QFT 3 : Problem Set 6

1.) Peskin & Schroeder 19.3 Computation of Anomaly coefficients

(a.)

A product r1 × r2 of SU(n) representations may be decomposed into its irreducible representation asfollows:

r1 × r2 =∑i

ri (1)

The representation matrices in the representation r1 × r2 are:

tar1×r2 = tar1 × θ1 + θ2 × tar2 (2)

Using this notation we may write the product of generators as:

tbr1×r2 tcr1×r2 = (tbr1 t

cr1 × θ2) + (θ1 × tbr2 t

cr2) + (tbr1 × t

cr2) + (tcr1 × t

br2) (3)

Using the above then it is easy to write the anti-commutator of two generators. In general we obtain thefollowing results:

{tbr1×r2 , tcr1×r2} = ({tbr1 , t

cr1} × θ2) + (θ1 × {tbr2 , t

cr2}) + 2 (tbr1 × t

cr2) + 2 (tcr1 × t

br2) (4)

tar1×r2{tbr1×r2 , t

cr1×r2} = (tar1{t

br1 , t

cr1} × θ2) + (θ1 × tar2{t

br2 , t

cr2}) + 2 (tar1 t

br1 × t

cr2) (5)

+2 (tar1 tcr1 × t

br2) + 2 (tbr1 × t

ar2 t

cr2) + 2 (tcr1 × t

ar2 t

br2) (6)

tr[tar1×r2{tbr1×r2 , t

cr1×r2}] = tr[tar1{t

br1 , t

cr1}] d2 + d1 tr[t

ar2{t

br2 , t

cr2}] (7)

where d1 and d2 are respectively the dimensions of the irreps r1 and r2. We have made use of the fact thattr[tar ] = 0 in obtaining the final line. We may now make use of the definition of an anomaly coefficientA(r) :

tr[tar {tbr , tcr}] =1

2A(r) dabc (8)

where dabc is a unique symmetric invariant for the SU(n) group, n ≥ 3. Thus we obtain:

A(r1 × r2) = d1A(r2) + d2A(r1) (9)

Now, the decomposition (1) implies that:

A(r1 × r2) =∑i

A(ri) (10)

Hence we obtain the relationship:

d1A(r2) + d2A(r1) =∑i

A(ri) (11)

(b.)

We know that the dimension of the antisymmetric 2-index tensor representation of SU(n) has dimension

da = n(n−1)2 and for SU(3) this number is da = 3. The generators of the antisymmetric rep of SU(3)

thus are antisymmetric 3× 3 Hermitian matrices which satisfy:

(tba)T = (tb3)∗ = −(tb3) (12)

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Now let us consider the generators of the representation 3:

tb3

= −(tb3)∗ (13)

Comparing (12) and (13), it is clear that in SU(3) the representation a is equivalent to 3.

We know that A(r) = −A(r) and that A(n) = 1. This implies that in SU(3) the representation a hasan anomaly coefficient given by:

A(a) = A(3) = −A(3) = −1 (14)

Let us now use the result of part (a) along with the decomposition n× n = s+ a so as to obtain:

2nA(n) = A(s) +A(a) (15)

∴ A(s) = 2 · 3 · 1− (−1) = 7 (16)

(c.)

Since Eqn. (15) is valid for any SU(n) group we know that, for the decomposition n× n = s+ a:

A(s) +A(a) = 2n (17)

Now let us consider the simplest decomposition of an SU(n) that contains an SU(3) subgroup. Thefundamental of SU(n) may be broken down into the fundamental of SU(3) plus (n− 3) singlets. Hencewe have the following decomposition (3 = fundamental of SU(3) and θ2 = singlet) :

n = 3 + (n− 3) θ2 (18)

n× n = (3× 3) + (n− 3)(3× θ2 + θ2 × 3) + (n− 3)2θ2 (19)

= s+ a (20)

3× 3 = 6 + 3 (21)

∴ a = 3 + (n− 3)3 (22)

Using the decomposition (18) we may now write down A(a) of SU(n) as follows:

A(a) = A(3) + (n− 3)A(3) = −1 + (n− 3) · 1 = n− 4 (23)

A(s) = 2n−A(a) = n+ 4 (24)

where we have made use of (22) and that A(3) = −A(3) = −1.

In order to obtain the anomaly coefficient for the j-index totally antisymmetric tensor rep of SU(n)we use (18) and hence find:

nj = (3 + (n− 3) θ2)j (25)

= 3j + (n− 3) (j 3j−1 × θ2) + ...+ (n− 3)j θ2 (26)

Note: j 3j−1 × θ2 ≡ 3j−1 × θ2 + 3j−2 × θ2 × 3 + ... j terms (27)

We are interested in the totally antisymmetric part of (26). Hence we need to check each irrep in (26) forits antisymmetric part. Note that there is only one independent component of a totally antisymmetrictensor in 3-dimensions, so that we may argue that the antisymmetric part of any of the terms in (26)that have three 3s is a singlet. Furthermore, there is no totally anti-symmetric part of any term withmore than three 3s. We now need to count the number of independent components of the terms witheither one or two 3s contributing to the totally antisymmetric tensor. Let’s consider the case with justone 3. In this case one of the j-indices of the tensor is taken by the 3, while the rest of the (j−1) indices

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come from the (n−3) singlets. The number of ways to obtain a totally antisymmetric tensor with (j−1)indices from an (n−3) dimensional matrix is n−3Cj−1. Next we consider the case with two 3s. This timewe need a (j − 2) index antisymmetric tensor from the (n− 3) singlets and the corresponding number isn−3Cj−2. Let us denote the totally antisymmetric j-index tensor rep of SU(n) by aj . Thus, excludingthe singlets (which yield zero for anomaly coefficients) we find that:

aj = n−3Cj−1 3 +n−3 Cj−2 3 (28)

∴ A(aj) = n−3Cj−1A(3) +n−3 Cj−2A(3) (29)

= n−3Cj−1 +n−3 Cj−2 (−1) (30)

=n− 2 j

j − 1n−3Cj−2 (31)

It is evident from (31) that for n = 2 j the result vanishes. This implies that the m-index totallyantisymmetric tensor representation of SU(2m) is always anomaly free.

2.) Bardeen Counterterm

(a.)

The given anomalous variation of an action is:

δS = 10C

∫Tr

[εL

((dAL)2 − i

2dA3

L

)− εR

((dAR)2 − i

2dA3

R

)](32)

The counterterm we need to consider is:

Sc = 5C

∫Tr [(dARAR +AR dAR)AL − (dALAL +AL dAL)AR] (33)

−i 5C

∫Tr

[A3RAL −A3

LAR +1

2(ARAL)2

](34)

We know that the one-form fields AR and AL transform as:

δ AL = d εL + i [εL , AL] (35)

δ AR = d εR + i [εR , AR] (36)

Since d εL is an exact form we obtain:

δ (dAL) = d (δ AL) = i d [εL , AL] (37)

δ (dAR) = d (δ AR) = i d [εR , AR] (38)

Using the above results we evaluate the variation of the counterterm Sc as follows:

δ Sc = 5C

∫Tr [(δ(dARAR) + δ(AR dAR))AL + (dARAR +AR dAR) δ AL − (L ↔ R)]

−i 5C

∫Tr

[δ(A3

RAL)− δ(A3LAR) +

1

2δ(ALAR)2

](39)

= 5C

∫Tr[(dARAR +AR dAR) d εL + i (dARAR +AR dAR) [εL , AL]

−i d [εL , AL]ALAR − dAL dεLAR − i dAL [εL , AL]AR

−d εL dALAR − i [εL , AL] dALAR − i AL[εL , AL]AR − (L ↔ R)]

−i 5C

∫Tr[A3R d εL + iA3

R [εL , AL] + ...]

(40)

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Let us now consider the first term in (40). Let us integrate by parts and notice that for AR sufficientlywell behaved at ∞ the boundary term doesn’t give any contribution. Also since dAR is an exact formd2AR = 0. Finally we use the cyclic property of the trace. A similar useful manipulation may be doneon other terms. Hence we obtain:

5C

∫Tr[(dARAR +AR dAR) d εL = 10C

∫Tr[εL (dAR)2] (41)

5C

∫Tr[dAL dεLAR] = −5C

∫Tr[εL dAR dAL] (42)

5C

∫Tr[dεL dALAR] = −5C

∫Tr[εL dAL dAR] (43)

Combining the above terms and dropping the terms that are order A3 or higher we get:

δ Sc = 10C

∫Tr

[εL

((dAR)2 +

1

2(dAR dAL + dAL dAR)

)− (L ↔ R)

](44)

δ (S + Sc) = 10C

∫Tr

[(εL − εR)

((dAL)2 + (dAR)2 +

1

2(dAR dAL + dAL dAR)

)](45)

Following the procedure outlined in [1] we arrive at the full variation of S + Sc:

δ(S + Sc) = 30C

∫Tr(εL − εR)

[1

4F 2V +

1

12F 2A −

1

6(FV A

2 +AFV A+A2 FV )− 1

6A4

](46)

FV = d(AL +AR) + i (A2L +A2

R) (47)

FA = d(AL −AR) + i (A2L −A2

R) (48)

A = AL −AR (49)

Apparent from the form of (46) is that δ(S + Sc) vanishes when εL = εR.

(b.)

Let us consider εL = −εR and AL = AR. Substituting these into (46) and (32) we obtain for the gaugevariations of S + Sc and S respectively:

δ(S + Sc) = 60C

∫Tr(εL)

[dAL + iA2

L

]2(50)

δ S = 20C

∫Tr

[εL

((dAL)2 − i

2dA3

L

)](51)

The above forms simplify if we neglect cubic and higher order gauge field terms:

δ(S + Sc) = 60C

∫Tr(εL) (dAL)2 = 3 δ S (52)

3.) Peskin & Schroeder 20.1 Spontaneous breaking of SU(5)

The covariant derivative acting on a scalar field Φ in the adjoint representation is given as:

(Dµ Φ)a = ∂µ Φa + g fabcAbµ Φc (53)

= ∂µ Φa + i g Abµ [tb ,Φa] (54)

where the ta s represent the generators of SU(5), and we have used [ta , tb] = i fabc tc. The gauge bosonmass comes from squaring the second term in (54) and tracing over the matrix indices. The relevantmass term thus looks like:

∆L = g2 Tr|[ta , < Φ >]|2(Aaµ)2 =1

2m2AA

2 (55)

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Now we have the following to ways that the SU(5) symmetry can be spontaneously broken:Case I:

< Φ > = A

1

11

1−4

(56)

Notice that [ta , < Φ >] is nonzero only when the generator ta has non-zero entries in the 5th rowand column, and there are 8 such generators. All the other generators of the original SU(5) gauge groupremain unbroken. This the residual gauge group is identified as SU(4) × U(1) which has (42−1)+1 = 16unbroken generators. In order to work out the mass of the gauge bosons corresponding to the brokengenerators, let us consider the following generator of SU(5):

t =1

2

−i

i

⇒ [t , < Φ >] =A

4

−5i

5i

(57)

⇒ Tr|[t , < Φ >]|2 =25

2A (58)

Hence the mass of the gauge bosons corresponding to the broken generators is m2A = 25 g2A2.

Case II:

< Φ > = B

2

22−3

−3

(59)

Here only those generators that carry entries only in the upper 3×3 or lower 2×2 diagonal blocks commutewith < Φ > reducing the residual gauge group to SU(3)× SU(2)× U(1) which has (32−1)+(22−1)+1 =12 unbroken generators. Thus 12 of the generators of SU(5) become massive. Once again we follow asimilar strategy to find the gauge boson mass as was done in Case I. We find that for the generatorscorresponding to the broken symmetry:

Tr|[t , < Φ >]|2 =25

2B (60)

Hence in this case there are 12 massive gauge bosons with mass m2A = 25 g2B2.

4.) Peskin & Schroeder 20.5 A model with two Higgs fields

(a.)

Here we have a model with two scalar fields φ1 and φ2 which transform as SU(2) doublets with Y = 12 :

L = (Dµφ†1) (Dµφ1) + (Dµφ

†2) (Dµφ2) (61)

where Dµ = ∂µ − i g Aaµ τa − i g′ Y Bµ (62)

τa =σa

2(63)

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The two scalar fields acquire parallel VEVs of the form:

< φ1 >=1√2

(0v1

); < φ2 >=

1√2

(0v2

)(64)

Then the relevant terms that give masses to the gauge fields involve squares of the kinetic terms of thescalars that contain its couplings to the gauge fields:

∆L =1

2

2∑i=1

(0 vi

)(g Aaµ τ

a + g′ Y Bµ) (g Aµ b τ b + g′ Y Bµ)

(0vi

)(65)

=v21 + v22

8

(g2 (A1

µ)2 + g2 (A2µ)2 + (g A3

µ − g′Bµ) (g Aµ 3 − g′Bµ))

(66)

It is evident from (66) that we obtain the following spectrum of gauge bosons:

W±µ =1√2

(A1µ ∓ i A2

µ) ; m2W =

g2

4(v21 + v22) (67)

Z0µ =

1√g2 + g′2

(g A3µ − g′Bµ) ; m2

Z =(g2 + g′

2)

4(v21 + v22) (68)

Aµ =1√

g2 + g′2(g A3

µ + g′Bµ) ; m2A = 0 (69)

The above result is same as is obtained in the one-Higgs case with the replacement v2 → (v21 + v22) .

(b.)

The most general potential for a model with two-Higgs doublets that has the discrete symmetry φ1 → −φ1,φ2 → −φ2 is:

V (φ1, φ2) = −µ21 φ†1 φ1 − µ2

2 φ†2 φ2 + λ1 (φ†1 φ1)2 + λ2 (φ†2 φ2)2

+λ3 (φ†1 φ1) (φ†2 φ2) + λ4 (φ†1 φ2) (φ†2 φ1) + λ5 ((φ†1 φ2)2 + (φ†2 φ1)2) (70)

We may use the freedom of SU(2) rotations on the Higgs fields and very near a stable minimum writethem as:

φi =1√2

(0θi

)(71)

where in general θi s are complex and at the minimum of the potential they assume real VEVs: < θi >= vi . In terms of the θi s the potential (70) takes the form:

V (θ1, θ2) = − µ21

2θ∗1 θ1 −

µ22

2θ∗2 θ2 +

λ14

(θ∗1 θ1)2 +λ24

(θ∗2 θ2)2

+λ34

(θ∗1 θ1) (θ∗2 θ2) +λ44

(θ∗1 θ2) (θ∗2 θ1) +λ54

((θ∗1 θ2)2 + (θ∗2 θ1)2) (72)

A minimum of (72) corresponds to:

∂ V

∂θi= 0 ;

∂2 V

∂θ∗i ∂θi> 0 ;

∂2 V

∂2θ∗i> 0 (73)

where all the above derivatives have been evaluated at the local minimum of the potential. The vanishingof the first derivative at the local minimum gives us the following two conditions:

v1

[λ1 v

21 + (λ5 +

λ3 + λ42

) v22 − µ21

]= 0 (74)

v2

[λ2 v

22 + (λ5 +

λ3 + λ42

) v21 − µ22

]= 0 (75)

(76)

6

Since if either v1 or v2 is zero these conditions should reduce to the case of a single Higgs field let usassume that both these VEVs are non-zero. Then we obtain the following conditions on the vi s and µi s:

v2i =µ2j [λ5 + 1

2 (λ3 + λ4)]− µ2i λi

[λ5 + 12 (λ3 + λ4)]2 − λ1λ2

> 0 ; i 6= j (77)

λ1 > 0 ; λ2 > 0 ; λ5 +λ42

< 0 (78)

where the conditions (78) were derived from the requirement that the potential (72) is bounded frombelow and a constraint on the mass term of the physical Higgs’ (from part (c.))

(c.)

In order to examine the Physical Higgs’ field let us write the two scalar fields as:

φi(x) =1√2

(φ+iφ0i

); < φ+i >= 0 ; < φ0i >= vi ; or φ0i = vi + hi ; (79)

⇒ φi(x) =1√2

(φ+i

vi + hi

); (80)

Using the above definition for the Higgs’ fields one may work out the algebra to simplify the potentialterm. Note that the terms quadratic in φi s are the ones that give a mass term to the Higgs’. We mayseparate out these terms and write the relevant part of the Lagrangian that gives mass to the Higgs’ as:

LV Q =1

2

(µ21 − λ1 v21 −

λ32v22

)φ+1 φ

−1 +

1

2

(µ22 − λ2 v22 −

λ32v12

)φ+2 φ

−2

− 1

2

(λ5 +

λ42

)v1 v2 (φ+1 φ

−2 + φ+2 φ

−1 ) (81)

Using conditions (74) and (75) we may simplify the above form for the mass term:

LV Q =1

2(λ5 +

λ42

) (v21 + v22)

[v2 φ

+1√

v21 + v22− v1 φ

+2√

v21 + v22

] [v2 φ

−1√

v21 + v22− v1 φ

−2√

v21 + v22

](82)

One may introduce the parameter β such that:

sinβ =v2√v21 + v22

; cosβ =v1√v21 + v22

; tanβ =v2v1

(83)

LV Q =1

2(λ5 +

λ42

) (v21 + v22) (sinβ φ+1 − cosβ φ+2 ) (sinβ φ−1 − cosβ φ−2 ) (84)

Equation (84) is a mass term. This is evident once we rewrite the scalar fields and the mass term is:

LV Q =1

2(λ5 +

λ42

) (v21 + v22)φ+ φ− (85)

φ+ = sinβ φ+1 − cosβ φ+2 = (φ−)∗ (86)

φ′+ = − cosβ φ+1 − sinβ φ+2 (87)

Thus it turns out that φ+ is the massive Higgs’ field while φ′+ corresponds to a massless N-G boson.

(d.)

The fundamental couplings of the Higgs’ to the quark fields is given as:

Lm = −λijd Qi

L · φ1 djR − λ

iju ε

abQi

Laφ†2b u

jR + h.c. (88)

7

We may use the familiar notation of (80) keeping only the terms involving couplings of φ+1 and φ+2 forthe coupling terms and only the ones involving vi s for the mass terms. Representing the quark fields by:

QiL =

(uiLdiL

)(89)

we may simplify (88) and obtain:

Lm+ = − 1√2

[λijd uiL φ

+1 d

jR − λ

iju d

i

Lφ−2 u

jR] + h.c. (90)

Lmv = − 1√2

[λijd di

L (v1 + h1) djR + λiju uiL (v2 + h∗2)ujR] + h.c. (91)

The standard way to diagonalize the mass terms is to introduce the change of variables and define theCKM matrices:

uiR → W iju ujR ; diR → W ij

d djR ; uiL → U iju ujL ; diL → U ijd djL ; VCKM = U†u Ud (92)

Using appropriate matrices to diagonalize the quark mass terms we find:

Lmv = −mid d

k

L dlR

(1 +

h1v1

)−mi

u ukL u

lR

(1 +

h∗2v2

)+ h.c. (93)

where we have used the following definitions:

Du(d) = U†u(d) λu(d)Wu(d) (94)

1√2Dklu(d) v1(2) = mk

u(d) δkl (95)

We may now rewrite the Higgs’-quark coupling terms using the above machinery as follows:

Lm+ = −mid V

ji ujLφ+1v1

diR +miu (V †)ji d

j

L

φ−2v2

uiR + h.c. (96)

Finally we use the field redefinitions (86) in order to obtain the couplings of the physical charged Higgs’to the mass eigenstates of the quarks:

Lm+ = − V ji√v21 + v22

(mid tanβ +

miu

tanβ

)ujL φ

+ diR

− (V †)ji√v21 + v22

(mid tanβ +

miu

tanβ

)dj

L φ− uiR (97)

Clearly the couplings depend only on the values of the quark masses and tanβ = v2v1

and on the elementsof the CKM matrix.

References

[1] O. Kaymakcalan et. al., Phys. Rev. D, 30, 594 (1984).

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