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.

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M =4G

22ρJμ

NC JNCμ

JμNC (q) = u γ μ

1

2cV − cAγ 5

( )u

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

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σ1 =0 1

1 0

⎛

⎝ ⎜

⎞

⎠ ⎟= τ 1

σ 2 =0 −i

i 0

⎛

⎝ ⎜

⎞

⎠ ⎟= τ 2

σ 3 =1 0

0 −1

⎛

⎝ ⎜

⎞

⎠ ⎟= τ 3

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τ + =1

2(τ 1 + iτ 2) =

0 1

0 0

⎛

⎝ ⎜

⎞

⎠ ⎟

τ − =1

2(τ 1 − iτ 2) =

0 0

1 0

⎛

⎝ ⎜

⎞

⎠ ⎟

We will do the same constructions we did last quarterfor 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:

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χL =ν

e−

⎛

⎝ ⎜

⎞

⎠ ⎟L

;τ + =0 1

0 0

⎛

⎝ ⎜

⎞

⎠ ⎟;τ − =

0 0

1 0

⎛

⎝ ⎜

⎞

⎠ ⎟

Jμ±(x) = χ Lγ μτ ±χ L

Jμ3(x) = χ Lγ μ

1

2τ 3χ L =

1

2ν Lγ μν L −

1

2e Lγ μeL

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

• 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.

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jμem = Jμ

3 +1

2jμ

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 T T3 Q Y

uL 1/2 1/2 2/3 1/3

dL 1/2 -1/2 -1/3 1/3

uR 0 0 2/3 4/3

dR 0 0 -1/3 -2/3

You get to verify the quark quantum numbers in HW3.

Note the difference in Y quantum numbers for left and right handed fermions of the same flavor.

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 .

• The two neutral currents B and W3 can, and do mix to give us the mass eigenstates of photon and Z boson.

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−ig J i( )

μWμ

i − i′ g

2JY

( )μBμBasic EWK interaction:

W3 and B mixing

• The physical photon and Z are obtained as:

• And the neutral interaction as a whole becomes:

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−ig J 3( )

μWμ

3 − i′ g

2JY

( )μBμ =

= −i gsinθW Jμ3 + ′ g cosθW

JμY

2

⎛

⎝ ⎜

⎞

⎠ ⎟Aμ

−i gcosθW Jμ3 − ′ g sinθW

JμY

2

⎛

⎝ ⎜

⎞

⎠ ⎟Z μ

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Aμ = Wμ3 sinθW + Bμ cosθW

Zμ = Wμ3 cosθW − Bμ sinθW

Constraints from EM

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ej em = e Jμ3 +

JμY

2

⎛

⎝ ⎜

⎞

⎠ ⎟= −i gsinθW Jμ

3 + ′ g cosθW

JμY

2

⎛

⎝ ⎜

⎞

⎠ ⎟

⇒ gsinθW = ′ g cosθW = e

⇒ ′ g =sinθW

cosθW

g

We now eliminate g’ and write the weak NC interaction as:

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−ig

cosθW

Jμ3 − sin2 θW jμ

em( )Z

μ ≡ −ig

cosθW

JμNC Z μ

Summary on Neutral Currents

€

jμem = Jμ

3 +1

2jμ

Y

JμNC = Jμ

3 − sin2 θW jμem

€

−ieQf γμ

Vertex Factors:

Charge of fermion

€

−ig

cosθW

γ μ 1

2(cV

f − cAf γ 5)

cV and cA differ according to Quantum numbers of fermions.

This thus re-expresses the “physical” currents for photon and Z in form of the “fundamental” symmetries.

Q, cV,cA

fermion Q cA cV

neutrino 0 1/2 1/2

e,mu -1 -1/2 -1/2 + 2 sin2W ~ -0.03

u,c,t +2/3 1/2 1/2 - 4/3 sin2W ~ 0.19

d,s,b -1/3 -1/2 -1/2 + 2/3 sin2W ~ -0.34

Accordingly, the coupling of the Z is sensitive to sin2W .You will verify this as part of HW3.

Origin of these values

The neutral current weak interaction is given by:

€

−ig

cosθW

ψ f γ μ

1

2(1− γ 5)T 3 − sin2 θWQ

⎛

⎝ ⎜

⎞

⎠ ⎟ψ f Z

μ

Comparing this with:

€

−ig

cosθW

γ μ 1

2(cV

f − cAf γ 5)

Leeds to:

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cVf = Tf

3 − 2Q f sin2 θW

cAf = Tf

3

Effective Currents

• In Chapter 12 of H&M, we discussed effective currents leading to matrix elements of the form:

€

MCC =4G

2J μ Jμ

*T

G

2=

g2

8MW2

€

M NC =4G

22ρJNCμ JNC

μ*T

ρG

2=

g2

8MZ2 cos2 θW

From this we get the relative strength of NC vs CC:

€

=MW

2

MZ2 cos2 θW

EWK Feynman Rules

€

−ieQf γμ

€

−ig

cosθW

γ μ 1

2(cV

f − cAf γ 5)

Photon vertex: Z vertex:

W vertex:

€

−ig

2γ μ 1

2(1− γ 5)

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:

• The Lagrangian in classical mechanics is given by:

L = T - V = Ekinetic - Epotential€

d

dt

∂L

∂ ˙ q

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂q= 0

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:

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L(q, ˙ q , t) → L(φ,∂μφ, xμ )

d

dt

∂L

∂ ˙ q

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂q= 0 →

∂

∂xμ

∂L

∂(∂μφ)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−

∂L

∂φ= 0

Let’s look at examples (1)

• Klein-Gordon Equation:

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∂∂xμ

∂L

∂(∂μφ)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−

∂L

∂φ= 0

L =1

2∂μφ∂ μφ −

1

2m2φ2

∂μ∂μφ + m2φ = 0

Note: This works just as well for the Dirac equation. See H&M.

Let’s look at examples (2)

Maxwell Equation:

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∂μ∂L

∂(∂μ Aν )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−

∂L

∂Aν

= 0

L = −1

4F μν Fμν − j μ Aμ

−∂L

∂Aν

= j μ

∂L

∂(∂μ Aν )=

∂

∂(∂μ Aν )−

1

4∂α Aβ −∂β Aα( ) ∂α Aβ −∂ β Aα

( ) ⎛

⎝ ⎜

⎞

⎠ ⎟=

= −1

2gαα gββ ∂

∂(∂μ Aν )∂α Aβ( )

2− ∂α Aβ∂β Aα( )( ) = −

1

22 ∂ μ Aν −∂ν Aμ( ) =

= −F μν

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∂μF μν = j μ=>

Aside on current conservation

• From this result we can conclude that the EM current is conserved:

• Where I used:

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∂∂μF μν = ∂ν j μ

∂ν ∂μ F μν = ∂ν ∂μ ∂ μ Aν −∂ν Aμ( ) =

= ∂μ∂μ∂ν Aν −∂ν ∂ν ∂μ Aμ = ∂μ∂

μ (∂ν Aν −∂μ Aμ ) = 0

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∂μ∂μ =∂∂

∂ Aν = ∂μ Aμ

Aside on mass term

• If we added a mass term to allow for a massive photon field, we’d get:

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(∂μ∂μ + m2)Aν = jν

This is easily shown from what we have done.Leave it to you as an exercise.