sign in sign up
<<
Home , Parity (physics)

Discrete Symmetries

• Parity • Parity Violation • Charge Conjugation • CP Violation • Reversal • The CPT Theorem • 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 . • − from v: s = v v • · P (s)= P (v v)=( v) ( v)= v v =+s · − · − · 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 P (p)= p − Vector P (v)= v − P (a)=+a

Physics 506A 4 - Discrete Symmetries Page 2 Parity in Physical Systems

ℓ Two-body systems have parity pApB( 1) • ℓ − P φ(12) = p1p2( 1) φ(12) − Intrinsically, particles and have opposite parity • Bound states like positronium e+ e and qq have parity of ( 1)ℓ+1. − − have a parity of ( 1), and this underlies the ∆ℓ = 1 • − ± in atomic transitions.

Note that parity is a multiplicative . • 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- have 1/2 and + parity and u-quarks have spin 1/2 and - parity

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

S L JP particle 0 0 0 0− π 0 1 0 1− ρ + 0 1 1 b1(1235)

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

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

In 1956, Yang and Lee realized that parity invariance had never been tested experimentally for weak interactions.

Wu’s experiment: recorded the direction of the emitted from a 60Co 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 5 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 νµ always emerge right-handed. • − − If parity were conserved by the , we would expect left-handed • pairs and right-handed pairs with equal probability (just as we observe with π0 2γ). → Assuming that are massless, • ALL neutrinos are left-handed ALL antineutrinos are right-handed

Physics 506A 4 - Discrete Symmetries Page 6 Charge Conjugation I

The charge conjugation , C, converts a particle to i ts . • C p = p | i | i In particular, C reverses every internal quantum number • (e.g. charge, /lepton number, strangeness, etc.).

C2 = 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 (π0, η, γ) are C eigenstates.

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

Physics 506A 4 - Discrete Symmetries Page 7 Charge Conjugation II

The has a C = 1 • − ff bound states have C =( 1)ℓ+s • − Charge conjugation is respected by both the strong and electromagnetic • interactions.

Example: the π0 (ℓ = s = 0 C = +1) can decay into 2γ but not 3γ • ⇒

C nγ =( 1)n γ | i − | i 0 0 C ˛π ¸ = ˛π ¸ ˛ ˛ π0 2γ is allowed (and observed) → π0 2γ is not allowed (and not observed < 3.1 10 8) → × −

Physics 506A 4 - Discrete Symmetries Page 8 G-Parity I

C- 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 . • | i | − i A 180 isospin gives 1 1 = eiπI2 1 1 . • ◦ | i | − i The charged are eigenstates under the G-parity, which combines C with • iπI2 a 180◦ isospin rotation: G = Ce G-parity is mainly used to examine decays to pions (which have G = 1). • − G nπ =( 1)n nπ | i − | i

Physics 506A 4 - Discrete Symmetries Page 9 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% −→ 4 πππ < 1.2 10− ×

Physics 506A 4 - Discrete Symmetries Page 10 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 (QFT) Look at a 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)+ νµ(L), but the νµ is still LH • − → − which does not occur in nature. ⇒ With C and P , though, we get a RH antineutrino. π µ (L)+ νµ(R) • − → − whiich is allowed in nature ⇒

Physics 506A 4 - Discrete Symmetries Page 11 CP Violation in the Sector I

Consider the neutral K0 (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

This section will focus on the CP aspects. The details on the time dependence of K0 K0 oscillations can be found in most textbooks.

Physics 506A 4 - Discrete Symmetries Page 12 CP Violation in the Kaon Sector II

Both K0 and K¯ 0 are pseudoscalar mesons, therefore P = 1. and the K0 and • − K¯ 0 are a particle-antiparticle pair. As a result, under CP , we have

0 0 0 0 CP ˛K ¸ = ˛K¯ ¸ CP ˛K¯ ¸ = ˛K ¸ ˛ − ˛ ˛ − ˛

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

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

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

Physics 506A 4 - Discrete Symmetries Page 13 CP Violation in the Kaon Sector III

K0 and the K0 are mass eigenstates and are each others antiparticles • K1 and the K2 are CP eigenstates • 0 1 0 1 ˛K ¸ = ( K1 + K2 ) ˛K ¸ = ( K1 K2 ) ˛ √2 | i | i ˛ √2 | i−| i 0 0 KS and the KL are the observed states and are nearly identical to the CP • 0 0 eigenstates (KS and KL are not antiparticles) 0 1 0 1 ˛KL¸ = ( K2 + ǫ K1 ) ˛KS ¸ = ( K2 ǫ K1 ) ˛ √1+ ǫ 2 | i | i ˛ √1+ ǫ 2 | i− | i | | | | Experimentally we observe • 10 KS ππ τ = 0.9 10 s → × − 7 KL πππ τ = 0.5 10 s → × − 0 + 0 However, we find about 1 KL π π in 440 KL decays → −

Physics 506A 4 - Discrete Symmetries Page 14 Other Tests of CP Violation

There are other CP -violating observables that have been measured in the kaon • sector. For example, there is an between the branching ratios of + + KL to π + e− +ν ¯e 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.

CP violation observed in the K and B mesons is not enough to explain the • domination of matter in the universe

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

Physics 506A 4 - Discrete Symmetries Page 15 Why study CP violation?

Sakarov pointed out that it is possible to start from a matter-antimatter • symmetric universe and end up in one that is asymmetric This requires that there be some process (or processes) that violates the CP symmetry.

The SM does not predict CP violation (it can accomodate no CPV or CPV). • However, the SM provides only one source of CP violation (CKM phase angle) which is only possible if there are more than 3 generations.

The currently observed SM CPV (using K and B mesons) is too small to • explain the matter-dominated universe.

There is another possible source of CPV within the SM in the neutrino sector • analoguous to the CKM matrix.

Beyond the SM: CP violation is ubiquitous in theories of New Physics • SUSY can provide enough CP violation to be observable at low . ⇒

Physics 506A 4 - Discrete Symmetries Page 16 Composition of the universe

The universe is more complicated and it is observed that only a small fraction (4%) of the universe is made of known matter.

Observations also show that there is more matter (dark matter) from un- known sources (21%).

In addition, there is another component (dark ) which we know even less about (75%).

Physics 506A 4 - Discrete Symmetries Page 17 CP Violation in τ decays

0 0 The goal is to search for direct CP violation in the decay τ h KS ` 0π ´ ντ − → − ≥ 0 + + 0 There is no charge asymmetry between τ π K ντ and τ π K ντ − → − → 0 0 Experimentally we observe the KS and KL mesons, which are linear combinations of K0 and K0 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 0 Γ `τ π KS ντ ´ Γ `τ π KS ντ ´ → − − → − AQ = + + 0 0 Γ `τ π KS ντ ´ + Γ `τ π KS ντ ´ → − → −

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

Physics 506A 4 - Discrete Symmetries Page 18 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 19 The CP T Theorem

The combination CPT is always conserved in any local quantum field theory. • CPT 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

See the PDG summary tables: • http://pdg.lbl.gov/2011/tables/rpp2011-conservation-laws.pdf Examples (fractional differences): M(K0) M(K0) < 8 10 19 − × − M(e+) M(e ) < 8 10 9 − − × − τ(µ+) τ(µ ) < 2 10 5 − − × − g(e+) g(e ) < 10 12 − − −

Physics 506A 4 - Discrete Symmetries Page 20 Lepton Number

There are 3 lepton numbers: Le, Lµ and Lτ • Le = +1 for e− and νe + Le = 1 for e and νe − + + + Conserved in the EM and Weak interactions γ e e and π µ νµ are • → − → allowed whereas µ+ e+γ is forbidden → Neutrino oscillations imply that lepton number is violated (at a very small • level)

See the PDG summary tables: • http://pdg.lbl.gov/2011/tables/rpp2011-conservation-laws.pdf Examples (fractional differences): Γ(µ e γ) < 10 11 − → − − Γ(Z0 e µ+) < 10 6 → − − Γ(τ e e e+) < 10 8 − → − − −

Physics 506A 4 - Discrete Symmetries Page 21 Physics 506A 4 - Discrete Symmetries Page 22 Physics 506A 4 - Discrete Symmetries Page 23 Physics 506A 4 - Discrete Symmetries Page 24 Physics 506A 4 - Discrete Symmetries Page 25