Quantum Field Theory

Quantum Field Theory

Set 3: solutions

After the correction of the three exercises there is a little review on totally antisymmetric tensors, i.e. , and on the contraction of tensors. Some properties that are used in the correction are listed.

Exercise 1

The Euler-Lagrange equations

∂µ

∂L

∂(∂µφ)

= ∂L

∂φ give

∂µ

∂

∂(∂µφ) 1

2∂ρφ∂σφη^ρσ

=∂µ(η^ρσδ^µ_ρ∂σφ)∂L

∂φ =∂µ∂^µφ= ∂V(φ)

∂φ . For the massiveλφ⁴theory one gets:

∂µ∂^µφ=−m²φ−λ 6φ³. Let us now consider the term

α(∂µφ ∂^µφ)²≡α(∂µφ ∂^µφ)(∂νφ ∂^νφ).

In order to compute the dimension in energy of the constantαlet us compute first the dimension of the fieldφ.

We recall that the dimension of the action is power of~ (depending on the spacetime dimension), therefore in natural units the action is dimensionless. From this one can extract the dimension of the Lagrangian densityL:

S= Z

dt d^D−1xL(~x, t)

[S] =E⁰, [dx] = [dt] =E⁻¹ ⇒ [L] =E^D

Therefore in 3+1 dimensions all the terms appearing in the Lagrangian density must have dimensionE⁴. Let us consider the kinetic term: knowing the dimension [∂_µ] = [_∂x^∂_µ] =E, we can extract the dimension ofφ:

[∂µφ ∂^µφ] = [∂µ]²×[φ]²=E²×[φ]² ⇒ [φ] =E

The fieldφ has dimension of a mass. One can repeat in arbitrary dimension and get [φ] =E^D−22 . Let us come back to the parameterα:

[L] =E⁴ ⇒ [α]×[∂µ]⁴×[φ]⁴=E⁴ ⇒ [α] =E⁻⁴ Finally let us compute the Euler-Lagrange equations:

∂_µ ∂L

∂(∂_µφ)

=∂_µ(∂^µφ+ 4α(∂_νφ ∂^νφ)∂^µφ) =−∂V

∂φ.

Exercise 2

The Lagrangian of the system is L=

Z

d³xL(x, t) L(x, t) = 1

2(∂µφ)(∂^µφ)−1

2m²φ²− λ 4!φ⁴. The conjugate momentum is

π(x, t) = δ L

δ(∂0φ) = ∂L(x, t)

∂(∂0φ(x, t)) =∂0φ(x, t).

The Hamiltonian reads H =

Z

d³xH(x, t) = Z

d³x[π∂0φ− L] = Z

d³x 1

2π²+1

2(∂iφ)(∂iφ) +1

2m²φ²+ λ 4!φ⁴

. Given two functionals ofφ, π

F[φ, π](t) = Z

f(φ(x, t), π(x, t))d³x, One defines the equal time Poisson brackets between the two as

{F(t), G(t)}= Z

δ F δπ(z, t)

δ G

δφ(z, t)− δ F δφ(z, t)

δ G δπ(z, t)

d³z, where as usual _δφ(z,t)^{δ F} = ^{∂ f(z,t)}_∂φ(z,t). In particular

{π(x, t), φ(y, t)}_t=

Z δ R

d³x₁π(x₁, t)δ³(x−x₁) δπ(z, t)

δ R

d³x₂φ(x₂, t)δ³(y−x₂) δφ(z, t)

−δ R

d³x1π(x1, t)δ³(x−x1) δφ(z, t)

δ R

d³x1φ(x2, t)δ³(y−x2) δπ(z, t)

! d³z

= Z

δ³(x−z)δ³(y−z)

d³z=δ³(x−y).

The equations of motion become:

φ˙ ={H, φ},

˙

π={H, π}, and therefore

φ(y, t) =˙ {H, φ(y, t)}= Z

d³x 1

2π²(x, t), φ(y, t)

= Z

d³x π(x, t){π(x, t), φ(y, t)}=π(y, t),

˙

π(y, t) ={H, π(y, t)}= Z

d³x 1

2(∂_iφ(x, t))²+1

2m²φ²(x, t) + λ

4!φ⁴(x, t), π(y, t)

=

= Z

d³x

∂iφ(x, t){∂iφ(x, t), π(y, t)}+m²φ(x, t){φ(x, t), π(y, t)}+ λ

3!φ³(x, t){φ(x, t), π(y, t)}

=− Z

d³x

∂iφ(x, t) ∂

∂xⁱδ³(x−y) + (m²φ(x, t) + λ

3!φ³(x, t))δ³(x−y)

=∂i∂iφ(y, t)−m²φ(y, t)− λ

3!φ³(y, t).

Substituting the former in the latter on can show the equivalence with the Lagrangian formalism:

∂_t²φ(y, t)−∂i∂iφ(y, t) =φ(y, t) =−m²φ(y, t)− λ

3!φ³(y, t).

Exercise 3

The Maxwell equations read

∇ ·~ E~ =ρ, ∇ ∧~ E~ +1 c

∂

∂ t B~ = 0,

∇ ·~ B~ = 0, ∇ ∧~ B~ −1 c

∂

∂ t E~ =

J~ c. One can also rewrite the latter expression in components; recalling that:

E~ = (E¹, E², E³), B~ = (B¹, B², B³), J~= (J¹, J², J³), ρ=J⁰,

∇~ = (∂₁, ∂₂, ∂₃), ∂

∂ t =∂₀, (1)

one obtains

∂iEⁱ =J⁰, ijk∂jE^k+1

c∂0Bⁱ= 0,

∂iBⁱ= 0, ijk∂jB^k−1

c∂0Eⁱ= Jⁱ c .

In defining four components quantities one must pay attention to the position of spatial indices; since these are lowered and raised with a metricη_µν = diag(1,−1,−1,−1), with Minkosky signature the spacial indices acquire a minus sign in the transition. For example:

V^µ= (V⁰, Vⁱ) = (V0,−Vi), ∂µ= (∂0, ∂i) = (∂⁰,−∂ⁱ).

The field strength F can be expressed in terms of the vector potentialAµ= (A0, Ai) as follows:

F_µν=∂_µA_ν−∂_νA_µ.

One should notice the antisymmetric nature of the tensorF for the exchange of the indices µ↔ν. In order to verify that this expression reflects the definition of F in terms of the physical fieldsE, B on can compute

F⁰ⁱ=∂⁰Aⁱ−∂ⁱA⁰=∂0Aⁱ+∂iA0= (∂0A~+∇A0)ⁱ=−Eⁱ, F¹²=∂¹A²−∂²A¹=−∂1A²+∂2A¹=−(∇ ∧~ A)~ ³=−B³, where we have lowered indices of derivatives to get the standard form as defined before.

In order to compute the equations of motion for the fieldA_µ one has to apply the Euler-Lagrange equation. This time the field with respect to which we differentiate carries an additional space-time index:

∂_µ

∂L

∂(∂_µA_ν)

= ∂L

∂A_ν.

The previous equation contains a free indexν, which selects the component of the fieldA_ν with respect to which one derives (therefore one has 4 ”independent” equations), and a summed indexµ. The equations read

∂µ

∂L

∂(∂_µA_ν)

=∂µ

∂

∂(∂_µA_ν)

−1

4(∂ρAσ−∂σAρ) (∂^ρA^σ−∂^σA^ρ)

=∂µ

∂

∂(∂_µA_ν)

−1

2(∂ρAσ−∂σAρ)∂^ρA^σ

=∂µ

∂

∂(∂_µA_ν)

−1

2(∂ρAσ−∂σAρ)∂αAβη^ραη^σβ

=∂_µ

−1

2 δ_ρ^µδ^ν_σ−δ^µ_σδ_ρ^ν

∂_αA_βη^ραη^σβ−1

2(∂_ρA_σ−∂_σA_ρ)δ^µ_αδ_β^νη^ραη^σβ

=−∂_µF^µν,

∂L

∂A_ν = ∂

∂ A_ν(−J^ρAρ) =−J^ρδ_ρ^ν=−J^ν, (2)

where we have derived using the relation

∂(∂µAν)

∂(∂ρAσ) =δ^ρ_µδ_ν^σ,

that is to say that one gets non-vanishing contribution only if the indices of derivative and of the vectorAmatch.

Otherwise the derivative gives zero since it’s like deriving a variable with respect to an independent one. Finally the equations of motion are given by

∂_µF^µν =J^ν.

At this point it’s straightforward to verify that one has obtained exactly the Maxwell equations: the component 0 reads

∂_µF^µ0=∂_iFⁱ⁰=J⁰=−∂_iF⁰ⁱ =∂_iEⁱ=ρ, while the componentiis

∂_µF^µi =∂₀F⁰ⁱ+∂_jF^ji=−∂0Eⁱ−∂_jjikB^k =Jⁱ,

where we have used the propertyF^mn=Fmn=−mnpB^p. However we immediately see that the Euler-Lagrange equations reproduce only the inhomogeneous Maxwell equations, that is to say the ones with a source on their l.h.s.. The homogeneous equations derive from the so calledBianchi identity:

_µνρσ∂^νF^ρσ= 2_µνρσ∂^ν∂^ρA^σ= 0,

since the tensor∂^ν∂^ρ is contracted with the total antisymmetric tensorµνρσ. Expanding the identity one gets:

0νρσ∂^νF^ρσ =0ijk∂ⁱF^jk=−ijk∂^ijklB^l=−2∂iBⁱ= 0

iνρσ∂^νF^ρσ =i0jk∂⁰F^jk+ 2ij0k∂^jF^0k

=ijk∂^0jklB^l+ 2ijk∂^j(−E^k) = 2∂0Bⁱ+ 2ijk∂jE^k= 0.

The Bianchi identities are relation encoded in the structure of the field strength and find their natural explication in the formalism of differential forms.

One can solve the latter equation for a simple external source. The simplest current one can think about is the one generated by a static pointlike charge. In general the current generated by a pointlike particle has the form

J^µ= eδ³(x−x(t)), e~v(t)δ³(x−x(t)) ,

however, since the particle doesn’t move, its velocity~vis null and the current has only the 0-component. Therefore J^µ=

eδ³(x), ~0

. For such a configuration one should expect to find the Coulomb potential generated by a charge e. Since the current is time independent we can look for a static solution. We start considering the equation for the scalar potentialA0(~x):

∂_iEⁱ=∂_i(−∂iA₀−∂₀Aⁱ) =−∇²A₀(x) =eδ³(x).

The solution of this Laplace equation can be easily obtained in Fourier transform; define A0(~x) =

Z ∞

−∞

d³p (2π)³

A(~˜ p)e^i~p·~x, δ³(~x) = Z ∞

−∞

d³p (2π)³e^i~p·~x. therefore the equation becomes

Z ∞

−∞

d³p (2π)³

A(~˜ p)∂_i∂_i+e e^i~p·~x=

Z ∞

−∞

d³p (2π)³

−A(~˜ p)|~p|²+e

e^i~p·~x= 0.

The latter expression states that the function ˜A(~p) −|~p|²+e

, thought as an element of the Hilbert space where the Fourier transform is defined, has vanishing scalar product with all the functionse^i~p·~xwhich is a complete basis in that space. Therefore it must be

A(~˜ p) = e

|~p|².

The solution in coordinate space is simply obtained by Fourier transforming the one in momentum space:

A₀(~x) =e Z ∞

−∞

d³p (2π)³

e^i~p·~x

|~p|² =e Z ∞

0

dp p² (2π)³

Z 1

−1

d(cosθ) Z 2π

0

dφe^ipxcosθ p²

= e 4π²

Z ∞

0

dp

e^−ipx−e^ipx

−ipx

= e 2π²

Z ∞

0

dp

sin (px) px

= e

2π²x Z ∞

0

dysiny y

| {z }

π/2

= e 4π

1

|~x|,

which is the usual Coulomb scalar potential generated by a static charge and gives an electric field E~ =−∇A~ 0(~x) = e

4π²

~ x

|~x|³.

The vector potential satisfies the homogeneous equation∇²A(~~ x)−∇(~ ∇ ·~ A(~~ x)) = 0. Thus, we can chooseA~= 0.

Antisymmetric tensors

Inddimension one can define a total antisymmetric tensor_{a1... ad} with the following properties:

• is antisymmetric for the exchange of any two near indices:

a₁..a_ia_i+1..a_d=−a₁..a_i+1a_i..a_d

• a₁... a_d is different from zero only if all the indices have different value. In particular 1 2...d= 1 a₁... a_d= (−1)ⁿ

wherenis the number of permutations of two near indices to reach the configuration 12...dfrom a1a2...ad. For example in three dimension one hasε_ijk which has entries:

ε₁₂₃= 1 ε₁₃₂=−1 ε₂₁₃=−1 ε₂₃₁= 1 ....

In four dimension one has_µνρσ. It’s interesting to note that, due to the antisymmetric property defined above, if one of the indices is temporal (that is to say equal to 0) all the other must be spacial (i, j, k) in order to have a non vanishing contribution. For example, in the 0-component of the Bianchi identity0νρσ∂^νF^ρσ, the summed indices can be only spacial, otherwiseis zero. Then

0νρσ→0ijk→εijk.

Contraction betweentensors can be expressed in terms of delta functions. The most frequent are the following:

εijkεimn=δimδkn−δinδkm, εijkεijm= 2δkm,

_µναβ^ρσαβ = 2 δ^ρ_µδ_ν^σ−δ^σ_µδ_ν^ρ , µναβ^ρναβ = (3!)δ^ρ_µ,

(3) Frequently one deals with symmetric or antisymmetric tensor, contracted among themselves or with some other generic tensor. A useful way to manipulate these expression is the following. First of all one can observe that every tensor (with two indices) can be written as symmetric part plus the antisymmetric one:

Tab=T_ab^S +T_ab^A, where

T_ab^S =T_ba^S = 1

2(Tab+Tba) T_ab^A=−T_ba^A= 1

2(Tab−Tba). Denoting byAmn andSmn an antisymmetric and a symmetric tensor respectively, one has:

Tr [A·S]≡AabSba=−AbaSab=−AabSba= 0,

where we have first used the (anti)symmetry property of the tensor and then renamed the indices to show that this contraction is equal to minus itself. More in general for a tensorTab one has:

Tr [A·T] =A_abT_ba=A_abT_ba^A Tr [S·T] =S_abT_ba=S_abT_ba^S. Only the part with the same symmetry property survives in the contraction.