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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
                 parity invariance had never been
                 tested experimentally for weak in-
                 teractions.




Physics 506A                              4 - Discrete Symmetries   Page 5
                          Madame Wu’s Experiment




         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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