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Phys624
Classical Field Theory
Homework 1
Homework 1 Solutions
Problem 1: Electromagnetic Field
The idea behind these problems is to “re-derive” some of the known results in electromagnetism using the classical field theory approach, i.e., with the Lagrangian
1
L = − F µν Fµν
4
(1)
Fµν = ∂µ Aν − ∂ν Aµ
(2)
where
and identifying the electric and magnetic fields as
E i = −F 0i ,
ijk B k = −F ij
(3)
(4)
For example, we already showed in lecture that Maxwell’s equations are simply the EulerLagrange equations.
a) Energy-momentum
Based on Noether’s theorem, construct the energy-momentum tensor for classical electromagnetism from the above Lagrangian.
Note that the usual procedure does not result in a symmetric tensor. To remedy that, we
can add to T µν a term of the form ∂λ K λµν , where K λµν is antisymmetric in its first two
indices. Such an object is automatically divergenceless, so
T̂ µν = T µν + ∂λ K λµν
(5)
is an equally good energy-momentum tensor with the same globally conserved energy and
momentum. Show that this construction, with
K λµν = F µλ Aν
(6)
leads to an energy-momentum tensor T̂ that is symmetric and yields the standard (i.e.,
known without using field theory) formulae for the electromagnetic energy and momentum
densities:
1
E2 + B2 ,
2
S = E×B
E =
1
(7)
(8)
Phys624
Classical Field Theory
Homework 1
Solution:
First, we calculate the energy-momentum tensor using
Tνµ =
δL
∂ν Aλ − δνµ L
δ(∂µ Aλ )
(9)
Expand the Lagrangian as
1
1
L = − F µν Fµν = − (∂ µ Aν ∂µ Aν − ∂ µ Aν ∂ν Aµ )
4
2
(10)
δL
= −F µλ
δ(∂µ Aλ )
(11)
1
T µν = −F µλ ∂ ν Aλ + η µν F ρσ Fρσ
4
(12)
we can calculate
Thus we get
where we have raised the ν index using metric η µν . This is obviously not symmetric under
exchange of µν indices. To make it a symmetric tensor, we add total derivative term:
∂λ K λµν = (∂λ F µλ )Aν + F µλ (∂λ Aν )
(13)
We know from equation of motion that ∂λ F µλ = 0. Therefore
1
T̂ µν = F µλ Fλ ν + η µν F ρσ Fρσ
4
(14)
which is manifestly symmetric in µν indices. Now we can express it in terms of physical
electric and magnetic fields. The energy density is given by
1
= T̂ 00 = F 0i Fi 0 + (2F 0i F0i + F ij Fij )
4
1 0i 0i 1 ij ij
F F + F F
=
2
4
1 ~ 2
~ 2)
=
(|E| + |B|
2
(15)
where in the last equality we used ijk ijl = 2δ kl . Similarly, the momentum density is given
by
~ i = T̂ 0i = F 0k F i = −E k kil B l = (E
~ × B)
~ i
S
k
(16)
b) Subtlety with going to Hamiltonian formalism
Exercises 2.4 and 2.5 of Lahiri and Pal.
Due to this subtlety, we will not quantize electromagnetic field to begin with (even though
historically it was the first QFT). We will return to this issue when we quantize the electromagnetic field later in the course.
2
Phys624
Classical Field Theory
Homework 1
Solution to Exercise 2.4
First, we need to find terms in the Lagrangian with time derivative of fields Aµ :
1
1
− F µν Fµν = − (2F 0i F0i + F ij Fij )
4
4
1 i 2
~ 0 )2 − 2Ȧi ∂ i A0 ] − 1 F ij Fij
=
[(Ȧ ) + (5A
2
4
(17)
The canonical momenta are
δL
=0
δ Ȧ0
(18)
δL
= Ȧi − ∂ i A0
δ Ȧi
(19)
Π0 ≡
Πi ≡
We can see that from the above equations we cannot solve for Ȧ0 . The reason is that there is
no term in the Lagrangian with time derivative of A0 . In other words, A0 is not a dynamical
field.
Solution to Exercise 2.5
Now, if we fix the gauge by choosing A0 = 0, and treat Ai as dynamical fields, we get
Πi ≡
δL
= Ȧi
i
δ Ȧ
(20)
Obviously, it can be inverted to solve for Ȧi .
Problem 2: Real, free scalar/Klein-Gordon Field
This is the simplest classical field theory and so the first one that we will quantize. For the
Lagrangian
L =
1 µ
1
(∂ φ) (∂µ φ) − m2 φ2 ,
2
2
(21)
where φ is a real-valued field,
(i) Show that the Euler-Lagrange equation is the Klein-Gordon equation for the field φ.
(ii) Find the momentum conjugate to φ(x), denoted by Π(x).
(iii) Use Π(x) to calculate the Hamiltonian density, H.
(iv) Based on Noether’s theorem, calculate the stress-energy tensor, Tνµ , of this field and the
conserved charges associated with time and spatial translations, i.e., the energy-momentum,
P µ , of this field.
(v) Using the Euler-Lagrange (i.e., Klein-Gordon) equation, show that ∂µ Tνµ = 0 for this
field. (Of course, this result was expected from Noether’s theorem.)
3
Phys624
Classical Field Theory
Homework 1
(vi) Finally, show that P 0 that you calculated above in part (iv) is the same as the total
Hamiltonian, i.e., spatial integral of H which you calculated above in part (iii).
We will determine eigenstates/values of this (total) Hamiltonian when we quantize the field.
And, P i can be interpreted as the (physical) momentum carried by the field (not to be
confused with canonical momentum!). This Pi will be used in interpreting the eigenstates of
the Hamiltonian of the quantized scalar field.
Solutions:
(i) Euler-Lagrange equation for φ,
δL
δL
= ∂µ
δφ
δ(∂µ φ)
(22)
−m2 φ = ∂µ (∂µ )φ
⇒ (∂ 2 + m2 )φ = 0
(23)
(24)
Substituting in the Lagrangian,
which is the Klein-Gordon equation.
(ii)
Π(x) =
δL
= φ̇
δ φ̇
(25)
(iii)
1
~ 2 + m2 φ2 ]
H = Πφ̇ − L = [φ̇2 + (5φ)
2
(26)
(iv)
δL ν
∂ φ − η µν L
δ(∂µ φ)
1
= ∂ µ φ∂ ν φ − η µν ∂ ρ φ∂ρ φ − m2 φ2
2
T µν =
(27)
The conserved charge is given by,
µ
P =
Z
d3 x T 0µ
(28)
(v) The divergence of the stress-energy tensor,
1
∂µ Tνµ = ∂µ (∂ µ φ∂ ν φ − η µν ∂ ρ φ∂ρ φ − m2 φ2 )
2
1 = ∂ 2 φ∂ ν φ + ∂ µ φ∂µ ∂ ν φ − ∂ ν ∂ ρ φ∂ρ φ − m2 φ2 )
2 ρ ν
2
ν
µ
ν
= ∂ φ∂ φ + ∂ φ∂ ∂µ φ − ∂ φ∂ ∂ρ φ + ∂ ν ∂ ρ φ ∂ρ φ − m2 φ∂ ν φ )
= (∂ 2 + m2 )φ∂ ν φ = 0
4
(29)
(30)
(31)
(32)
Phys624
Classical Field Theory
Homework 1
Therefore, if the field satisfies its equation of motion (the Klein-Gordon equation in this
case), the stress-energy tensor is conserved. Therefore, Noether current conservation relies
on the equations of motion which are satisfied for a classical field.
(vi) Using the expression above for P µ , we get
Z
Z
1 2
2
2 2
0
3
~
(33)
P = d x [φ̇ + (5φ) + m φ ] = d3 x H
2
i
P =
Z
d3 x φ̇∂ i φ
(34)
Problem 3: Scale invariance
Exercise 2.10 of Lahiri and Pal.
The transformations involve a simultaneous re-scaling of the coordinates and the fields, hence
the name “scale invariance” given to this symmetry.
Solution:
The transformations in Lahiri and Pal and those in Peskin and Schroeder follow different
conventions. They are potentially quite confusing, so it is a good idea to keep one convention
handy. We will use the Lahiri and Pal notation here.
The transformation is
x → x0 = bx
φ(x)
φ(x) → φ0 (x0 ) =
b
(35)
(36)
It is important to note that in this convention, the argument of the field (in the right-most
expression) does not change with the transformation. As a reference for this convention, one
can remember that the scalar transforms like φ(x) → φ(x) under a Lorentz transformation.
The infinitesimal version of the transformation is given by
x → (1 + )x
φ(x) → 1/(1 + )φ(x)
δxµ = xµ
δφ = −φ
(37)
(38)
Again, remember that for Lorentz transformations on a scalar field, δφ would be zero in this
convention.
The Lagrangian is given by,
1
L = ∂µ φ∂ µ φ − λφ4
2
(39)
Under the transformation, ∂ → 1b ∂. Therefore, the transformed Lagrangian becomes,
11
λ
∂µ φ∂ µ φ − 4 φ4
4
b 2
b
1
= 4L
b
Lb =
5
(40)
Phys624
Classical Field Theory
Homework 1
We can now look at the transformation of the action under this symmetry,
Z
Z
Z
4
4 4
d x L → b d x L b = d4 x L
(41)
Therefore, the action is invariant under this symmetry. The Noether current is,
∂L
δφ − T µν δxν
∂(∂µ φ)
∂L
∂L ν
µν
=
δφ −
∂ φ − g L δxν
∂(∂µ φ)
∂(∂µ φ)
ρ
4
ν
µν 1
= −φ∂µ φ − ∂µ φ∂ φ − g ( ∂ρ φ∂ φ − λφ ) xν
2
J µ =
(42)
(43)
(44)
Problem 4: Complex scalar/Klein Gordon field coupled to electromagnetism (Scalar electrodynamics)
Exercises 2.9 (b) and (c) of Lahiri and Pal. Neglect the potential term, V φ† φ , given in
the Lagrangian in exercise 2.3 of Lahiri and Pal for these problems.
The free complex Klein-Gordon field was discussed in lecture. In particular, it was already
shown that the Euler-Lagrange equation is the Klein-Gordon equation (exercise 2.3 of Lahiri
and Pal) and the conserved current corresponding to the transformation φ → eiα φ was
already calculated [exercise 2.9 (a) of Lahiri and Pal] so that that there is no need to do it
again here. This field is a simple generalization of the case of a real field so that it will be
the second field to be quantized.
The purpose of exercises 2.9 (b) and (c) in Lahiri and Pal is to study the addition of an
interaction of this field with the electromagnetic field. We will return to quantization of this
theory later in the course.
Solution:
b) The Lagrangian we are asked to assume is
−1
Fµν F µν + (∂ µ − iqAµ )φ† [(∂µ + iqAµ )φ] − m2 φ† φ
4
The infinitesimal transformations are,
L=
δφ = −iqθ φ
†
†
δφ = iqθ φ
(45)
(46)
(47)
The Noether current,
∂L
∂L
δφ +
δφ†
∂(∂µ φ)
∂(∂µ φ† )
= (∂ µ − iqAµ )φ† (−iqθ φ) + [(∂µ + iqAµ )φ] (iqθ φ† )
= −iqθ φ ∂ µ φ† − φ† ∂µ φ − 2iqAµ φφ†
θJ µ =
6
(48)
(49)
(50)
Phys624
Classical Field Theory
Homework 1
c) The Euler-Lagrange equation for Aµ
∂L
∂L
=
∂
ν
∂(Aµ )
∂(∂ν Aµ )
(51)
Note that
∂(Fαβ F αβ )
= (gαν gβµ − gαµ gβµ )F αβ + (δνα δµβ − δµα δµβ )Fαβ
∂(∂ ν Aµ )
= 4Fνµ
(52)
∂ν Fµν = −iqφ† [(∂µ + iqAµ )φ] + iqAµ φ (∂ µ − iqAµ )φ†
(53)
Therefore,
The term on the right hand side is simply the Noether current derived above.
∂ν F νµ = j µ
7
(54)