Quantum Field Theory

Quantum Field Theory

Set 9: solutions

Exercise 1

In addition to the usual transformation properties under the Poincar´e group (that define scalars, vectors, spinors, ...), any given field can have non trivial transformation properties under the action of additional symmetries. In the present case we consider a pair of Lorentz scalarsφ1andφ2, which are assumed to transform in the representation j= 1/2 of an SU(2) group. This isospin symmetry acts on fields as

Φ(x)−→Φ⁰(x⁰) =UΦ(x), Φ^†(x)−→Φ^0†(x⁰) = Φ^†(x)U^†,

φa(x)−→φ⁰_a(x⁰) =U_a^bφb(x), φ^∗a(x)−→φ^0∗a(x⁰) =φ^∗b(x)(U^†)_b^a,

whereU is aSU(2) element and we have defined the complex conjugate fields with the index up to recall that they transform with the hermitian conjugate matrix (they are a row instead that a column). The Lagrangian density

L=∂µΦ^†∂^µΦ−V Φ^†Φ

is invariant because the two doublets Φ and Φ^† are contracted in such a way to form an invariant underSU(2) transformations (that is to say an object transforming in thej = 0 representation):

Φ^†Φ−→Φ^0†Φ⁰ = Φ^†U^†U

|{z}

1

Φ,

∂µΦ^†∂^µΦ−→∂µΦ^0†∂^µΦ⁰=∂µ(Φ^†U^†)∂^µ(UΦ) =∂µΦ^†U^†U

|{z}

1

∂^µΦ,

where we have used the definition of SU(2) matrices. Since the invariance of the Lagrangian density has been proved for a generic potential being function only of theSU(2) invariant combination Φ^†Φ, one can go on and compute the Noether’s current associated to this symmetry. Using the exponential representation of theSU(2) matrices in terms of the generators (which we showed to be the Pauli matrices divided by two in the j = 1/2 representation)

φ⁰_a(x⁰) =

e^{iαi σi2 b}

a φb(x)'φa(x) +iαⁱ σⁱ

2 ^b

a

φb≡φa(x) + ∆_φaⁱ(x)αⁱ, φ^0∗c(x⁰) =φ^∗b(x)

e^{−iαi σi2c}

b 'φ^∗c(x)−iαⁱφ^∗b(x) σⁱ

2 ^c

b

≡φ^∗c(x) + ∆_φ∗^ci(x)αⁱ,

where we have defined ∆φ and ∆φ^∗ as the variations of φ and φ^∗, respectively. Note that in this case there is no variation of the positionx, so the total variation of the fields coincides with the variation at fixed coordinate, hence the use of the name ∆(x) instead ofε(x), see Solution8. The Noether’s current then reads

J^{i µ}= ∂L

∂(∂_µφ_a)∆_φaⁱ + ∂L

∂(∂_µφ^∗a)∆_φ∗^ai=i∂^µφ^∗a σⁱ

2 ^b

a

φ_b−i∂^µφ_cφ^∗b σⁱ

2 ^c

b

= i

2∂^µφ^∗a(σⁱ)_a^bφb− i

2φ^∗b(σⁱ)_b^c∂^µφc= i

2∂^µΦ^†σⁱΦ− i

2Φ^†σⁱ∂^µΦ.

One has to notice the similarity of this expression with the Noether’s current associated to theU(1) symmetry:

the previous current is its obvious generalization and the only difference is the presence of the generators between the fields in the appropriate representation. Another point to be stressed is the complete independence of the result here obtained from the explicit form of the potentialV, since this doesn’t contain derivative of the fields.

Let us now consider another set of three Lorentz scalar fieldsA~ = (A1, A2, A3) and impose that under the action ofSU(2) they transform in the representation j= 1:

Φ(x)−→Φ⁰(x⁰) =UΦ(x), Φ^†(x)−→Φ^0†(x⁰) = Φ^†(x)U^†, A(x)~ −→A(x~ ⁰) =R[U]^(j=1)A.~

Notice that the representationj= 1 ofSU(2) consists of rotations acting on a three dimensional vector space (and coincides withSO(3) rotations) therefore we have collectively denoted the fieldsA_i with a vectorA. The matrix~ R[U]^(j=1)is the representative ofU in the representationj= 1, that is to say the element ofSO(3) associated to U. We add to the previous Lagrangian density a kinetic term for the fieldsAand an interaction term:

L˜=∂_µΦ^†∂^µΦ−V Φ^†Φ +1

2∂_µA^T∂^µA+λA_iΦ^†σⁱΦ.

The last term of the r.h.s. is called aninteraction term since it connects the equations of motion for the different fields:

φ_a : φ^∗a = −∂V

∂φa

+λA_iφ^∗c(σⁱ)_c^a, φ^∗c: φc = − ∂V

∂φ^∗c +λAi(σⁱ)_c^aφa, Ai: Ai = λΦ^†σⁱΦ.

One has now to check whether the part of the Lagrangian involvingAis invariant underSU(2) transformations:

∂_µA^T∂^µA−→∂_µA^0T∂^µA⁰=∂^µA^T(R[U]^(j=1))^TR[U]^(j=1)

| {z }

1

∂_µA=∂_µA^T∂^µA,

because R[U]^(j=1) is a matrix of SO(3). The interaction term is more subtle (from now on we will omit the notation (j= 1)):

A⁰_iΦ^0†σⁱΦ⁰=R[U]_i^jAjΦ^†(U^†σⁱU)Φ.

In order to complete the proof of the invariance of L under SU(2), one should recall the definition of adjoint representation ofSU(2), that is to say the representation acting on the algebra itself. In Solution5 we have shown that, given the space of the hermitian traceless 2×2 matricesM (which coincides with the Lie algebra ofSU(2)), the action of an elementV ofSU(2), defined by

M =xiσⁱ−→ VMV^† =M⁰=x⁰_iσⁱ,

induces on this space a representation which turns out to be thej = 1 representation, since x⁰_i =R[V]_i^jx_j. In particular the Pauli matrices themselves transform as a triplet:

σ^k =δ_i^kσⁱ−→ V(δ^k_iσⁱ)V^†= (R[V]_i^jδ_j^k)σⁱ=R[V]_i^kσⁱ In the case at hand we haveU^†σⁱU =R[U^†]_lⁱσ^l, and collecting all together we obtain:

A⁰_iΦ^0†σⁱΦ⁰−→ R[U]_i^jR[U^†]_lⁱ

| {z }

δ^j_l

AjΦ^†σ^lΦ =AiΦ^†σⁱΦ,

where we have used the fact that (R[U^†]^(j=1)R[U]^(j=1))_i^m=R[U^†U]_i^m=R[1]_i^m=δ_i^m. This completes the proof of invariance under theSU(2) symmetry of the Lagrangian density ˜L.

Exercise 2

We want to show the invariance under Lorentz transformations of the measure over momentum space _(2π)^d33^k2k₀, where k0 ≡ω(~k) =

q

|~k|²+m² is the energy associated to a given particle of mass m. There are two ways to achieve the result: the first consists in checking explicitly the invariance performing a Lorentz transformation on momenta; however we first prove it performing a manipulation. The measure can be rewritten as

d³k

(2π)³2k₀ = d³k

(2π)³dk0δ(k²−m²)θ(k0) = d⁴k

(2π)³δ(k²−m²)θ(k0). (1)

In order to convince oneself that this is true, one can consider a test functionf of momenta and integrate it over k0:

Z d³k

(2π)³dk0δ(k²₀−|~k|²−m²)θ(k0)f(~k, k0) =

Z d³k (2π)³dk0



 δ(k0+

q

|~k|²+m²)

2|k0| +δ(k0− q

|~k|²+m²) 2|k0|



θ(k0)f(~k, k0), where we have used the well known relation for theδfunction: given a functiong(x) which vanishes in the points {x1, ..., xn}, then

δ(g(x)) =

n

X

i=1

δ(x−xi) 1

|g⁰(xi)|.

In the present case the equation (k0)²− |~k|²−m²= 0 admits two opposite solutions, hence the two terms in the parenthesis, but the theta function gets rid of the second one since in that casek0<0. Finally, integrating onk0

one gets the initial measure:

Z d³k

(2π)³dk₀δ(k0− q

|~k|²+m²)

2|k0| θ(k₀)f(~k, k₀) =

Z d³k (2π)³2

q

|~k|²+m² f(~k,

q

|~k|²+m²).

Notice that the integrated function depends only on~k (and the measure we are considering is defined on three- momenta). We have formally extended it to be a function of~k, k0 but this is only a trick because theδ function forces the variables to be related by themass shell condition k²=m².

The form of the measure we got allows us to show immediately the invariance under a Lorentz transformations.

• d⁴k is invariant since the jacobian determinant of the change of variables is 1.

• δ(k²−m²) is a function of the scalark²=kµk^µ and therefore it’s itself invariant.

• The theta function is not a priori invariant under Lorentz transformations: it is so only if the sign ofk₀⁰ is the same of that ofk₀. Lorentz transformations in general don’t preserve the sign of the 0-component of a four vector: for example ifv_µ= (1,0,0,2), then one can easily find a boost in the third direction that makes v₀⁰ <0:

v₀⁰ =γv₀−βγv₃= (1−2β)γ =⇒ β≥ 1 2. However one has to recall that the mass-shell conditionk0=

q

|~k|²+m² makeskµ a timelike fourvector (a fourvector in which the 0-component is larger than the modulus of its spatial threevector), that is to say a vector which lies inside the future lightcone centered in the origin. Transformations of the orthochronous Lorentz group, defined by the condition Λ⁰₀ >0 (those non containing the time reversal) send the future lightcone in itself and therefore if the four vectorkµ have positive 0-component, the same will be fork_µ⁰. One can prove this explicitly. The transformed 0-component of k^µ is k⁰⁰ = Λ⁰₀k⁰ +P

iΛⁱ₀kⁱ, with k⁰ ≥ pP

i(kⁱ)², due to the mass-shell condition. The defining relation ηµνΛ^µ_ρΛ^ν_σ = ηρσ implies Λ⁰₀ >

pP

i(Λⁱ₀)². Moreover, sinceP

iΛⁱ₀kⁱ≥ −pP

i(kⁱ)²pP

i(Λⁱ₀)², then:

(k⁰)⁰≥Λ⁰₀k⁰− s

X

i

(kⁱ)² s

X

i

(Λⁱ₀)²≥



Λ⁰₀− s

X

i

(Λⁱ₀)²



k⁰>0 implying that the sign ofk⁰⁰ is the same as the one ofk⁰.

Therefore the theta function will be left invariant in the distribution (1).

One can also check it explicitly performing Lorentz transformations: clearly the measure _(2π)^d3₃^k_2k

0 is invariant under space rotations sinced³kis so andk0 is a scalar underSO(3). Therefore only pure boosts are left to check.

Consider then a boost in the direction~nwith rapidityη(recall that the rapidity is defined asη= tanh⁻¹(β)): we decompose the spatial momentum~k in its longitudinal (i.e. along~n) and transverse (i.e. orthogonal to~n) parts:

~k=~k_T +~k_L. Then the transformed quantities are:

k⁰₀=k0cosh (η) +|~kL|sinh (η),

~k⁰_L=k0~nsinh (η) +~kLcosh (η),

~k⁰_T =~k_T.

Note that the direction of~kL is fixed to be parallel to ~n, therefore one can remove the symbol of vector and consider only the modulus kL. One can as well decompose the differential d³k −→ d²kTdkL; therefore the measure transforms as:

d²k_T⁰ = d²kT, dk⁰_L = ∂k⁰_L

∂kL

dkL= ∂

∂kL

(k0sinh (η) +kLcosh (η))dkL= ∂

∂kL

( q

k²_L+|~kT|²+m² sinh (η) +kLcosh (η))dkL

=





kL

q

k²_L+|~kT|²+m²

sinh (η) + cosh (η)



dkL= 1 k0

(kLsinhη+k0coshη)dkL =k⁰₀ k0

dkL.

At the end the ratio _(2π)^d33^k2k₀ is invariant also under Lorentz boosts.

Finally one can consider the distributiond³kδ³(~k). The fastest way to see that it is invariant is to see it as the result of an integration overk0:

d³kδ³(~k) = Z

dk⁰d³k δ⁴(k) = Z

d⁴k δ⁴(k) (2)

The expression on the right-hand side is Lorentz invariant becaused⁴k⁰=|J(Λ)|d⁴k andδ⁴(k⁰) =δ⁴(k)|J(Λ)|⁻¹, whereJ(Λ) is the determinant of the Jacobian of the Lorentz transformation.

Finally, the distribution (2π)³k⁰δ³(~k) is Lorentz-invariant because it’s obtained by dividing the two invariant distributionsd³kδ³(~k) and _(2π)^d3k_3k0

Exercise 3

To begin with, we derive formally the expression of the Noether’s current for a general Lagrangian density L[φa, ∂µφa] invariant under a Lie group defined by the following transformations

x^µ−→x^0µ=f^µ(x)'x^µ−^µ_iαⁱ,

φa(x)−→φ⁰_a(x⁰) =D[φ]a(f⁻¹(x⁰))'φa(x) +αⁱε[φ]ai(x)'φa(x⁰) +αⁱ∆[φ]ai(x⁰).

The action written in terms of the transformed fields is Z

Ω⁰

L[φ⁰_a, ∂_µ⁰φ⁰_a](x⁰)d⁴x⁰' Z

Ω⁰

L[φa, ∂µφa](x⁰) + ∂L

∂ φ⁰_a(x⁰)∆ai(x⁰)αⁱ+ ∂L

∂(∂_µ⁰ φ⁰_a)(x⁰)∂_µ⁰(∆ai(x⁰)αⁱ) +O(α²)

d⁴x⁰

= Z

Ω

L[φa, ∂_µφ_a](x⁰) + ∂L

∂ φ⁰_a(x⁰)∆_ai(x⁰)αⁱ+ ∂L

∂(∂_µ⁰ φ⁰_a)(x⁰)∂_µ⁰(∆_ai(x⁰)αⁱ) +O(α²)

∂x

∂x⁰

−1

d⁴x

= Z

Ω

L[φa, ∂µφa](x)−α^iµ_i∂µL[φa, ∂µφa](x) + ∂L

∂ φ_a(x)∆ai(x)αⁱ+ ∂L

∂(∂_µφ_a)(x)∂µ(∆ai(x)αⁱ) +O(α²)

∂x⁰

∂x

d⁴x '

Z

Ω

L[φa, ∂µφa](x)−α^iµ_i∂µL[φa, ∂µφa](x) + ∂L

∂ φ_a(x)∆ai(x)αⁱ+ ∂L

∂(∂_µφ_a)(x)∂µ(∆ai(x)αⁱ)

(1−∂µ(^µ_iαⁱ))d⁴x.

Note that in last two steps we have traded derivatives w.r.t. primed fields with derivatives w.r.t. untransformed fields in terms that are linear inα, since the difference betweenφandφ⁰ is itselfO(α), so the error we are doing isO(α²). If the parametersαⁱ are constant, then, neglecting higher order terms, one gets

Z

Ω⁰

L[φ⁰_a, ∂_µ⁰φ⁰_a]d⁴x⁰ ' Z

Ω

L[φa, ∂µφa]d⁴x+αⁱ Z

Ω

∂L

∂ φa

(x)∆ai(x) + ∂L

∂(∂µφa)∂µ∆ai(x)−∂µ(L(x)^µ_i(x))

d⁴x.

The request that the transformations define a symmetry translates in the equation

∂L

∂ φa

(x)∆ai(x) + ∂L

∂(∂µφa)∂µ∆ai(x)−∂µ(L(x)^µ_i(x)) = 0 (3) that, adding a subtracting the term∂µ(∂L/∂(∂µφa)) ∆ai(x), can be rewritten as

∂_µ

∂L

∂(∂µφa)∆_ai− L^µ_i

+ ∂L

∂φa

−∂_µ ∂L

∂(∂µφa)

∆_ai= 0.

The quantity in first parentheses is the Noether current, and its divergence vanishes if the fields satisfy the equations of motion.

Let us now consider the following construction:

• Consider the same (global) symmetry defined at the beginning of the exercise.

• Define the following transformation:

x^µ−→x^0µ =x^µ−^µ_i(x)βⁱ(x),

φ_a(x)−→φ⁰_a(x⁰)'φ_a(x⁰) + ∆[φ]_ai(x⁰)βⁱ(x⁰),

where ∆_aiand^µ_i are the ones given by the symmetry transformation at the beginning of the exercise. Note that the parametersβⁱ depend explicitly onx. In general this transformation willnot define a symmetry of the theory, in the sense that if the theory is invariant under the global transformation, it is not said that it’s invariant under its local version (while the converse is always true, since the local symmetry contains the global one as a particular case).

• Compute the variation of the action to first order inβⁱ(x):

Z

Ω⁰

L[φ⁰_a, ∂_µ⁰φ⁰_a]d⁴x⁰− Z

Ω

L[φa, ∂µφa]d⁴x '

Z

Ω

−β^iµ_i∂µL[φa, ∂µφa]− L[φa, ∂µφa]∂µ(^µ_iβⁱ) + ∂L

∂ φa

∆aiβⁱ+ ∂L

∂(∂µφa)∂µ(∆aiβⁱ)

(x)d⁴x

= Z

Ω

∂L

∂ φ_a∆aiβⁱ+ ∂L

∂(∂_µφ_a)∂µ(∆aiβⁱ)

(x)d⁴x (the first two terms are a total derivative).

Now plug in the condition for invariance under global transformations, Eq.(3):

Z

Ω⁰

L[φ⁰_a, ∂_µ⁰φ⁰_a]d⁴x⁰− Z

Ω

L[φa, ∂µφa]d⁴x

= Z

Ω

− ∂L

∂ ∂_µφ_a(∂µ∆ai)βⁱ+∂µ(L^µ_i)βⁱ+ ∂L

∂(∂_µφ_a)∂µ(∆aiβⁱ)

d⁴x

= Z

Ω

∂L

∂(∂_µφ_a)∆_ai− L^µ_i

∂_µβⁱd⁴x

= Z

Ω

j_i^µ∂µβⁱd⁴x

where we made use of integration by parts in some steps. This suggests a simple way to find the Noether’s current: when a Lagrangian has a global symmetry, one can simply consider the parameters of the trans- formation as local instead of global (that is, dependent onxinstead of constant), and evaluate the change in the action (which is not 0 in general since the local transformation is not a symmetry): this change is proportional to the space-time derivative of the parameters (since for constant parameters it has to vanish), and the coefficient is the current associated to the global transformation.

We can check how this construction works in the explicit example of a U(1) transformation applied to a free massive complex field with Lagrangian

L=∂µφ^†∂^µφ−m²φ^†φ.

The transformation is simply:

x^µ −→x^0µ=x^µ,

φ(x)−→φ⁰(x) =U(α)φ(x)≡e^iαφ(x), φ^†(x)−→φ^0†(x) =φ^†U^†(α)≡φ(x)^†e^−iα.

Let us apply the procedure and consider the same transformation with parameter α→ β(x). In this case the symmetry is internal (i.e. it only changes fields, not the space-time), so one can consider the variation in the Lagrangian instead of the variation of the action (i.e. the measure remains the same):

L[φ⁰_a, ∂_µ⁰φ⁰_a]− L[φa, ∂µφa] =φ^†(∂µU^†)U(∂^µφ) +φ^†(∂µU^†)(∂^µU)φ+ (∂µφ^†)U^†(∂^µU)φ.

Considering that∂µU =iU∂µβ, and∂µU^†=−iU^†∂µβ, then, at first order inβ, one gets L[φ⁰_a, ∂_µ⁰φ⁰_a]− L[φ_a, ∂_µφ_a] =−i(∂_µβ)φ^†∂^µφ+i(∂_µβ)∂^µφ^†φ=∂_µβJ^µ, hence

J^µ=i(∂^µφ^†φ−φ^†∂^µφ).

One can recognize the usualU(1) Noether’s current for a complex scalar field.

Exercise 4

Let’s first introduce some notation for delta functions, used also in next exercise.

Z

d³x e^i~k·~x= (2π)³δ³(~k), Z

d³k e^i~k·~x= (2π)³δ³(~x), Z

d³x δ³(~x) = Z

d³k δ³(~k) = 1.

When~k can assume only discrete values~kn= 2π~n/L, the Dirac delta becomes a Kronecker delta, since it has to give 1 when summed,notwhen integrated, and the integral on momenta becomes a sum over numbers. Basically the discrete case can be deduced from the continuous one making the following formal replacements.

δ³(~k) −→

L 2π

3

δ_~³_n,~0, Z

d³k −→

2π L

3

X

~n∈Z³

, and in particular

Z

d³x e^i~k·~x −→

Z

d³x e^i~kn·~x= (2π)³ L

2π 3

δ_~n,~³₀. Now one can compute explicitly the required expression:

Z

d³x X

~n∈Z³

1

L^3/2φn(t)∂ie^i2πL~n·~x X

m∈~ Z³

1

L^3/2φm(t)∂ie^{i2πLm·~~ x}= 1 L³

i2π L

² X

~ n, ~m∈Z³

~

n·m φ~ n(t)φm(t) Z

d³x e^{i2πL(~m+~n)·~x}

= 1 L³

i2π L

2

X

~ n, ~m∈Z³

~

n·m φ~ n(t)φm(t)L³δ³_{~n+m,~~ 0}= 2π

L 2

X

~n∈Z³

|~n|²|φn(t)|².

Exercise 5

The first commutator is [H, a(~p)] =

Z d³k (2π)³ω(~k)h

a^†(~k)a(~k), a(~p)i

=

Z d³k

(2π)³ω(~k) a^†(~k)h

a(~k), a(~p)i +h

a^†(~k), a(~p)i a(~k)

=

Z d³k

(2π)³ω(~k)

0−(2π)³δ³(~p−~k)a(~k)

=−ω(~p)a(~p).

Analogously, H, a^†(~p)

=

Z d³k (2π)³ω(~k)h

a^†(~k)a(~k), a^†(~p)i

=

Z d³k

(2π)³ω(~k) a^†(~k)h

a(~k), a^†(~p)i +h

a^†(~k), a^†(~p)i a(~k)

=

Z d³k

(2π)³ω(~k)

a^†(~k)(2π)³δ³(~p−~k) + 0

= +ω(~p)a^†(~p).