Quantum Field Theory: Example Sheet 1 Solution prepared by Peng Zhao

1. Decoupled harmonic oscillator

If we express the displacement as a sine series Fourier expansion of the form
r ∞
2X  nπx 
y(x, t) = qn (t) sin ,
a n=1 a

then its time and spatial derivatives are

r ∞ r ∞
∂y 2X  nπx  ∂y 2 X nπ  nπx 
= q̇n (t) sin , = qn (t) cos .
∂t a n=1 a ∂x a n=1 a a

Note that sin(nπx) and cos(mπx) form an orthonormal basis for L2 ([−1, 1]). Hence, for example
Z a  2 ∞  ∞ ∞
2 a  nπx 
Z
∂y X  mπx  X X
dx = q̇m (t)q̇n (t) dx sin sin = (q̇m (t)q̇n (t)δmn ) = q̇n2 (t).
0 ∂t m,n=1
a 0 a a m,n=1 n=1

The Lagrangian becomes

∞ 
" 2 2 #
Z a   
σ ∂y T ∂y X σ T  nπ 2 2
L= dx − = q̇n2 − qn .
0 2 ∂t 2 ∂x n=1
2 2 a

The Euler-Lagrange equation associated with this Lagrangian is

T  nπ 2
q̈n = qn .
σ a
We recognize this as an infinite set (one for each n) of decoupled harmonic oscillator with frequencies
r  
T nπ
ωn = .
σ a
2. The Klein-Gordon equation is Lorentz-invariant
Under a Lorentz transformation xµ → Λµν xν , a scalar field transforms as φ(x) → φ(y) = φ(Λ−1 x). We find

ηµν ∂ µ ∂ ν φ(x) → ηµν ∂ µ ∂ ν φ(Λ−1 x) = ηµν (Λ−1 )µρ (Λ−1 )νσ ∂ ρ ∂ σ φ(y) = ηρσ ∂ ρ ∂ σ φ(y).

Thus the Klein-Gordon equation ∂µ ∂ µ φ + m2 φ2 = 0 is invariant under Lorentz transformations.

3. Complex scalar field
The Lagrangian for a complex field φ(x) is given by
λ ∗ 2
L = ∂µ φ∗ ∂ µ φ − m2 φ∗ φ − (φ φ) ,
2
where we treat φ(x), φ∗ (x) independently. The equation of motion for φ is obtained by varying w.r.t. φ∗

∂µ ∂ µ φ + m2 φ + λ|φ|2 φ = 0.

The equation for φ∗ is obtained by varying w.r.t. φ and is the complex conjugate of the equation above.
Under the infinitesimal transformation

δφ = iαφ, δφ∗ = −iαφ∗ ,

the Lagrangian changes to first order in α as

δL = iα∂µ φ∗ ∂ µ φ − iα∂µ φ∗ ∂ µ φ − iαm2 |φ|2 + iαm2 |φ|2 = 0.

The Noether current associated with this transformation is

∂L ∂L ∗
jµ = δφ + δφ = iα(φ∂ µ φ∗ − φ∗ ∂ µ φ).
∂φ ∂φ∗
We can verify that it is conserved using the field equation:

∂µ j µ = iα(φ∂µ ∂ µ φ∗ − φ∗ ∂µ ∂ µ φ) = iα(−m2 φ − λ|φ|2 φ + m2 φ + λ|φ|2 φ) = 0.

1
4. Lagrangian with so(3) internal symmetry
The Lagrangian for a triplet of real fields φa (a = 1, 2, 3)
1 1
L= ∂µ φa ∂ µ φa − m2 φa φa
2 2
is invariant under SO(3) rotation because it is composed of inner product of fields. Suppose we rotate the
fields along the na vector infinitesimally by θ, then φa → θa + θabc nb φc . Then to first order in θ,

φa φa → φa φa + 2θabc nb φa φc = φa φa ,
∂µ φa ∂ µ φa → ∂µ φa ∂ µ φa + 2θabc nb ∂µ φa ∂ µ φc = ∂µ φa ∂ µ φa .

The terms linear in θ vanish because contracting an antisymmetric tensor with a symmetric tensor gives 0.
The Noether current associated with the infinitesimal symmetry is

∂L
jµ = δφa = abc nb ∂ µ φa φc ,
∂(∂µ φa )

which gives rise to conserved charges

Z Z
Qa = d3 x j 0 = d3 x abc nb φ̇a φc .

We verify this directly using the field equations ∂µ ∂ µ φa + m2 φa = 0 and throwing away surface terms:
Z Z Z
Q̇a = d x abc nb (φ̈a φc + φ̇a φ̇c ) = d x abc nb (∇ φa − m φa φc ) = d3 x abc nb (−∇φa · ∇φa ) = 0.
3 3 2 2

5. Lorentz transformation
Under a Lorentz transformation xµ → Λµν xν , the spacetime interval transforms as

ηµν dxµ dxν → ηµν Λµρ Λνσ dxρ dxσ .

The Minkowski metric is preserved if and only if ηρσ = ηµν Λµρ Λνσ .
Under an infinitesimal transformation Λµν = δ µν + ω µν , the Minkowski metric transforms as

ηρσ → ηµν (δ µρ + ω µρ )(δ νσ + ω νσ ) = ηρσ + ηρν ω νσ + ηµσ ω µρ + O(ω 2 ) = ηρσ + ωρσ + ωσρ + O(ω 2 ).

Thus the transformation is Lorentz if and only if ωρσ = −ωσρ .

An antisymmetric 4 × 4 tensor has 6 independent components, namely,
 0 1   0 1   0 1
 0  0  0 
K1 = −1 0 0 , K2 = −1 0 0 , K3 = 0
0 , J1 = 0
0 1 , J2 = 0 −1
0 , J3 = 0 1
−1 0 .
0 0 −1 0 −1 0 1 0 0

More compactly, (Ji )jk = ijk and (Ki )jk = ηij η0k − ηik η0j . The Ji , Ki correspond to spatial rotation and
Lorentz boost along the xi -axis. A rotation through an infinitesimal angle θ along the x3 -axis is given by
0 
µ µ 0 −θ
ω ν = θ(J3 ) ν = θ 0
.
0

Note that we have used the metric to raise an index. It generates a finite rotation by the exponential map
0
1  0  ! 1 
1 2 2 1 3 3 1 −θ − θ
cos θ − sin θ
exp(ω) = 1+θJ3 + θ J3 + θ J3 +· · · = 1 + θ
+ 2!
θ
− 2!
+· · · = sin θ cos θ
.
2! 3! 1 0 1 0

1
Similarly, a boost with rapidity β along the x is given by
0 β 
ω µν = β(K1 )µν = β 0
0
.
0

It generates a finite boost by the exponential map

 
0 β2
1  β   cosh β sinh β 
1 1 2!
exp(ω) = 1 + βK1 + β 2 K12 + β 3 K13 + · · · = 1
1 + β 0
0
+ β2
2!
= sinh β cosh β
1
.
2! 3! 1 0 0 1
0

2
6. Angular momentum of a field
Recall that a scalar field φ(x) transforms under a Lorentz transformation xµ → Λµν xν as φ(x) → φ(Λ−1 x).
Infinitesimally, Λµν = δ µν + ω µν and

φ(x) → φ(xµ − ω µν xν ) = φ(x) − ω µν xν ∂µ φ(x).

Similarly, ∂µ φ(x) → ∂µ φ(x) − ω σρ xρ ∂σ ∂µ φ(xν ). The Lagrangian changes by a surface term:

∂L ∂L ∂L µ ν ∂L
δL = δφ + δ∂µ φ = − ω ν x ∂µ φ − ω σ xρ ∂σ ∂µ φ(xν )
∂φ ∂(∂µ φ) ∂φ ∂(∂µ φ) ρ
 
µ ν ∂L ∂L
= −ω ν x ∂µ φ + ∂µ ∂ρ φ = −ω µν xν ∂µ L = −∂µ (ω µν xν L),
∂φ ∂(∂ρ φ)

where in the last step we commuted ∂µ past ω µν xν because ∂µ (ω µν xν ) = ω µν δ νµ = 0. The Noether current
associated with this transformation is
∂L ∂L
jµ = − ω ρ xν ∂ρ φ + ω ρν xν δ µρ L = −ω ρν T µρ xν , T µρ = ∂ρ φ − δ µρ L.
∂(∂µ φ) ν ∂(∂µ φ)

For spatial rotation along the xi -axis, ωjk = ijk The conserved angular momentum is
Z Z Z
1
Qi = d3 x j 0 = − d3 x ijk T 0j xk = ijk d3 x xj T 0k − xk T 0j .


2

For Lorentz boost along the xi -axis, ωjk = ηij η0k − ηik η0j . The conserved charge is
Z Z
Qi = d3 x(ηij η0k − ηik η0j )T 0j xk = d3 x T 0i x0 − T 00 xi .

Because Qi is conserved, Q̇i = 0. It follows that

Z Z Z Z
d 3 00 i d
d xT x = d x T x = d x T + t d3 x Ṫ 0i .
3 0i 0 3 0i
dt dt

The LHS gives the velocity of the center of energy density. The first term on the RHS side is the conserved
linear momentum along the xi -axis and so the second term is 0. Hence this equation says that the center of
energy travels at a constant speed.
7. EM tensor of an EM field
Under the gauge transformation Aµ → Aµ + ∂µ ξ, the field strength tensor transforms as

Fµν = ∂µ Aν − ∂ν Aµ → (∂µ Aν + ∂µ ∂ν ξ) − (∂ν Aµ + ∂ν ∂µ ξ) = Fµν .

1
Hence L = − Fµν F µν is gauge-invariant.
4
Under an infinitesimal spacetime translation xµ → xµ + µ , the potential transforms as Aµ → Aµ − ν ∂ν Aµ .
Thus the Noether current associated with the spacetime translational symmetry is

∂L
T µν = − ∂ν Aρ − δ µν L = −F µρ ∂ν Aρ − δ µν L.
∂(∂µ Aρ )

If we lower an index using the metric, we obtain a 02 tensor

Tµν = −Fµρ ∂ν Aρ − ηµν L.

However,
the first term is neither symmetric nor gauge-invariant! To salvage this situation, consider an-
other 02 tensor known as the Belinfante-Rosenfeld tensor

Θµν = Tµν − Fρµ ∂ ρ Aν = Fµρ F ρν − ηµν L.

This is clearly symmetric and gauge-invariant. In addition, it is trace-free:

Θµµ = −Fµρ F µρ − δ µµ L = 0.

3
8. Massive photon
The Proca Lagrangian describing a massive vector field is
1 1
L = − Fµν F µν + m2 Cµ C µ .
4 2
The equation of motion is
∂ν F µν = m2 C µ .
If m 6= 0, then we can apply ∂µ to both sides to get the Lorentz gauge condition

0 = ∂µ ∂ν F µν = m2 ∂µ C µ .

Since the gauge fixing condition always hold, the theory is not gauge-invariant (one can also see from the
action that Cµ → Cµ + ∂µ ξ is not a symmetry). However, if we treat ξ as an independent scalar field, then
the new Lagrangian
1
L = − Fµν F µν + m2 (Cµ + ∂µ ξ)(C µ + ∂ µ ξ)
4
is invariant under the gauge transformation Cµ → Cµ + ∂µ Λ and ξ → ξ − Λ. We can always recover
the original Lagrangian by choosing the gauge Λ = ξ. Thus we explicit see that ξ corresponds to the
longitudinal mode of the massive photon. Had we started with a massless field Cµ and gauge it with the
longitudinal mode ξ, then Cµ absorbs ξ and adds a mass term in the unitary gauge. This is the Higgs
mechanism.
To quantize this Lagrangian we need secondary class constraints. Note that ∂ν F 0ν = m2 C 0 implies

∂i (∂ 0 C i − ∂ i C 0 ) = m2 C 0 .

This allows us to express C 0 in terms of C i . The momenta conjugate to C i is

∂L
Πµ = = ∂µ C0 − ∂0 Cµ = Fµ0 .
∂(∂ 0 C µ )
Antisymmetry of Fµν implies that Π0 = 0. The Hamiltonian density, in terms of C0 , Ci and Πi , is
1 1 1 1
H = Πi Ċ i − L = Πi (∂ i C0 − Πi ) + F0i F 0i + Fij F ij − m2 C0 C 0 − m2 Ci C i
2 4 2 2
1 1 1 1
= − Πi Πi + Πi ∂ i C0 + (∂i Cj − ∂j Ci )∂ i C j − m2 C0 C 0 − m2 Ci C i .
2 2 2 2

9.Dil atation
Consider the Lagrangian for a real scalar field in n + 1 dimensional spacetime
Z  
n+1 1 µ 1 2 2 p
S= d x ∂µ φ∂ φ − m φ − gφ .
2 2
Under the dilatation (scaling transformation) x → λxµ , the scalar field and its derivatives transform as

φ(x) → λ−D φ(λ−1 x), ∂µ φ → λ−D−1 ∂µ φ.

The kinetic term transforms as

Z Z   Z
n+1 1 µ n+1 n+1 1 −2D−2 −2D+n−1 1
∂µ φ∂ µ φ dn+1 x ∂µ φ∂ µ φ.


d x ∂µ φ∂ φ → λ d x λ =λ
2 2 2
n−1
It is invariant if λ−2D+n−1 = 1, or D =. For n = 3, D = 1. The potential terms transform as
2
Z   Z  
1 2 2 1 −(n−1)+n+1 2 2
dn+1 x m φ + gφp → dn+1 x λ m φ + gλ−p(n−1)/2+n+1 φp .
2 2

To be invariant, the φ2 term must vanish, i.e., m = 0. Thus a scale-invariant theory cannot have a mass
2(n + 1)
term. We also need −p(n − 1)/2 + n + 1 = 0, or p = . For n = 3, p = 4. Hence the scale-invariant
n−1
Lagrangian in 3 + 1 dimensional spacetime is
1
L= ∂µ φ∂ µ φ − gφ4 .
2

4
Infinitesimally, λ = 1 +  and φ(x) → λ−1 φ(λ−1 x) ≈ (1 − )φ(x − x) ≈ (1 − )φ(x) − xν ∂ ν φ(x). The
Lagrangian changes by a surface term as L → L − ∂µ (xµ L). By Noether’s theorem, the conserved current
associated to scale-invariance is
∂L
Dµ = δφ + xµ L = −∂ µ φ(φ + xν ∂ ν φ) + xµ L.
∂(∂µ φ)