Quantum Field Theory

Quantum Field Theory

Set 16: solutions

Exercise 1

When we deal with the quantization of massless vectors we have to define the physical states of the theory and in addition the physical observables. These can be defined as all the operatorsO that, applied to a physical state, still give a physical state. This translates into a condition involving the commutator [O, L], whereL ≡q^ρaρ(q).

Indeed:

L|physi= 0, O|physi=|si,

L|si=LO|physi= [L, O]|physi+OL|physi= [L, O]|physi

If we want |sito be a physical state we don’t need to impose the vanishing of the commutator: is sufficient to require:

[L, O]∼L . We want to show that this is the case for the momentum Pν = R

d³x T0ν, the Noether charge associated to translations. Since Pν is a Noether charge we know that it can be regarded as the generator of translations, therefore we expect to find [Pµ, L]∼∂µL.

We now check that this is indeed the case. First we compute the conserved momentum Pµ, recalling that the photon field can be written in Fourier space as:

A_µ(x) = Z

dΩ_~kh

a_µ(~k, t) +a^†_µ(−~k, t)i e^i~k·~x.

From the Gupta-Bleuler Lagrangian we getT^0ν =−∂⁰A^ρ∂^νA_ρ+^g0ν₂ ∂^αA^ρ∂_αA_ρ. Let’s first focus on the energy, which receives contribution from both terms:

P₀=1 2

Z

d³x(−∂⁰A^ρ∂⁰A_ρ+∂ⁱA^ρ∂_iA_ρ) = Z

dΩ_q~ q⁰ 4

a_ρ(−~q, t)a^ρ(~q, t)−a_ρ(−~q, t)a^ρ†(−~q, t)−a^†_ρ(~q, t)a^ρ(~q, t) +a^†_ρ(~q, t)a^ρ†(−~q, t) + Z

dΩq~

qⁱqi

4q⁰

aρ(−~q, t)a^ρ(~q, t) +aρ(−~q, t)a^ρ†(−~q, t) +a^†_ρ(~q, t)a^ρ(~q, t) +a^†_ρ(~q, t)a^ρ†(−~q, t)

=

− Z

dΩ_~q q0 a^†_ν(~q, t)a^ν(~q, t),

where we have usedqⁱqi=−qⁱqⁱ=−q₀²and dropped infinite constants as usual. For the spatial components, only the first term in the energy-momentum tensor gives a non vanishing contribution and we have:

Pⁱ= Z

d³x(−∂⁰A^ρ∂ⁱA_ρ) = Z

dΩ~q

qⁱ 2

aρ(−~q, t)a^ρ(~q, t) +aρ(−~q, t)a^ρ†(−~q, t)−a^†_ρ(~q, t)a^ρ(~q, t)−a^†_ρ(~q, t)a^ρ†(−~q, t)

=

− Z

dΩ~q qⁱ a^†_ν(~q, t)a^ν(~q, t),

where we have noticed that two of the four terms vanish by antisymmetry for~q→ −~q. Thus finally:

Pµ=− Z

dΩ~k kµa^†_ν(~k)a^ν(~k), L(~q) =qρa^ρ(~q).

Note that the sign is correct, since for the transverse componentsa^⊥_ρ there is a +. Note also thatP_µis independent of time, so it is without loss of generality that we dropped the symboltin last equation.

Hence:

[Pµ, L(~q)] =− Z

dΩ~k kµq^ρh

a^†_ν(~k), aρ(~q)i

a^ν(~k) =− Z

d³k kµq^ρaρ(~k)δ³(~k−~q) =−qµq^ρaρ(~q) =−qµL(~q).

Then, definingL(q) =L(~q)e^−iq0t, we get:

L(x) =∂^µA⁻_µ(x) = Z

dΩ_~q e^−iqxL(~q) =⇒ [L(x), P_ν] =i∂_νL(x).

Exercise 2

The EOM is:

A_v−(1−ξ)∂_ν∂_µA^µ =−J_ν which can be cast in the form:

ΠνµA^µ(x) =−Jν(x) where:

Πµν ≡(ηµν−(1−ξ)∂ν∂µ) One can solve formally forAµ in terms of (Π⁻¹)^µν:

A^µ(x) =−(Π⁻¹)^µνJν(x).

At this point it is more convenient (but not compulsory) to work in momentum space:

A˜^µ(k) =− Π˜^−1µν

J˜ν(x) where:

Π˜_νµ=k²

η_νµ−(1−ξ)k_νk_µ k²

.

In order to find Π˜⁻¹µν

, it is useful to notice that ˜Πνµ can be split into a sum of orthogonal projectors P_L^µν, P_T^µν:

P_L^µν = k^µk^ν

k² , P_T^µν =η^µν−k^µk^ν k²

(PT)^µ_α(PT)^αν= (PT)^µν, (PL)^µ_α(PL)^αν= (PL)^µν, (PL)^µ_α(PT)^αν = 0, (PL)^µν+ (PT)^µν =η^µν Π˜_νµ=k²(P_T)_νµ+k²ξ(P_L)_νµ.

Therefore, it is easy to check that Π˜⁻¹µν

is simply given by the following expression:

Π˜⁻¹µν

= 1

k²(PT)^µν+ 1 k²

1

ξ(PL)^µν = 1 k²

η^µν+1−ξ ξ k^µk^ν

.

Forξ= 0, this expression is not well defined. This is expected, because in the presence of gauge-invariance the longitudinal part ofA^µ is not physical, therefore it cannot be determined from the EOM.

The choiceξ= 1 appears particularly simple:

Π˜^−1µν

= 1 k²η^µν

2

The Green function of the theory in momentum space is given by:

G˜^µν(k) =η^µν 1 (2π)²

1 k²

In order to find its expression in coordinate space, one has to take its anti Fourier-transform

G^µν(x) =η^µν 1 (2π)²

Z

d⁴ke^−ikx 1

k² =η^µν 1 (2π)²

Z

dk⁰d³k e^{−ik0t+i~k~x˙} 1 k²₀− |~k|²

The results depend on the convention used to go around the poles atk⁰=±|~k|in the complexk₀-plane.

Going above both poles corresponds to the retarded Green function G^µν_R. Indeed, for t < 0, the e^−ik0t factor is exponentially suppressed for Imk0 >0, therefore one can apply the Cauchy theorem for a closed integration contour contained in the upper complex plane. This does not embrace any pole, therefore the result of the integration is 0. Conversely, fort >0, the closed contour goes in the lower half plane, and it contains both poles.

By the Cauchy theorem, the result of the integration is:

G^µν_R(x) =η^µν Z

dk⁰d³k e^{−ik0t+i~k~x˙} 1

k₀²− |~k|² =η^µνθ(t) 1 (2π)²

Z

d³k2πi 1 2|~k|e^i~k~x˙

e^i|~k|t−e^−i|~k|t

The integration ind³kcan be performed by going to polar coordinates as usual:

G^µν_R(x) =η^µνθ(t) i 4π

Z ∞

0

d|~k|

Z 1

−1

dcosθ|~k|e^i|~k|rcosθ

e^i|~k|t−e^−i|~k|t

=

=−η^µνθ(t) 1 4πr

Z ∞

0

d|~k|

e^i|~k|r−e^−i|~k|r e^i|~k|t−e^−i|~k|t

=

=η^µνθ(t) 1 8πr

Z ∞

−∞

d|~k|

e^i|~k|(t−r)+e^{−i|~k|(t−r)}

Where in the last passage we omitted terms that would give rise toδ(t+r), which is 0 becauset+r >0. The final result is:

G^µν_R (x) =η^µνθ(t) 1

4πrδ(t−r)

Instead, the Feynman propagation is defined by the prescription k² →k²+i for →0⁺. This corresponds to going above the pole atk⁰=|~k|and below the pole atk⁰=−|~k|. Indeed:

k²₀− |~k|²+i∼

((k0+i)²− |~k|², k⁰>0 (k₀−i)²− |~k|², k⁰<0

Therefore, fort >0 (t <0) the closed contour embraces the pole atk⁰=|~k| (k⁰=−|~k|). Thus:

G^µν_F (x) =−η^µν i (2π)³θ(t)

Z d³k 1

2|~k|e^{−i|~k|t+i~k~x˙} −η^µν i

(2π)³θ(−t) Z

d³k 1

2|~k|e^{i|~k|t+i~k~x˙}

The Feynman prescription therefore implies the propagation of positive frequency in the future and negative frequencies in the past. It will be useful in the formalism of Feynman diagrams used in the perturbation theory for QFT. Performing the integration over angular variables one arrives at

G^µν_F (x) = −iη^µν (2π)²r

Z ∞

0

d|~k|sin(|~k|r)

θ(t)e^−i|~k|t+θ(−t)e^i|~k|t .

3

Making use of the formulae, 1

2i Z ∞

0

he^ik|(r−t)−e^−ik(r+t)i

= lim

→0

r

r²−t²+ 2it = r

r²−t² −rπi·sgn(t)δ(t²−r²), 1

2i Z ∞

0

h

e^ik(r+t)−e^−ik(r−t)i

= lim

→0

r

r²−t²−2it= r

r²−t² +rπi·sgn(t)δ(t²−r²), one obtains the final expression for the Feynman propagator in theξ= 1 gauge,

G^µν_F (x) = iη^µν

(2π)²(t²−r²)−η^µν

4πδ(t²−r²).

Exercise 3

The solution to the free Klein-Gordon-Fock equation is just a plain wave, ψ∝e^{−iEt+i~k·~x},

but the notion of energy in relativistic physics differs from that of the classical one, namely E=

q

m²+~k².

However, in the non-relativistic limitkmthey match in the following way,

∆E≡E_cl=E−m≈ ~k² 2m. Then it’s natural to decompose the full relativistic wave-function in this way,

ψ=e^−imtψ ,˜

where ˜ψis the non-relativistic part of the wave-function. Plugging this ansatz into the Klein-Gordon-Fock equation one arrives at

e^−imt ∂²

∂t²−2im∂

∂t−m²− ∇²+m²

ψ˜= 0. Next,

∂²

∂t²

ψ˜∼∆E²ψ˜m∂

∂t

ψ˜∼∆Emψ ,˜

thus, the first term in the parentheses is negligible compared to the second one. Since the third and the fifth terms are cancelled, the Klein - Gordon - Fock equation reduces to the Schroedinger equation,

i∂

∂t

ψ˜=−∇² 2m

ψ .˜

4