf1 and sinf2-sinf1.
• In summary, application of the symmetry
  operations yield

52
Wavefunctions Bases for C3v (c)

• In matrix notation
• with the characters ?(E)2, ?(C3)-1,
  ?(sv)0.
• It is the 2nd-order E irrep.
• Thus the coordinates x and y, and correspondingly
  the orbitals px and py transform according to the
  E irrep.

53
Projection Operators (a)

• To construct the SALCs, one employs projector
  operators.
• This tool requires the knowledge of the entire
  matrices for each symmetry operation.
• A partial information is the knowledge of the
  matrix characters, which is more practical,
  though less complete.
• Take an orthonormal set of ni functions fik
  k1,, ni which is a basis for the ith
  ni-dimensional irrep of a group of order h.
• The operation of any symmetry element on any
  function of the basis is
• This operation defines the s,t matrix element

54
Projection Operators (b)

• Multiply by the matrix element
• and sum over all group elements
• The last interchange of summation is since the
  functions fik are independent of R.
• Due to the GOT the inner summation is
• Only common irreps (ij) survive, and only
  corresponding elements (ss', tt') contribute

55
Projection Operators (c)

• The projection operator is defined
• When applied on an arbitrary function fit this
  operator projects out the component fjs' of the
  function
• The rest is erased.
• If the original function happens not to include a
  component tt', or is not of the jth irrep basis
  defining the projection operator, then the result
  vanishes.

56
Projection Operators (d)

• As for any projector operator, these operators
  are orthonormal in the sense that they are alien
  to each other, and that once is enough
• If the element is diagonal t't, the application
  of Pjtt on fis yields
• From an arbitrary function fis this operator
  projects out fjt
• a function with the symmetry of the specific
  irrep.
• Hence, by applying the entire nj diagonal
  projectors on an arbitrary function, the entire
  basis for the jth irrep is generated.

57
Incomplete Projection Operators

• Since usually the apparent information about the
  irreps are the character tables, the utilization
  of the projectors is not practical for
  multi-dimensional irreps.
• Take the diagonal projector operator and sum over
  the entire matrix elements

58
Constructing SALCs by Projection Operators (a)

• Example Application of C3v group elements on
  various functions

59
Constructing SALCs by Projection Operators (b)

• Application of the projection operator PE, of the
  two-dimensional irrep E of C3v group on the
  arbitrary function xzyzz2
• The two functions xz and yz that form a basis for
  the two-dimensional irrep E have been projected
  out of the arbitrary function xzyzz2.
• The term z2 which does not contribute to the
  irrep has been erased.
• This property of the projectors looks like a
  miracle.

60
Constructing SALCs by Projection Operators (c)

• A separate application of the complete projection
  operators PE11 and PE22 on the arbitrary function
  xzyzz2 yields correspondingly the two
  projections xz and yz.
• (Homework show it).
• This is obvious because the incomplete projection
  operator PE is derived by adding the two
  individual complete projection operators PE11 and
  PE22
• Hence the sum of their projections is the result
  of the application of the incomplete operator.

61
Constructing SALCs by Projection Operators (d)

• To construct the SALCs one refers to
  symmetry-equivalent atoms those which are
  interchanged by the symmetry operations.
• e.g. six carbon atoms of benzene, six hydrogen
  atoms of benzene, two hydrogen atoms of water,
  six fluor atoms of SF6.

62
Constructing MO-LACO by Projection Operators

• Molecular orbitals (MOs) should correspond to the
  molecular symmetry, in the sense that they form
  bases for the molecular irreps.
• Often MOs are approximately expressed as
  functions of the supposedly known AOs. (?? ???
  ?????? ????, ?? ???? ????? ???)
• A convenient way is to construct the MOs as
• Linear combinations (LCs) of corresponding
  atomic orbitals (AOs).
• Each atom participating in the molecular
  structure contributes its own AOs to the overall
  MO-LCAOs.
• Our concern is to find the algorithm to build
  these MOs that satisfy the appropriate molecular
  symmetry.

63
SALCs of 1-Dimensional Irreps of D2h (C2H4) (a)

• For a one-dimensional irrep the character and the
  full matrix are the same, so the complete and
  incomplete projection operators bear out the same
  results.
• e.g. Ethylene transforms according to the C2h
  group.
• There are four symmetry-equivalent hydrogen
  atoms, each with a 1-s atomic orbital.
• Using the LCAO-MO approximation,
• it is assumed that each hydrogen atom
• is connected just to its adjacent carbon
• atom by a sigma bond.

64
SALCs of 1-Dimensional Irreps of D2h (C2H4) (b)

• The basis functions (currently the four sigma
  bonds) are analyzed according to how they are
  affected by the appropriate symmetry elements.
• C2(z) yields the following vector with the proper
  matrix
• Obviously, since C2(z) takes each sigma bond into
  another one, its character must vanish.
• Any basis function that is dislocated contributes
  null to the character.

65
SALCs of 1-Dimensional Irreps of D2h (C2H4) (c)

• Since only the operations E and s(xy) leave the
  sigma bonds
• the reducible representation is
• Reduction to irreps AgB1gB2uB3u
• The projection procedure may be applied to any
  one of the basis members.

66
SALCs of 1-Dimensional Irreps of D2h (C2H4) (d)

• Sigma bond s1 is taken as a candidate
• Likewise

67
SALCs of 2-Dimensional Irreps of D4h (PtCl4-2) (a)

• In the LCAOMO approximation Pt tetra-chloride
  anion (PtCl4-2)
• has 4 sigma bonds (likewise, cyclobutadiene).
• The symmetry group is D4h with 16 elements.
• Irreps A1gB1gEu

68
SALCs of 2-Dimensional Irreps of D4h (PtCl4-2) (a)

• For the two 1-dimensional irreps application of
  the projection operations result in
• For the Eu irrep only the operations E, C42, i,
  sh have nonzero characters
• The second member of the set must be orthogonal
  to s1-s3 which is the wanted SALC.
• Applications of the elements E, C2, 2C'2, i, sh,
  sv yield (s1-s3).
• The other symmetry elements C4, C''2, S4, sd
  yield (s2-s4).

69
SALCs of 2-Dimensional Irreps of D4h (PtCl4-2) (b)

• Since only two orthogonal functions constitute
  the basis for the 2-dimensional irrep Eu, they
  are (in normal form)
• Homework
• (1) Obtain the reducible matrices of the 16
  symmetry elements on the basis of the four sigma
  bonds.
• (2) Reduce this representation into the
  appropriate irreps.
• (3) Apply each symmetry operation on one of
  the bonds.
• (4) Construct SALCs that span each irrep.