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 J NC J NC
2
1
NC
J  (q)  u   cV  c A  5 u
2
 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
1  
1
0
 2  
i
1
 3  
0
1
 1
0
i
  2
0 
0 
  3
1
0 1
1
   (1  i 2 )  

2
0 0
0 0
1
   (1  i 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:
 
0 1
0 0
 L     ;   
;   

e L
0 0
1 0
J  (x)  L      L
1
1
1
J  (x)  L    3  L  L   L  eL   eL
2
2
2
3
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
em
j
1 Y
 J  j 
2
3
• 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

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:
g Y 
igJ  W   i J  B
2
i 
i
• 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
A  W  sin W  B cos W
Z   W 3 cosW  B sin W
• And the neutral interaction as a whole
becomes:
g Y 
3 
3

igJ
W

i
J B

2


Y 

J
 igsin W J 3  gcosW  A 
2 

Y 

J
igcos W J 3  gsin W  Z 
2 

Constraints from EM
ej em
Y 
Y 


J
J
 eJ 3    igsin W J 3  gcos W  
2  
2 

 gsin W  gcos W  e
sin W
 g
g
cosW
We now eliminate g’ and write the weak NC interaction as:

g
g
3
2
em

NC 
i
J

sin

j
Z

i
J


W  
 Z
cosW
cosW
Summary on Neutral Currents
1 Y
j  J  j 
2
NC
3
2
em
J  J  sin W j 
em
3
This thus re-expresses the “physical” currents for
photon and Z in form of the “fundamental” symmetries.

Vertex Factors:
ieQ f 

Charge of fermion
g
 1
f
f 5
i
 (cV  c A  )
cosW 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 sin2W ~ -0.03
u,c,t
+2/3
1/2
1/2 - 4/3 sin2W ~ 0.19
d,s,b
-1/3
-1/2
-1/2 + 2/3 sin2W ~ -0.34
Accordingly, the coupling of the Z is sensitive to sin2W .
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
 f    (1  )T  sin W Q f Z
2

cosW
Comparing this with:
Leeds to:

g
 1
f
f 5
i
 (cV  c A  )
cosW 2
c  T  2Q f sin W
f
V
c T
f
A
3
f
3
f
2
Effective Currents
• In Chapter 12 of H&M, we discussed effective
currents leading to matrix elements of the
form:
4G  *T
J J
2
G
g2

2
8M
2
W
M CC 
4G
2 J NC J NC  *T
2
G
g2


2
2
8M
cos
W
2
Z
M NC 
From this we get the relative strength of NC vs CC:
2
M

  2 W2
M Z cos W
EWK Feynman Rules
Photon vertex:
ieQ f 
Z vertex:
g
 1
i
 (cVf  c Af  5 )
cosW 2


W vertex:
g 1
i
 (1  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 qÝ 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 (x,t), and
thus a Lagrange density, and covariant
derivatives:
L(q, qÝ,t)  L(, , x  )




d L L

L
L
0

0


 


dt qÝ q
x   (  )  
Let’s look at examples (1)
• Klein-Gordon Equation:



L
L


0




x   (   )  
1
1 2 2

L      m 
2
2

2
   m   0
Note: This works just as well for the Dirac equation. See H&M.
Let’s look at examples (2)
Maxwell Equation:
 L  L
 
 ( A ) 
 A  0
   



1 
=>
L   F F  j  A

4
L

 j
A
L
  1
 
  

  A    A  A   A 

 ( A )  ( A )  4
 F
 

j
2
1  

1
 
 
 g g

A


A

A


2

A


A 



 
   
2
 (  A )
2
 F 
Aside on current conservation
• From this result we can conclude that the EM
current is conserved:
  F

  F


  j

     A   A











   A      A   ( A   A )  0
• Where I used:
     

 A    A

Aside on mass term
• If we added a mass term to allow for a
massive photon field, we’d get:


(  m )A  j
2
This is easily shown from what we have done.
Leave it to you as an exercise.
