Neutrino interactions with quarks For right now lets assume that the - PowerPoint PPT Presentation

Heidi Schellman Northwestern Neutrino interactions with quarks For right now lets assume that the proton (which like mass eigenstates) and the W boson (which wants weak eigenstates) agree that a d quark is a d quark and an s quark is an s quark. Note that the W boson treats a quark just like an electron, there is no charge suppression. June 2010 HUGS 1

Heidi Schellman Northwestern Charge conservation means that some weak interactions can’t happen. µ µ ν νµ µ W W u d d u ν µ → W + µ − and ν µ → W − µ + so ν µ d L → µ − u L and ν µ u R → µ − d R ν µ u L → µ + d L and ν µ d R → µ + u R June 2010 HUGS 2

Heidi Schellman Northwestern telling d L from u R μ− μ + d u ν μ ν μ J = 1 J = 0 u d μ− μ + u d ν μ ν μ J = 0 J = 1 d u June 2010 HUGS 3

Heidi Schellman Northwestern telling d L from u R μ− μ + d u ν μ ν μ J = 1 J = 0 u d The ν µ d → uµ − scattering cross section is isotropic because a left-handed neutrino interacting with a left-handed d-quark has total spin 0. This will also be true for ν µ d → uµ + which has a right handed anti-neutrino scattering off of a right handed u quark. June 2010 HUGS 4

Heidi Schellman Northwestern μ− μ + u d ν μ ν μ J = 0 J = 1 d u This is not true for the ν µ u → dµ + cross section which has a right-handed anti-neutrino interacting with a left-handed u quark or for νu → dµ − which has a left handed neutrino scattering off of a right handed d quark. In this case forward scattering is allowed but going completely backwards is not. There is a spin suppression factor of ((1 + cos θ ) / 2) 2 = (1 − y ) 2 . y = ( P · q ) = E lab − E ′ = (1 − cos θ cm ) lab P · E lab 2 . June 2010 HUGS 5

Heidi Schellman Northwestern How do you measure your observables Neutrinos are invisible, so instead of measuring k , you measure E ′ k and ν = E had then work backwards to the neutrino energy E k and hope it is coming in at an angle you understand. June 2010 HUGS 6

Heidi Schellman Northwestern June 2010 HUGS 7

Heidi Schellman Northwestern June 2010 HUGS 8

Heidi Schellman Northwestern Where neutrinos and muons come from June 2010 HUGS 9

Heidi Schellman Northwestern Strangeness and Charm Of course, we already know that the weak interactions have different ideas about what an s and d quark are than the mass eigenstates. The weak s ′ quark is a superposition of the d and s quark mass eigenstates. So a weak interaction involving an ” s ” quark can produce either a u or a c quark. An initial state u quark decays weakly to a W + d ′ weak eigenstate. You actually observe a superposition of 3 mass eigenstates - which you can distinguish. | W + d ′ > = V du | W + d > + V su | W + s > + V bu | W + b > June 2010 HUGS 10

Heidi Schellman Northwestern Unitarity check In the limit where E >> m b , the 3 diagrams with different mass eigenstates will add incoherently since they have different final states. |M| 2 ∝ | U ud | 2 + | U us | 2 + | U ub | 2 = 1 if there are no other unknown flavors. Interestingly, this check was off by around 1% until the early 2000’s. June 2010 HUGS 11

Heidi Schellman Northwestern However if s ≃ m 2 c you run into mass suppression effects and some of the diagrams are suppressed. In that case you can revert to the 2 × 2 formalism using only the u, d, s, c quarks where        d ′ cos θ C sin θ C  d     s ′ − sin θ C cos θ C s where the Cabibbo angle θ C is such that V ud ≃ V cs = cos θ C ≃ 0 . 973 and V us = − V dc = sin θ C ≃ 0 . 227 Normally the b and t quark masses are too big to be important in deep inelastic scattering so the Cabibbo formalism is sufficient for us. June 2010 HUGS 12

Heidi Schellman Northwestern One can then write the cross sections for all the neutrino processes as follows d 2 σ G 2 F M 4 s W dxdy ( ν µ e → µ − ν e ) = π ≡ σ 0 ( M 2 W + Q 2 ) 2 d 2 σ | U ud | 2 σ 0 xd ( x ) dxdy ( νd → µ − u ) = d 2 σ | U ud | 2 σ 0 xu ( x )(1 − y ) 2 dxdy ( νu → µ − d ) = d 2 σ dxdy ( νu → µ + d ) | U ud | 2 σ 0 xu ( x )(1 − y ) 2 = d 2 σ dxdy ( νd → µ + u ) | U ud | 2 σ 0 xd ( x ) = d 2 σ | U us | 2 σ 0 xs ( x ) dxdy ( νs → µ − u ) = d 2 σ dxdy ( νs → µ + u ) | U us | 2 σ 0 xs ( x ) = June 2010 HUGS 13

Heidi Schellman Northwestern Cross section scales with s ≡ 2 ME lab June 2010 HUGS 14

Heidi Schellman Northwestern ν + s → µ − + c → µ + + ν µ + X June 2010 HUGS 15

Heidi Schellman Northwestern Charm production is suppressed by the need to have a 1.7 GeV quark in the final state d 2 σ | U cs | 2 σ 0 xs ( x ) ∗ f ( m c ) dxdy ( νs → µ − c ) = d 2 σ dxdy ( νs → µ + c ) | U cs | 2 σ 0 xs ( x ) ∗ f ( m c ) = d 2 σ | U cd | 2 σ 0 xs ( x ) ∗ f ( m c ) dxdy ( νd → µ − c ) = d 2 σ dxdy ( νd → µ + c ) | U cd | 2 σ 0 xs ( x ) ∗ f ( m c ) = f ( m c ) is the possible suppression factor for charm production. June 2010 HUGS 16

Heidi Schellman Northwestern If we take proton and neutron targets and the parton probabilities we get: dσ σ 0 x [ d ( x ) + s ( x ) + | U ud | 2 u ( x )(1 − y ) 2 ] dxdy ( νp → µ − X ) = dσ dxdy ( νp → µ + X ) σ 0 x [ d ( x ) + s ( x ) + | U ud | 2 u ( x )(1 − y ) 2 ] = dσ σ 0 x [ u p ( x ) + s ( x ) + | U ud | 2 d p ( x )(1 − y ) 2 ] dxdy ( νn → µ − X ) = dσ dxdy ( νn → µ + X ) σ 0 x [ u p ( x ) + s ( x ) + | U ud | 2 d p ( x )(1 − y ) 2 ] = If you take σ ν p + σ ν n + σ ν p + σ ν n you get something close to σ 0 x ( q ( x ) + q ( x ))(1 + (1 − y ) 2 ) where q are all of the quarks. If you take σ ν p + σ ν n − ( σ ν p + σ ν n ) you get something close to σ 0 x ( q ( x ) − q ( x ))(1 − (1 − y ) 2 ) . June 2010 HUGS 17

Heidi Schellman Northwestern If you study the y dependence, you can solve for u ( x ) , d ( x ) , s ( x ) , u ( x ) , d ( x ) , s ( x ) . June 2010 HUGS 18

Heidi Schellman Northwestern Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino Anti-Neutrino 2 x=0.015 1.5 1 0.5 2 x=0.045 1.5 1 0.5 2 x=0.125 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1/E d 2 � /dxdy (x 10 -38 cm 2 /GeV) 1.5 1 0.5 2 x=0.175 1.5 1 0.5 x=0.275 1.5 1 0.5 x=0.35 1.5 1 0.5 0.4 x=0.55 0.2 0.3 x=0.65 0.2 0.1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) (E=150 GeV) June 2010 HUGS 19

Heidi Schellman Northwestern Cross sections for neutrino data taken on an iron (30 n and 26 p) target. Note that the anti-neutrino data show a much stronger y dependence than the neutrino data, as you would expect from the quark contents. June 2010 HUGS 20

Heidi Schellman Northwestern x The parton distribution functions xu ( x ) , xd ( x ) , ... derived from these data. June 2010 HUGS 21