Physics 222 UCSD/225b UCSB
Lecture 9
Weak Neutral Currents
Chapter 13 in H&M.
Weak Neutral Currents
•  “Observation of neutrino-like interactions
without muon or electron in the gargamelle
neutrino experiment” Phys.Lett.B46:138-140,1973.
•  This established weak neutral currents.
4G
M=
2 ρJ µNC J NCµ
2
1
NC
J µ (q) = u γ µ (cV − c A γ 5 ) u
2
ρ allows for different coupling
from charged current.
cv = cA = 1 for neutrinos, but
not for quarks.
Experimentally: NC has small right handed component.
EWK Currents thus far
•  Charged current is strictly left handed.
•  EM current has left and right handed component.
•  NC has left and right handed component.
=> Try to symmetrize the currents such that we get one
SU(2)L triplet that is strictly left-handed, and a singlet.
€
Reminder on Pauli Matrices
0
σ1 = 
1
0
σ2 = 
i
1
σ3 = 
0
1
 = τ1
0
−i
 = τ2
0
0
 = τ3
−1
 0 1
1
τ + = (τ1 + iτ 2 ) = 

2
 0 0
 0 0
1
τ − = (τ1 − iτ 2 ) = 

2
 1 0
We will do the same
€ constructions we did last quarter
for isospin, using the same formalism. Though, this time,
the symmetry operations are identified with a “multiplet of
weak currents” . The states are leptons and quarks.
Starting with Charged Current
•  Follow what we know from isospin, to form
doublets:
ν 
 0 1
0 0
χ L =  −  ;τ + = 
;τ − = 

e  L
 0 0
1 0
±
µ
J (x) = χ L γ µ τ ± χ L
1
1
1
J (x) = χ L γ µ τ 3 χ L = ν L γ µν L − eL γ µ eL
2
2
2
3
µ
We thus have a triplet of left handed currents W+,W-,W3 .
€
Hypercharge, T3, and Q
•  We next take the EM current, and decompose it such
as to satisfy:
Q = T3 + Y/2
j
em
µ
1 Y
= J + jµ
2
3
µ
•  The symmetry group is thus: SU(2)L x U(1)Y
•  And the generator of Y must commute with the
generators Ti, i=1,2,3
€ of SU(2)L .
•  All members of a weak isospin multiplet thus have
the same eigenvalues for Y.
Resulting Quantum Numbers
Lepton T T3 Q Y
ν
1/2 1/2 0 -1
e -L
1/2 -1/2 -1 -1
e -R
0
0
-1 -2
Quark
uL
dL
uR
dR
T
1/2
1/2
0
0
T3
1/2
-1/2
0
0
Q
2/3
-1/3
2/3
-1/3
Note the difference in Y quantum numbers for
left and right handed fermions of the same flavor.
You get to verify the quark quantum numbers in HW3.
Y
1/3
1/3
4/3
-2/3
Now back to the currents
•  Based on the group theory generators, we
have a triplet of W currents for SU(2)L and a
singlet “B” neutral current for U(1)Y .
Basic EWK interaction:
−ig( J
i µ
)
g′ Y µ
W − i ( J ) Bµ
2
i
µ
•  The two neutral currents B and W3 can, and
do mix to give
€ us the mass eigenstates of
photon and Z boson.
W3 and B mixing
•  The physical photon and Z are obtained as:
3
Aµ = W µ sin θW + Bµ cos θW
Z µ = W µ3 cosθW − Bµ sin θW
•  And the neutral interaction as a whole
becomes:
g′ Y µ
3 µ
3
€
−ig( J
)W
µ
−i
J )
(
2
Bµ =
Y 

J
= −i gsin θW J µ3 + g′ cosθW µ  A µ
2

Y 

J
−i gcos θW J µ3 − g′ sin θW µ  Z µ
2

Constraints from EM
ej em
Y 
Y 


J
J
= e J µ3 + µ  = −i gsin θW J µ3 + g′ cos θW µ 
2
2


⇒ gsin θW = g′ cos θW = e
sin θW
⇒ g′ =
g
cosθW
We now eliminate g’ and write the weak NC interaction as:
€
g
g
3
2
em
µ
NC µ
−i
J
−
sin
θ
j
Z
≡
−i
J
(
µ
W µ )
µ Z
cos θW
cosθW
Summary on Neutral Currents
1 Y
j = J + jµ
2
NC
3
2
em
J µ = J µ − sin θW j µ
em
µ
3
µ
This thus re-expresses the “physical” currents for
photon and Z in form of the “fundamental” symmetries.
Vertex Factors:
€
−ieQ f γ
µ
Charge of fermion
€
g
µ 1
f
f 5
−i
γ (cV − c A γ )
cos θW
2
cV and cA differ according to
Quantum numbers of fermions.
Q, cV,cA
fermion Q
cA
cV
neutrino 0
1/2
1/2
e,mu
-1
-1/2
-1/2 + 2 sin2θW ~ -0.03
u,c,t
+2/3
1/2
1/2 - 4/3 sin2θW ~ 0.19
d,s,b
-1/3
-1/2
-1/2 + 2/3 sin2θW ~ -0.34
Accordingly, the coupling of the Z is sensitive to sin2θW .
You will verify this as part of HW3.
Origin of these values
The neutral current weak interaction is given by:
1

g
5
3
2
−i
ψ f γ µ  (1− γ )T − sin θW Qψ f Z µ
2

cos θW
Comparing this with:
€
Leeds to:
€
g
µ 1
−i
γ (cVf − c Af γ 5 )
cos θW
2
f
V
3
f
f
A
3
f
2
c = T − 2Q f sin θW
c =T
€
Effective Currents
•  In Chapter 12 of H&M, we discussed effective
currents leading to matrix elements of the
form:
4G µ *T
J Jµ
2
G
g2
=
2
8M
2
W
M CC =
4G
2 ρJ NCµ J NC µ *T
2
G
g2
ρ
=
2
2
8M
cos
θW
2
Z
M NC =
From this we get the relative strength of NC vs CC:
2
M
W
€ρ =
M Z2 cos2 θW
EWK Feynman Rules
Photon vertex:
−ieQ f γ
Z vertex:
g
µ 1
−i
γ (cVf − c Af γ 5 )
cos θW
2
µ
€
W vertex:
g µ1
−i
γ (1− γ 5 )
2 2
Chapter 14 Outline
•  Reminder of Lagrangian formalism
–  Lagrange density in field theory
•  Aside on how Feynman rules are derived from
Lagrange density.
•  Reminder of Noether’s theorem
•  Local Phase Symmetry of Lagrange Density leads to
the interaction terms, and thus a massless boson
propagator.
–  Philosophically pleasing …
–  … and require to keep theory renormalizable.
•  Higgs mechanism to give mass to boson propagator.
Reminder of Lagrange Formalism
•  In classical mechanics the particle equations
of motion can be obtained from the Lagrange
equation:
 
d ∂L ∂L
=0
 −
dt  ∂q˙  ∂q
•  The Lagrangian in classical mechanics is
given by:
L = T - V = Ekinetic - Epotential
€
Lagrangian in Field Theory
•  We go from the generalized discrete
coordinates qi(t) to continuous fields φ(x,t),
and thus a Lagrange density, and covariant
derivatives:
L(q, q˙,t) → L(φ,∂µ φ, x µ )




d ∂L ∂L
∂
∂L
∂L
=0→
=0

 −
 −
dt  ∂q˙  ∂q
∂x µ  ∂ (∂µ φ )  ∂φ
Let’s look at examples (1)
•  Klein-Gordon Equation:


∂
∂L
∂L
=0

 −
∂x µ  ∂ (∂ µ φ )  ∂φ
1
1 2 2
µ
L = ∂µ φ∂ φ − m φ
2
2
µ
2
∂µ∂ φ + m φ = 0
Note: This works just as well for the Dirac equation. See H&M.
Let’s look at examples (2)
Maxwell Equation:
 ∂L  ∂L
∂µ 
=0
 −
 ∂ (∂ µ Aν )  ∂Aν
µν
µ
1 µν
L = − F Fµν − j µ Aµ
=>
µ
4
∂L
−
= jµ
∂Aν
∂L
∂  1
α β
β α 
=
− (∂α Aβ − ∂ β Aα )(∂ A − ∂ A ) =

∂ (∂µ Aν ) ∂ (∂µ Aν )  4
∂ F
€(
=j
2
1 αα ββ
∂
1
µ ν
ν µ
=− g g
∂
A
−
∂
A
∂
A
=
−
2
∂
A
−
∂
A )=
(
(
(
α β)
α β β α)
2
∂ (∂ µ Aν )
2
= −F µν
)
Aside on current conservation
•  From this result we can conclude that the EM
current is conserved:
∂ν ∂ µ F
µν
µ
= ∂ν j
∂ν ∂ µ F
µν
= ∂ν ∂ µ (∂ A − ∂ A
µ
ν
ν
µ
)=
= ∂µ∂ µ∂ν Aν − ∂ν ∂ ν ∂ µ A µ = ∂µ∂ µ (∂ν Aν − ∂µ A µ ) = 0
µ
•  Where I used:
∂µ∂ = ∂ν ∂
ν
ν
∂ν A = ∂ µ A
µ
Aside on mass term
•  If we added a mass term to allow for a
massive photon field, we’d get:
µ
2
ν
(∂ µ∂ + m )A = j
This is easily shown from what we have done.
Leave it to you as an exercise.
ν