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Lecture 7: Symmetries II

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Lecture 7: Symmetries II Lecture 7: Symmetries II • Charge Conjugation • Time Reversal • CPT Theorem • Baryon & Lepton Number • Strangeness • Applying Conservation Laws Useful Sections in Martin & Shaw: Sections 3.1-3.4, 5.3, 5.4, 5.6 C-Parity (charge conjugation) Changes particle to anti-particle (without affecting linear or angular momentum) Electromagnetism is obviously symmetric with respect to C-parity (flip the signs of all charges and who would know?). It turns out that the strong force is as well. But, again, not the weak force (otherwise there would be left-handed anti-neutrinos... there aren’t !) For particles with distinct anti-particles (x = e-, p, π-, n, ...) a state characterized C ∣x, ψ> = ∣ x, ψ > ! where∣x, ψ > ≡ by a particle x and a wave function ψ For particles which no not have distinct anti-particles (y = γ, π0, ...) where C is a ''phase factor" of ±1 ∣ ∣ y C y, ψ > = Cy y, ψ > ! (like for parity) used to determine C-conservation in interactions For multi-particle systems ∣ ∣ | C x, ψ1; x, ψ2 > = x, ψ1; x, ψ2 > = ± x, ψ1; x, ψ2 > depending on whether the system is symmetric or antisymmetric under the operation Example: consider a π+π- pair in a state of definite orbital angular momentum L C ∣ π+π-, L > = (-1)L ∣ π+π-, L > since interchanging π+ and π- reverses their relative position vector in the spatial wave function Example: Experimentally, π0 → γ + γ but never π0 → γ + γ + γ How is this reconciled with the concept of C-parity ?? using a similar argument 0 0 C ∣ π > = C ∣ π > as for parity, π° Cγ = -1 C = +1 π° ∣ ∣ C ∣ γγγ = C C C ∣ γγγ C γγ > = CγCγ γγ > > γ γ γ > ! 2 ∣ = C 2 C ∣ γγγ = Cγ γγ > ! γ γ > ! = C ∣ = ∣ γγ > ! γ γγγ > ! = -∣ γγγ > ! Ah! So we can never get π0 → γ + γ + γ if C-parity is conserved !! Time Reversal In analogy with parity, we could try t → -t However, note that if we start with ∂ H Ψ(x,t) = i ℏ Ψ(x,t) ∂t ∂ Now apply t -t H (x,-t) = - i ℏ (x,-t) → Ψ ∂t Ψ But we want the Schrodinger equation to be invariant! Wigner, This can be patched up by taking the complex conjugate 1931 (but without touching the phase of Ψ, otherwise t doesn’t get reversed): ∂ T [H Ψ(x,-t)] = T [i ℏ Ψ(x,-t) ] ∂t ∂ H Ψ*(x,-t) = i ℏ Ψ*(x,-t) ∂t Both Ψ(x,t) and Ψ*(x,-t) satisfy the same equation. Thus we define T S(x,t) = S*(x, -t) Note that if Ψ ≡ Αϕ1(x,t) + Βϕ2(x,t) T Ψ = TΑϕ1(x,t) + TΒϕ2(x,t) = Α*{ Tϕ1(x,t) } + Β*{ Tϕ2(x,t) } ≠ Α{ Tϕ1(x,t) } + Β{ Tϕ2(x,t) } ''anti-linear" So the operator is not Hermitian and the eigenvalues are not real! Hermitian ⇒ ∫ Ψα* (Ο Ψβ) = ∫ (Ο Ψα)* Ψβ so ∫ Ψn* (G Ψn) = gn ∫ Ψn* Ψn = gn ∫ (G Ψn)* Ψn = gn* ∫ Ψn* Ψn = gn* ⇒ gn= gn* so gn must be real So, unlike other symmetries, time-reversal does not give rise to real conserved quantities (i.e. no conservation laws per se) In practice this is difficult to test (how do you ''reverse time" in an experiment?) It’s more useful to consider the combination T P : Note that P p = -p and P L = P (x × m dx/dt) T P p = p = x × m dx/dt T P L = - (x × m dx/dt) = -L (assume this holds for spin) the spin-averaged a(pa, sa) + b(pb, sb) → aʹ(paʹ, saʹ) + bʹ(pbʹ, sbʹ) apply rates of these should thus be the same a(p , -s ) + b(p , -s ) ← aʹ(p ʹ, -s ʹ) + bʹ(p ʹ, -s ʹ) T P ⇒ a a b b a a b b under T P symmetry ''Principle of Detailed Balance" (confirmed in various EM & strong interactions) Charge Conjugation -X +X Charge Conjugation Parity -X +X -X +X Flip orientation in time Charge Conjugation Parity -X +X -X +X +X -X Flip Switch orientation coordinate in time definitions Charge Conjugation Parity -X +X -X +X +X -X Flip Switch orientation coordinate in time definitions Charge Conjugation Parity -X +X -X +X -X+X Flip Switch orientation coordinate in time definitions Charge Conjugation Parity -X +X -X +X -X +X Flip Switch orientation coordinate in time definitions Charge Time Conjugation Parity Reversal -X +X -X +X -X +X Flip Switch orientation coordinate in time definitions Charge Time Conjugation Parity Reversal -X +X -X +X -X +X -X +X Flip Switch Run movie orientation coordinate backwards in time definitions in time CPT CPT Theorem (independently discovered by Pauli, Luders and Bell and Schwinger) States that if a quantum field theory is invariant under Lorentz transformation, then C P T is an exact symmetry !! 1) Integer spin particles obey Bose-Einstein statistics (bosons) 2) 1/2 - spin particles obey Fermi-Dirac statistics (fermions) 3) Particles and antiparticles must have identical masses & lifetimes 4) All internal quantum numbers of antiparticles are opposite to those of the corresponding particles ( Note that if, for example, CP is violated, then T must be violated ) Baryon and Lepton Number It is an empirical observation that the number of baryons (fermions with masses ≥ the proton mass) minus the number of antibaryons is conserved in all reaction thus far observed. Thus we define the ''baryon number" B ≡ (# baryons) - (# antibaryons) as a conserved quantity NO Experimental The same has been assumed to be true for Leptons. evidence that this is the case!!! However, there is also a form of the rule that seems to operate which relates to individual lepton ''families" or ''generations" : e- µ- τ- L νe νµ ντ } } } } HOT OFF THE PRESS! These can be violated by L L L ''neutrino oscillations" e µ τ Among other things, conservation of B and L means that protons and electrons don’t decay (so matter is stable) and baryons don’t mix with leptons. ⇒ GUT models therefore predict these laws to break down at some point Symmetry Summary: Transformation Symmetry Type Conserved Quantity translation global, continuous linear momentum rotation global, continuous angular momentum time global, continuous energy (Lorentz global, continuous CM velocity) (''space-time") rotation in ''isospin-space" global, continuous isospin additive (''internal") electromagnetic scalar/vector potential local, continuous, gauge charge - (global, continuous, gauge) baryon number - (global, continuous, gauge) lepton number parity global, discreet ''P-value" charge conjugation global, discreet ''C-value" multiplicative time reversal global, discreet none! In Addition: Local, Lorentz-Invariant, Quantum Field Theories ⇒ CPT Strangeness π- K0 π+ π- Λ p π- Originally found in cosmic ray cloud chambers in 1947, Then in bubble chambers at the Brookhaven Cosmotron in 1953 The new particles were always produced in pairs (''associated production"), suggesting a new conserved quantity ⇒ ''strangeness" so define SΛ ≡ -1 and SK° = +1 The cross-section for production indicates a strong interaction, however the decay timescale is much longer than expected for a strong decay: -15 8 -23 tS ~ 1 fm / c = (10 m) / (3x10 m/s) = 10 s but the observed decays took 10-10s (practically forever!) It can be shown that this is about what you’d expect for a weak decay: - First consider: n → p + e + νe from Fermi Golden Rule: 1/t ∝ ρ (density of final states) 2 2 dNe ~ pe dpe dNν ~ pν dpν since the distributions for 2 2 - ⇒ dN = dN dN ~ p dp p dp e and νe are unconstrained e ν e e ν ν in a 3-body decay and, assuming the proton gets basically no kinetic energy, - the energy, E , available for the e & νe is determined. ≃ If E ≡ Ee + Eν ⇒ pν E - Ee ≃ then, for a given value of Ee , dpν dE 2 2 Thus, dρ = dN/dE = dN/dpν ~ pe (E -Ee) dpe E and, in the limit E ≫ m , 2 2 e e ρ ~ ∫ Ee (E -Ee) dEe 0 ⇒ τ ∝ E-5 ≃ For β-decay, E mn - mp ~ 1.3 MeV and the neutron lifetime is ~ 1000 s so take an In case of the strange particles: mΛ = 1116 MeV m = 498 MeV average value K° of E~750 MeV Thus, we’d expect a weak decay timescale of order t = 1000 s (1.3/750)5 = 10-11s ( which is certainly alot closer W to what is actually observed! ) Interpretation: S is conserved by the strong interaction, which is why these particles are produced in pairs and why the individual particles cannot undergo strong decay to non-strange products. However, S is not conserved by the weak interaction, which eventually does allow the Λ and K0 to decay ! For “1st-order" Weak Interactions: ΔS = 0, ±1 Applying Conservation Laws: Example: In the following pairs of proposed reactions, determine which ones are allowed and the relevant force at work π - + p → Σ0 + η0 π - + p → Σ0 + K0 Σ - → π - + n Σ - → π - + p Interaction: strong weak charge: -1 + 1 = 0 + 0 -1 + 1 = 0 + 0 -1 = -1 + 0 -1 = -1 + 1 lepton 0 + 0 = 0 + 0 0 = 0 + 0 number: 0 + 0 = 0 + 0 0 = 0 + 0 baryon 0 + 1 = 1 + 0 0 + 1 = 1 + 0 1 = 0 + 1 1 = 0 + 1 number: strangeness: 0 + 0 = -1 + 0 0 + 0 = - 1 + 1 -1 + 1/2 = 0 - 1/2 -1 + 1/2 = 0 + 0 Isospin (I3) : Example: Κ- π- Identify the neutral particle X Κ- → π- + X Lepton number: 0 = 0 + ? ⇒ L = 0 Baryon number: 0 = 0 + ? ⇒ B = 0 Spin: 0 = 0 + ? ⇒ J = 0 Candidates: π0, Κ0, Κ0, η0 But, m < m - m X Κ π X = π0 (X≠Κ,η0) + - X → π + π π- π+ Lepton #: ? = 0 + 0 ⇒ L = 0 - X Baryon #: ? = 0 + 0 ⇒ B = 0 π Spin: ? = 0 + 0 ⇒ J = 0 π- Y (as before) Candidates: π0, Κ0, Κ0, η0 But, m > m + m p X π π (X≠π) & < 10-18 s (cτ < 0.3 nm) (X≠η) - τη η Y → π + p Lepton #: ? = 0 + 0 ⇒ L = 0 0 0 X = K or K Baryon #: ? = 0 + 1 ⇒ B = 1 Spin: ? = 0 + 1/2 ⇒ J = 1/2 strong Candidates: n , Λ, Σ0, Ξ0 π- +p → X + Λ interaction But, m > m + m Y π p (Y≠n) Strangeness: 0 + 0 = ? + -1 ⇒ S = +1 Σ0 → Λ + γ (Y≠Σ0) ΔS=0,±1 (weak decay) (Y≠Ξ0) X = K0 Y = Λ .
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