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
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
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
Symmetry Operations (energy invariance)
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
Symmetry Operations (energy invariance)
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
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.
Label electrons ‘a’ and ‘b’
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)
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 states are ‘a’:
(φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]
φ0(|αβ>-|βα>)
(φ1+φ2)(|αβ>-|βα>)
He atom. S2 group.
Label electrons ‘a’ and ‘b’
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)
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’
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.
Label electrons ‘a’ and ‘b’
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)
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 states are ‘a’:
(φ1-φ2)[|αα>,(|αβ>+|βα>),|ββ>]
φ0(|αβ>-|βα>)
(φ1+φ2)(|αβ>-|βα>)
He atom. S2 group.
Label electrons ‘a’ and ‘b’
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)
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’
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
Nuclear spin statistical weights for H2O
The total internal wavefunction ΦH2O has B1 or B2 symmetry
Γint
Γint
para
paraorthoortho
States with higher nuclear spin statistical weight called ortho
int
Allsymmetriespossible
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
Nuclear spin statistical weights for H2O
The total internal wavefunction ΦH2O has B1 or B2 symmetry
Γint
Γint
para
paraorthoortho
States with higher nuclear spin statistical weight called ortho
int
Allsymmetriespossible
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
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
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
+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
+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
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
+s for even J, -a for odd Jin ground ev state
ψevr
C16O2
Nuclear spin state has symmetry +s
ψint Symmetry has to be +s or –s.
Thus levels with odd J are missing.
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
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
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,….
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
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)
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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