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Perturbation Theory Anomaly
Zhiguang Xiao
June 14, 2019
Contents
1 Two dimensional axial anomaly
2 Four Dimensional Axial Anomaly
3 Gauge anomaly
4 Appendix: An integral
Pertuabation theory of anomaly• Classically, the global symmetry ⇒ conserved currents ∂µJµ = 0 and
conserved charge Q =∫
dx3J0, Q = 0.• Quantum mechanically, the classical current may not be conserved —
Quantum anomaly.We first look at a two dimensional massless QED as an example:
L = ψ(iD/)ψ − 14 (Fµν)
2, Dµ = ∂µ + ieAµ, µ = 0, 1.
[ψ] = 1/2, [A] = 0, [e] = 1.Dirac spinor, two components: ψ =
(ψ+
ψ−
), Dirac matrix: 2× 2 matrices,
γµ, γν = 2gµν .
γ0 =
(0 −ii 0
), γ1 =
(0 ii 0
), γ5 = γ0γ1 =
(1 00 −1
)Global symmetry:
ψ → eiαψ, vector current jµ = ψγµψ.
ψ → eiαγ5ψ, axial vector current jµ5 = ψγµγ5ψ.
Two dimensional QED• Fermionic part in another way:
L ∼ψ†(γ0γ0D0 + γ0γ1D1)ψ
∼ψ†+i(D0 + D1)ψ+ + ψ†−i(D0 −D1)ψ−
• Define x± = x0 ± x1, ∂± = ∂∂x± . Free EOM, no Aµ ,
i(∂0 ± ∂1)ψ± = i∂±ψ± = 0, ψ± = ψ(x∓) .
ψ+: right-moving fermion. ψ−: left-moving fermion. ψ− and ψ+ do notmix in the Lagrangian.• The left and right fermion currents seems to conserve separately,
jµ± = ψγµ(1± γ5
2 )ψ, ∂µjµ± = 0
• The vector and axial vector currents are not independent:γµγ5 = −ϵµνγν , ϵ01 = 1, jµ5 = −jνϵµν .
Two dimensional QED: Vacuum Polarization Diagram
In the m→ 0 limit
iΠµν2 = −i e2
(4π)d/2 tr[1]∫ 1
0dx(−gµν(1 −
d2)Γ(1 − d
2 )
∆1−d/2
−(−q2gµν + 2qµqν)x(1 − x)Γ(2 − d
2 )
(−x(1 − x)q2)2−d/2
)
= −i(q2gµν − qµqν) 2e2
(4π)d/2 tr[1]∫ 1
0dx x(1 − x)
Γ(2 − d2 )
(−x(1 − x)q2)2−d/2
d→2−−−→ −i(q2gµν − qµqν)2e2
4π· 2 ·
−1q2 , Finite
= i(
gµν −qµqν
q2
) e2
π≡ i(
gµνq2 − qµqν)Π(q2) , Π(q2) =
1q2
e2
π.
Dimension for spinor rep. tr[1] = 4 for d=4, and tr[1] = 2 for d=2.
Two dimensional QED: Vacuum Polarization Diagram
• The denominator q2(1−Π) = q2 − e2
π. [e] = 1.
• There is a pole at q2 = e2
π, photon developes a mass.
Two dimensional QED: Vacuum Polarization Diagram
In the background Aµ field, expectation value for thevector current at one-loop:
⟨jµ(q)⟩ =∫
d2xeiq·x⟨jµ(x)⟩ one loop−−−−−→ −ie∫
d2x d2y eiq·xψγµψ(x)ψγνψ(y)Aν(y)
=1ie (iΠ
µν(q))Aν(q) = −(gµν −qµqνq2 ) · e
πAν(q) .
conserved: qµ⟨jµ(q)⟩ = 0.The axial vector current:
⟨jµ5(q)⟩ = −ϵµν⟨jν(q)⟩
=eπϵµν
(Aν(q)−
qνqλq2 Aλ
)qµ⟨jµ5(q)⟩ = e
πϵµνqµAν(q)
x space−−−−−→ ∂µjµ5 =e
2π ϵµνFµν
So the axial vector is not conserved.
Two dimensional QED: Vacuum Polarization Diagram
The problem comes from the regularization of the vacuum polarizationdiagram.• B is finite. Can unambiguously determined by the low energy structure of
the theory— residue of the q2 pole.• A is logrithmically divergent: Dimensional regularization automatically
subtract the divergent to have A = B. — Ward Id. satisfied.• Other regularization to have A = 0, then qµ⟨jµ5⟩ = 0, but
qµ⟨jµ⟩ = eπ
qνAν(q) — not gauge invariant.• It is impossible to regularize 2-d QED — both gauge inv. and axial current
conserved.
Two dimensional QED: Axial Vector Current Operator Eq.Another point of view:• Classical EOM, ∂/ψ = −ieA/ψ , ∂µψγ
µ = ieψA/ .Classically
∂µjµ = ∂µ(ψγµψ) = ieψA/ψ − ieψA/ψ = 0
∂µjµ5 = ∂µ(ψγµγ5ψ) = ieψA/γ5ψ − ieψA/γ5ψ = 0
• Operator products at one point in QFT is always singular, needsregulariztion.Operator product expansion: O1(x)O2(0)→
∑n Cn
12(x)On(0)
G12 = ⟨O1(x)O2(0)ϕ(y1) . . . ϕ(ym)⟩ =∑
nCn
12(x)⟨On(0)ϕ(y1) . . . ϕ(ym)⟩
In the limit x→ 0, the coefficients may be divergent. eg. Free masslessfermions
T(ψ(x + ϵ)ψ(x)) =⟨0|T(ψ(x + ϵ)ψ(x))|0⟩+ : ψ(x + ϵ)ψ(x) :
=
∫ ddk(2π)d
ie−ik·ϵ
k/+∑n=0
1n!
: ((ϵ · ∂)nψ(x))ψ(x) :
in d = 2,∫ ddk(2π)d
ie−ik·(y−z)k/k2 =− ∂/y
∫ ddk(2π)d
e−ik·(y−z)
k2 = −i
4π∂/y log(y − z)2
=−i
2πγµ(y − z)µ(y − z)2
Two dimensional QED: Axial Vector Current Operator Eq.d=2
T(ψ(x + ϵ)Γψ(x)) =− i2π
tr(γµϵµΓ)(ϵ)2 + regular terms
Regularize ψγµγ5ψ by separating the 2 ψ operators by ϵ. And insert a wilsonline to make it gauge invariant. Then take symmetric ϵ→ 0 limit.
jµ5 = symm limϵ→0
ψ(x +
ϵ
2 )γµγ5 exp
[− ie
∫ x+ ϵ2
x− ϵ2
dz ·A(z)]ψ(x− ϵ
2 )
Symmetric limit ϵ→ 0: symm limϵ→0
ϵµ
ϵ2
= 0, symm lim
ϵ→0
ϵµϵν
ϵ2
= 1
d gµν .
Two dimensional QED: Axial Vector Current Operator Eq.Keep to ϵ order:
∂µjµ5 =symm limϵ→0
(∂µψ)(x +
ϵ
2)γµγ5 exp
[− ie
∫ x+ ϵ2
x− ϵ2
dz · A(z)]ψ(x −
ϵ
2)
+ ψ(x +ϵ
2)γµγ5 exp
[− ie
∫ x+ ϵ2
x− ϵ2
dz · A(z)]∂µψ(x −
ϵ
2)
+ ψ(x +ϵ
2)γµγ5[−ieϵν∂µAν(x) exp
[− ie
∫ ϵ2
− ϵ2
dz · A(x + z)]]ψ(x −
ϵ
2)
=symm lim
ϵ→0
ψ(x +
ϵ
2)ieA/(x +
ϵ
2)γ5ψ(x −
ϵ
2)
−ψ(x +ϵ
2)γ5(−ieA/(x −
ϵ
2))ψ(x −
ϵ
2) + ψ(x +
ϵ
2)γµγ5[−ieϵν∂µAν(x)]ψ(x −
ϵ
2)
=symm limϵ→0
ieψ(x +
ϵ
2)ϵµ∂µA/(x)γ5ψ(x −
ϵ
2)
−ieψ(x +ϵ
2)γµγ5ϵν∂µAν(x)ψ(x −
ϵ
2)
=symm limϵ→0
ieψ(x +
ϵ
2)γνγ5ϵµ(∂µAν(x)− ∂νAµ(x))ψ(x −
ϵ
2)
OPE=====symm lim
ϵ→0
e2π
tr(γαϵαγνγ5)
(ϵ)2 ϵµFµν = symm limϵ→0
e2π
2εανϵα(ϵ)2 ϵµFµν
=e
2πεµνFµν
Anomaly in two dimensional QED: Non-conservation ofFermion numbers
For conserved currents: jµ, jµ5.• 0 =
∫dx1∂µjµ =
∫dx1∂0(j0
L + j0R) + j1|∞−∞ = NL + NR,
∆NL +∆NR =∫
dtNL + NR = 0Conservation of total left and right-moving charges.• 0 =
∫dx1∂µjµ5 =
∫dx1∂0(j0
R − j0L) = NR − NL,
∆NR −∆NL =∫
dtNR − NL = 0Conservation of the difference of the left and right-moving charges.• The left and right-moving charges are conserved separately,
∆NL = ∆NR = 0.Anomalous nonconservation of the axial current:
∂µjµ5 =e
2π εµνFµν =
e4π∂µ(ε
µνAν)
∆(NR −NL) =
∫dx0dx1∂µjµ5 =
e4π
∮∂
dxµAµ
• If Aµ fall off sufficiently rapidly at the infinity, the integral vanishes —charge conservation still preserved.• There can be situations where the integral not vanishing — charge
conservation is violated.
Anomaly in two dimensional QED: Non-conservation ofFermion numbers
Eg.• A background A1(t), varies very slowly. Finite spatial size L, with periodic
boundary condition.• This background can not be removed by gauge transformation with
periodic boundary condition , 1e∂1α ∼ A1(t), α(x + L) = α(x).
• Large gauge transformation: α(x) = 2nπL x, eiα(0) = eiα(L),
A1 → A1 − 2nπeL . A1 ∈ [0, 2π
eL ).• The wilson line is non-trivial, exp−ie
∫ L0 dx1A1(t) = 1, which is gauge
invariant — can not be removed by gauge trans.• The Hamiltonian:
H = −i∫
dx : ψγ1D1ψ :=
∫dx
−i :ψ†+(∂1 − ieA1)ψ+ : +i :ψ†−(∂1 − ieA1)ψ− :
• For constant A1, Eigenvalues and eigenfunctions of the covariant
derivative:
i(∂1 − ieA1)eiknx =− (kn − eA1)eiknx, eikn(x+L) = eiknx
⇒ kn =2πnL , n = −∞, . . . ,∞
Anomaly in two dimensional QED: Non-conservation ofFermion numbers
• Single particle state:
ψ+,n =eiknx, En = +(kn − eA1) ,
ψ−,n =eiknx, En = −(kn − eA1) .
• Vacuum: all the negative energy level are filled.
ψ+ =
∞∑En>0
ane−iEnt+iknx +
∞∑En<0
b†ne−iEnt+iknx
ψ− =
∞∑En>0
cne−iEnt+iknx +
∞∑En<0
d†ne−iEnt+iknx
Different kn’sa†n|Ω⟩, En > 0, kn > eA1: generate one right-moving fermion;b†n|Ω⟩, En < 0, kn < eA1: generate one right-moving hole, antifermion;c†n|Ω⟩, En > 0, kn < eA1: generate one left-moving fermion;d†n|Ω⟩, En < 0, kn > eA1: generate one left-moving hole, antifermion;
Anomaly in two dimensional QED: Non-conservation ofFermion numbers
• Adiabatically change A1, the fermionenergy level shift slowly.
ψ+,n =eiknx, En = +(kn − eA1) ,
ψ−,n =eiknx, En = −(kn − eA1) .
• If adiabatically change ∆A1 = 2πeL ,
for ψ+, En → En−1;for ψ−, En → En+1.occupation number on these level notchanged
• One left-moving fermion appear from the vacuum, and one right-movingfermion disappears from the vacuum.• The wilson line also does not change.
exp−ie∫ L
0 dx1A1(t) → exp−ie∫ L
0 dx1A1(t)− i2π• Anomalous nonconservation eq:
∆(NR −NL) =
∫d2x e
2π ϵµνFµν =
∫dt dx e
π∂0A1 =
eπ
L(−∆A1) = −2 .
Anomaly in Four Dimensional massless QEDMassless Four dimensional QED:
L = iψD/ψ − 14FµνFµν , Dµψ = (∂µ + ieAµ)ψ
γµ =
(0 σµ
σµ 0
), σµ = (1, σi), σµ = (1,−σi), γ5 = iγ0γ1γ2γ3 =
(−1 00 1
)Global symmetry:
ψ → eiαψ, vector current jµ = ψγµψ.
ψ → eiαγ5ψ, axial vector current jµ5 = ψγµγ5ψ.
∂µjµ5 =symm limϵ→0
ieψ(x +
ϵ
2 )γνγ5ϵµFµν(x)ψ(x−
ϵ
2 )
We then need to look at the operator product expansion of ψ(x + ϵ2 )ψ(x−
ϵ2 ),
with nonzero background Aµ, not as a quantum field.Vertex iL ∼ −e
∫dx4iψA/ψ, only fermions are contracted.
Anomaly in Four Dimensional massless QEDLeading order:
ψ(y)ψ(z) =∫ d4k
(2π)4 e−ik·(y−z) ik/k2 = −∂/(
i4π2
1(y − z)2 ) =
i2π2
γα(y − z)α(y − z)4
tr(γνγ5γα) = 0.Next to leading order: in-going state A(p) =
∫d4xeip·xA(x).
⟨ψ(y)ψ(z)⟩ =⟨ψ(y)ψ(z)∫
d4x(−ie)ψ(x)A/(x)ψ(x)⟩
=− ie∫ d4k
(2π)4d4p(2π)4
∫d4xe−ik·(x−z)e−i(k+p)·(y−x)
(ik/+ p/
(k + p)2 A/(x) ik/k2
)=− ie
∫ d4k(2π)4
d4p(2π)4 eik·ze−i(k+p)·y
(ik/+ p/
(k + p)2 A/(p) ik/k2
)⟨ψ(x +
ϵ
2)γµγ5ψ(x −
ϵ
2)⟩
=− ie∫ d4k
(2π)4d4p(2π)4 eik·ϵe−ip·(x− ϵ
2 )tr(−
i(k/+ p/)A/(p)ik/γµγ5
(k + p)2k2
)=− ie
∫ d4kd4p(2π)8 eik·ϵe−ip·x
(−4iϵαβγµ(k + p)αAβkγ
(k + p)2k2
)
=− 4e∫ d4kd4p
(2π)8 eik·ϵe−ip·x(ϵαβγµpαAβkγ(k + p)2k2
)tr(γ5γαγβγγγδ) = −4iϵαβγδ
Anomaly in Four Dimensional massless QEDFor ϵ→ 0, the most divergent part is from large k integral, k + p→ k
⟨ψ(x +ϵ
2)γµγ5ψ(x −
ϵ
2)⟩
=− 4e∫ d4kd4p
(2π)8 eik·ϵe−ip·x(ϵαβγµpαAβkγ(k + p)2k2
)
→− 4eϵαβγµ∫ d4p
(2π)4 e−ip·xpαAβ(p)∫ d4k
(2π)4 eik·ϵ(
kγk4
)=− 4eϵαβγµi∂α
∫ d4p(2π)4 e−ip·xAβ(p)
∫ d4k(2π)4 eik·ϵ
(kγk4
)=eϵαβγµiFαβ(
ϵγ
4π2ϵ2)
where ∫ d4k(2π)4 eik·ϵ
(kγk4
)=− i∂ϵγ
∫ d4kE(2π)4 e−ikE·ϵE
(i
k4E
)
=− i∂ϵγ−i ln(−ϵ2)
16π2 = −ϵγ
8π2ϵ2
∂µjµ5 =symm limϵ→0
ieψ(x +
ϵ
2)γνγ5ϵµFµν(x)ψ(x −
ϵ
2)
=symm limϵ→0
ieϵµFµν(x)(eϵαβγν iFαβ(
ϵγ
4π2ϵ2))
= −e2
16π2 ϵαβµνFαβFµν
This is know as Adler-Bardeen-Jakiw anomaly.
4D-Anomaly in massless QED: perturbation calculation,triangle diagram
Matrix element: of ∂µj5µ with two photon, look first at∫d4xe−iq·x⟨p, k|j5µ(x)|0⟩ = (2π)4δ(4)(p + k− q)ϵ∗ν(p)ϵ∗λ(k)Mµνλ(p, k)
Leading order diagrams:
the first one:
the second: (p, ν)↔ (k, λ).
4D-Anomaly in massless QED: perturbation calculation∂µ the amplitude ∼ iq· the triangle diagrams: usingqµγµγ5 = (l/+ p/− l/+ k/)γ5 = (l/+ p/)γ5 + γ5(l/− k/)The first one:
ie2∫
d4l(2π)4 tr
[((l/+ p/)γ5 + γ5(l/− k/)
) i(l/− k/)(l− k)2 γ
λ il/l2 γ
ν i(l/+ p/)(l + p)2
]=e2
∫d4l
(2π)4 tr[γ5 (l/− k/)
(l− k)2 γλ l/
l2 γν + γ5γλ
l/l2 γ
ν (l/+ p/)(l + p)2
]=e2
∫d4l
(2π)4 tr[γ5 (l/− k/)
(l− k)2 γλ l/
l2 γν − γ5 l/
l2 γν (l/+ p/)(l + p)2 γ
λ]
l−k→l======e2
∫d4l
(2π)4 tr[γ5 l/
l2 γλ l/+ k/(l + k)2 γ
ν − γ5 l/l2 γ
ν (l/+ p/)(l + p)2 γ
λ]
• Anti-symmetric under (p, ν)↔ (k, λ). Cancels with the second diagram.• However, the integral is linearly divergent, the shift of integration variable
may not be valid.• The integral should be regularized first and then the integration variable
can be shifted.• The shift may not be consistent with the regularization.
4D-Anomaly in massless QED: perturbation calculationDimenaional regularization: How to define γ5? ’t Hooft and Veltmanγ5 = iγ0γ1γ2γ3, anti-commutes with γµ, µ = 0, 1, 2, 3, but commutes withothers.
• All external momenta pµ, kµ are physical four components vector, onlyinternal momentum l is in d dimensions: l = l∥ + l⊥.• qµγµγ5 = (l/∥ + p/− l/∥ + k/)γ5 = (l/+ p/)γ5 + γ5(l/− k/)− 2γ5l/⊥• The first two terms cancel. The last term gives
• Feynmann parameterization:l→ l + (kx− py), l⊥ not change.denominator → [l2 + k2x(1− x) + p2y(1− y) + 2xyp · k]3 = (l2 +∆)3,numerator → 2tr(l→ l + (kx− py)) , the 2 from Feynmannparameterization. And the overall integral
∫ 10 dx
∫ x0 dy.
• l/⊥ can only be paired with another l⊥ to give non-zero integral.l/⊥l/⊥ = l2⊥ → d−4
d l2• tr(γ5γαγβγγγδ) = −4iϵαβγδ, ϵαβγδ lαlβ = 0. One lα → 0• Only terms like tr(γ5l/⊥l/⊥k/p/γλγν) are left.
4D-Anomaly in massless QED: perturbation calculationNumerator:
− 4tr(γ5l/⊥(l/⊥ + k/(x− 1)− p/y)γλ(l/⊥ + k/x− p/y)γν(l/⊥ + p/(1− y) + k/x))
→− 4tr(γ5l/⊥l/⊥(γλ
(k/xγνp/(1− y)− p/yγνk/x
)+
(k/(x− 1)γλγνp/(1− y)− p/yγλγνk/x
)−
(k/(x− 1)γλp/yγν + p/yγλk/xγν
))→− 4tr(γ5l/⊥l/⊥k/γλp/γν)
→− 4d− 4d l2tr(γ5k/γλp/γν)
∫ddl
(2π)dl/⊥l/⊥
(l2 −∆)3 =d− 4
di
(4π)d/2d2
Γ(2− d2 )
Γ(3)∆2−d/2 →−i
2(4π)2
The second graph, (p, ν)↔ (k, λ), the same as the first graph.
4D-Anomaly in massless QED: perturbation calculationThe total amplitude:
⟨p, k|∂µjµ5(0)|0⟩ =− e2
2π2 ϵανβλkαpβϵ∗ν(p)ϵ∗λ(k)
=− e2
2π2 ϵανβλ(−ipα)ϵ∗ν(p)(−ikβ)ϵ∗λ(k)
=− e2
16π2 ⟨p, k|ϵανβλFανFβλ(0)|0⟩
The same ABJ anomaly equation as before.
4D-Anomaly in massless QED: Linear divergenceLet’s look at the correlation function: j5µ(q) =
∫d4xeiqxj5µ(x),
jν(p) =∫
d4xe−ipxjν(x).
iMµλν5 (q, k, p)(2π)4δ(q− p− k) = ⟨Ω|j5µ(q)jλ(k)jν(p)|Ω⟩
iMµλν5 =−
∫d4l
(2π)4 tr[γµγ5 i(l/− k/)
(l− k)2 γλ il/
l2 γν i(l/+ p/)(l + p)2 + γµγ5 i(l/− p/)
(l− p)2 γν il/
l2 γλ i(l/+ k/)(l + k)2
]
−qµMµλν5 =−
∫d4l
(2π)4 tr[γ5 (l/− k/)
(l− k)2 γλ l/
l2 γν − γ5 l/
l2 γν (l/+ p/)(l + p)2 γ
λ]+ (p, ν)↔ (k, λ)
=− 4iϵµλρν∫
d4l(2π)4
[ kµlρ(l− k)2l2 +
pµlρl2(l + p)2 −
pµlρ(l− p)2l2 −
kµlρl2(l + k)2
]• ∫ d4l
(2π)4kµlρ
(l−k)2l2 ∼ kµkρ, contracted with ϵµλρν , appears to vanish.
4D-Anomaly in massless QED: Linear divergence
pνMµλν5 =
∫d4l
(2π)4 tr[γµγ5
( (l/− k/)(l− k)2 γ
λ l/l2 −
(l/− k/)(l− k)2 γ
λ (l/+ p/)(l + p)2 )
− l/l2 γ
λ (l/+ k/)(l + k)2 +
(l/− p/)(l− p)2 γ
λ (l/+ k/)(l + k)2
)]=4iϵµρλν
∫d4l
(2π)4
[ (l− k)ρlν(l− k)2l2 −
(l− k)ρ(l + p)ν(l− k)2(l + p)2
− lρ(l + k)νl2(l + k)2 +
(l− p)ρ(l + k)ν(l− p)2(l + k)2
]• Shift l→ l + k in the first term, and l→ l− p + k in the fourth term, then
all terms cancel.• The above integrals are linear divergent, shift integration variable is not
valid.• Consider ∆(a) =
∫∞−∞ dx[f(x + a)− f(x)], where f→ c1 as x→∞ and
f→ c2 as x→ −∞ ; c1, c2 are constants
∆(a) =∫ ∞−∞
dx[af′(x) + a2
2 f′′(x) + . . .] = a[f(∞)− f(−∞)] = a(c1 − c2)
4D-Anomaly in massless QED: Linear divergenceIn four dimension: consider integral
∆α =
∫d4k(2π)4 (F
α[k + a]− Fα[k]) wick rot.======= i
∫d4kE
(2π)4 (Fα[kE + a]− Fα[kE])
For linear divergence: limkE→∞ Fα(kE) = A kαEk4
E. Expand around a = 0
∆α(aµ) =i∫
d4kE
(2π)4
[aµ∂µFα[kE] +
12aµaν∂µ∂νFα[kE] + . . .
]=iaµ
∫d3Sµ(2π)4 Fα[kE]
The higher derivative terms vanish too quickly such that the integral over theinfinite surface → 0 after using Gauss theorem.The infinitesimal surface area of the dSµ is normal to the 3-sphere at kE →∞,d3Sµ = k2kµdΩ3.
∆α(aµ) =iaµ lim|k|→∞
∫dΩ3
(2π)4 Akµkαk2 = iaµ lim
|k|→∞
∫dΩ3
(2π)4 Ak2δµα
4k2
=i
32π2 Aaα
∆α is finite.
4D-Anomaly in massless QED: Linear divergence
pνMµλν5 =4iϵµρλν
∫d4l
(2π)4
[ (l− k)ρlν(l− k)2l2 −
(l− k)ρ(l + p)ν(l− k)2(l + p)2
− lρ(l + k)νl2(l + k)2 +
(l− p)ρ(l + k)ν(l− p)2(l + k)2
]=4iϵµρλν
∫d4l
(2π)4
[ −kρlν(l− k)2l2 −
−(k + p)ρlν(l− k)2(l + p)2
+kρlν
l2(l + k)2 +−(p + k)ρlν
(l− p)2(l + k)2
]The first term and the third term: l→ l + k, aν = −kν , contract with ε, notcontribute, or depend only on k, contracted with ϵ and gives zero.The second term can be obtained from the fourth term by l→ l− k + p,aµ = p− k,
4iϵµρλν(p + k)ρFν(lE) = 4iϵµρλν (p + k)ρlν(l− k)2(l + p)2
k→∞−−−−→ 4iϵµρλν(p + k)ρlνl4
Thus,
pνMµλν5 =
14π2 ϵ
µρλνkρpν = 0
Not as we expected: The vector current is not conserve, but the axial currentseems conserved.
4D-Anomaly in massless QED: Linear divergenceThe problem: the initial internal integral momentum is chosen by hand, can beshift arbitrarily. We can make an arbitrary shift lµ → lµ + b1p + b2k for the firstdiagram and lµ → lµ + b1k + b2p for the second to keep Bose symmetry.
−qµMµλν5 =− 4iϵµλρν
∫d4l
(2π)4
[ kµlρ(l− k)2l2 +
pµlρl2(l + p)2−
pµlρ(l− p)2l2 −
kµlρl2(l + k)2
]lµ→lµ+b1p+b2k============
lµ→lµ+b2p+b1k− 4iϵµλρν i
32π2
((k + p)µ(b1p + b2k)ρ−(p + k)µ(b2p + b1k)ρ
)=
14π2 ϵ
µλρνkµpρ(b1 − b2)
4D-Anomaly in massless QED: Linear divergenceFor vector current: l1 = l + b1p + b2k, l2 = l + b1k + b2p.The first term = -The third term(l→ l + (b2 − b1)(k− p)− k,The second term = -The fourth term(l→ l + (−1 + b2 − b1)(k− p),
pνMµλν5 =4iϵµρλν
∫d4l
(2π)4
[ −kρl1,ν(l1 − k)2l21
− −(k + p)ρl1,ν(l1 − k)2(l1 + p)2
+kρl2,ν
l22(l2 + k)2 +−(p + k)ρl2,ν
(l2 − p)2(l2 + k)2
]→4iϵµρλν i
32π2
[−kρ(b2 − b1)(k− p)ν + (k + p)ρ(−1 + b2 − b1)(k− p)ν
]=− ϵµρλν 1
8π2
[kρpν(b2 − b1)− 2kρpν(−1 + b2 − b1)
]=− 1
8π2 ϵµρλνkρpν(2 + b1 − b2)
4D-Anomaly in massless QED: Linear divergence
−qµMµλν5 =
14π2 ϵ
µλρνkµpρ(b1 − b2)
pνMµλν5 =− 1
8π2 ϵµρλνkρpν(2 + b1 − b2)
• If b1 − b2 = 0, the axial current is conserved but the vector current is not.• If b1 − b2 = −2, the vector current is conserved but the axial current is
not, and this is consistent with what we expected.
with Non-abelian gauge fields: Linear divergenceLet’s look at the correlatoin function for most general current:
j5a,µ(q) =∫
d4xeiqxja,5µ(x) =∫
d4xeiqxψiγµγ5Ta
ijψj,
jb,ν(p) =∫
d4xe−ipxjb,ν(x) =∫
d4xeiqxψiγνTb
ijψj.
Ta is the generator for all the symmetries, not necessarily irreducible, includingglobal and local symmetry, G1 ⊗G2 ⊗ . . . .
iMµλν5,abc(q, k, p)(2π)
4δ(q− p− k) = ⟨Ω|ja,5µ(q)jb,λ(k)jc,ν(p)|Ω⟩
iMµλν5,abc =−
∫d4l
(2π)4 tr[γµγ5Ta i(l/− k/)
(l− k)2 γλTb il/
l2 γνTc i(l/+ p/)
(l + p)2
+ γµγ5Ta i(l/− p/)(l− p)2 γ
νTc il/l2 γ
λTb i(l/+ k/)(l + k)2
]
with Non-abelian gauge fields: Linear divergenceShift arbitrarily: lµ → l1 = lµ + b1p + b2k for the first diagram andlµ → l2 = lµ + b1k + b2p for the second to keep Bose symmetry.
−qµMµλν5,abc =− 4iϵµλρν
∫d4l
(2π)4
[( kµl1,ρ(l1 − k)2l21
+pµl1,ρ
l21(l1 + p)2
)tr(TaTbTc)
−( pµl2,ρ(l2 − p)2l22
+kµl2ρ
l22(l2 + k)2
)tr(TaTcTb)
]=− 4iϵµλρν i
32π2
((k + p)µ(b1p + b2k)ρtr(TaTbTc)
−(p + k)µ(b2p + b1k)ρtr(TaTcTb))
=1
8π2 ϵµλρνkµpρ(b1 − b2)tr(TaTb,Tc)
with Non-abelian gauge fields: Linear divergence
pνMµλν5,abc =4iϵµρλν
∫d4l
(2π)4
[( (l1 − k)ρl1,ν(l1 − k)2l21
− (l1 − k)ρ(l1 + p)ν(l1 − k)2(l1 + p)2
)tr(TaTbTc)
+(− l2,ρ(l2 + k)ν
l22(l2 + k)2 +(l2 − p)ρ(l2 + k)ν(l2 − p)2(l2 + k)2
)tr(TaTcTb)
]=4iϵµρλν
∫d4l
(2π)4
[( −kρl1,ν(l1 − k)2l21
− −(k + p)ρl1,ν(l1 − k)2(l1 + p)2
)tr(TaTbTc)
+( kρl2,ν
l22(l2 + k)2 +−(p + k)ρl2,ν
(l2 − p)2(l2 + k)2
)tr(TaTcTb)
]→4iϵµρλν i
32π2
[(− kρ(b1p + b2k)ν + (k + p)ρ(p + b1p + b2k)ν
)tr(TaTbTc)
+(kρ(b1k + b2p)ν − (k + p)ρ(−p + b1k + b2p)ν
)tr(TaTcTb)
]=− 1
8π2 ϵµρλνkρpν
[(1− b2)tr(TaTbTc) + (1 + b1)tr(TaTcTb)
]
with Non-abelian gauge fields: Linear divergence
−qµMµλν5,abc =
18π2 ϵ
µλρνkµpρ(b1 − b2)tr(TaTb,Tc)
pνMµλν5,abc =− 1
8π2 ϵµρλνkρpν
[(1− b2)tr(TaTbTc) + (1 + b1)tr(TaTcTb)
]• Vector current conserved:b1 = −b2 = −1,
−qµMµλν5,abc = − 1
4π2 ϵµλρνkµpρtr(TaTb,Tc)
• Axial current conserved: b1 = b2,
pνMµλν5 = − 1
8π2 ϵµρλνkρpνtr(TaTb,Tc)
• Symmetric: Chiral current γ5 → 1−γ5
2 , q→ −q, q + k + p = 0,qµMµλν
L,abc(q, k, p) = − 12 qµMµλν
5,abc(q, k, p)
qµMµλνL,abc(q, k, p) = kµMνµλ
L,abc(q, k, p) = pµMλνµL,abc(q, k, p)
Condition: −(1− b2) = b1 − b2, −(1 + b1) = b1 − b2, ⇒ b1 = −b2 = − 13
qµMµλν5,abc =
124π2 ϵ
µλρνkµpρtr(TaTb,Tc)
4D-Anomaly in massless QED:Path integralThe Axial vector Ward Id. Consider the functional integral: More general, nfermions, ψn, with continuous non- Abelian or abelian gauge symmetry, e.g.
Z =
∫ ∏n[Dψn][Dψn] exp
[i∫
d4xψi(iD/)ijψj],Dµ = ∂µ − igAa
µta
• The classical conserved current of the global axial transformation.Change variable :(Aµ not change), t: Can be any global axial symmetrygenerator, [t, ta] = 0.
ψ(x) → ψ′(x) = eiα(x)γ5tψ = (1 + iα(x)tγ5)ψ(x),
ψ(x) → ψ′(x) = ψeiα(x)tγ5= ψ(x)(1 + iα(x)tγ5) ,
The global transformation is a symmetry,∫d4xL[ψ′,A] =
∫d4x[L[ψ,A]− ∂µα(x)jµ5]
=
∫d4x[L[ψ,A] + α(x)∂µjµ5] .
for ψ satisfies EOM, since α(x) arbitrary, we have ∂µjµ5 = 0.• Ward Id. From 0 = Zψ′ − Zψ
0 =δZ ?==
∫[Dψ][Dψ]
∫d4xα(x)∂µjµ5eiS =
∫d4xα(x)⟨Ω|∂µjµ5|Ω⟩
4D-Anomaly in massless QED:Path integralWe have assumed that the measure is invariant under the symmetrytransformation. For a general transformation:
ψ(x) →U(x)ψ(x) =∫
yU(x)δ(x − y)ψ(y) ≡
∫yU(x, y)ψ(y),
ψ(x) →ψ(x)γ0U†(x)γ0 =
∫yψ(y)δ(x − y)γ0U†(y)γ0 ≡
∫yψ(y)U(x, y).
U(x, y) = U(x)δ(x − y) , U(x, y) = δ(x − y)γ0U†(y)γ0
[Dψ][Dψ] → [Dψ][Dψ] det[U(x, y)]−1 det[U(x, y)]−1
• Vector like symmetry: U(1) symmetry, U = eiα(x)t,γ0U†γ0 = U† = e−iα(x)t.
[U(x, y)]−1 = U(x, y),detU(x, y) det U(x, y) = 1
4D-Anomaly in massless QED:Path integral• Axial vector like symmetry: U(x) = eiα(x)tγ5 ,γ0U†γ0 = eiα(x)tγ5 .
[dψ(x)][dψ(x)]→[dψ(x)][dψ(x)] det[eiα(x)γ5tδ(x− y)]−2
=[dψ(x)][dψ(x)] exp(−2iTr(γ5α(x)tδ(x− y))
=[dψ(x)][dψ(x)] exp∫
d4x iα(x)A(x),
A(x) =− 2tr(γ5t)δ(x− x)
detM = expTr lnM is used. Tr is over both lorentz spinor indices,flavor indices and x. tr does not include the trace over x.• Ward id.
Z[A] =
∫ ∏n[Dψn][Dψn] exp
[i∫
d4xψi(iD/)ijψj + α(∂µJµ +A)]
0 = δZ[A] =
∫[Dψ][Dψ]
∫d4xα(x)(∂µjµ5 +A)eiS =
∫d4xα(x)⟨Ω|∂µjµ5 +A|Ω⟩
• Spinor space trace: trγ5 = 0. Seems to be invariant.However,
tr(α(x)δ(x− y)) =∫
d4xα(x)δ(x− x)→∞.
We need regularization first.
4D-Anomaly in massless QED:Path integralRegularize the δ(x− x) in a gauge invariant manner:
AM(x) = −2[tr(γ5tf(D/2
x/M2))δ4(x− y)]
y→x
f(0) = 1, f(∞) = 0, sf ′(s)→ 0 at s→ 0 and s→∞.
At M→∞, AM → A(x), f(s) a smooth function, droping from 1 to 0 for sfrom 0→∞, e.g. e−s or 1
s+1 .Fourier transform δ4(x− y) to momentum space:
A(x) =− 2∫
d4k(2π)4
[tr(γ5tf(D/2
x/M2))e−ik·(x−y)]y=x
=− 2∫
d4k(2π)4
[tr(γ5tf([−ik/+ D/x]
2/M2))]
rescale k→Mk==========− 2M4
∫d4k(2π)4
[tr(γ5tf([−ik/+ D/x/M]2))
]=− 2M4
∫d4k(2π)4
[tr(γ5tf
(−k2 − 2k ·Dx
M +(D/x
M
)2))]
only 4γterm=========
1/M4 term− 2M4
∫d4k(2π)4
[tr(γ5t1
2 f′′(−k2)(D/x
M
)4)]
Higher terms vanish as M→∞. We have usedD/(e−ik·x . . . ) = e−ik·x(−ik/+ D/)(. . . ).
4D-Anomaly in massless QED:Path integralWick rotation, k→ iκ, k2 → −κ2, this is equivalent to working in Euclideanspace from the begining∫
d4kf′′(−k2) =i∫
d4κf′′(κ2) = i∫ ∞
0dκ2π2κ3f′′(κ2)
=iπ2∫ ∞
0ds sf′′(s) = −iπ2
∫ ∞0
ds f′(s) = iπ2 .
The trace: tr(γ5γµγνγργδ) = −4iϵµνρδ
tr(γ5t(D/x)4) =− 4itr(tϵµνρδDµDνDρDδ) = −itr(tϵµνρδ[Dµ,Dν ][Dρ,Dδ])
=ig2tr(tϵµνρδFµνFρδ) = ig2ϵµνρδFbµνFc
ρδtr(ttbtc)
M4 cancelsA =
g2
16π2 ϵµνρδFb
µνFcρδtr(ttbtc)
So the anomalous Ward id. ∂µj5µ = − g2
16π2 ϵµνρδFb
µνFcρδtr(ttbtc) = −A
4D-Anomaly in massless QED:Path integralAnother way to look at the path integral: go to Euclidean space path integral:• x0 = −ix4 , xi = xi; ∂0 = i∂4 = i∂/∂x4, ∂i = ∂i;
Scalar: iS = i∫
d4x(∂µϕ∂µϕ−m2ϕ2) =∫
d4x(−∂µϕ∂µϕ−m2ϕ2) = −SE
• A0 = iA4, Ai = Ai; F0i = ∂0Ai − ∂iA0 = i(∂4Ai − ∂iA4) = iF4i,Fij = Fij.Action: iS ∼ i
∫d4x(− 1
4 FµνFµν) = −∫
d4x( 14 Fµν Fµν)
ϵαβγδFαβFγδ = −iϵαβγδFαβFγδ, ϵ1234 = ϵ0123 = 1.• γ0 = γ4, γi = iγi; γµ = 㵆, γ5 = −γ1γ2γ3γ4; γ0γµγ0 = 㵆,γ4γ4γ4 = γ4†, γ4γiγ4 = −γi†.• Eucl: ψ and ψ are defined to be two independent spinors, not ψ = ψ†γ0 in
Minkowski. We integrate over ψ and ψ not over their complex conjugate.• Under SO(4) trans: ψ′E = e 1
2 Σµνωµν
ψE, ψ′†E = ψ†Ee− 12 Σµνω
µν ,Σµν = 1
4 (γµγν − γν γµ).SO(4) inv.: ψ†EψE. We define ψM = iψE to be transform the same as ψ†E,then ψψ is Lorentz invariant in Eucl..• Action: iS ∼ i
∫d4x ψ(i∂/−m)ψ → −
∫d4x ψ(i∂/+ im)ψ
4D-Anomaly in massless QED:Path integralWe can define the measure more carefully• Covariant derivative in Eucl.:iD/ = i∂/+ gA/ata is hermitian, has real
eigenvalues.
(iD/)ϕm = λmϕm,
∫d4xEϕ
†m(x)ϕn(x) = δmn
Completeness:∑
k ϕk(x)ϕ†k(y) = δ4(x− y)I. I unit operator in spinorspace and flavor space.• If Aµ = 0, free Dirac theory, λ2
m = k2, in Eucl. For large k, and fixed Aµ,these eigenvalues gives asymptotic eigenvalues.• Expand ψ and ψ using the basis of the eigenstates of iD/
ψ(x) =∑
mamϕm(x), ψ(x) =
∑m
amϕ†m(x)
am and am are anti-commuting coefficients.• the functional measure
[Dψ][Dψ] = Πmdamdam,
4D-Anomaly in massless QED:Path integralThe transformation ψ′(x) = eiαγ5tψ(x) = (1 + iα(x)γ5t)ψ(x) ,ψ′(x) = ψ(x)eiαγ5t = ψ(x)(1 + iα(x)γ5t)
a′m =∑
n
∫d4xϕ†m(x) expiα(x)γ5tϕn(x)an = (eC)mnan ,
a′m =∑
n
∫d4xanϕ
†n(x) expiα(x)γ5tϕm(x) = an(eC)nm
Cnm =
∫d4xϕn(x)(iα(x)γ5t)ϕm(x), J = eC = I + C + O(C2)
[Dψ′][Dψ′] =Πmda′mda′m = ΠmdamdamJ−2,
J−2 =det[eC]−2 = exp[−2Tr log[eC]] = exp[−2∑
nCnn].
Note:∫d4xϕn(x)(iα(x)γ5t)2ϕm(x) =
∫d4x
∫d4yϕn(x)(iα(x)γ5t)δ4(x − y)(iα(y)γ5t)ϕm(y)
=∑
l
∫d4xϕn(x)(iα(x)γ5t)ϕl(x)
∫d4yϕl(y)(iα(y)γ5t)ϕm(y)
=(C · C)nm
4D-Anomaly in massless QED:Path integralRegularize: The same as before,
∫d4x iα(x)A(x) = −2iTr(γ5α(x)tδ(x− y))
−2TrC =− 2∑
ni∫
d4xϕ†n(x)α(x)γ5tϕn(x) ≡∫
d4xα(x)A(x)
=− 2 limM→∞
∑n
i∫
d4xϕ†n(x)α(x)γ5tϕn(x)f(λ2
nM2 )
=− 2 limM→∞
∑n
i∫
d4xϕ†n(x)α(x)γ5tf(−D/2
M2 )ϕn(x)
4D-Anomaly in massless QED:Path integralUsing the completeness relation: ∑n ϕn(x)ϕ†n(y) = δ4(x − y)I.
limM→∞
∑n
i∫
d4xϕ†n(x)α(x)γ5tf(−D/2
M2 )ϕn(x)
= limM→∞
i∫
d4xα(x)tr(γ5t(
f(−D/2
M2
)δ4(x − y)
)∣∣∣y→x
)= lim
M→∞i∫
d4x∫ d4k
(2π)4 α(x)tr(γ5t(
f(−D/2
M2
)eik·(x−y)
)∣∣∣y→x
)= lim
M→∞i∫
d4x∫ d4k
(2π)4 α(x)tr(γ5t(
eik·(x−y)f(−(ik/+ D/)2
M2
))∣∣∣y→x
)= lim
M→∞i∫
d4x∫ d4k
(2π)4 α(x)tr(γ5t(
f( k2 − 2ik · D − D/2
M2
)))Expand f
===========keep 4γ terms
limM→∞
i∫
d4x∫ d4k
(2π)4 α(x)tr(γ5t 1
2
(f′′( k2
M2
)(−D/2
M2
)2))k→kM======i
∫d4xα(x)tr(γ5tD/4)
12
∫ d4k(2π)4 f′′(k2)
=i2
∫d4xα(x)(−4ϵαβγδ)(− g2
4)tr(tFαβFγδ)(
116π2 ))
=i2
∫d4xα(x)
(g2
16π2 ϵαβγδtr(tFαβFγδ)
)
4D-Anomaly in massless QED:Path integral
[Dψ′][Dψ′] =ΠmdamdamJ−2
=Πmdamdam exp
−i
∫d4xα(x)
(g2
16π2 ϵαβγδtr(tFαβFγδ)
)Minkowski========Πmdamdam exp
−i
∫id4xα(x)
(g2
16π2 ϵαβγδtr(itFαβFγδ)
)=Πmdamdam exp
i∫
d4xα(x)(
g2
16π2 ϵαβγδtr(tFαβFγδ)
)In Minkowski:
Z =
∫[dψ][dψ] exp
[iS + i
∫d4xα(x)
(∂µjµ5 +
g2
16π2 ϵαβγδtr(tFαβFγδ)
)]Anomalous axial current eq:
∂µjµ5 = − g2
16π2 ϵαβγδtr(tFαβFγδ)
This can be easily generalized to d = 2n dimensions.
4D-Anomaly in massless QED:Path integralWe have seen the trace over x, y, spinor and flavor index (Euclidean space)
limM→∞
Tr(∫
d4xϕ†m(x)γ5tf(−D/2
M2
)ϕn(x)) =
g2
32π2
∫d4xϵαβγδtr(tFαβFγδ)
For any nonzero eigenvalue λk,• iD/ϕk = λkϕk• there is another ϕ−k = γ5ϕk,
iD/ϕ−k = iD/γ5ϕk = −γ5iD/ϕ = −λkϕk
• ∫d4xϕ†−kϕk =
∫d4xϕ†kγ
5ϕk = 0• For U(1) generator t, [t, ta] = 0, we can choose tϕk = tkϕk and then
tϕ−k = tkϕ−k.• For α(x) =const., Ckk ∼
∫d4xϕ†kγ
5f(λ2k/M2)tkϕk = 0 for λk = 0.
• Only for λk = 0, zero modes, there are not such cancellations. We canchoose eigenfunctions both for γ5 and iD/ for zero modes. SinceiD/ϕ0 = iD/γ5ϕ0 = 0, γ5( 1±γ5
2 )ϕ0 = ±( 1±γ5
2 )ϕ0.
iD/φu = 0, γ5φu = φu, u = 1, . . .n+, tφu = t+u φu;
iD/ψv = 0, γ5ψv = −ψv, v = 1, . . .n−, tψv = t−v ψv.∫d4xφ†u1φu2 = δu1u2 ,
∫d4xψ†v1ψv2 = δv1v2 ,
∫d4xψ†vφu = 0.
4D-Anomaly in massless QED:Path integralSince f(0) = 1, the trace
Tr(γ5tf(D/2/M2)δ4(x− y))
= limM→∞
∑n
i∫
d4xϕ†n(x)γ5tf(−D/2
M2 )ϕn(x)
=
∫d4x
[∑u
t+u (φ†u(x)φu(x))−∑
vt−v (ψ†v(x)ψv(x))
]=∑
ut+u −
∑v
t−vfor t=1
=========t+=1,t−=1
n+ − n−
So, we have
n+ − n− =g2
32π2
∫d4xϵµνρδFb
µνFcρδtr(tbtc)
• This is a special case of Atiyah-Singer index theorem.• The left hand is an integer. The right hand side can not be changed
smoothly, and only depends on the topology of the gauge field.• Difference of left-handed and right-handed zero modes of the Covariant
Dirac operator iD/ (Eucl.)— algebraic index;Topological invariant numbers — topology index;Atiyah-singer index theorem relate the algebraic index and the topologicalindex.
Global Symmetry in QCDQCD, SU(3) gauge theory: quarks, uc, dc, cc, sc, tc, bc, have color indicesc = 1, 2, 3. Gauge field: Aa
µ, a = 1, 2, . . . , 8. We discuss only the lightest u andd here.
L =(u d
)(iD/ 00 iD/
)(ud
)−(u d
)(mu 00 md
)(ud
)• If mu = md = 0, we have chiral symmetry
U(2)L ×U(2)R = U(1)L ×UR(1)× SU(2)L × SUR(2) for uL = 1−γ5
2 u,uR = 1+γ5
2 u.
QL ≡(
ud
)L→ UL
(ud
)L, QR ≡
(ud
)R→ UR
(ud
)R,
Conserved currents:
jµL = QLγµQL, jµR = QRγ
µQR,
jµaL = QLγ
µτaQL, jµaR = QRγ
µτaQR,
Combinations give baryon number and isospin vector currents forUL = UR:
jµ = QγµQ, jµa = QγµτaQ,and axial current:
jµ5 = Qγµγ5Q, jµ5a = Qγµγ5τaQ,
Global Symmetry in QCDSSB of the symmetry: quark and anti-quark have strong interaction, thevacuum may contain a condensate of quark-antiquark pairs with zeromomentum and angular momentum.• The v.e.v ⟨0|QQ|0⟩ = ⟨0|QLQR + QRQL|0⟩ = 0.• Inv. under UL = UR, not inv. under UL = UR. The axial transformation
symmetries are spontaneously broken.• There should be 4 goldstones corresponding to the four axial symmetry
generators.• Low energy strong interaction have three light pseudo-scalars, πa, isospin
triplet. Can be regarded as the goldstones, generated by the three isospinvectors ⟨0|jµ5a|πb(p)⟩ = −ipµfπδabe−ip·x, fπ: pion decay constant.• ∂µjµ5a = 0, p2 = 0, this is the Goldstone theorem.
• With quark masses, only conserve the vector symmetry: axial current nolonger conserved. EOM:
iD/Q = mQ , ⇒ ∂/Q = −imQ + igA/aTaQ
−iD∗µQγµ = Qm, ⇒ Q←−∂/ = iQm− igQA/aTa
m =
(mu 00 md
).
• Current conservation:
∂µjµ5a =∂µ(Qγµγ5τaQ)
=(iQm− igQA/bTb)γ5τaQ− (Qγ5τa(−imQ + igA/bTbQ))
=iQm, τaQ
∂µ⟨0|jµ5a(0)|πb(p)⟩ = −p2fπδab = ⟨0|iQm, τaγ5Q|πb(p)⟩.The right hand : ⟨Ω|iQjγ5Qi|πb⟩ = −(τ b)ijM2
tr[m, τaτ b] =12δ
ab(mu + md) ⇒ m2π = (mu + md)
M2
fπ• M estimated to be ∼ 400 MeV, mπ ∼ 140 MeV, mu + md ∼ 10MeV≪ 300 MeV, the effective mass of the quark from the SSB. Isospin is anapproximated symmetry.
Anomaly of Chiral currents• For axial isospin current: in the massless limit, there is an anomaly
∂µjµ5a = − g2
16π2 ϵµνρδFb
µνFcρδtr(τatbtc)
τa is the flavor SU(2) symmetry generator, tc is the color SU(3) matrix.tr(τatbtc) = tr(τa)tr(tbtc)) = 0• Axial U(1) current: for nf = 2 here
∂µjµ5 = − g2nf
32π2 ϵµνρδFc
µνFcρδ
So the Axial isospin singlet current is not conserved. However, the righthand side is a total derivative. There are field configurations that givenontrival space-time integration of the right hand side. Thus, thissymmetry is not the real good symmetry of QCD. So, no goldstone isassociated with this symmetry.
Isospin current AnomalyAxial isospin currents have anomaly associated with the couplings of quarks tothe electromagnetism.
∂µjµ5a = − e2
16π2 ϵµνρδFµνFρδtr(τaQ2), Q =
( 23 00 − 1
3
).
∂µjµ53 = − e2
32π2 ϵµνρδFµνFρδ.
Since jµ53 annihilates/creates a π0, this contributes to π0 → 2γ.• Consider matrix element: ⟨p, k|jµ53(q)|0⟩ = ϵ∗νϵ
∗λMµνλ(p, k).
• Symmetry: interchange (p, ν) and (k, λ), pνMµνλ = kλMµνλ = 0.• The most general: with p2 = k2 = 0
Mµνλ =qµϵνλαβpαkβM1 + (ϵµναβkλ − ϵµλαβpν)kαpβM2
+ [(ϵµναβpλ − ϵµλαβkν)kαpβ − ϵµνλσ(p− k)σp · k]M3
• Using q = p + k, red terms vanish; and using q2 = 2p · kiqµMµνλ =iq2ϵνλαβpαkβM1 − iϵµνλσqµ(p− k)σp · kM3
=iq2ϵνλαβpαkβ(M1 + M3).
Vanishes in the limit q2 → 0.• Compared with axial anomaly: iqµMµνλ = − e2
4π2 ϵνλαβpαkβ . There must
be a 1q2 pole in M1 or M3.
Isospin current Anomaly
The current create a π0 goldstone and then decay to two γ, giving a 1q2 pole.
Amplitude iM(π0 → 2γ) = iAϵ∗νϵλϵνλαβpαkβ . Using⟨0|jµ5a|πb(p)⟩ = −ipµfπδabe−ip·x This contributes to
Mµνλ ∼ (iqµfπ)i
q2 (iAϵ∗νϵ∗λϵνλαβpαkβ)
Contributes to M1 = − 1q2 fπA. Compared with the anomaly equation
iqµMµνλ = − e2
4π2 ϵνλαβpαkβ ., we have A = e2
4π21
fπ .
This amplitude can be used to calculate the decay width of π0 → 2γ
Γ(π0 → 2γ) = 12mπ
18π
12∑pols
|M(π0 → 2γ)|2 = A2 m3π
64π
=α2
64π3m3
f2π
.
Anomalous variation of the effective actionConsider the generating functional:
Z[χ, χ] =∫
[dA][dψ][dψ] expi∫
d4xLgauge[A] + LM[ψ,Dψ, ψ,Dψ] + ψ(x)χ(x) + χ(x)ψ(x)
=
∫[dA] expi(Sgauge + W[χ, χ,A])
eiW[χ,χ,A] =
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ] + ψ(x)χ(x) + χ(x)ψ(x)
• W is just taking A as external fields— invariant under the gaugetransformation. All the gauge fixing procedure and Faddeev-Popovquantization can be done after integration over fermions.• Anomaly of the global symmetry is only caused by fermion integration,
thus only appear in W[χ, χ,A].• If there is no anomaly, W[χ, χ,A] is invariant under the original axial
transformaion symmetry. ψ → ψ′ = eiαγ5tψ, α const
W[χ, χ,A] =
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ] + ψ(x)χ(x) + χ(x)ψ(x)
=
∫[dψ′][dψ′] expi
∫d4xLM[ψ′,Dψ′, ψ′,Dψ′] + ψ(x)eiαγ5tχ(x) + χ(x)eiαγ5tψ(x)
No anomaly========= W[χ′, χ′,A]
Anomalous variation of the effective actionWith anomaly, Slavnov-Taylor i.d.
eiW[χ,χ,A] =
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ] + ψ(x)χ(x) + χ(x)ψ(x)
=
∫[dψ′][dψ′] expi
∫d4xLM[ψ′,Dψ′, ψ′,Dψ′] + ψ(x)eiαγ5tχ(x) + χ(x)eiαγ5tψ(x)
=
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ]
+ ψ(x)(χ(x) + δχ(x)) + (χ(x) + δχ(x))ψ(x) + αA(x)
0 =
∫d4x←−δ W[χ, χ,A]
δχ(x) δχ(x) +∫
d4xδχ(x)δW[χ, χ,A]
δχ(x) + α
∫d4xA
=
∫d4x⟨ψ⟩χiαγ5tχ+
∫d4xχtγ5iα⟨ψ⟩χ + α
∫d4xA
We have defined: Ψ = ⟨ψ⟩χ = δW[χ,χ,A]δχ
, Ψ = ⟨ψ⟩χ =←−δ W[χ,χ,A]
δχ
Define effective action: Γ[Ψ, Ψ,A] = W[χ, χ,A]−∫Ψχ−
∫χΨ.
We then have δΓ[Ψ,Ψ,A]
δΨ(x) = −χ(x),←−δ Γ[Ψ,Ψ,A]δΨ(x) = −χ(x).∫
d4xΨ(x)γ5t iδΓ[Ψ, Ψ,A]
δΨ(x)+
∫d4x i←−δ Γ[Ψ, Ψ,A]
δΨ(x) tγ5Ψ(x) = −∫
d4xA
Anomalous variation of the effective actionFor α(x), χ′(x) = χ(x) + α(x)∆χ,
eiW[χ,χ,A] =
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ] + ψ(x)χ(x) + χ(x)ψ(x)
=
∫[dψ′][dψ′] expi
∫d4xLM[ψ′,Dψ′, ψ′,Dψ′] + ψ(x)eiαγ5tχ(x) + χ(x)eiαγ5tψ(x)
=
∫[dψ][dψ] expi
∫d4xLM[ψ,Dψ, ψ,Dψ] + ψ(x)(χ(x) + α(x)∆χ(x))
+ (χ(x) + α(x)∆χ(x))ψ(x) + α(x)∂µJµ5 + αA(x)
∂µ⟨Jµ5(x)⟩χ,χ,A = −A(x)−←−δ W[χ, χ,A]
δχ(x) ∆χ(x)−∆χ(x)δW[χ, χ,A]
δχ(x)
Gauge anomalyWe have studied the global symmetry anomaly:• The global symmetry is different from the gauge symmetry. [t, ta] = 0.
Fermions are coupled to the gauge theory by vector current.• We regularize the UV divergence respecting the gauge symmetry.• If the Gauge fields couples to the chiral current, like GWS model, there
could be gauge anomaly — We don’t want this to happen. This causes thetheory not well-defined. In reality the gauge anomaly must be cancelled.
Chiral fermionsMassless fermions: Helicity = Chirality for fermions, Helicity= − Chirality forantifermions. Σi = 1
2
(σi 00 σi
)
ψ(x) =∑
s
∫d3p√
(2π)32Ep[u(p, s)e−ip·xb(p, s) + v(p, s)eip·xd†(p, s)] ,
Σ · pu(p, h) = hu(p, h) , Σ · pv(p, h) = −hv(p, h)
h u(p, h) = Σ · p u(p, h) = γ5
2 u(p, h) , h v(p, h) = −Σ · p v(p, h) = −γ5
2 v(p, h).
• Dirac fermion: ψ = ψL + ψR, ψL,R = 1∓γ5
2 ψ
• For Weyl rep.: γµ =
(0 σµ
σµ 0
), γ5 =
(−1 00 1
). ψ =
(ψLψR
).
L = ψi(x)i∂/ψi(x) = ψ†Liiσ · ∂ψLi + ψ†Riiσ · ∂ψRi
• The left- and right-Chirality fermions can be assigned to differentrepresentions.
ψLi → eiαataLiψL, ψR → eiαata
RψR, Aaµ → Aa
µ +1g Dµα
a.
• Transform the right-Chirality fermion fields to left-Chirality fermion fields:
ψ′Li = σ2ψ∗Ri, ψ′†Li = ψTRiσ
2.
Lorentz transform: exp i2Σ
µνϵµν
Σµν = − i4 [γ
µ, γν ] =
((σµν)β
α 00 (σµν)β
)α ,
σµν = − i4 (σ
µσν − σν σµ)σµν = − i
4 (σµσν − σνσµ)
σµν† = σµν
using σ2σµ = σµ∗σ2 and σµ = 1,−σi, σ2σµν∗ = σµνσ2.Right-handed fermions h = 1/2 → Right-handed anti-fermions h = 1/2Left-handed antifermions h = −1/2 → Left-handed fermions h = −1/2.• Lagrangian:∫
d4xψ†Riiσ · ∂ψRi =
∫d4x(ψ′TLiσ
2)iσ · ∂(σ2ψ′∗Li)
=
∫d4x(ψ′TLi )i(σT) · ∂(ψ′∗Li)
int. by part==========
interchage
∫d4xψ′†Liiσ · ∂ψ′Li
∫d4xψ†Riiσ · (∂ − igAata
r )ψRi =
∫d4x(ψ′Li
T)i(σT) · (∂ − igAatar )(ψ
′Li∗)
=
∫d4xψ′Li
†iσ · (∂ + igAa(tar )
T)ψ′Li =
∫d4xψ′Li
†iσ · (∂ − igAa(tar ))ψ
′Li
Chiral Fermions• Dirac fermion ψ: rep r, then ψ → ψL: r; ψ′L(x) belongs to the conjugate
rep r.ψL ⊕ ψ′L : r⊕ r, Real rep.• Dirac mass: Can not be chiral, left right fermion should transform in the
same rep, ψL → eiαatarψL, ψR → eiαata
rψR, ψ′L → eiαatarψ′L
mψiψi = m(ψ†RiψLi + ψ†LiψRi) = −m(ψ′TLiσ2ψLi + ψ†Liσ
2ψ′∗Li)
• Most general Mass term: ∆L = MijψTLiσ
2ψLj + h.c. Mij symmetric. (sinceσ2 is anti-symmetric and ψ are grassman field.)• If gauge invariant, strictly real rep. For intrinsic Chiral theory, no gauge
invariant mass term.
Gauge anomaly• The current: jµa = ψγµ
(1−γ5
2
)ψ is coupled the the gauge bosons.
• There could be axial current anomaly in one-loop: If regularize as previous,
∂µ⟨p, ν, b; k, λ, c|jµa|0⟩ = g2
8π2 ϵανβλpαkβAa,b,c,
Aabc =tr[tatb, tc].
• We have made the transformation of ψR → ψL. tr is the trace over all leftfermions.• Current conservation is violated — Gauge symmetry is broken, Ward id is
not true any more, unphysical state may not be cancelled in the S-matrix,violate unitarity — the theory is not well-defined.• Realistic chiral gauge theory, the gauge symmetry must be anomaly free.Aabc = tr[tatb, tc] = 0
Chiral Gauge anomalyAabc = tr(tatb, tc) gauge invariant.e.g.• SU(2) chiral gauge theory. Fermions are in fundamental rep. ta = σa
2 ,a = 1, 2, 3.
Aabc = tr(tatb, tc) = 12tr(taδbc) = 0
No anomaly.• U(1) chiral gauge theory. U(1) charge Q. We need to look atAabc ∼ tr(Q3).• For real Rep R, equivalent to its conjugate rep R. ta
R ∼ taR = −(ta
R)T by
unitary trans:Aabc = tr[(−ta
R)T(−tb
R)T, (−tc
R)T] = −tr[tb
R, tcRta
R] = −Aabc = 0
Thus, for real representation , the anomaly cancels.Dirac Fermions, real rep.• For SU(n), n > 3, there is a unique symmetric invariant dabc using
fundamental rep tan: ta
n, tbn = 1
nδab + dabctc
n.For each rep r, we can define anomaly coefficient A(r):tr(tatb, tc) = 1
2 A(r)dabc.For r, A(r) = −A(r).• Coupling to gravity: A ∼ tr[ta
R], only for U(1), the anomaly can benon-zero.
Gauge anomaly cancellation in Electro-Weak theoryStandard model: SUc(3)× SU(2)×UY(1) gauge theory.GWS: we work in the basis of the gauge bosons before the symmetry breaking.Triangle Diagrams:Aabc = tr(TaTb,Tc). ta SU(3)c generator. τa: SU(2)generator.
• SU(3)-SU(3)-SU(3) color alone, SU(3)-G-G, SU(3)-SU(3)-G: gaugefields couples to vector current. Left- and right-handed fermions in thesame rep. — NO anomaly.• SU(2)-SU(2)-SU(2) ∼ tr(τaδbc), No anomaly.• SU(3)-SU(2)-SU(2) ∼ tr[ta]tr[τ b, τ c] = 0,
SU(2)-SU(3)-SU(3) ∼ tr[τa] = 0,SU(2)-U(1)-U(1) ∼ tr[τa] = 0.SU(2)-SU(3)-U(1) ∼ tr[τa]tr[ta] = 0.• U(1)-SU(3)-SU(3) ∼ 1
2δab ∑
q Yq
Gauge anomaly cancellation in Electro-Weak theoryleft-handed Q T3 Y right-handed Q T3 Y
QL =
(uLdL
) 23− 1
3
12− 1
2
16
uRdR
23− 1
3
00
23− 1
3
EL =
(νLeL
)0−1
12− 1
2− 1
2(νR)not consider
eR
0−1
00
0−1
• U(1)-SU(3)-SU(3) ∼ 12δ
ab ∑q Yq,
For one generation of quarks uL, dL, u∗R, d∗R:∑q Yq = 2× 1
6 + (− 23 ) +
13 = 0.
• U(1)-SU(2)-SU(2) ∼ 12δ
ab ∑fL YfL,
Only for left-handed fermions: QL, EL,∑
fL YfL = 3× 16 −
12 = 0
• U(1)-U(1)-U(1): Sum over all left- and (right-handed∗) fermions,
tr[Y3] =2× (−12 )
3 + 13 + 3×[2× (
16 )
3 − (23 )
3 + (13 )
3]
=− 14 + 1 + 3× [
14× 27 −
827 +
127 ] = 0
• U(1)-G-G ∼ trY
trY =2× (−12 ) + 1 + 3×
[2× (
16 )− (
23 ) + (
13 )
]= 0
So for one generation of fermions, the anomaly cancels. For eachgeneration, the anomaly cancels separately.
Gauge anomalyWe define: W[Aa] ≡ W[0, 0,Aa] = Γ[0, 0,A], we include g into Aµ,Dµ → ∂µ − iAµ(x).Then, if there is no gauge anomaly, under gauge transformationAaµ → A′aµ = Aa
µ + Dµαa, W[A′] = W[A]
If gauge symmetry has anomaly,∫[dψ][dψ]→
∫[dψ′][dψ′] =
∫[dψ][dψ]ei
∫αa(x)Aa .
eiW[A′] =
∫[dψ][dψ] expiSM[ψ,D′ψ, ψ,D′ψ]
=
∫[dψ′][dψ′] expiSM[ψ′,D′ψ′, ψ′,D′ψ′]
gauge inv.=========
∫[dψ][dψ] expiSM[ψ,Dψ, ψ,Dψ] + i
∫d4xαa(x)Aa
⇒ W[A′] =W[A] +
∫d4xαa(x)Aa
0 =
∫d4x δW[A]
δAaµ(x)
δAaµ(x)−
∫d4xαa(x)Aa =
∫d4x δW[A]
δAaµ(x)
Dµαa(x)−∫
d4xαa(x)Aa
=−∫
d4xDµδW[A]
δAaµ(x)
αa(x)−∫
d4xαa(x)Aa
So Dµ δW[A]δAa
µ(x) = −Aa ⇒ Dµ⟨Jµa ⟩A = (∂µδba + CacbAc
µ)⟨Jµb (x)⟩A = −Aa
Gauge anomaly
∂µδW[A]
δAaµ(x)
+ CadeAdµ(x)
δW[A]
δAeµ(x)
= −Aa
Take two times δδA , then A→ 0
∂µδ2W[A]
δAaµ(x)δAb
ν(y)+ Cabeδ
4(x − y) δW[A]
δAeν(x)
+ CadeAdµ(x)
δ2W[A]
δAeµ(x)δAb
ν(y)= −
δAa
δAbν(x)
(∂µ
δ3W[A]
δAaµ(x)δAb
ν(y)δAbρ(z)
+ Cabeδ4(x − y) δ2W[A]
δAeν(x)δAc
ρ(z)
+ Caceδ4(x − z) δ2W[A]
δAeρ(x)δAb
ν(y)
)∣∣∣A=0
= −δ2Aa
δAbν(y)δAc
ρ(z)
∣∣∣A=0
i∂µ⟨Jµa (x)Jνb (y)Jρc (z)⟩
=− Cabeδ4(x − y)⟨Jνe (x)Jρc (z)⟩ − Caceδ
4(x − z)⟨Jρe (x)Jνb (y)⟩+δ2Aa
δAbν(y)δAc
ρ(z)
∣∣∣A=0
=− (Cabcδ4(x − y)ΠνρM − δ4(x − z)ΠρνM ) +
δ2Aa
δAbν(y)δAc
ρ(z)
∣∣∣A=0
Reference: Weinberg II, arXiv:0802.0634
Appendix:An integralIn Euclidean space, x2 = r2∫
dxd|x|−αeik·x
=
∫(rd−1dr)(sin θ)d−2dθdΩd−2r−αeikr cos θ
=Ωd−2
(( ∫ π/2
0dθ +
∫ π
π/2dθ
)(sin θ)d−2
∫ ∞0
(rd−α−1dr)eikr cos θ)
=Ωd−2
(∫ π/2
0dθ(sin θ)d−2
∫ ∞0
(rd−α−1dr)(eikr cos θ + e−ikr cos θ))
r cos θ→r=======Ωd−2
(∫ π/2
0dθ(sin θ)d−2(cos θ)α−d
)(∫ ∞0
(rd−α−1dr)(eikr + e−ikr))
=Ωd−2
(∫ π/2
0dθ(sin θ)d−2(cos θ)α−d
)(∫ ∞0
(2rd−α−1 cos kr)dr)
Using integral: (Ωd−2 = 2π(d−1)/2
Γ( d−12 )
)
∫ π/2
0dθ sinα θ cosβ θ =
Γ( 1+α2 )Γ( 1+β
2 )
2Γ( 2+α+β2 )
,
∫ ∞0
dr(rα cos(kr)
)= −
Γ(1 + α) sin(απ2 )
|k|α+1
Appendix:An integral
∫dxd|x|−αeik·x
=− 2π(d−1)/2
Γ( d−12 )
Γ( d−12 )Γ( 1+α−d
2 )
2Γ(α2 )2Γ(d− α) sin( (d−α−1)π
2 )
|k|d−α
=−π(d−1)/2Γ( 1+α−d
2 )
Γ(α2 )
2Γ(d− α) sin( (d−α−1)π2 )
|k|d−α
Using the Γ function relation
Γ(1− z)Γ(z) = π
sin(πz) , Γ(z)Γ(z + 1/2) = 21−2z√πΓ(2z)
we have Γ(d− α) = Γ( d−α2 )Γ( d−α+1
2 )
21+α−dπ1/2 , Γ( 1+α−d2 )Γ( 1−α+d
2 ) = πsin(π(1+α−d)/2) ,∫
dxd|x|−αeik·x = −2π(d−1)/2Γ( d−α
2 )
21+α−dπ1/2Γ(α2 )
π sin( (d−α−1)π2 )
sin( (1+α−d)π2 )|k|d−α
=πd/2Γ( d−α
2 )
2α−dΓ(α2 )
1|k|d−α =
πα/2Γ( d−α2 )
Γ(α2 )
( 4π|k|2
)(d−α)/2
(d→ α limit) = πα/2
Γ(α2 )
(Γ(
d− α2 ) + ln
( 4π|k|2
))
Appendix:An integralAnother more effective derivation: |x|−α scales as λ−α,
∫dxd|x|−αeik·x should
scale as |k|α−d. Thus ∫dxd|x|−αeik·x = cd,α|k|α−d
Next, determine cd,α. Using the fourier transform
F [e−x2] =
∫ddx
(2π)d/2 e−x2e−ik·x =
e−k2/4
2d/2 ,
and ∫ddkF [f]kF [g]k =
∫ddxf(x)g(x) .
Thus ∫ddx|x|−αe−x2
=
∫ddkF [|x|−α]kF [e−x2
]k
=1
(2π)d/2
∫ddk cd,α|k|α−d e−k2/4
2d/2
=cd,α
(2π)d/22d/2
∫ddk |k|α−de−k2/4
Appendix:An integralFrom Ωd−1 = 2πd/2
Γ(d/2)∫ddx|x|αe−x2
=
∫rd−1dr dΩd−1 rαe−r2
= Ωd−1
∫ ∞0
dr rα+d−1e−r2
=Ωd−1
∫ ∞0
dr2
2 (r2)(α+d−2)/2e−r2
=2πd/2
Γ(d/2)Γ(α+d
2 )
2
=πd/2Γ(α+d
2 )
Γ(d/2)
Thus
cd,α =πd/2Γ(−α+d
2 )
Γ(d/2)(2π)d/22d/2Γ(d/2)
2απd/2Γ(α/2)
=2d−απd/2Γ( d−α
2 )
Γ(α/2)
