Vacuum Polarization

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Vacuum Polarization Vacuum Polarization Vacuum Polarization e- e- Thus, we never actually ever see a ''bare" charge, only an effective charge shielded by polarized virtual electron/positron pairs. A larger charge (or, equivalently, α) will be seen in interactions involving a high momemtum transfer as they probe closer to the central charge. - q ⇒ ''running coupling constant" In QED, the bare charge of an electron is actually infinite !!! Note: due to the field-energy near an infinite charge, the bare mass of the electron (E=mc2) is also infinite, but the effective mass is brought back into line by the virtual pairs again !! Symmetry Translational Invariance Space-Time Symmetries F m1 m2 x1 x2 | | If the force does not change as a function of position, then force felt at x : F = m x 2 2 ¨2 recoil felt at x : F = -m x 1 1 ¨1 subtracting: m x + m x = 0 2 ¨ 2 1¨1 d/dt [ m x + m x ] = 0 2 ˙2 1 ˙1 m2 v2 + m1 v1 = constant Translational Invariance ⇔ Conservation of Linear Momentum Time Invariance Consider a system with total energy 1 assume this basic E = m x2 + V 2 description also holds at other times dE dV dx = m x x + dt dx dt dV = m x x + x dx but dV/dx = -F = -mx (Newton’s 2nd law) dE ⇒ = m x x - m x x = 0 dt ⇒ E = constant Time Invariance ⇔ Conservation of Energy Gauge Invariance in EM ''local" Gauge Invariance in Electromagnetism: symmetry A → A + ∇χ(x,t) Φ → Φ - ∂ χ(x,t) ∂t E = -∇Φ - ∂ A → -∇[Φ - ∂ χ(x,t) ] - ∂ [A + ∇χ(x,t)] ∂t ∂t ∂t = -∇Φ - ∂ A = E ∂t B = ∇ × A → ∇ × [A + ∇χ(x,t)] = ∇ × A = B Gauge Invariance ⇔ Conservation of Charge Necessity of Charge Conservation (Wigner, 1949) To see this, assume charge were not conserved So a charge could And destroyed here, be created here by with the output of inputing energy E some energy Eʹ PoP ! q PoP ! x1 x2 - - E = qΦ(x1) Eʹ = qΦ(x2) - - - Thus we will have created an overall energy Eʹ E = q { Φ(x2) Φ(x1) } So, to preserve energy conservation, if Φ is allowed to vary as a function of position, charge must be conserved Noether’s Theorem Noether’s Theorem Continuous Symmetries ⇔ Conserved ''Currents" (Emmy Noether, 1917) Importance of Gauge Invariance Gauge symmetry from another angle... Take the gauge transformation of a wavefunction to be Ψ → eiqθ Ψ where θ is an arbitrary ''phase-shift" as a function of space and time Say we want the Schrodinger equation to be invariant under such a transformation ∂ Ψ i 2 = 2m ∇ Ψ clearly we’re in trouble ! ∂t Consider the time-derivative for a simple plane wave: Ψ = Aei(px-Et) Ψ → Aei(px-Et+qθ) ∂/∂t Ψ = i ( -E + q ∂θ/∂t ) Ψ Note that if we now introduce here’s the problem! an electric field, the energy level gets shifted by -qΦ ∂/∂t Ψ = i ( -E + qΦ + q ∂θ/∂t ) Ψ But we can transform Φ → Φ - ∂θ/∂t, thus cancelling the offending term! (a similar argument holds for the spatial derivative and the vector potential) Gauge invariance REQUIRES Electromagnetism !! Gauge Invariance and Gravity Another example... Special Relativity: Invariance with respect to reference frames moving at constant velocity ⇒ global symmetry Generalize to allow velocity to vary arbitrarily at different points in space and time (i.e. acceleration) ⇒ local gauge symmetry Require an interaction to make this work GRAVITY! Chicken or egg ? All known forces in nature are consequences of an underlying gauge symmetry !! or perhaps Gauge symmetries are found to result from all the known forces in nature !! Symmetry and pragmatism Pragmatism: Symmetries (and asymmetries) in nature are often clear and can thus be useful in leading to dynamical descriptions of fundamental processes True even for ''approximate" symmetries ! Meson Nonets For ''pre-1974" hadrons, the following symmetries were also observed Q = I3 + (B+S)/2 thus, define ''Hypercharge" as Gell-Mann - Nishijima Formula Y ≡ B + S Mesons Y Y Κ0 1 Κ+ Κ*° 1 Κ*+ (498) (494) (896) (892) 0 0 - (135) π η (547) + - (769) ρ ω (782) + π π I ρ ρ I (140) ηʹ (140) 3 (769) ϕ (769) 3 (958) (1019) - *- Κ -1 Κ0 Κ -1 Κ*0 (494) (498) Note the presence (892) (896) of both particles Parity - ( Spin ) 0 nonet and antiparticles 1- nonet Baryon Octet & Decuplet Baryons Y Y n 1 p Δ- Δ° 1 Δ+ Δ++ (940) (938) (1232) (1232) (1232) (1232) Σ0 (1193) Σ- Σ+ I Σ*- Σ*ο Σ*+ I (1197) Λ (1116) (1189) 3 (1387) (1384) (1383) 3 Ξ- -1 Ξ0 Ξ*- -1 Ξ*ο (1321) (1315) (1535) (1532) Ω- (1672) Note antiparticles ( SpinParity ) 1/2+ octet are not present 3/2+ decuplet Inelastic Scattering Inelastic Scattering: Evidence for Compositeness 3-Quark Model Consider a 3-component ''parton" model where the constituents have the following quantum numbers: Y Y 1 1 s d u I -1 1 I3 -1 1 3 u d s -1 -1 ''quarks" ''anti-quarks" Quarks and Mesons • Mesons are generally lighter than baryons, suggesting they contain fewer quarks • Also, the presence of anti-particles in the meson nonets suggests they might be composed of equal numbers of quarks and anti-quarks (so all possible combinations would yield both particles and anti-particles) • Further, if we assume quarks are fermions, the integer spins of mesons suggest quark-antiquark pairs We can add quarks and anti-quarks quantum numbers together graphically by appropriately shifting the coordinates of one ''triangle" with respect to the other: Y d s u s 1 d d u u d u u d I -1 s s 1 3 s u -1 s d Building Baryons Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building 3-quark states Start with 2: The Decuplet Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building ddd ddu duu uuu 3-quark states dds uus Now add a 3rd: uds dss uss The baryon decuplet !! and the Ω- sealed the Nobel prize ⇒ sss Building Baryons Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building 3-quark states Start with 2: The Decuplet Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building ddd ddu duu uuu 3-quark states dds uus Now add a 3rd: uds dss uss The baryon decuplet !! and the Ω- sealed the Nobel prize ⇒ sss Coping with the Octet Y But what about the octet? n 1 p (940) (938) It must have something to do with spin... (in the decuplet they’re all parallel, 0 (1193) Σ- Σ Σ+ I here one quark points the other way) (1116) 3 (1197) Λ (1189) We can ''chop off the corners" by J=1/2 0 artificially demanding that 3 identical Ξ- -1 Ξ quarks must point in the same direction (1321) (1315) But why 2 states in the middle? ddd ddu duu uuu ways of getting spin 1/2: dds uus ↑ ↓ ↑ ↓ ↑ ↑ ↑ ↑ ↓ uds u d s u d s u d s 0 Σ dss uss these ''look" pretty much the same as far as the strong J=3/2 force is concerned (Isospin) Λ sss Coping with the Octet Y Charge: n 1 p (940) (938) d + d + u = 0 0 (1193) - Σ Σ+ I Σ (1116) 3 u = -2d (1197) Λ (1189) J=1/2 0 d + u + u = +1 Ξ- -1 Ξ (1321) (1315) d + 2(-2d) = +1 ddd ddu duu uuu -3d = +1 dds uus d = -1/3 & u= +2/3 uds dss uss u + d + s = 0 J=3/2 s = -1/3 sss Quark Questions So having 2 states in the centre isn’t strange... Y but why there aren’t more states elsewhere ?! n 1 p (940) (938) i.e. why not ↑ ↓ ↑ and ↑ ↑ ↓ ??? 0 u u s u u s (1193) + - Σ Σ I Σ (1116) 3 (1197) Λ (1189) We can patch this up again by altering J=1/2 the previous artificial criterion to: - 0 Ξ -1 Ξ The lowest energy state (1321) (1315) ''Any pair of similar quarks must is be in identical spin states" ddd ddu duu uuu Not so crazy ⇒ lowest energy states of simple, 2-particle systems tend to be dds uus ( ''s-wave" (symmetric under exchange) ) uds What happened to the Pauli Exclusion Principle ??? dss uss Why are there no groupings suggesting J=3/2 qq, qqq, qqqq, etc. ?? sss What holds these things together anyway ?? Colour Pauli Exclusion Principle ⇒ there must be another quantum number which further distinguishes the quarks (perhaps a sort of ''charge") Perhaps, like charge, it also helps hold things together ! We see states containing up to 3 similar quarks ⇒ this ''charge" needs to have at least 3 values (unlike normal charge!) Call this new charge ''colour," and label the possible values as Red, Green and Blue We need a new mediating boson to carry the force between colours (like the photon mediated the EM force between charges) call these ''gluons" Gluons In p-n scattering, u and d quarks appear to swap places.
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