# Discrete Symmetries

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```					               Discrete Symmetries
• Parity
• Parity Violation
• Charge Conjugation
• CP Violation
• Time Reversal
• The CP T Theorem
• Lepton number

Physics 506A       4 - Discrete Symmetries   Page 1
Parity
• A parity transformation, P , inverts every spatial coordinate: P (t, x) = (t, −x)
P 2 = I, and therefore the eigenvalues of P are ±1.

• Ordinary vector v. P (v) = −v .

• Scalar from v: s = v · v
P (s) = P (v · v) = (−v) · (−v) = v · v = +s

• Cross product of two vectors: a = v × w
P (a) = P (v × w) = (−v) × (−w) = v × w = +a

• Scalar from a and v: p = a · v
P (p) = P (a · v) = (+a) · (−v) = −a · v = −p

Scalar                  P (s) = +s
Pseudoscalar            P (p) = −p
Vector                  P (v) = −v
Pseudovector            P (a) = +a

Physics 506A                               4 - Discrete Symmetries                          Page 2
Parity in Physical Systems
• Two-body systems have parity pA pB (−1)ℓ
P φ(12) = p1 p2 (−1)ℓ φ(12)

• Intrinsically, particles and antiparticles have opposite parity
Bound states like positronium e+ e− and mesons qq have parity of (−1)ℓ+1 .

• Photons have a parity of (−1), and this underlies the ∆ℓ = ±1 selection rule
in atomic transitions.

• Note that parity is a multiplicative quantum number.
This is true for all discrete symmetries.
Continuous symmetries have additive quantum numbers.

Physics 506A                              4 - Discrete Symmetries                        Page 3
Example: uu mesons
By conventions: u-quarks have spin 1/2 and + parity and
u-quarks have spin 1/2 and - parity

Parity of a uu meson is P = pu pu (−1)ℓ
The intrinsic spin (S) of the uu meson is 0 or 1
but may have any orbital angular momentum (L) value.

S     L       JP           particle
0     0       0−              π0
1     0       1−              ρ0
0     1       1+           b1 (1235)

See the PDG for a table of the quantum numbers of the low-mass mesons.

Physics 506A                           4 - Discrete Symmetries                    Page 4
Parity in the Standard Model

• It has long been realized that par-
ity is a respected symmetry of the
strong and electromagnetic interac-
tions.
• In 1956, Yang and Lee realized that
tested experimentally for weak in-
teractions.

Physics 506A                              4 - Discrete Symmetries   Page 5

Wu recorded the direction of the emitted electron from a 60 Co β-decay when the
nuclear spin was aligned up and down
The electron was emitted in the same direction independent of the spin.
−→ Parity is not conserved in the weak interations

Physics 506A                           4 - Discrete Symmetries                       Page 6
Parity Violation in π Decay
• Consider the weak decay π + → µ+ + νµ .
Since the π is spin-0 and the µ and ν emerge back-to-back in the CM frame, the
spins of the µ and ν must cancel.

• Experiments show that every µ+ is left-handed, and therefore every νµ is also
left-handed.

• Similarly, in π − decay, both the µ− and ν µ always emerge right-handed.

• If parity were conserved by the weak interaction, we would expect left-handed
pairs and right-handed pairs with equal probability (just as we observe with
π 0 → 2γ).

• Assuming that neutrinos are massless,
ALL neutrinos are left-handed
ALL antineutrinos are right-handed

Physics 506A                               4 - Discrete Symmetries                          Page 7
Helicity
• Only the z-component of angular momentum can be speciﬁed
⇒ align the z-axis with the direction of motion of a particle.

• Deﬁne the helicity of a particle by:       h = ms /s

−→                     −→
⇒                   ⇐
h = +1                     h = −1

• For a spin-1/2 particle the helicity can be either +1 or -1.

• Helicity is not Lorentz invariant unless the particle is massless

Physics 506A                                4 - Discrete Symmetries                  Page 8
Charge Conjugation I
• The charge conjugation operator, C, converts a particle to i ts antiparticle.
C |p = |p

• In particular, C reverses every internal quantum number
(e.g. charge, baryon/lepton number, strangeness, etc.).

• C 2 = I implies that the only allowed eigenvalues of C are ±1.

• Unlike parity, very few particles are C eigenstates.
Only particles that are their own antiparticles (π 0 , η, γ) are C eigenstates.

For example,
˛ +¸ ˛ −¸
C ˛π   = ˛π
C |γ = − |γ

Physics 506A                                 4 - Discrete Symmetries                           Page 9
Charge Conjugation II
• The photon has a C = −1

• f f bound states have C = (−1)ℓ+s

• Charge conjugation is respected by both the strong and electromagnetic
interactions.

• Example: the π 0 (ℓ = s = 0 ⇒ C = +1) can decay into 2γ but not 3γ

C |nγ = (−1)n |γ
C ˛π 0 = ˛π 0
˛ ¸ ˛ ¸

π 0 → 2γ is allowed (and observed)
π 0 → 2γ is not allowed (and not observed < 3.1 × 10−8 )

Physics 506A                              4 - Discrete Symmetries                         Page 10
G-Parity I
• C-symmetry is of limited use.
⇒ Most particles are not C eigenstates

• The C operator converts π + to π − .

• These two particles have isospin assignments | 1 1 and | 1 − 1 .

• A 180◦ isospin rotation gives | 1 1 = eiπI2 | 1 − 1 .

• The charged pions are eigenstates under the G-parity, which combines C with
a 180◦ isospin rotation:        G = CeiπI2

• G-parity is mainly used to examine decays to pions (which have G = −1).
G |nπ = (−1)n |nπ

Physics 506A                               4 - Discrete Symmetries                    Page 11
G-Parity II
G-Parity of a few mesons

Particle      JP         I      G    Decay
ρ(770)        1−        1       +1    2π
ω(783)        1−        0       -1    3π
φ(1020)       1−        0       -1    3π
f (1270)      2+        0       +1    2π

For example, the ρ(770) has G = 1 which means it should only decay to an even
number of pions. Experimentally we ﬁnd that

ρ −→         ππ        100%
πππ        < 1.2 × 10−4

Physics 506A                          4 - Discrete Symmetries                      Page 12
CP Symmetry
The combination of C and P (and time reversal T ) have special signiﬁcance.
The violation of CP is the reason we live in a matter universe
CP T is required to be conserved in Quantum Field Theory (QFT)
Look at a pion decay example:

• In the pion decay π + → µ+ (R) + νµ (L), the νµ is always left-handed (LH)

• Under C, this becomes π − → µ− (R) + ν µ (L), but the ν µ is still LH
⇒ which does not occur in nature.

• With C and P , though, we get a RH antineutrino. π − → µ− (L) + ν µ (R)
⇒ whiich is allowed in nature

Physics 506A                               4 - Discrete Symmetries                       Page 13
CP Violation in the Kaon Sector I
¯
• Consider the neutral kaons K 0 (ds) and K 0 (sd)
These particles can mix via a second-order weak interaction:

s                    d                    s               d
u
W
u          u                              W       W

W                                        u
d                     s                    d               s

Physics 506A                              4 - Discrete Symmetries                       Page 14
CP Violation in the Kaon Sector II
¯
• Both K 0 and K 0 are pseudoscalar mesons, therefore P = −1. and the K 0 and
¯
K 0 are a particle-antiparticle pair.
As a result, under CP , we have
˛K = − ˛K 0   ¯          ¯
˛ 0¸
˛K = − ˛K 0
˛ ¸         ˛ 0¸      ˛ ¸
CP                     CP

`˛ 0 ¸ ˛ 0 ¸´ √
• Deﬁning               |K1      =        ˛K − ˛K¯    / 2
`˛ 0 ¸ ˛ 0 ¸´ √
|K2      =        ˛K + ˛K¯    / 2

we have                    CP |K1           =    + |K1
CP |K2           =    − |K2

• If CP is conserved, then
|K1 can only decay to 2π (CP = +1)
|K2 can only decay to 3π (CP = −1).

Physics 506A                             4 - Discrete Symmetries                      Page 15
CP Violation in the Kaon Sector III
• We observe the KS and the KL :

KS → ππ           τ = 0.9 × 10−10 s
KL → πππ             τ = 0.5 × 10−7 s
0
• K 0 and the K are mass eigenstates and are each others antiparticles

• KS and the KL are CP eigenstates (have different masses) and are not
antiparticles

Physics 506A                              4 - Discrete Symmetries                       Page 16
CP Violation in the Kaon Sector IV
• Start out with a beam of K 0 (π − p → K 0 Λ). This will be a superposition of K1
and K2 (CP eigenstates):

˛ 0¸               √
˛K = (|K1 + |K2 ) / 2

• The K1 component of the beam will decay away over a few centimeters,
thereby leaving a nearly pure beam of K2 .
⇒ expect to see only 3π decays

• Experimentally, we ﬁnd that about 1 in 440 decays is to 2π.
⇒ the KL has a small mixture of K1 :          q
|KL = (|K2 + ǫ |K1 ) / 1 + |ǫ|2

Physics 506A                               4 - Discrete Symmetries                        Page 17
Other Tests of CP Violation
• There are other CP -violating observables that have been measured in the kaon
sector. For example, there is an asymmetry between the branching ratios of
KL to π + + e− + νe versus π − + e+ + νe
¯

• Within the last few years, the BaBar and Belle experiments have measured CP
violation in the B-meson sector.

• CP violation should also be observable in the D-meson (charm) sector, though
this will be a small effect that will be very difﬁcult to measure.

• There have been earches in the charged lepton sector
At UVic, we are studying the τ − → π − KS ν decay with the BaBar data

• With the observation that neutrino has mass, it is expected that we will
observe CP violation in the neutrino sector

Physics 506A                               4 - Discrete Symmetries                          Page 18
Why Measure CP Violation?
• The universe contains much more matter than antimatter.
⇒ This requires CP violation.

• Since the SM provides only one source of CP violation (CKM phase)
The SM CP V appears to be inadequate to account for this asymmetry.

• However the SM may have another source if there is CP V in the neutrino
sector

• CP violation is ubiquitous in theories of New Physics (ie, beyond SM).
⇒ SUSY can provide enough CP violation to be observable at low energies.

Physics 506A                             4 - Discrete Symmetries                       Page 19
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CP Violation in τ decays
0
The goal is to search for direct CP violation in the decay τ − → h− KS ≥ 0π 0 ντ
`     ´

There is no charge asymmetry between τ − → π − K 0 ντ and τ + → π + K 0 ν τ
0      0
Experimentally we observe the KS and KL mesons, which are linear combinations
of K 0 and K 0 states.
0      0
Earlier experiments have shown that the KS and KL are not exact CP eigenstates.
As a result the charge asymmetry is
0
Γ τ + → π + KS ν τ − Γ τ − → π − KS ντ
0
`               ´   `                ´
AQ = ` +             0
´   `            0
´              (1)
Γ τ → π + KS ν τ + Γ τ − → π − KS ντ

Theory predicts AQ = (3.3 ± 0.1) × 10−3

Little can be learnt from this CP asymmetry, but this measurement provides a
valuable calibration point in CP searches and any deviation from the Standard
Model value would be evidence for new physics.

Physics 506A                           4 - Discrete Symmetries                         Page 23
Time Reversal Symmetry
• Time reversal symmetry, as you might guess, reverses the time component:

T (t, x) = T (−t, x)

• Although we expect the weak interaction to violate T , direct T violation has
not been deﬁnitively observed yet.

• Experimentally, one tries to measure the rate of a reaction in both directions
A + B → C + D but this is not so easy

Physics 506A                               4 - Discrete Symmetries                         Page 24
The CP T Theorem
• The combination CP T is always conserved in any local quantum ﬁeld theory

• CP T violation in essentially synonymous with a violation of Lorentz
invariance

• CP T symmetry mandates that particles and antiparticles must have certain
identical properties, such as the same mass, lifetime, charge, and magnetic
moment

Physics 506A                              4 - Discrete Symmetries                       Page 25
Experimental tests of CP T
Citation: S. Eidelman et al. (Particle Data Group), Phys. Lett. B 592, 1 (2004) (URL: http://pdg.lbl.gov)

CPT INVARIANCE
(m W + ? m W ? ) / m average                                                 ? 0:002 0:007
(m e + ? m e ? ) / m average                                                 <8    10?9 , CL = 90%
qe + + q e ? e                                                              <4 10?8
(g e + ? g e ? ) / gaverage                                                  (? 0:5 2:1) 10?12
( + ? ? ) / average                                                          (2 8) 10?5
(g + ? g ? ) / g average                                                     (? 2:6 1:6) 10?8
(m + ? m ? ) / m average                                                     (2 5) 10?4
( + ? ? ) / average                                                          (6 7) 10?4
(m K + ? m K ? ) / m average                                                 (? 0:6 1:8) 10?4
( K + ? K ? ) / average                                                     (0:11 0:09)% (S = 1.2)
K !               rate di erence/average                                    (? 0:5 0:4)%
K !             0 rate di erence/average                                f ] (0:8 1:2)%
in K  0 ? K 0 mixing
real part of                                                       (2:9 2:7) 10?4
imaginary part of                                                  (0:02 0:05) 10?3
m K 0 ? m K 0 / m average                                             g] <10?18 , CL = 90%
(? K 0 ? ? K 0 )/m average                                                (8 8) 10?18
phase di erence 00 ? +?                                                   (0:2 0:4)
Re( 2 +? + 3 00 )? 2
3
1         L                                             (? 3 35) 10?6
ACPT (K          ) in D 0 ! K ? + , D 0 ! K + ?                           0:008 0:008
m p ?m p /m p                                                         h] <1:0 10?8 , CL = 90%
q q q
( mpp { mp )/ mpp                                                         (? 9 9) 10?11
p
qp + qp e                                                             h] <1:0 10?8 , CL = 90%
( p + p) p                                                                (? 2:6 2:9) 10?3
(m n ? m n )/ m n                                                         (9 5) 10?5
(m ? m ) m                                                                (? 0:1 1:1) 10?5 (S = 1.6)
( ? )/                                                                    ? 0:001 0:009
( + ? ?) / +                                                              (? 0:6 1:2) 10?3
( + + ?)                  +                                               0:014 0:015
(m ? ? m + ) / m ?                                                        (1:1 2:7) 10?4
( ? ? +) / ?                                                              0:02 0:18
( ? + +) / ?                                                              +0:01 0:05
(m ? ? m + ) / m ?                                                        (? 1 8) 10?5
( ? ? +) / ?                                                              ? 0:002 0:040

HTTP://PDG.LBL.GOV                                      Page 11                             Created: 6/17/2004 19:40

Physics 506A                                                                                                                4 - Discrete Symmetries   Page 26
Lepton Number
• There are 3 lepton numbers: Le , Lµ and Lτ
Le = +1 for e− and νe
Le = −1 for e+ and ν e

• Conserved in the EM and Weak interactions γ → e+ e− and π + → µ+ ν µ are
allowed whereas µ+ → e+ γ is forbidden

• BaBar (UVic group) put a new limit on the τ + → µ+ γ branching fraction
(10−8 )

• Neutrino oscillations imply that lepton number is violated (at a very small
level)

Physics 506A                               4 - Discrete Symmetries                        Page 27
Experimental tests of Lepton Number
Citation: S. Eidelman et al. (Particle Data Group), Phys. Lett. B 592, 1 (2004) (URL: http://pdg.lbl.gov)

TESTS OF NUMBER CONSERVATION LAWS

LEPTON FAMILY NUMBER
Lepton family number conservation means separate conservation of each of Le , L ,
L .

?(Z ! e        )/?total                                                 i ] <1:7 10?6 , CL = 95%
?(Z ! e        )/?total                                                 i ] <9:8 10?6 , CL = 95%
?(Z !          )/?total                                                 i ] <1:2 10?5 , CL = 95%
limit on ? ! e ? conversion
( ? 32 S ! e ? 32 S) /                                                <7     10?11 , CL = 90%
( ? 32 S ! 32 P )
( ? Ti ! e ? Ti) /                                                    <4:3     10?12 , CL = 90%
( ? Ti ! capture)
( ? Pb ! e ? Pb) /                                                    <4:6     10?11 , CL = 90%
( ? Pb ! capture)
limit on muonium ! antimuonium conversion Rg =                               <0:0030,    CL = 90%
GC / GF
?( ? ! e ? e )/?total                                                   j ] <1:2 10?2 , CL = 90%
?( ? ! e ? )/?total                                                          <1:2     10?11 , CL = 90%
?( ? ! e ? e + e ? )/?total                                                  <1:0     10?12 , CL = 90%
?( ? ! e ? 2 )/?total                                                        <7:2     10?11 , CL = 90%
?( ? ! e ? )/?total                                                          <2:7     10?6 , CL = 90%
?( ? ! ? )/?total                                                            <1:1     10?6 , CL = 90%
?( ? ! e ? 0 )/?total                                                        <3:7     10?6 , CL = 90%
?( ? ! ? 0 )/?total                                                          <4:0     10?6 , CL = 90%
?( ? ! e ? K 0 )/?total
S                                                             <9:1     10?7 , CL = 90%
?( ? ! ? K 0 )/?total
S                                                             <9:5     10?7 , CL = 90%
?( ? ! e ? )/?total                                                          <8:2     10?6 , CL = 90%
?(  ? ! ? )/?total                                                           <9:6     10?6 , CL = 90%
?( ? ! e ? 0 )/?total                                                        <2:0     10?6 , CL = 90%
?( ? ! ? 0 )/?total                                                          <6:3     10?6 , CL = 90%
?( ? ! e ? K (892)0 )/?total                                                 <5:1     10?6 , CL = 90%
?( ? ! ? K (892)0 )/?total                                                   <7:5     10?6 , CL = 90%
?( ? ! e ? K (892)0 )/?total                                                 <7:4     10?6 , CL = 90%
?( ? ! ? K (892)0 )/?total                                                   <7:5     10?6 , CL = 90%
?( ? ! e ? )/?total                                                          <6:9     10?6 , CL = 90%
?( ? ! ? )/?total                                                            <7:0     10?6 , CL = 90%
?( ? ! e ? e + e ? )/?total                                                  <2:9     10?6 , CL = 90%
?( ? ! e ? + ? )/?total                                                      <1:8     10?6 , CL = 90%
HTTP://PDG.LBL.GOV                                      Page 12                             Created: 6/17/2004 19:40

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Summary
• The 3 discrete symmetries P , C, and T are respected
individually by the strong and electromagnetic forces.

• Parity violation is the signature of the weak interaction.

• The weak interaction also violates CP , as a result of a complex
phase in the CKM matrix.

• The combined symmetry of CP T is obeyed by all local
quantum ﬁeld theories.

Physics 506A                          4 - Discrete Symmetries                 Page 33
The Quark Model
• Sticks quarks and antiquarks together like Lego.

• Makes no attempt to account for the complexities of low-energy QCD other
than to hope that it treats all members of a multiplet in the same way.

• With a handful of experimental inputs, can be used to obtain ballpark
estimates for hadron masses and magnetic moments.

Physics 506A                              4 - Discrete Symmetries                        Page 34
Mesons
• There are 9 distinct ﬂavor combinations that can be contructed from {u, d, s}
u ¯¯                              ¯
and {¯, d, s}. Group theoretically, 3 ⊗ 3 = 8 ⊕ 1, but we usually lump the 1
with the 8 to create a meson nonet.

• Once we include spin, we get a pseudoscalar meson nonet (π, K, η) and a
vector meson nonet (ρ, K ∗ , ω, φ).

• The vector mesons are heavier, and this can be modeled by the meson mass
formula:
(S1 · S2 )
M = m1 + m2 + A
m1 m2
With 4 parameters (mu = md = 310 MeV, ms = 483 MeV,
A = (2mu )2 160 MeV), we can predict 18 meson masses to about 1%.

Physics 506A                               4 - Discrete Symmetries                       Page 35
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Baryons
• There are 27 distinct ﬂavor combinations that can be constructed from three
copies of {u, d, s}. Group theoretically, 3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10.

• One of the baryon octets contains the p, n, Σ, Λ, Ξ. The baryon decuplet
contains the ∆, Σ ∗ , Ξ ∗ , Ω. The other octet and the singlet contain less familiar
baryons.

• As with the mesons, we can construct a baryon mass formula in terms of the
effective masses of the quarks (different from those used for mesons) ’ and
their relative spins.

• No evidence for particles with more than 3 quarks

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