Quantum Field Theory

Quantum Field Theory

Set 8: solutions

Exercise 1

The usual classical mechanics can be regarded as a 0 + 1 dimensional field theory. The coordinates that describe a classical system are functions of time and therefore can be thought of as fields:

3 + 1 0 + 1

~

x, t t

φ_a(~x, t) ~q_a(t)

where ~q_a(t) are coordinates and can be in general a three-dimensional vector (note that the three dimensions are those of the physical space where the system evolves). Since the classical mechanics is a trivial example of field theory, one can directly apply the formulas obtained from the Noether theorem and get several currents corresponding to different symmetries. For example a Lagrangian of the form

L=X

a

ma

2 |~q˙a|²−X

a6=b

V(|~qa−~qb|)

has the following symmetries:

• time translation: t⁰=t−α, ~q_a⁰(t⁰) =~qa(t);

• collective coordinates translation: t⁰=t, ~q_a⁰ =~qa+~α;

• coordinates rotation: t⁰=t, q⁰ⁱ_a=Rⁱ_jq^j_a. In the case of time translation one gets

t⁰=t−α =⇒ = 1,

~

q_a⁰(t⁰)−~qa(t) = 0 =⇒ ~εa = 0,

~

q_a⁰(t)−~qa(t) =α~q˙a =⇒ ∆~a= ˙~qa,

where we have used the vector notation for εa and ∆a since there is one of them for any component of the coordinate vector. And hence the current is

J⁰=X

a

∂ L

∂~q˙_a ·∆~a−L=X

a

~

πa·~q˙a−L=H.

The Noether current associated to the invariance under time translation coincides with the Hamiltonian and the condition of null divergence is∂₀J⁰= ^∂H_∂t = 0, so the Hamiltonian is conserved.

Analogously one can consider traslations along some direction ˆn:

t⁰ =t =⇒ = 0,

~

q_a⁰(t⁰)−~q_a(t) =αˆn =⇒ ~ε_a= ˆn,

~

q_a⁰(t)−~qa(t) =αˆn =⇒ ∆~a = ˆn, J⁰=X

a

∂ L

∂~q˙_a ·∆~a=X

a

~

πa·ˆn=P~·n.ˆ

Therefore the invariance under translation along some direction implies the conservation of the conjugate momen- tum in that direction.

Finally, consider the invariance under rotation around a direction (take thez direction for concreteness):

t⁰=t =⇒ = 0,

q^0x_a(t⁰) =q_a^x(t) +αq^y_a(t) =⇒ q^0x_a(t)−q_a^x(t) =αq_a^y(t) =⇒ ∆^x_a =q^y_a(t), q^0y_a(t⁰) =−αq_a^x(t) +q_a^y(t) =⇒ q^0y_a(t)−q_a^y(t) =−αq^x_a(t) =⇒ ∆^y_a =−q^x_a(t), q^0z_a(t⁰) =q^z_a(t) =⇒ ∆^z_a= 0,

J⁰=X

a

∂ L

∂q˙^x_a∆^x_a+ ∂ L

∂q˙a^y

∆^y_a

=X

a

(π^x_aq_a^y−π^y_aq_a^x) =X

a

(~πa∧q~a)^z.

Therefore the invariance under rotation around some direction implies the conservation of the angular momentum in that direction.

Exercise 2

The Noether theorem states the following: given a theory described by the Lagrangian densityL[φa, ∂µφa], where φa is a generic field, and a symmetry of this Lagrangian acting on coordinates and fields as follows (with some transformation labelled by a set of parameters{αⁱ})

x^µ −→x^0µ'x^µ−^µ_iαⁱ,

φa(x)−→φ⁰_a(x⁰) =D[φ]a(x)'φa(x) +αⁱεai(x)'φa(x) +αⁱεai(x⁰) 'φ_a(x⁰) +α^iµ_i∂_µφ_a(x⁰) +αⁱε_ai(x⁰)

≡φa(x⁰) +αⁱ∆a i(x⁰), then the four vector defined by

J_i^µ =X

a

∂L

∂(∂µφa)∆a i−^µ_iL

satisfies∂µJ_i^µ= 0 when one uses the equation of motion for the field φa. For clearness, note that the easiest way to compute ∆ is

εa i(x⁰)αⁱ=φ⁰_a(x⁰)−φa(x),

∆_{a i}(x⁰)αⁱ =φ⁰_a(x⁰)−φ_a(x⁰),

∆a i(x)αⁱ=φ⁰_a(x)−φa(x).

Note also that when the symmetry is such thatx⁰ =x, then ∆ai =εai. Consider the explicit example of a free complex scalar fieldφ=φ1+iφ2. The Lagrangian density reads

L=∂_µφ^†∂^µφ−m²φ^†φ.

It’s straightforward to show that the transformation given in the text is a symmetry of the theory: the transformed Lagrangian is in fact

L⁰=∂µφ^†e^−iα∂^µφe^iα−m²φ^†e^−iαφe^iα,

which is equal toLbecause αis a globalparameter (it’s not a function of coordinates).

To find the Noether current it is useful to write the transformation at the infinitesimal level:

x⁰ =x =⇒ ^µ= 0,

φ⁰(x⁰) =e^iαφ(x)∼(1 +iα)φ(x) =⇒ ε_φ=iφ,

φ^0†(x⁰) =e^−iαφ^†(x)∼(1−iα)φ^†(x) =⇒ ε_φ^†=−iφ^†.

Here the indecesa, iappearing in the general definition and labeling respectively the fields and the parameters of the transformation assume the values

a=φ, φ^†,

i= 1, since only the phaseαis present.

The Noether current reads

J^µ= ∂L

∂(∂µφ)∆φ+ ∆_φ†

∂L

∂(∂µφ^†) =∂^µφ^†(iφ) + (−iφ^†)∂^µφ, and the divergence gives

−i∂µJ^µ=φ^†φ−φ^†φ=m²φ^†φ−φ^†(m²)φ= 0,

where in the last equality we have used the equations of motion forφandφ^†, namelyφ=m²φandφ^†=m²φ^†. This is an important feature of the Noether current: its divergence vanishes only if one makes explicit use of the equations of motion for the fields. The above behavior is somehow expected because in the derivation of the current one neglects terms that are proportional to the Euler-Lagrange equation:

Z

d⁴x⁰ L[φ⁰](x⁰)− Z

d⁴xL[φ](x)∼ Z

d⁴x αⁱ∂µJ_i^µ+ Z

d⁴x

∂µ

∂L

∂(∂µφa)− ∂L

∂φa

∆aiαⁱ.

If the transformations define a symmetry the l.h.s. has to be identically zero and therefore one deduces the vanishing of the divergence of the current once the Euler-Lagrange equation are satisfied.

Exercise 3

In order to build an operatorW^µtransforming as a vector under Lorentz transformation we must use a contraction of the known covariant tensors: η^µν, J^µν, P^µ, and ^µναβ. In this way its square will be automatically a scalar (invariant under Lorentz transformations):

J^µν, W²

= 0 (1)

The Casimir operator must be also invariant under translations:

P^µ, W²

(2) We can start by trying with W^µ ≡ J^µνP_ν, but one can explicitly check that [W^µ, P^ν] 6= 0. Let us try with W^µ≡¹₂^µναβJ_ναP_β:

[W^µ, P^γ] = 1

2^µναβ[J_ναP_β, P^γ] = 1

2^µναβJ_να[P_β, P^γ] +^µναβ[J_να, P^γ]P_β= 1

2i^µναβ(δ^γ_αP_ν−δ_ν^γP_α)P_β= 0 (3) where we used the anty-symmetricity of thetensor and [P^µ, P^ν] = 0. The operator W² is then a Casimir, but what does it correspond to physically? Let us take a massive representation in the subspace of the rest frame:

P^µ= (M,0,0,0). We have:

W⁰= 1

2^0ναβJναPβ=1

2^ijkJijPk = 0 (4)

Wⁱ= 1

2^iναβJναPβ= 1

2^ijkJijM =S^kM (5)

whereS^k≡ ¹₂^ijkJ_ij is the spin operator. The CasimirW²is then proportional to the spin of the particle S².

Exercise 4

Given a symmetry acting on coordinates and fields as follows 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),

Noether’s current is defined as

J_i^µ= ∂L

∂(∂_µφ_a)∆ai−^µ_iL,

satisfying∂µJ_i^µ= 0. As a consequence, one defines the conserved Noether’s charge associated to the symmetry Qi=

Z

d³x J_i⁰= Z

d³x πa∆ai−⁰_iL

, Q˙i = 0,

whereπais the momentum conjugate to the fieldφa. One can use this charge to define a generator of the symmetry transformations in the following sense:

δ0φ(x) =φ⁰(x)−φ(x) = ∆ai(x)αⁱ={Qi, φa(x)}αⁱ.

Indeed substituting the expression for Noether’s charge and recalling the definition of the Poisson brackets ( see the solutions of Set3) one gets

{Qi, φa(x)}= Z

d³y {πb(y), φa(x)}∆bi(y) +πb(y){∆bi(y), φa(x)} −⁰_i {L(y), φa(x)}

= Z

d³y

δ_abδ³(x−y)∆_bi(y) +π_b(y)δ³(x−y)∂∆_bi

∂ πa

(x)−⁰_iδ³(x−y)∂L

∂ πa

(x)

= ∆_ai(x) +π_b(x)∂(ε_bi[φ] +^µ_i∂_µφ_b)

∂ πa

(x)−⁰_i ∂L

∂ ∂0φb

∂(∂₀φ_b)

∂πa

(x)

= ∆ai(x) +πb(x)⁰_i∂(∂0φb)

∂ πa

(x)−⁰_iπb(x)∂(∂0φb)

∂πa

(x) = ∆ai(x),

where we have used the fact that in the Hamiltonian formalism the independent variables areφ_a and π_a, while the spatial derivatives of the field,∂_iφ_a, are considered non independent.

The Noether’s charges, as functions of the fields and their conjugate momentum, can be thought of as generators in the Poisson brackets formalism, in complete analogy with analytical mechanics. The labelgenerators has however a deeper meaning. Since the transformations reflect the action of an element of the Lie symmetry group, one can regard this charges as a realization of the Lie algebra on the space of fields. The operation under which these generators must be closed are the Poisson brackets.

Consider the explicit example of translations:

x^µ−→x^0µ=x^µ−a^µ=x^µ−δ_ν^µa^ν,

φ_a(x)−→φ⁰_a(x⁰) =φ_a(x) =φ_a(x⁰) +a^ν∂⁰_νφ_a(x⁰).

The associated Noether’s current is calledEnergy-Momentum Tensor and can be computed as usual:

T^µ_ν = ∂L

∂(∂µφa)∆_aν −^µ_νL

= ∂L

∂(∂µφa)∂_νφ_a−δ_ν^µL.

The Noether’s charge in the present case is a four vector:

Pν= Z

d³x πa∂νφa−δ⁰_νL

= R

d³x(πa∂0φa− L) =H forν = 0, Rd³x(πa∂iφa) forν=i.

The zero component of the Noether’s charge corresponds to the energy of the system, therefore the complete four vector is the conserved four momentum of the system. Finally the infinitesimal transformation can be obtained using the Poisson brackets formalism:

{H, φ_a(x)}= ˙φ_a(x), {P_i, φ_a(x)}=

Z

d³y{π_b(y)∂_iφ_b(y), φ_a(x)}= Z

d³y δ³(x−y)δ_ab∂_iφ_b(y) =∂_iφ_a(x),

which confirms the interpretation of Pi as spatial momentum. One can repeat the procedure for the Lorentz transformations, assuming the theory to be Poincar´e invariant, (and for simplicity assume again scalar fields):

x^µ −→x^0µ= Λ^µ_νx^ν'x^µ−w^µ_νx^ν=x^µ−w^αβ(η_βνδ^µ_α−η_ανδ^µ_β)x^ν, φ_a(x)−→φ⁰_a(x⁰) =φ_a(x) =φ_a(x⁰) +w^αβ(η_βνδ_α^µ−η_ανδ^µ_β)x^0ν∂_µ⁰φ_a(x⁰).

The associated Noether’s current is:

M_αβ^µ = ∂L

∂(∂µφa)∆_aαβ−^µ_αβL

= ∂L

∂(∂µφa)(x_α∂_βφ_a−x_β∂_αφ_a)−(x_αδ^µ_β−x_βδ^µ_α)L

= x_αT^µ_β−x_βT^µ_α,

whereT^µ_β is the energy momentum tensor associated to spacetime translation invariance. The conservation of the current reads:

∂_µM_αβ^µ =T_αβ−T_βα= 0,

so the invariance under Lorentz transformations, together with the one under translations, implies the symmetry of the energy momentum tensor in the two indices (for a scalar theory). Notice in fact that we have assumed that the theory is Poincar´e invariant, so the energy momentum tensor is by itself conserved. The Noether’s charge in the present case is a Lorentz tensor:

Q_αβ= Z

d³x x_αT⁰_β−x_βT⁰_α

= Z

d³x x_α(π_a∂_βφ_a−δ_β⁰L)−x_β(π_a∂_αφ_a−δ_α⁰L) .

To see the meaning of this charge one can compute the Poisson bracket of itsij−component with the fieldφ_a(x):

{Qij, φ_a(x)}= Z

d³y(y_i{πb(y), φ_a(x)}∂_jφ_b(y)−y_j{πb(y), φ_a(x)}∂_iφ_b(y)) = (x_i∂_j−x_j∂_i)φ_a(x).

This charge is thus the angular momentum tensor of the system.

A note on functional derivatives

Consider a functionalF[φ] defined as

F[φ] = Z

d⁴x f(φ, ∂φ, x).

Given a small variation of the field,φ(x)−→φ(x) +δφ(x), then the change inF is δF =

Z

d⁴x δF

δφ(x)δφ(x) +O(δφ²), which generalizes the usual definition in the discrete case

δF =X

i

∂F

∂qi

δq_i+O(δq_i²).

In terms of the integrand, the variationδF is δF =

Z d⁴x

∂f(x)

∂φ(x)δφ(x) + ∂f(x)

∂(∂_µφ(x))∂_µδφ(x)

, which, up to boundary terms, can be rewritten as

δF = Z

d⁴x

∂f(x)

∂φ(x)−∂_µ ∂f(x)

∂(∂µφ(x))

δφ(x).

In the previous equations we have simply renamedf(φ, ∂φ, x)≡f(x). Note that the symbol∂stands forordinary derivative. Identifying this equation and the second, one gets the definition of functional derivativeδas

δF

δφ(x) = ∂f(x)

∂φ(x)−∂_µ ∂f(x)

∂(∂_µφ(x)). An example in which this definition is used is the Euler-Lagrange equation

δS

δφ(x)= 0 ⇐⇒ ∂L(x)

∂φ(x)−∂µ

∂L(x)

∂(∂_µφ(x))= 0.

In particular one can considerF[φ] =φ(y), so thatf(x) =φ(x)δ⁴(x−y) and δφ(y)

δφ(x) =∂(φ(x)δ⁴(x−y))

∂φ(x) =δ⁴(x−y), which is the generalization of the discrete relation∂qi/∂qj =δ_jⁱ.

In the hamiltonian formalism time is distinguished from the spatial coordinates, so one has integrals over d³x rather than overd⁴x. Given a functionalF of fields and conjugate momenta

F[φ, π](t) = Z

d³x f(φ, ∂iφ, π, x, t), its change due to variationsδφ(x, t) andδπ(x, t) is

δF(t) = Z

d³x

δF(t)

δφ(x, t)δφ(x, t) + δF(t)

δπ(x, t)δπ(x, t)

+O(δφ², δπ²)

= Z

d³x

∂f(x, t)

∂φ(x, t)δφ(x, t)−∂_i ∂f(x, t)

∂(∂_iφ(x, t))δφ(x, t) +∂f(x, t)

δπ(x, t)δπ(x, t)

, so that

δF(t)

δφ(x, t) = ∂f(x, t)

∂φ(x, t)−∂i

∂f(x, t)

∂(∂_iφ(x, t)), δF(t)

δπ(x, t) = ∂f(x, t)

∂π(x, t).

Note that in the last steps we have neglected boundary terms and used the shorthand notationf(φ, ∂iφ, π, x, t)≡ f(x, t). Particularly relevant cases areF[φ, π](t) =π(y, t), corresponding tof(x, t) =π(x, t)δ³(x−y), which yields

δπ(y, t)

δπ(x, t) =∂(π(x, t)δ³(x−y))

∂π(x, t) =δ³(x−y), andF[φ, π](t) =φ(y, t), orf(x, t) =φ(x, t)δ³(x−y), which gives

δφ(y, t)

δφ(x, t) =∂(φ(x, t)δ³(x−y))

∂φ(x, t) =δ³(x−y).

Given now two functionals

A[φ, π](t) = Z

d³x a(φ, ∂iφ, π, x, t), B[φ, π](t) =

Z

d³x b(φ, ∂iφ, π, x, t),

their Poisson brackets (at constant time) are defined starting from the definition of functional derivative as {A[φ, π](t), B[φ, π](t)} =

Z d³x

δA(t) δπ(x, t)

δB(t)

δφ(x, t)− δB(t) δπ(x, t)

δA(t) δφ(x, t)

= Z

d³x

∂a(x, t)

∂π(x, t)

∂b(x, t)

∂φ(x, t)−∂_i ∂b(x, t)

∂(∂iφ(x, t))

− ∂b(x, t)

∂π(x, t)

∂a(x, t)

∂φ(x, t)−∂_i ∂a(x, t)

∂(∂iφ(x, t))

.

For the particular caseA=π(z, t),B =φ(w, t), one finds {π(z, t), φ(w, t)}=

Z

d³xδπ(z, t) δπ(x, t)

δφ(w, t) δφ(x, t) =

Z

d³x δ³(x−z)δ³(w−x) =δ³(z−w).