Vacuum Polarization

Vacuum Polarization

e- e-

Thus, we never actually ever see a ''bare" charge, only an effective charge shielded by polarized virtual / 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 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 assume this basic E = 1 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" , 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 - ( ) 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- 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

''" ''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 - Σ (1193) 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 ↑ ↑ ↓ ??? u u s 0 u u s - Σ (1193) 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 mediated the EM force between charges) call these ''" Gluons

In p-n scattering, u and d quarks appear to swap places. But their colours must also swap (via interactions). ⇒ This suggests that an exchange-force is involved...

But then we run into trouble while trying to conserve charge at an interaction vertex:

R G (a quark-screw!)

⇒ the only way out is to attribute colour to the gluons as well. For the above case, the gluon would have to carry away RG quantum numbers Flux Tubes

We could explain only having the quark combinations seen if we only allowed ''colourless" quark states involving either colour-anticolour, all 3 colours (RGB) , or all 3 anticolours.

(hence the analogy with ''colour", since white light can be decomposed into either red, green & blue or their opposites - cyan, magenta & yellow)

If the carriers of the force (the gluons) actually carry colour themselves, the field lines emanating from a single quark will interact:

''flux tube" q q q

* formally still just a hypothesis (calculation is highly non-perturbative) Confinement

For this configuration, the field strength (flux of lines passing through a surface) does not fall off as 1/r2 any more ⇒ it will remain constant. The field energy will thus scale with the length of the string and so as L → ∞ then E → ∞ Clearly we can’t allow this!! Can be stopped by terminating field line on another colour charge

⇒ Ah! So only colourless states have finite energy !

''Confinement" PoP ! q q q q q q q q q q

''fragmentation" Asymptotic Freedom

Getting very close to a quark: RB

q q q q q

RG So, on average, the colour is ''smeared" out q into a sort of ''fuzzy ball"

Thus, the closer you get, the less colour charge you see enclosed within a Gaussian surface. So, on distance scales of ~1 fm, quarks move around each other freely ⇒ ''Asymptotic Freedom" Where Are The Coupling Constants Running ??? 1974 ABBA e- e+ The November Revolution

Come The Revolution... (1974)

SPEAR (Richter et al.) The Psi

e+ π+ - ( ) π Ψʹ3.7 → Ψ3.1

e- The J

Brookhaven (Ting et al.) the J/Psi

J/Ψ CC ''Charmonium" Not entirely unexpected ⇒ had been some hints of needing another quark (Glashow, Bjorken, ...) (“GIM” mechanism)

nicer parallel with the (then known) : e µ u c νe νµ d s There are different energy states of Charmonium, in analogy to atom:

There are different energy states of Charmonium, in analogy to hydrogen atom: Hydrogen Splittings

Hydrogen

Different mechanisms which separate energy levels:

n - basic Bohr energy level ∝ α2

''Fine Structure" - spin-orbit coupling ∝ α4

''Hyperfine Structure" - magnetic moment coupling (very small in hydrogen) ∝ α4 Charmonium &

Charmonium & Positronium Energy Levels

Note in charmonium the various splittings are all of the same order ≃ ⇒ so αS ~ 1 (best fit actually gives αS 0.2) Naked Charm

''Naked" Charm: Naked Bottom Top Production Discovery of the Tau

''Meanwhile, back at the lab, Martin was about to make another unexpected discovery..."

e+ e- → µ+ e- + ''missing energy" number violation??

e+ e- → τ+ + τ- - e + νe + ντ + µ + νµ + ντ

mτ = 1.78 GeV ! Tau Decay

τ decays weakly and we would naively expect the lifetime to be roughly ≃ 5 5 -6 Tτ (mµ/mτ) Tµ = (0.106/1.78) (2.2x10 s) = 1.6x10-12s However The τ is heavy enough to decay into hadrons as well as muons and (where as muon can only decay to electrons)

s d τ- W- - τ- W- - Κ π u u

ντ ντ

“Cabibo-mixing" (only counts as 1) Tau Decay

So there is a relative phase-space factor of roughly:

µ-decay τ-decay electron electron muon quarks 1 versus 1 + 1 + 1

3 quark colours !!

Experiment: ≃ -12 -13 3.0 0.1 x10-13s Tτ (1.6x10 s) (1/5) = 3.2x10 s ± Reality of Quarks: R

But is this real ? fragmentation

e+ µ+ e+ q hadrons Consider: versus

- - - q e µ e hadrons Cross-sections for both processes should be basically the same except for an additional phase-space factor for the number of different quarks and different colour states that can be produced

(e+e- hadrons) (e+e- qq ) R ≡ σ → = σ → σ (e+e- → µ+µ-) σ (e+e- → µ+µ-) For the CM energies we will look at, only the 5 lightest quarks can be produced.

= N [ (q /e)2 + (q /e)2 + (q /e)2 + (q /e)2 + (q /e)2 ] ⇒ R C u d s c b

= NC [ 4/9 + 1/9 + 1/9 + 4/9 + 1/9 ] = (11/9) NC ≃ (in fact, higher order corrections suggest a better estimate of R (11/9) NC (1+αS/π) ) Measurement of R

⇒ NC = 3 !! Pi-Zero Decay

More Evidence for Colour: π0 → γ + γ u γ 0 π u + u γ u γ 0 π u + u γ u γ 0 π u In this case, the amplitudes γ add coherently and calculation u yields: N 2 α2 m 3 C π +... if there are Γ (π0 → 2γ) = = 7.73 (N /3)2 (3 ) 64π3 f 2 C other colours π

(''pion decay constant" fπ = 92.4 MeV from charged pion decay rate)

Experimentally ⇒ Γ = 7.7 ± 0.6 eV NC = 2.99 ± 0.12 Parity Violation

Parity Violation in Weak Interactions First suggested in 1956 by Lee & Yang based on review of kaon decay modes

Directly observed by Wu et al. in 1957 from the decay 60 60 - Co → Ni* + e + νe γ (1.173 MeV) + γ (1.332 MeV)

e- nuclear spins aligned by cooling to 0.01 oK P in a magnetic field 60Co 60Co (degree of polarisation determined from the anisotropy of γ-rays) e-

Should be the same under parity transformation, but fewer electrons are actually seen going forward !

Garwin, Lederman & Weinrich (1957)

νµ

π+

e+ precess ν µ+ µ polarised (polarised) muons νe Neutrino Helicity

Also, in 1958, Goldhaber et al. measured the helicity of the neutrino:

- 152 152 e + Eu(J=0) → Sm*(J=1) + νe

152Sm(J=0) + γ

events were chosen with the final states collinear

⇒ γ and νe travel in opposite directions, so helicity of the neutrino is found from that of the gamma

⇒ all neutrinos are left-handed ! Neutral Kaons

Kaons: Ko = ds Ko = sd (S = +1) (S = -1)

u But S is not conserved d s in weak interactions so Ko W+ W- Ko Ko-Ko mixing can occur: s d u

o o o We can thus define ⎜K1 〉 = 1/√2 ( ⎜K 〉 + ⎜K 〉 ) two orthogonal mixtures: o o - o ⎜K2 〉 = 1/√2 ( ⎜K 〉 ⎜K 〉 )

o o o - o Note: C P ⎜K1 〉 = + ⎜K1 〉 and C P ⎜K2 〉 = ⎜K2 〉

o + - o o o + - o o o o K1 → π π ; π π K1 → π π π ; π π π o + - o o o o o + - o o K2 → π π π ; π π π K2 → π π ; π π Allowed Forbidden CP Violation

Experimentally, 2 kaon states are observed with different lifetimes:

o + - o o ≃ -11 KS → π π ; π π τ 9x10 s

o + - ο o o o ± ± ≃ -8 KL → π π π ; π π π ; π lepton ν (ν) τ 5x10 s

o o o o So we associate KS ⇔ K1 and KL ⇔ K2

However, in 1964, Christenson, Cronin, Fitch & Turlay discovered

o + - KL → π π (branching ratio ~ 2x10-3) CP Experiment

beam lead-glass cuts collimator out

30 GeV protons KS+KL KL

18 m

steel target magnets sweeps out charged particles

CM of π+π- pair

K beam )θ L direction CP Violating Term

o 2 o - o ⎜KS 〉 = 1/√1+ ε ( ⎜K1 〉 ε ⎜K2 〉 )

o 2 o o ⎜KL 〉 = 1/√1+ ε ( ε ⎜K1 〉 + ⎜K2 〉 )

where ε ≡ small complex number parameterizing the size of the CP violation (experimentally, ε ≃ 2.3x10-3 )

What does this mean??

Reason for antimatter assymmetry ??

Perhaps we can learn more from studying CP violation in other particle systems... Sakarov Conditions

Matter-Antimatter Asymmetry Revisited:

Sakarov Conditions (1967)

!!! 1) Baryon Number Violation (GUTs) allows baryons and anti-baryons to appear and disappear independently of each other Establishes 2) CP Violation Asymmetry so the rate of appearance/disappearance of baryons is different from anti-baryons

3) Non-Equilibrium Conditions Locks In since equilibrium would then tend to Asymmetry ''average-out" any asymmetry Weak Coupling & the W Mass

Recall that the ''matrix element" for scattering from a Yukawa potential is

2 2 2 〈 Ψf ⎪V⎪ Ψo〉 = g /(q +M )

- In the Fermi theory of β decay, this is what essentially becomes GF or, more precisely, 2 2 2 2 2 GF/√2 = g /(q +M ) = 4παW/(q +M )

2 σ ∝ GF and the relatively small value of GF characterizes the fact that the weak interaction is so weak

We can get this small value either by making αW small or by making M large

So what if we construct things so αW = α ??? ⇒ UNIFICATION !!

Assuming M ≫ q2 , M = √ 4π √2 α / GF

α = 1/137 -5 -2 GF = 10 GeV ⇒ M ~ 100 GeV u hadrons p u d νe, νµ, ντ - W u e-, µ-, τ- p u hadrons d

Stochastic Cooling

Electron Cooling Weak Coupling & the W Mass

Recall that the ''matrix element" for scattering from a Yukawa potential is

2 2 2 〈 Ψf ⎪V⎪ Ψo〉 = g /(q +M )

- In the Fermi theory of β decay, this is what essentially becomes GF or, more precisely, 2 2 2 2 2 GF/√2 = g /(q +M ) = 4παW/(q +M )

2 σ ∝ GF and the relatively small value of GF characterizes the fact that the weak interaction is so weak

We can get this small value either by making αW small or by making M large

So what if we construct things so αW = α ??? ⇒ UNIFICATION !!

Assuming M ≫ q2 , M = √ 4π √2 α / GF

CERN, 1983 α = 1/137 -5 -2 M = 80 GeV !! GF = 10 GeV ⇒ M ~ 100 GeV W

Electroweak Interlude

A Brief Theoretical Interlude (electroweak theory... at pace!!) Weak Isospin

But how can this be the ''same" force when the W’s are charged and the photon certainly isn’t !? Is there a way we can ''bind up" the W’s along with a neutral exchange particle to form a ''triplet" state (i.e. like the pions) ??

Well, like with the pions, we seem to have a sort of ''Weak" Isospin since the weak force appears to see the following left-handed doublets u c t νe νµ ντ ( e )L ( µ)L ( τ)L ( d)ʹ L ( sʹ)L ( b)ʹ L (3) as essentially two different spin states: IW = ± 1/2 (like p-n symmetry)

+ Thus, in νe e The W must carry + (3) the process W away +1 units of IW

so let’s symbolically denote W+ ≡ ⎪↑〉 and, similarly, W- ≡ ⎪↓〉 ’ o If IW = 1 for the W s then, similar to the π , there is also a neutral state: o W = 1/√2 ( ⎪↑〉 - ⎪↓〉 ) (which completes the triplet) The Higgs

There is, however, another orthogonal state: 1/√2 ( ⎪↑〉 + ⎪↓〉 ) If we ascribe this to the photon, then perhaps we might expect to see weak ''neutral currents" associated with the exchange of a W o with a similar mass to the W ± so we’d have a nice ''single package" which describes EM and weak forces!

Hold on... any simple symmetry is obviously very badly broken ⇒ the photon is massless and the W’s are certainly not! The photon is also blind to weak isospin and also couples to right-handed leptons & quarks as well

Assume the symmetry was initially perfect and all states were massless Then postulate that there exists some overall (non-zero) ''field" which couples to particles and gives them additional virtual loop diagrams : (kind of like an ''aether" which produces a sort of ''drag") but in the limit of zero momentum transfer (rest mass), so represent as ⇒ Higgs Mechanism Mixing: the Photon & Z

Further suppose that this field is blind to weak isospin and, thus, allows for it’s violation.

This would allow the neutral weak isospin states to mix ⇒ like with the mesons (the W± are charged and cannot mix)

We will call the ''pure," unmixed states Wo and Γ And we will call the physical, mixed states Zo and γ and Couplings

Think about mathematically introducing this Higgs coupling by applying some ''mass-squared" operator to the initial states (since mass always enters as the square in the )

G G M2 W± = W± + W W W± G G G G M2 Wο = Wο + W W Wο + W Γ Γ G G G G M2 Γ = Γ + Γ Γ Γ + Γ W Wo

where the right-most terms represent the weak isospin - violating terms

’ Assume couplings to W s are all the same (GW) but coupling to Γ may be different (GΓ )

± 2 2 For the W the mass would then simply be given by MW = GW (where G2 contains the coupling plus a few other factors) For the latter 2 equations, we can think of M2 as an operator which yields the mass-squared, M2 , for the coupled state: 2 o 2 o M W = GW W + GW GΓ Γ 2 2 ο M Γ = GΓ Γ + GW GΓ W Massless Photon / Massive Z

GW G Γ o From the second of these: Γ = 2 2 W (M -GΓ )

2 2 G G 2 o 2 o W Γ o Substituting into the first: M W = G W + W W 2 2 (M -GΓ )

4 - 2 2 2 2 - 2 2 2 2 M M GΓ = M GW GW GΓ + GW GΓ 2 2 - 2 - 2 M ( M GΓ GW ) = 0

2 2 2 2 ⇒ M = 0 or M = GW + GΓ

2 2 2 2 Thus, associate Mγ = 0 and MZ° = GW + GΓ

Note also that MZ° > ΜW Weinberg Angle & Z Mass

We can parameterize the γ as a mixture of Wo and Γ as follows:

- o γ ≡ Γ sinθW W cosθW θW ≡ ''Weinberg Angle"

2 2 2 - o Thus, applying M : M γ = M (Γ sinθW W cosθW) = 0

2 ο - 2 o - 0 = ( GΓ Γ + GW GΓ W ) sinθW (GW W GW GΓ Γ) cosθW

o - 2 Coefficient of W ⇒ GW GΓ sinθW GW cosθW = 0 2 - Coefficient of Γ ⇒ GΓ sinθW GW GΓ cosθW = 0

''unification condition" ''anomaly condition" tan θW = GΓ / GW 2 2 2 2 2 2 Q + 3 Q = 0 MZ° /MW = (GW + GΓ )/GW = 1/cos θW Σ l Σ q (leptons) (quarks) is satisfied separately ⇒ MZ° = MW/cosθW for each generation Neutral Current Event

Neutral Current Event (Gargamelle Bubble Chamber, CERN, 1973)

- νµ π

p Z Discovery

From comparing neutral and charged current rates

2 ⇒ sin θW = 0.226

MW = 80 GeV

⇒ MZ° = 91 GeV (predicted)

Z° → e+ e-

MZ° = 91 GeV (observed!!) Flavour-changing neutral currents While we’re here... u u pre-ABBA weak doublet = = d cosθ + s sinθ ( d´) ( C C)

So, consider the coupling to the Z0 :

(d cos + s sin ) u θC θC

Z0 Z0

+

u (d cosθC + s sinθC)

Probability ∝ product of wave functions:

2 2 uu + (dd cos θC + ss sin θC) + (sd + ds ) sinθC cosθC

ΔS = 0 ΔS = 1 “Flavour-Changing Neutral Currents” ⇒ never seen! GIM mechanism

Postulate 2 doublets: (Glashow, Iliopolis & Maiani: “GIM” mechanism)

u u c c = s cos - d sin = d cos + s sin θC θC ( d´) ( θ C θ C) & ( s´) ( )

(d cos + s sin ) u θC θC

Z0 Z0

+

u (d cosθC + s sinθC)

(s cos - d sin ) c θC θC

Z0 Z0

+ +

- c (s cosθC d sinθC)

2 2 uu + cc + (dd+ss)cos θC + (ss+dd) sin θC) + (sd + ds - sd - sd) sinθC cosθC

ΔS = 0 ΔS = 1 Resonant Cross Section

formation ''rate" (recall Γ = ℏ/τ) Blam ! of initial state prob for decay to Transition dP 0 particular final state Rate W = Γ ( dN ) given the total number of available states dP dE = 0 Γ ( dE ) ( dN ) dP 1 Γ = f

2 2 dE 2π (E-E0) + Γ /4

- 2 - dN 1 V q dq dΩ 1 = 3 But recall that (dE ) ((2 π ) dE )

W = vB σ / V V q2 -1 ( 2 π2 v ) π Γ0 Γf

σ = 2 2 2 q (E-E0) + Γ /4 The Z Resonance

Thus, for the production of Z0 near resonance and the subsequent decay to some final state ''X" :

2 0 + - 0 12π MZ Γ(Ζ → e e ) Γ(Ζ → X) (e+e- X) = σ → E2 (E2 - M 2)2 + M 2 2 CM [ Z Z Γ ] CM Ζ

since Γ(Ζ0 → e+e-) can be related by time-reversal to Γ(e+e- →Ζ0)

Peak of resonance ⇒ MZ

Height of resonance ⇒ product of branching ratios

0 + - 0 - (Ζ → e e ) (Ζ → X) Br(Ζ0 → e+e ) Br(Ζ0 → X) = Γ Γ ΓΖ ΓΖ Z Decay: Generation Limit

Results:

0 MZ = 91.188 ± 0.002 GeV Γ(Ζ → hadrons) = 1.741 ± 0.006 GeV Γ = 2.495 ± 0.003 GeV 0 + - Z Γ(Ζ → l l ) = 0.0838 ± 0.0003 GeV 1.741 + (3 x 0.0838) = 1.9924 ≠ 2.495 !! So what’s left ???

''Invisible modes"

Neutrinos !!

(limit for light, ''active" neutrinos) End To Generation Game

An End To The Generation Game ???

(not necessarily a bad thing!) Pi in the Sky

''π In The Sky" p

Blam !

π0 γ + - π π N γ µ 2 N ∼ + - e µ µ

+ e e-

νµ νe ν ν µ µ νe νµ

SuperK SuperK Muon-Neutrino SuperK Electron-Neutrino SuperK Mu-Nu Disappearance

Neutrinos oscillate! or maybe decay...

⇒ νµ ’s disappear in an energy-dependant way

They have mass!! The Standard Solar Model (p-p chain)

2 + - 2 p+p → H+e +νe p+e +p → H+νe 99.7% 0.23%

2H+p → 3He+γ

84.92% 15.08% ~10-5% 3 3 3 + He+ He → α+2p He+p → α+e +νe

3He+α → 7Be+γ

15.07% 0.01%

7 - 7 7 8 Be+e → Li+γ+νe Be+p → B+γ

7 8 + Li+p → α+α B → 2α+e +νe Chlorine Experiment

Homestake Gold Mine (South Dakota)

37 37 - νe + Cl = Ar + e

(Ray Davis) Chlorine Results

What’s SNU ? 1 SNU = 10-36 captures per second per target nucleus

ν Reactions in SNO

- CC ν e + d ⇒ p + p + e

-ν e only -Gives νe energy spectrum well -Weak direction sensitivity ∝ 1-1/3cos(θ)

NC ν x + d ⇒ p + n +ν x

-Measures total 8B ν flux from Sun -Equal cross section for all ν types

ES - - ν x + e ⇒ ν x + e

-Low Statistics

-Mainly sensitive to νe,, some sensitivity to νµ and ντ -Strong direction sensitivity The Basic Maths:

In terms of the Standard Solar model, we see:

ΦSNO = Φ(νe) = 0.35

If this is due to flavour conversion, SuperK would see: - ΦSK(predicted) = 0.35 + (1 0.35) (1/6.5) = 0.45

electron non-electron relative sensitivity neutrinos neutrinos of ES reaction to neutral current

ΦSK (measurement) = 0.451 !!! SSM

total (unconstrained

Φ CC spectrum)

June 2001 April 2002 Sept 2003 May 2008 (indirect) (direct) (salt - unconstrained) (indep. NCD measurement) SK

ΦES

SNO SNO

ΦES ΦCC

SNO

Φ NCD SNO

ΦNC SNO

Φ Salt ΦSSM

MINOS (θ23)