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Ψ e-v-r-ns 0. Neglect Ortho-para mixing. Neglect Cent. Dist Coriolis. Born-Opp approx. ψ elec ψ vib ψ rot ψ ns. Harmonic oscillator. Rigid rotor. Uncoupled spins. Molecular Orbitals. (espin and Slater determinants). X i Y i Z i.
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Ψe-v-r-ns0 Neglect Ortho-para mixing Neglect Cent. Dist Coriolis Born-Opp approx ψelec ψvib ψrot ψns Harmonic oscillator Rigid rotor Uncoupled spins Molecular Orbitals (espin and Slater determinants)
XiYi Zi Space-fixed origin i = 1 to ℓ Molecule-fixed origin XiYi Zi X0Y0 Z0 i = 2 to ℓ translation rovibronic Nuclear-fixed origin + Born-Opp. Approx. ξjηjζj ξiηiζi i = 2 to N j = N+1 to ℓ electronic rovibration Molecule-fixed axes nuclear CofM origin θφχxiyizi rotation vibration
More changes in ‘coordinates’ (actually in momenta) occur as conjugate momenta in Hrv θφχ PθPφPχ Jx = sinχ Pθ – cscθ cos Pφ + cotθ cos P χ χ χ χ Jy = cosχ Pθ + cscθ sin Pφ – cotθ sinχ P χ Jz = P = -iħd/d χ χ
The Angular Momentum Operators J2 = Jx2 + Jy2 + Jz2 [J2,Jz ] = [J2,Jζ ] = 0. Simultaneous eigenfunctions of the three operators are the rotation matrices Dm,k(J)(θφχ) Eigenvalues are J(J+1)ħ2, mħ, and kħ Page 241 J2 Jζ Jz
Final coordinate change ℓ Δαi = ∑mi-½αi,r Qr 3N Cartesian disp. Coords Δx1,Δy1,…,ΔzN 3N-6 Vibrational Normal Coordinates 3 translational and 3 rotational ‘normal coordinates’ complete 3N coords See Eq.(10-134) in ℓ- matrix chosen to diagonalize force constant matrix.
Zeroth order rot-vib Hamiltonian Neglect anharmonicity in VN, neglect vibrational angular momenta neglect dependence of μαβ on Qr: Hrv0 = ½ Σμααe Jα2 + ½ Σ(Pr2 + λrQr2) r α rigid rotor + harmonic oscillator Sum of 3N-6 terms (At equilibrium axes are principal axes so off-diagonal elements of μ vanish)
The Harmonic Oscillator Evib = Ev1+ Ev2 + … + Ev3N-6 ωe Evr/hc = (vr + ½) Φvr = Nvr Hvr( γr½Qr) exp(-γrQr2/2) H0 = 1 H1 = 2(γr½Qr ) H2 = 4(γr½Qr )2 – 2 H3 = 8(γr½Qr )3 -12γr½Qr
Rotation and vibration wavefunctions Chapter 11 Symmetry classification of wavefunctions Chapter 12 Can also go to my website ‘downloads’ www.chem.uni-wuppertal.de/prb P-everything.ONE.pdf P-everything.TWO.pdf P-everything.THREE.pdf
1 version E (12) (12)*
Ka Kc symm even even A1 even odd B1 odd even B2 odd odd A2 Φrot = |JKaKc> Φvib = |v1 v2 v3> Q1 and Q2 are of A1 symmetry and Q3 is B2 v3 symm even A1 odd B2
1 version E (23) Rxπ Ryπ (23)*
Pure rotation (ΔK = 3) transition in H3+ v2(E’)=1, J=4,K=1 A2’’ Ground vib state, J=4,K=3 A2’’ J=3,K=0 A2’ Allowed rotation-vibration transition μA is A1” Forbidden rotation transition since H3+ has no dipole moment
Pure rotation (ΔK = 3) transition in H3+ v2(E’)=1, J=4,K=1 A2’’ Hrv’ Ground vib state, J=4,K=3 A2’’ J=3,K=0 A2’ Allowed rotation-vibration transition Forbidden rotation transition steals intensity from rv transition 13
Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Identical nuclei-----------Permute identical nuclei Parity conservation-----Inversion (p,q) (-p,-q)E* or P Reversal symmetry-----Time reversal (p,s) (-p,-s) T Ch. conj. Symmetry-----Particle antiparticle C Symmetry Operations (energy invariance) What about the symmetry operations in black?
Uniform Space ----------Translation Symmetry Operations (energy invariance) Separate translation… Translational momentum Ψtot = Ψtrans Ψint int = rot-vib-elec.orb-elec.spin-nuc.spin
Uniform Space ----------Translation Isotropic Space----------Rotation Symmetry Operations (energy invariance) Group K(spatial) of all rotations about all axes. Irred. reps give ang mom quantum no. ~ Ψint = Ψr-v-eoΨesΨns For singlet states N=J N J F
Using J as generic angular momentum quantum number Group K(spatial)of all rotations in space has irreducible representations D(J) of dimension (2J+1) X (2J+1). Classifying Ψ in the group K(spatial) gives the overall rotational quantum number labels J and mJ. Transform as the mJth row of the irreducible representation D(J). H has symmetry D(0); Only states of same J interact. Hamiltonian matrix is block-diagonal in J (as well as In the MS group symmetry species). Dipole moment has symmetry D(1) Transitions with ΔJ > 1 or J=0 → 0 are forbidden
Uniform Space ----------Translation Isotropic Space----------Rotation Identical electrons-------Permute electrons Symmetry Operations (energy invariance) For n-electrons use group of n! electron permutations
ELECTRON PERMUTATION SYMMETRY The electron permutation group or “Symmetric Group” Sn (Contains all n! permutations of the coordinates and spins of the electrons in an n-electron atom or molecule). This is a symmetry group of the electronic (orbital plus spin) Hamiltonian
Since Horb-spin commutes with any electron permutation, has to Ψorb-spin transform as an irrep of Sn For the LiH2 molecule (5 electrons), the electron permutation group is the Symmetric Group S5 So you would think there are 7 different possible symmetries for Ψorb-spin
Writing permutations as the product of pair exchanges 123456 ↓(34) 124356 ↓(23) 134256 ↓(12) 234156 123456 ↓(45) 123546 ↓(34) 124536 (345) (1234) (34)(45) = (345) (12)(23)(34) = (1234)
Writing permutations as the product of pair exchanges 123456 ↓(34) 124356 ↓(23) 134256 ↓(12) 234156 123456 ↓(45) 123546 ↓(34) 124536 (345) (1234) (34)(45) = (345) An EVEN permutation (12)(23)(34) = (1234) An ODD permutation
An odd permutation is one that involves an odd number of pair exchanges. Examples: (12), (12)(345)=(12)(34)(45), (1234)=(12)(23)(34) An even permutation has an even number: (12)(34), (123)=(12)(23), (12345)=(12)(23)(34)(45) The S5 group for LiH odd
Mother Nature seems to have designed our Universe in such a way that electronic wavefunctions change sign if a pair of electrons are exchanged (or permuted). This means that electronic spin-orbital functions have to transform as the irrep having -1 for odd permutations
For the LiH2 molecule (5 electrons) Ψorb-spin transforms as D(0) of S5 Slater determinant ensures antisymmetry. Ψorb-spin is “symmetrized”
The so-called Pauli Exclusion Principle, “discovered” by Stoner, Phil. Mag., 48, 719 (1924) (see ) is a result of electronic states having to transform as the anti-symmetric representation. http://www.jstor.org/stable/27757517 The PEP is sometimes called a ukase We can use the He atom as an example to explain this.
He atom. S2 group. E (ab) s 1 1 a 1 -1 Label electrons ‘a’ and ‘b’ Orbital state: φ0 = a(1s)b(1s) symmetry is ‘s’ φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s) Symmetry of φ0 is ‘s’ Symmetry of φ1 + φ2 is ‘s’ Symmetry of φ1 - φ2 is ‘a’ Spin states: |αα>, |αβ>+|βα>, |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’ φ0(|αβ>-|βα>) Allowed state is ‘a’: φ0 Excluded state is ‘s’: [|αα>,(|αβ>+|βα>),|ββ>]
He atom. S2 group. E (ab) s 1 1 a 1 -1 Label electrons ‘a’ and ‘b’ Orbital states: φ0 = a(1s)b(1s) φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s) Symmetry of φ0 is ‘s’ Symmetry of φ1 + φ2 is ‘s’ Symmetry of φ1 - φ2 is ‘a’ Spin states: |αα>, |αβ>+|βα>, |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’ φ0(|αβ>-|βα>) Allowed states are ‘a’: (φ1+φ2)(|αβ>-|βα>) (φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]
[|αα>,(|αβ>+|βα>),|ββ>] (|αβ>-|βα>) (φ1-φ2) He atom. S2 group. E (ab) s 1 1 a 1 -1 Label electrons ‘a’ and ‘b’ Orbital states: φ0 = a(1s)b(1s) φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s) Symmetry of φ0 is ‘s’ Symmetry of φ1 + φ2 is ‘s’ Symmetry of φ1 - φ2 is ‘a’ Spin states: |αα>, |αβ>+|βα>, |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’ [|αα>,(|αβ>+|βα>),|ββ>] φ0 Excluded states are ‘s’: (φ1+φ2) ‘paronic states’
Fermi-Dirac statistics Mother Nature has designed our Universe in such a way that the wavefunctions describing a system of particles having half-integral spin are changed in sign if a pair of such particles are exchanged (or permuted). Particles having half-integral spin are called fermions.
Bose-Einstein statistics Mother Nature has also designed our Universe in such a way that the wavefunctions describing a system of particles having integral spin are unchanged if a pair of such particles are exchanged (or permuted). Particles having integral spin are called bosons.
The connection between spin and permutation symmetry is empirical There is no proof of why there has to be this connection between spin and permutation symmetry. See the book review by Wightman, Am. J. Phys. 67, 742 (1999), and references therein. So maybe it isn’t true? And it should be experimentally tested
He atom. S2 group. E (ab) s 1 1 a 1 -1 Label electrons ‘a’ and ‘b’ Orbital states: φ0 = a(1s)b(1s) φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s) Symmetry of φ0 is ‘s’ Symmetry of φ1 + φ2 is ‘s’ Symmetry of φ1 - φ2 is ‘a’ Spin states: |αα>, |αβ>+|βα>, |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’ φ0(|αβ>-|βα>) Allowed states are ‘a’: (φ1+φ2)(|αβ>-|βα>) (φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]
[|αα>,(|αβ>+|βα>),|ββ>] (|αβ>-|βα>) (φ1-φ2) He atom. S2 group. E (ab) s 1 1 a 1 -1 Label electrons ‘a’ and ‘b’ Orbital states: φ0 = a(1s)b(1s) φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s) Symmetry of φ0 is ‘s’ Symmetry of φ1 + φ2 is ‘s’ Symmetry of φ1 - φ2 is ‘a’ Spin states: |αα> |αβ>+|βα> |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’ [|αα>,(|αβ>+|βα>),|ββ>] φ0 Excluded states are ‘s’: (φ1+φ2) ‘paronic states’
Exchange Symmetric (Paronic) States of He Singlet and triplet spin functions interchanged. Relativistic calc of energies. G.W.F.Drake, Phys. Rev. A39, 897 (1989). (φ1+φ2)(|αβ>-|βα>) (φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>] φ0(|αβ>-|βα>)
Experiment to look for paronic state of He Atomic beam fluorescence experiment does not see transition from paronic state. Thus violation of PEP < 5 10–5. 36 Deilamian et. al., Phys. Rev. Lett. 74, 4787 (1995)
Law of Nature: The Symmetrization Postulate The states that occur in nature are symmetric with respect to identical boson exchange and antisymmetric with respect to identical fermion exchange. This applies to nuclei and electrons. It fixes the symmetry of Ψint with respect to electron permutation and identical nuclear permutations. where Ψoverall = ΨtransΨint and Ψint = Ψevrns
Symmetrization All basis functions should be symmetrized in Sn Ψeorb-spin ψvib ψrot ψns Ψint0 = Electron permutation has no effect on these functions Symmetrized in Sn Use Slater determinant for spin-orbit basis functions. This ensures symmetrization of to the antisymmetric irrep of Sn Ψint0
Symmetrization All basis functions should be symmetrized in the MSG Ψeorb-spin ψvib ψrot ψns Ψint0 = Each is symmetrized in the MSG Ψint0 Only need construct that satisfy the Symmetrization Principle Get nspin statistical weights. We use H2O as an example
The water molecule Protons are fermions 1 2 C2v(M) ψevrψns Ψint= Rules of statistics apply only to ψint ?What is the symmetry of ψint
The water molecule Protons are fermions 1 2 C2v(M) ψevrψns Ψint= Rules of statistics apply only to ψint ψint can only be B1 or B2
Rotational energy levels of H2O An asymmetric top Levels labelled JKaKc b a +c Γevr In vibronic ground state
C2v(M) ψevrψns Ψint= ψint can only be B1 or B2 Ψevr can be A1, A2, B1 or B2 I, mI 1H spin functions are α = |½,½>, andβ =|½,-½>. 16O spin function is δ= |0,0>. Four nuclear spin functions in total for H2O: α α δ α β δ βα δ β β δ A1 A1 + B2 A1
Nuclear spin functions for H2O Symmetrized spin functions A1 A1 A1 B2 N.B. Nuclear spin functions always have + parity
Nuclear spin statistical weights for H2O int The total internal wavefunction ΦH2O has B1 or B2 symmetry All symmetries possible Γint para Γint ortho para ortho
Nuclear spin statistical weights for H2O int The total internal wavefunction ΦH2O has B1 or B2 symmetry All symmetries possible Γint para Only B1 or B2 allowed Γint ortho para ortho States with higher nuclear spin statistical weight called ortho
Nuclear spin statistical weights for H2O int The total internal wavefunction ΦH2O has B1 or B2 symmetry All symmetries possible Γint para Only B1 or B2 allowed as parity is – or + Γint ortho para ortho States with higher nuclear spin statistical weight called ortho
Ortho and para H2O Levels labelled JKaKc Γns B2 3A1 b a +c Γrve Γrve
Ortho and para H2O Levels labelled JKaKc Γns B2 3A1 b a +c Γrve Γint Γrve B2 B1 B1 B2
Intensity alternation for H2O para-para ortho-ortho Γint=B2(+) 1 Γint=B1(-) 2 Γns = 3A1 Γns = B2