Ψ elec ψ vib ψ rot ψ ns Born-Opp approx Neglect Cent. Dist Coriolis Neglect Ortho-para mixing Molecular Orbitals Harmonic oscillator Rigid rotor Uncoupled

ψelec ψvib ψrot ψns

Born-Oppapprox

NeglectCent. DistCoriolis

NeglectOrtho-para

mixing

MolecularOrbitals

Harmonicoscillator

Rigidrotor

Uncoupledspins

Ψe-v-r-ns0

(espin and Slater determinants)

XiYi Zi

X0Y0 Z0XiYi Zitranslation

ξiηiζi

i = 1 to ℓ

i = 2 to ℓ

i = 2 to N j = N+1 to ℓ

rovibronic

Space-fixed origin

Molecule-fixed origin

ξjηjζjNuclear-fixed origin +Born-Opp. Approx.

rovibration electronic

θφχ xiyizirotation vibration

Molecule-fixed axesnuclear CofM origin

More changes in ‘coordinates’ (actually in momenta)

occur as conjugate momenta in Hrv

Jx = sinχ Pθ – cscθ cos Pφ + cotθ cos P

Jy = cosχ Pθ + cscθ sin Pφ – cotθ sinχ P

Jz = P = -iħd/d

θφχ

PθPφPχ

χ χ

χχ

χ

χχ

The Angular Momentum Operators

J2 = Jx2 + Jy

2 + Jz2

[J2,Jz ] = [J2,Jζ ] = 0.

Simultaneous eigenfunctions of the threeoperators are the rotation matrices

Dm,k(J)(θφχ)

Eigenvalues are J(J+1)ħ2, mħ, and kħJ2 Jζ Jz

Page 241

Final coordinate change

Δαi = ∑mi-½ αi,r Qrℓ

3N Cartesiandisp. CoordsΔx1,Δy1,…,ΔzN

3N-6 VibrationalNormal Coordinates

3 translational and 3 rotational ‘normal coordinates’ complete 3N coords

ℓ- matrix chosen to diagonalize force constant matrix.

See Eq.(10-134) in

Neglect anharmonicity in VN, neglect vibrational angular momentaneglect dependence of μαβ on Qr:

Hrv0 = ½ Σ μαα

e Jα2 + ½ Σ (Pr

2 + λrQr2)

α r

rigid rotor + harmonic oscillator

Zeroth order rot-vib Hamiltonian

(At equilibrium axes are principal axes so off-diagonal elements of μ vanish)

Sum of 3N-6 terms

The Harmonic Oscillator

Evib = Ev1 + Ev2

+ … + Ev3N-6

Evr /hc = (vr + ½)

Φvr = Nvr

Hvr( γr

½Qr) exp(-γrQr2/2)

H0 = 1H1 = 2(γr

½Qr )H2 = 4(γr

½Qr )2 – 2H3 = 8(γr

½Qr )3 -12γr½Qr

ωe

www.chem.uni-wuppertal.de/prb

Rotation and vibration wavefunctionsChapter 11

Symmetry classification of wavefunctionsChapter 12

Can also go to my website ‘downloads’

P-everything.ONE.pdf

P-everything.TWO.pdf

P-everything.THREE.pdf

http://www.chem.uni-wuppertal.de/prb

E

(12)

(12)*

1 version

Φrot = |JKaKc>

Ka Kc symm even even A1

even odd B1

odd even B2

odd odd A2

Φvib = |v1 v2 v3>

Q1 and Q2 are of A1 symmetry and Q3 is B2

v3 symm even A1

odd B2

E

(23)

(23)*

1 version

Ryπ Rx

π

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

Forbidden rotation transitionsince H3

+ has no dipole momentμA is A1”

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

Forbidden rotation transitionsteals intensity from rv transition

Hrv’

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


Group K(spatial) of all rotations about allaxes. Irred. reps giveang mom quantum no.

Ψint = Ψr-v-eo ΨesΨns

N

J

F

For singletstates N=J

~

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 asIn the MS group symmetry species).

Using J as generic angular momentum quantum number

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


For n-electronsuse group of n!electron permutations

The electron permutation group or “Symmetric Group” Sn

(Contains all n! permutations of the coordinates and spins of

the electrons in an n-electronatom or molecule).

ELECTRON PERMUTATION SYMMETRY

This is a symmetry group of theelectronic (orbital plus spin) Hamiltonian

For the LiH2 molecule (5 electrons), the electron permutation group is the Symmetric Group S5

Since Horb-spin commutes with anyelectron permutation, Ψorb-spin

has to

transform as an irrep of Sn

So you would think there are 7 differentpossible symmetries for Ψorb-spin

123456↓(45)

123546↓(34)

124536

(345)

(34)(45) = (345)

123456↓(34)

124356↓(23)

134256↓(12)

234156

(1234)

(12)(23)(34) = (1234)

Writing permutations as the product of pair exchanges

123456↓(45)

123546↓(34)

124536

(345)

(34)(45) = (345)

123456↓(34)

124356↓(23)

134256↓(12)

234156

(1234)

(12)(23)(34) = (1234)

Writing permutations as the product of pair exchanges

An EVEN permutation

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 functionshave 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-spinis “symmetrized”

The so-called Pauli ExclusionPrinciple, “discovered” by Stoner,Phil. Mag., 48, 719 (1924)(see )is a result of electronic stateshaving to transform as theanti-symmetric representation.

We can use the He atom as anexample to explain this.

http://www.jstor.org/stable/27757517

The PEP is sometimes called a ukase

http://www.jstor.org/stable/27757517

He atom. S2 group.

Label electrons ‘a’ and ‘b’

E (ab)s 1 1a 1 -1

Orbital state: 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’

Allowed state is ‘a’: φ0(|αβ>-|βα>)

symmetry is ‘s’

[|αα>,(|αβ>+|βα>),|ββ>] φ0Excluded state is ‘s’:

He atom. S2 group.


E (ab)s 1 1a 1 -1

Orbital states: φ0 = a(1s)b(1s)

φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s)




Allowed states are ‘a’:

(φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]

φ0(|αβ>-|βα>)

(φ1+φ2)(|αβ>-|βα>)

He atom. S2 group.


E (ab)s 1 1a 1 -1


φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s)




Excluded states are ‘s’: φ0 [|αα>,(|αβ>+|βα>),|ββ>]

(φ1-φ2)

[|αα>,(|αβ>+|βα>),|ββ>]

(|αβ>-|βα>) ‘paronic states’ (φ1+φ2)

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 describinga 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.

There is no proof of why there hasto 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

The connection between spin and permutation symmetry is empirical

He atom. S2 group.


E (ab)s 1 1a 1 -1


φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s)




Allowed states are ‘a’:

(φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]

φ0(|αβ>-|βα>)

(φ1+φ2)(|αβ>-|βα>)

He atom. S2 group.


E (ab)s 1 1a 1 -1


φ1 = a(1s)b(2p) and φ2 = a(2p)b(1s)



Spin states: |αα> |αβ>+|βα> |ββ> symmetry is 3’s’ |αβ>-|βα> symmetry is ‘a’

Excluded states are ‘s’: φ0 [|αα>,(|αβ>+|βα>),|ββ>]

(φ1-φ2)

[|αα>,(|αβ>+|βα>),|ββ>]

(|αβ>-|βα>) ‘paronic states’ (φ1+φ2)

Exchange Symmetric (Paronic) States of He

Singlet and triplet spinfunctions interchanged.Relativistic calc of energies.

G.W.F.Drake,Phys. Rev. A 39, 897 (1989).

(φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]

φ0(|αβ>-|βα>)

(φ1+φ2)(|αβ>-|βα>)

Experiment to look for paronic state of He

Atomic beam fluorescence experiment does not see transitionfrom paronic state. Thus violation of PEP < 5 10–5.

Deilamian et. al., Phys. Rev. Lett. 74, 4787 (1995) 36

Law of Nature: The Symmetrization PostulateThe 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 permutationand identical nuclear permutations.

where Ψoverall = ΨtransΨint

and Ψint = Ψevrns

Ψeorb-spin ψvib ψrot ψnsΨint0 =

Symmetrization

All basis functions should be symmetrized in Sn

Symmetrized in Sn

Use Slater determinant forspin-orbit basis functions. This ensures symmetrization of to the antisymmetric irrep of Sn

Electron permutationhas no effect on thesefunctions

Ψint0

Ψeorb-spin ψvib ψrot ψnsΨint0 =

Symmetrization

All basis functions should be symmetrized in the MSG

Each is symmetrized in the MSG

Only need construct that satisfy the SymmetrizationPrinciple

Ψint0

Get nspin statistical weights. We use H2O as an example

The water molecule

1 2

C2v(M)ψevrψnsΨint

=

Rules of statistics applyonly to ψint

Protons arefermions

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1 ?What is the symmetry of ψint

The water molecule

1 2

C2v(M)ψevrψnsΨint

=

Rules of statistics applyonly to ψint

Protons arefermions

ψint can only be B1 or B2

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

LevelslabelledJKaKc

ab

+c

Γevr

Rotational energy levels of H2O An asymmetric top

In vibronicground state

C2v(M) ψevrψnsΨint =

ψint can only be B1 or B2

Ψevr can be A1, A2, B1 or B2

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

1H spin functions are α = |½,½>, and β =|½,-½>.

16O spin function is δ = |0,0>.Four nuclear spin functions in total for H2O:

α α δ α β δ β α δ β β δ

I, mI

A1

A1 + B2

A1

Nuclear spin functions for H2O

B2

A1

A1

A1

Symmetrized spinfunctions

N.B. Nuclear spin functions always have + parity

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Nuclear spin statistical weights for H2O

The total internal wavefunction ΦH2O has B1 or B2 symmetry

Γint

Γint

para

paraorthoortho

int

Allsymmetriespossible

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1



Γint

Γint

para

paraorthoortho

States with higher nuclear spin statistical weight called ortho

int


Only B1 or B2 allowed

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1



Γint

Γint

para

paraorthoortho

States with higher nuclear spin statistical weight called ortho

int


Only B1 or B2 allowedas parity is – or +

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Γ

LevelslabelledJKaKc

ab

+c

Ortho and para H2O

Γrve

B2 3A1Γns

Γ

LevelslabelledJKaKc

ab

+c

Ortho and para H2O

Γrve

B2 3A1Γns

Γint B2 B1 B1 B2

Intensity alternation for H2Oortho-ortho para-para

1

2

Γint=B2(+)

Γint=B1(-)

Γns = B2

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Γns = 3A1

The (H2O)3 Molecule

Two rotational transitions in thespectrum of (D2O)3

Two rotational transitions in thespectrum of (D2O)3

The Water Trimer

The equilibrium structureThe equilibrium structure

The intensity pattern76:108:54:11for (D2O)3 showsthat there is tunnelingbetween 48 minima onpotential surface.JACS 115,11180 (1993)JACS 116, 3507 (1994)

hCNPI = 8640

MS Group = G48

51

Applying the Symmetrization Principlecan lead to “missing levels.”

The ozone molecule

1 2

3

C2v(M) E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Nuclear spin statistical weights for 16O3

In 16O3, all nuclei have I = 0.

Thus, there is only one spin function of A1 symmetry

Missinglevels

A1 A1 A1

A2 A1 A2

B1

B2

intThe total internal wavefunction ΦO3 has A1 or A2 symmetrySince 16O nuclei are bosons.

int

int E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

rve ns int

Γ

LevelslabelledJKaKc

ab

+c

Ozone missing levels

Γrve

A1Γns

+s -s -a+a

+s for even J, -a for odd Jin ground ev state

ψevr

12C16O2

Nuclear spin state has symmetry +s

ψint Symmetry has to be +s or –s.

Thus levels with odd J are missing.

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

+s -s -a+a


ψevr

12C16O2




E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

+s -s -a+a


ψevr

C16O2




Exchange Antisymmetric States of CO2

For the CO2 molecule states with odd values of J are missingbecause 16O nuclei are Bosons. Mazzotti et al., Phys. Rev. Lett. 86, 1919 (2001) looked for transition:

Sensitivityshows thatprobability < 10–11

(0001)-(0000)R(25) calc at2367.265 cm-1

12C has spin 0

Spin symmetry Ag

Int symmetry Ag or Au

gns = 1

missing levels

12C60

62

Ortho-para mixing andortho-para transitions

Pages 407-9,413 and 473

Γ

LevelslabelledJKaKc

ab

+c

Ortho and para H2O

Γrve

B2 3A1Γns

B2 B1 B1 B2

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Γint

Γ

LevelslabelledJKaKc

ab

+c

Ortho and para H2O

Γrve

B2 3A1Γns

B2 B1 B1 B2

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

Γint

Ortho-para transition in H216O

Transition (b) steals (or borrows) intensity from transition (a)

A1 B2 B2(+)

B2 A1 B2(+)

B1 A1 B1(-)

Para state

Ortho state

Ortho state

Γrve Γns Γint

101

212

202

o-p transition

States with same F and parity can be mixed by Hhfs

Φo

189-191

The H2 molecule E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

+s -s -a+a

J=1 (-a) I=1: αα,αβ+βα,ββ (+s) F=0,1,2 (-a)

φnsψint

J=0 (+s) I=0: αβ-βα (+a) F=0 (+a)

Ortho and para states have opposite parity and sowithin the ground state there cannot be ortho-paramixing and ortho/para nature is a conserved quantity.Can keep para (i.e., J=0) H2 in a bottle.

Rotational levels in theground vibronic state:odd J (-a), even J (+s)

φevr

evr symmetry int symmetryns symmetry

Σu+

Σg+

Σu-

Σg-

The H2 molecule E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

+s -s -a+a

Rotational levels in theground vibronic state:odd J (-a), even J (+s)

J=1 (-a) I=1: αα,αβ+βα,ββ (+s) F=0,1,2 (-a)

φnsψint

J=0 (+s) I=0: αβ-βα (+a) F=0 (+a)

F=0 to 1 hfs component of J=0 to 1 is

forbidden since <φev|μz|φev> = 0 no dipole moment(+a)

φevr

Σu+

Σg+

Σu-

Σg-

evr symmetry int symmetryns symmetry

1Σu+

excited electronic state

J=1 (-s) I=0: αβ-βα (+a) F=1 (-a)

J=1 (-a) I=1: αα,αβ+βα,ββ (+s) F=0,1,2 (-a)

J=0 (+s) I=0: αβ-βα (+a) F=0 (+a)

1Σg+

ground electronic state

Hns

para

para

ortho

~10-10 D. Dodelson. J. Phys. B., At. Mol. Phys., 19, 2871 (1986)

N.B.

o-pmixing

a ‘u’ state

a ‘g’ state

H2+

g-u mixingby Fermicontact op.in Hhfs

g-u mixing gives N = 1 0 rotational transitions in H2+

For v=19N=1 0μ ~ 0.42 D

Predicted in CPL 316, 266 (2000). Effect of mixing with continuum is dominant effect. Observation reported in PRL 86, 1725 (2001).

ortho para

For v=18N=1 0μ ~ 0.02 D

KaKc

e e Ag

o o B2g

e o B1g

o e B3g

Γrot (14)(23)(56)* = p(14)(23)(56)* R0 i

Point Group operation i commutes with Hrve Thus g and u are good rovibronic labelsbut not good rovibronic-nuclear spin labels

The point groupoperation “ ”i

Gives g/u symmetry

72

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

Do they all commute with the Hamiltonian?Do they all commute with the Hamiltonian?

IdenticalParticles

The Breakdown of Symmetry

We have looked at the breakdown of the Symmetrization Principle

Experiment to look for paronic state of He

Breakdown of the Pauli Exclusion Principle

Atomic beam fluorescence experiment does not see transitionfrom paronic state. Thus violation of PEP < 5 10–5.

Phys. Rev. Lett. 74, 4787 (1995)

Exchange Antisymmetric States of CO2

For the CO2 molecule states with odd values of J

are missing because 16O nuclei are Bosons.

Phys. Rev. Lett. 86, 1919 (2001) looked for transition:

Sensitivityshows thatprobability < 10–11

(0001)-(0000)R(25) calc at2367.265 cm-1

Parity Symmetry Breakdown P (= E*)

• We have considered the Electromagnetic Force in setting up the molecular Hamiltonian. The effect of the Gravitational Force is negligible, the Strong Force does not affect electrons, but the Weak Force has interesting consequences.

• Including the Weak Force gives the Electroweak Hamiltonian HEW which does not commute with P. Because [P,HEW] = there is Parity Violation. As a result:

•Parity forbidden transitions become (very weakly) allowed.

•Enantiomers have (very slightly) different energies.

A Parity-Violating Transition in Cesium

7S(+) 6S(+) transition has been observed in Cesium.

The electric dipole transition moment is about 10-10 D.

HEW mixes 10-11 of P(-) states into these S(+) states.

Phys. Rev. Letters 61, 310 (1988), Can. J. Phys. 77, 7 (1999)

The electric dipole moment operator has (–) parity.Thus it only connects states of opposite parity.

Result agrees with Standard Model of fundamental PhysicsTo better than 1%, and it constrains alternative theories. Symmetry breakdown is studied in order to test these alternative theories…String Theory, Supersymmetry,….

Darquié et al, Chirality, 22, 870 (2010)

PRL 83, 1554 (1999)

PARITY VIOLATION FOR ENANTIOMERS

A molecule with a PV shift of ~ 0.1 Hz

See also: PCCP 15, 10952 (2013)

The Violation of CP Symmetry

In 1957 after Parity violation had been observed by

Wu, it was thought that CP was a “good” symmetry.

But CP violation was observed in the decay of K mesons in 1964Some CP violation can be accommodated in the Standard Model

We should all be grateful for CP violation since it is probablythe reason for matter-antimatter asymmetry in the Universe.

CP violation allows nuclear decays to occur with a very slightasymmetry in the production of particles and antiparticles.As the Universe cooled such processes would compete withmatter-antimatter annihilation to produce M-A asymmetry.

The Violation of CP Symmetry (cont)

By observations of the cosmic radiation background

(the 3 K remnant radiation from the Big Bang)

it is found that the Universe has:

Matter = 10–9 Photons

Thus as Universe cooled there were (109 + 1) particles of matter for every 109 particles of antimatter; a very small asymmetry.

~

Violation of Time Reversal (T) Symmetry

implies CP violation (assuming CPT)

Weak Force breaks T-Symmetry. Extent can be calculated

using the Standard Model. But is it correct?

The dipole moment of the electron. Within the Standard Model: de (Standard Model) 10–38 e cm

Spectroscopy experiment on 174YbF in an electric field.Splitting of E(F=1,mF=±1) level of N=0, v=0, X2+ state.

Current null result is at a precision that leads to an upper limitof |de| 10.5 × 10–28 e.cm. Nature 473, 493 (2011).Several extended theories predict values in this range.

http://dx.doi.org/10.1088/1367-2630/14/10/103051 See also

http://dx.doi.org/10.1088/1367-2630/14/10/103051

Recent (Jan 2014) lower value for the upper limit

Using Thorium monoxide an upper limit of

8.7 x 10-29 e.cm with 90% confidence

has been obtained.

See DOI:10.1126/science.1248213.

See also wikipedia on electric dipole moment of the electron

Violation of CPT Symmetry

Theoretical Physicists really believe in this even after

the violations of P and CP symmetry have been found.

It leads to the result that a measurement of T-violation

yields the extent of CP violation.

A spectroscopy experiment could determine the extent of its violation (if any). Anti-hydrogen consists of an anti-proton anda positron. The positronic 2S-1S spectral line in anti-hydrogen should occur at a different wavenumber from the 2S-1S electronic line in normal hydrogen if there is CPT violation.

Science, 298, 1327 (2002)

High energy physics http://lhc.web.cern.ch/lhc/

Nature,483,439 (2012)

http://lhc.web.cern.ch/lhc/



Research on Symmetry: Where are we going?

“Hidden Symmetry” of the

initially created Universe?

One Force?

P, C, CP and T Symmetry? Symmetrization Principle?

“Broken Symmetry” of thestructured cold universe?

Four Forces. Extent of breakdown of P, C, CP or T Symmetry? Is there CPT Symmetry? Symmetrization Principle?

Atomic and Molecular Spectroscopy can play a part

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Ψ elec ψ vib ψ rot ψ ns Born-Opp approx Neglect Cent. Dist Coriolis Neglect Ortho-para mixing Molecular Orbitals Harmonic oscillator Rigid rotor Uncoupled