Solutions - PS1

Solutions - PS1

Questions on the solutions can be sent to [email protected].

1

1 a): Recall the definition of the field strength:

F iµν = ∂µAiν − ∂νAiµ + gf ijkAjµA

kν . (1.1)

Then,

δF iµν = ∂µδAiν − ∂νδAiµ + gf ijk(δAjµ)Akν + gf ijkAjµδA

kν , (1.2)

where we used the Leibniz property of the variational operator δ, and the fact that δ commutes with ∂µ.The quantity δAµ transforms under the adjoint representation of the group. To see this, recall that undera group action U ∈ G, Aµ transforms as

Aµ → A′µ = UAµU−1 − i

g

(∂µU

)U−1 . (1.3)

The final term on the left is independent of Aµ, and is known as the Maurer-Cartan form. With this,δAµ, which is the difference of two gauge fields, transforms as

δAµ → δA′µ = U(δAµ)U−1 . (1.4)

The Maurer-Cartan form cancels, and we see that δAµ transforms in the adjoint representation. Recallingthat the generators of the adjoint transformation are given by (T iadjoint)

jk = −if ijk, we have

DµδAiν = ∂µδA

iν + gf ijkAjµδA

kν . (1.5)

Substituting this into (1.2), we obtain the Palatini identity:

δF iµν = DµδAiν −DνδA

iµ . (1.6)

1 b): The operator form of Fµν is

Fµν =i

g[Dµ, Dν ] . (1.7)

1

The Jacobi identity,

[Dµ, [Dν , Dρ]] + [Dν , [Dρ, Dµ]] + [Dρ, [Dµ, Dν ]] = 0 , (1.8)

then implies that

[Dµ, Fνρ] + [Dν , Fρµ] + [Dρ, Fµν ] = 0 . (1.9)

Now, the complete antisymmetrisation of a tensor Bµνρ is given by

B[µνρ] =1

6

(Bµνρ −Bµρν +Bρµν −Bρνµ +Bνρµ −Bνµρ

). (1.10)

If, however, Bµνρ = −Bµρν initially, then the above formula becomes

B[µνρ] =1

3

(Bµνρ +Bρµν +Bνρµ

). (1.11)

Let’s now define Bµνρ = [Dµ, Fνρ]. Clearly, Bµνρ = −Bµρν , and the Jacobi identity in (1.9) is simply thestatement that

B[µνρ] = 0 . (1.12)

One way to ‘pick out’ the antisymmetric part of any tensor is to contract it with the epsilon symbol.Since we are in 4 dimensions (each index µ runs over 4 values), the epsilon symbol has 4 indices. As such,we can rewrite (1.12) as

εµνρσBνρσ = 0 =⇒ [Dν , εµνρσFρσ] = 0 =⇒ [Dν , F

µν ] = 0 , (1.13)

where we used the definition

Fµν =1

2εµνρσFρσ . (1.14)

Finally, note that

[∂µ, X]φ = (∂µX)φ+X∂µφ−X∂µφ = (∂µX)φ =⇒ [∂µ, X] = ∂µX (1.15)

for any operator X and field φ. Thus,

0 = [Dµ, Fµν ] = [∂µ − igAµ, Fµν ]

= ∂µFµν − ig[Aµ, F

µν ]

=(∂µF

µν i)T i − igAjµFµν k[T j , T k]

=(∂µF

µν i + gf ijkAjµFµν k)T i

=(DµF

µν i)T i , (1.16)

2

where we used the fact that Fµν is in the adjoint representation. (1.16) implies that DµFµν i = 0, as the

generators T i are linearly independent.

1 c): We need to compute the variation of F iµνFµν i. First, note that

F iµνδFµν i =

1

2F iµνε

µνρσδF iρσ =1

2F iρσε

ρσµνδF iµν =1

2F iρσε

µνρσδF iµν = Fµν iδF iµν . (1.17)

Therefore,

δ(F iµνF

µν i)

= F iµνδFµν i + Fµν iδF iµν = 2Fµν iδF iµν . (1.18)

Using the Palatini identity (1.6) in question 1 a), we have

δ(F iµνF

µν i)

= 2Fµν iδF iµν

= 4Fµν iDµδAiν

= 4Fµν i(∂µδA

iν + gf ijkAjµδA

kν

)= 4∂µ

(Fµν iδAiν

)− 4(δAiν)∂µF

µν i + gf ijkFµν iAjµδAkν

= 4∂µ(Fµν iδAiν

)− 4(δAiν)DµF

µν i

= 4∂µ(Fµν iδAiν

), (1.19)

where we used the Bianchi identity DµFµν i = 0 we derived in question 1 b).

2

2 a): T i ba is Hermitian, so (T i ba )∗ = T i ab . Using this, and the transformations of ϕa and its conjugate(given in the problem sheet), we find that the fields

φa1 =1

2

(ϕa + ϕ∗a

), φa2 = − i

2

(ϕa − ϕ∗a

)(2.1)

transform as

δφa1 =i

2δθi(T i ba ϕb − (T i ba )∗ϕ∗b

), (2.2)

δφa2 =1

2δθi(T i ba ϕb + (T i ba )∗ϕ∗b

). (2.3)

We can invert (2.1) to give

ϕa = φa1 + iφa2 , ϕ∗a = φa1 − iφa2 . (2.4)

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Substituting this into the transformations, we find

δφa1 = −δθi(

Im(T i ba )φb1 + Re(T i ba )φb2), (2.5)

δφa2 = δθi(

Re(T i ba )φb1 − Im(T i ba )φb2). (2.6)

We can package φa1 and φa2 into a 2n component vector φa:

φa =

(φa1φa2

)(2.7)

which will transform under the real representation as

δφa = −δθi(

Im(T i ba )φb1 + Re(T i ba )φb2−Re(T i ba )φb1 + Im(T i ba )φb2

). (2.8)

We know, however, that

δφa = iδθi(T irealφ)a . (2.9)

where T ireal are the generators in the real representation. From (2.8), we can read off these generators to be

T ireal = i

(Im(T i ba ) Re(T i ba )−Re(T i ba ) Im(T i ba )

), (2.10)

which are purely imaginary, antisymmetric 2n× 2n matrices.

2 b): Define Sij = (T iφ, T jφ), where T i is purely imaginary and antisymmetric, and φ is real.

Symmetric: Since (u, v) = (v, u), Sij = (T iφ, T jφ) = (T jφ, T iφ) = Sji.

Real: (Sij)∗ = (T i∗φ, T j∗φ) = (−T iφ,−T jφ) = Sij , as T i is imaginary.

Since Sij is symmetric and real, it is diagonalisable with

S = ODOT , (2.11)

where O is the orthogonal matrix constructed by the orthonormal eigenvectors of S, and D is the diagonalmatrix of the eigenvalues of S. The mass squared matrix M2 is proportional to S, so if

M2(T iφ) = 0 , T iφ 6= 0 , (2.12)

for some values of i, then ϕi = T iφ satisfies

Sϕi = 0 . (2.13)

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This means that each ϕi is a zero eigenvector of S, and consequently, each non-zero T iφ corresponds toa single zero eigenvalue in the mass squared matrix. The unique one-to-one correspondence between anon-zero T iφ and a zero eigenvalue is due to the diagonalisability of S.

3

3 a): The complex doublet is given by

ϕ =

(φ4R3 + iφ4R4

φ4R1 + iφ4R2

). (3.1)

Using the results in question 2, we find that the real fields φ1 and φ2 are

φ1 =

(φ4R3

φ4R1

), φ2 =

(φ4R4

φ4R2

), (3.2)

and the 4-component real field φ is then

φ =

(φ1φ2

)=

φ4R3

φ4R1

φ4R4

φ4R2

. (3.3)

This transforms under the real representation with generators T ireal given in (2.10). What we want, how-ever, are the generators for the field

φ4R =

φ4R1

φ4R2

φ4R3

φ4R4

. (3.4)

This can be obtained from φ by a transformation

φ4R =

0 1 0 00 0 0 11 0 0 00 0 1 0

φ := Mφ , (3.5)

which means that the generators for φ4R, T i4R, can be written in terms of the generators for φ, T ireal, as

T i4R = MT irealM−1 . (3.6)

We now list the generators T i in the complex representation:

T 1 =1

2

(0 11 0

), T 2 =

1

2

(0 −ii 0

), T 3 =

1

2

(1 00 −1

), T 4 =

1

2

(1 00 1

). (3.7)

We can now obtain the desired T i4R by applying (2.10) to obtain T ireal and then converting using (3.6).

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As an explicit example, let’s consider i = 1. Here,

Im(T 1) =

(0 00 0

), Re(T 1) =

1

2

(0 11 0

), (3.8)

so (2.10) gives

T 1real =

i

2

0 0 0 10 0 1 00 −1 0 0−1 0 0 0

. (3.9)

Finally, (3.6) tells us that

T 14R =

i

2

0 1 0 00 0 0 11 0 0 00 0 1 0

0 0 0 10 0 1 00 −1 0 0−1 0 0 0

0 0 1 01 0 0 00 0 0 10 1 0 0

=i

2

0 0 0 10 0 −1 00 1 0 0−1 0 0 0

. (3.10)

(Mathematica is useful in computing these matrix products quickly.)

3 b): We will derive the general form of the vector mass matrix (given in the problem sheet) using grouptheory only, and assuming an unbroken symmetry given by (T 3 + T 4)φ = 0. From this, we have

[T 3 + T 4, T iT j ]φ = (T 3 + T 4)T iT jφ+ T iT j(T 3 + T 4)φ = (T 3 + T 4)T iT jφ . (3.11)

Furthermore,[T 4, T iT j ] = [T 4, T i]T j + T i[T 4, T j ] = 0 , (3.12)

as T 4 generates the Abelian U(1)Y group, so [T 3 + T 4, T iT j ]φ = [T 3, T iT j ]φ. Using this, we have

Kij = gigj(φ, [T3, T iT j ]φ)

= gigj(φ, (T3 + T 4)T iT jφ)

= −gigj((T 3 + T 4)φ, T iT jφ)

= 0 , (3.13)

where in the second line we used the fact that the generators are antisymmetric. Defining P ij =[T 3, T iT j ] = [T 3, T i]T j + T i[T 3, T j ], we obtain the identity

Kij = gigj(φ, Pijφ) = 0 . (3.14)

i = j = 1: P 11 = [T 3, T 1]T 1 + T 1[T 3, T 1] = iT 2T 1 + iT 1T 2, so

K11 = 2ig22(φ, T 2T 1φ) = 2(M2)12gen = 0 , (3.15)

where we used the fact that (φ, T iT jφ) = −(T iφ, T jφ) = −(T jφ, T iφ) = (φ, T jT iφ), and the definition of

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the vector matrix.

i = 1, j = 2: P 12 = i(T 2T 2 − T 1T 1), so K12 = i((M2)22gen − (M2)11gen

)= 0, meaning that (M2)22gen =

(M2)11gen.

For i = 1, j = 3, we find that P 13 = iT 2T 3, so (M2)23gen = 0. For i = 1, j = 4, we have P 14 = iT 2T 4, so(M2)24gen = 0. For i = 2, j = 3, P 23 = −iT 1T 3, so (M2)13gen = 0. Lastly, for i = 2, j = 4, P 24 = −iT 1T 4, so(M2)14gen = 0.

Unfortunately, we cannot use the same trick to deal with the (M2)33gen, (M2)34gen, and (M2)44gen components,

as P 33, P 34, and P 44 are trivially zero. What we know so far from our group theoretic discussion is then

(M2)gen =1

4

g22v

2 0 0 00 g22v

2 0 00 0 x y0 0 y z

, (3.16)

where we parameterised (M2)11gen as g22v2/4.

There is one last thing that we can do to constrain the bottom 2 × 2 matrix. We know that there isan unbroken symmetry generated by T 3 + T 4, so Goldstone’s theorem and the Higgs mechanism tells usthat there must be a single massless vector corresponding to this unbroken symmetry. This means that(M2)gen must have a single zero eigenvalue that must occur in the bottom 2× 2 matrix. Another way ofsaying this is that the determinant of the bottom 2× 2 matrix must vanish:

det

(x yy z

)= 0 =⇒ y = ±xz . (3.17)

Choosing the negative root and parameterising the free parameters (x, z) as x = g22u2 and z = g21u

2, wefind that the generic form of the vector mass matrix is

(M2)gen =1

4

g22v

2 0 0 00 g22v

2 0 00 0 g22u

2 −g1g2u0 0 −g1g2u g22u

2

. (3.18)

4

For this question, we need to recall Goldstone’s theorem, which states that each broken generator of aglobal symmetry corresponds to a massless mode. It is also useful to write out the generators of SO(3)in matrix form for clarity. We are given that (T i)jk = −iεijk, with i, j, k ∈ {1, 2, 3}, and so

T 1 = i

0 0 00 0 −10 1 0

, T 2 = i

0 0 10 0 0−1 0 0

, T 3 = i

0 −1 01 0 00 0 0

. (4.1)

4 a): φ =

000

.

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Here, T iφ = 0 trivially, so there are no broken generators, and the stability group is still SO(3). ByGoldstone’s theorem, all 3 components of the fluctuations ϕ where φ = φ+ ϕ are massive.

4 b): φ = v

100

.

Here, T 1φ = 0, but

T 2φ = −iv

001

, T 3φ = iv

010

. (4.2)

It is clear that there exists no non-zero a, b ∈ C such that (aT 2 + bT 3)φ = 0, as T 2φ and T 3φ are linearlyindependent vectors in C3. Therefore, T 2 and T 3 are independently broken. The only unbroken generatoris T 1. Exponentiating T 1, we find

exp(iθT 1) =

1 0 00 cos θ sin θ0 − sin θ cos θ

∈ SO(2) . (4.3)

As such, the stability group generated by T 1 is simply SO(2). By Goldstone’s theorem, there will be 1massive mode corresponding to the unbroken T 1 generator, and 2 massless modes corresponding to thebroken T 2 and T 3 generators.

4 c): φ = v√2

110

.

Here, (T 1+T 2)φ = 0, and the broken generators are T 1−T 2 and T 3. Again, the stability group generatedby T 1 + T 2 is SO(2). By Goldstone’s theorem, there will be 1 massive mode corresponding to T 1 + T 2,and 2 massless modes corresponding to T 1 − T 2 and T 3.

4 d): φ = v√2

1−10

.

The analysis is the same as 4 c) with the role of T 1 + T 2 exchanged with T 1 − T 2.

4 e): Consider the most general form of the vacuum expectation value

φ =

xyz

∈ C3 . (4.4)

If x = y = z = 0, then there are no broken generators and the stability group is SO(3). If φ 6= 0, then forsome a, b, c ∈ C, we have

(aT 1 + bT 2 + cT 3)φ = i

bz − cycx− azay − bx

. (4.5)

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For the combination aT 1 + bT 2 + cT 3 to be an unbroken generator, we require the RHS of (4.5) to vanish.This requires solving the following 3 linear, simultaneous equations for the parameters a, b, c:

bz = cy , cx = az , ay = bx . (4.6)

Without loss of generality, let’s assume x 6= 0 and a 6= 0. Then, the last two equations can be solved by

c =az

x, b =

ay

x. (4.7)

With this expressions for c and b, we find that the first equation bz = cy is trivially satisfied:

bz − cy =ayz

x− azy

x= 0 . (4.8)

We have now found a unique solution for the unbroken generator, and it is given by

a(T 1 +

y

xT 2 +

z

xT 3)

(4.9)

Since a 6= 0, we can ignore the overall factor of a. The broken generators are formed by linearly inde-pendent combinations of T i that are orthogonal to the unbroken generator, and since there’s only oneunbroken generator, there must be two broken one. After a bit of algebra, we find that one particularbasis for the broken generators is

Unbroken: T 1 = T 1 +y

xT 2 +

z

xT 3 (4.10)

Broken: T 2 =y

xT 1 − T 2 , T 3 =

z

xT 1 +

yz

x2T 2 −

(1 +

y2

x2

)T 3 . (4.11)

Let’s just check that this works for the cases we talked about. For 4 b), we had x = v, y = 0, z = 0, sothe unbroken generator is T 1 = T 1, and the broken generators are T 2 = T 2, and T 3 = T 3. For 4 c), wehad x = v/

√2, y = x, z = 0, so T 1 = T 1 + T 2, T 2 = T 1 − T 2, and T 3 = −2T 3. Finally, for 4 d), we had

x = v/√

2, y = −x, z = 0, so T 1 = T 1 − T 2, T 2 = −(T 1 + T 2), and T 3 = −2T 3.

5

5 a): Let M ∈ GL(n,C) and define K = M †M . Clearly, K† = (M †M)† = K, so K is Hermitian, whichmeans that it has a set of non-zero eigenvectors ei, i ∈ {1, . . . , n}, with real eigenvalues Kei = λiei,λi ∈ R. The eigenvectors are orthonormal with respect to the Cn inner product

(ei, ej) =

n∑α=1

(eαi )∗eαj = δij , (5.1)

where eαi is the αth component of ei. By orthonormality,

λi = (ei,Kei) = (ei,M†Mei) = (Mei,Mei) = ||Mei||2 ≥ 0 , (5.2)

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where we used the definition of the Hermitian conjugate: (u,Mv) = (M †u, v). This shows that theeigenvalues are strictly positive unless Mei = 0, which is a case we will ignore, and we shall write themas λi = ω2

i , with ωi ∈ R \ {0}.

To diagonalise K, construct the matrix U = (e1, e2, . . . , en), where we treat ei as column vectors. Byorthonormality of the eigenvectors, we see that

U †U = UU † = 1n , (5.3)

so U is unitary. Therefore,

K = K(UU †) = (KU)U † . (5.4)

By defining the diagonal matrix

D2 = diag(ω21, ω

22, . . . , ω

2n) , (5.5)

we have

KU = K(e1, e2, . . . , en) = (ω21e1, ω

22e2, . . . , ω

2nen) = UD2 . (5.6)

Thus,

K = UD2U † . (5.7)

5 b): Define H = UDU †. To compute H−1, note that since ωi 6= 0,

Dm = diag(ωm1 , ωm2 , . . . , ω

mn ) (5.8)

for any m ∈ Z. This, combined with the unitarity of U then implies that

H−1 = UD−1U † . (5.9)

Now, define U = MH−1 = MUD−1U †. Then,

U †U = UD−1U †M †MUD−1U †

= UD−1U †(UD2U †

)UD−1U †

= 1n , (5.10)

where we used the result in 5 a) that K = M †M = UD2U †. This means that U is unitary. With thesedefinitions, we can now define the matrix V = UU = MUD−1, which is unitary because it is the productof unitary matrices. Then,

V †MU = D−1U †M †MU = D−1U †(UD2U †

)U = D . (5.11)

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