Introduction to the

Lecture 8: Quantisation and Feynman Rules

Quantisation of Gauge Fields

problem with gauge fields: Given the field equation:

M µν A = J ν where M µν ∂µgµν ∂ν ∂µ µ ≡ − µν we see that because M ∂ν = 0, M is not invertible. The problem can be solved by using the fact that not all degrees of freedom for Aµ are physical (). This can be seen by applying a gauge transformation to Aµ:

A′µ = Aµ + ∂ν Λ ∂ A′µ = ∂ Aµ + Λ → µ µ

µ We can now choose Λ such that ∂µA = 0 which removes one degree of freedom from the vector field Aµ. The gauge function Λ is not completely determined; there is another gauge freedom

Λ Λ+Λ′ → such that Λ′ = 0. Then we have

Aµ A′′µ = Aµ + ∂µΛ+ ∂µΛ′ → µ µ µ ′ where ∂ Λ can be used to remove ∂µA and the term ∂ Λ can be used to remove another degree of freedom, e.g. A0 = 0 . Thus, Aµ now only has two degrees of freedom; the other two can be “gauged away.”

The mode expansion for the 4-component gauge field is

3 d3k A (x)= a (k)εr (k)e−ik·x + a†(k)εr∗(k)eik·x µ (2π)3 r µ r µ Z r=0 X  ∗ Notice that Aµ(x)= Aµ(x); the gauge fields are real-valued.

Appyling the gauge condition:

µ rµ ∂µA =0= kµε =0 ⇒ = k ε =0 A0 =0= εµ = (0, ε) ⇒ · ⇒ ) This is the transversality condition; it is consistent with the observation that EM radiation is tranversly polarised. µ 0 Note: kµεr = 0 is manifestly covariant whereas A = 0 is not.

1 By choosing a reference frame kµ = ω(1, 0, 0, 1), the polarisation vectors read:

ε1µ = (0, 1, 0, 0) 2µ Linearly polarised ε = (0, 0, 1, 0))

Using a basis change, one obtains the polarisation vectors which correspond to circularly polarised light:

±µ 1 µ µ 1 ε = (ε1 ε2 )= (0, 1, i, 0) ∓√2 ± ∓√2 ±

± εµ are the helicity eigenstates of the .

± Note: ε1,2,εµ correspond to the two observable degrees of freedom of the gauge field Aµ.

µ The quantisation of gauge fields is non-trivial because ∂ Aµ = 0 cannot be implemented at the operator level due to a contradiction with the canonical commutation relations. This issue is solved by the Gupta-Bleuler formalism (see Rel. QFT notes for more detail):

only states which correspond to transverse (ε±/ε1,2) are relevant for • .

unphysical degrees of freedom do not contribute in the scattering ( matrix) • S− elements.

The Gupta-Bleuler formalism works for any U(1) gauge but fails for Non-Abelian (this was later solved by Fedeev and Popov in 1958 -see Modern QFT).

Feynman Rules and Feynman Diagrams

The of a theory are determined by the propagation of the fields and the between them. To start, we shall mention some points:

Symmetries and the particle content determine the Lagrangian • The terms in the Lagrangian define propagation and of the particles (or • field quanta).

The quantum field theory foralism leads to computational rules to evaluate the matrix • S− elements: = i f where is the scattering operator (see further lectures). Sif h |S| i S b b

2 :

The is the Green’s function for the inhomogeneous field equation. i) propagator ( + m2)φ(x)= J(x) where J(x) is the source term that creates the inhomogeneity required for this definition; this result follows from the inhomogeneous Lagrangian: = + J(x)φ(x). L LKG

4 φ(x)= φ0(x)+ i d y G(x y)J(y) − Z b ( + m2)G(x y)= iδ(x y) − − − We use a Fourier ansatz: b

d4k G(x)= G(k)e−ik·x (2π)4 Z b d4k δ(x)= e−ik·x (2π)4 Z to find k i G(k)= ⇔ k2 m2 + iǫ −

ii) propagator (i∂/ m)S(x y)= iδ(x y) − − − b d4k S(x)= S(k)e−ik·x (2π)4 Z where S(k) (S(k)) is a matrixb in . ≡ αβ

k k/ + m i S(k)αβ = i 2 2 = α β ⇔ k m + iǫ! k/ m αβ − αβ  − 

iii) Gauge propagator 1 g + 1 ∂ ∂ Dνρ(x)= gρ δ(x) − µν − λ µ nu − ν   k  b i kµkν Dµν (k)= gµν + 1 λ ν µ ⇔ k2 + iǫ − k2 iǫ − −   3 where λ is a gauge fixing term:

λ = 1 Feynman Gauge λ =0 LandauGauge

µν µν In the Landau gauge, D (k) obeys transversality condition, kµD (k)=0.

Interaction Vertices

Derivation of Feynman Rules Each term in a Lagrangian that contains products of µ fields, ϕ1,...,ϕ ; ϕ φ,ψ,A , leads to an n- vertex: N j ∈  φ3(x3) φ4(x4)

φ1(x1) δ δ 4 Vφ1...φn (x1,...,xn)= i d x ∼ δφ1(x1) ··· δφn(xn) L  Z  φ2(x2) φn(xn) We desire a momentum space representation Fourier Transform Noether’s theorem → translational⇒ invariance ⇒ Poincar´einvariance ⇒ and momentum are conserved:

δ(p1 + + p ) overall ··· n A propagator (up to a minus sign) is the inverse two-vertex:

−1 Pφφ† = Vφφ† − k2→k2−iǫ   1 = − φ † ! 2 2 − φ k k iǫ → −

Note: The iǫ term is called the ‘Feynman prescription’ (or simply the ‘iǫ prescription’) and it ensures causality.

4 Recipe for deriving Feynman rules: Rather than actually performing the Fourier trans- form and the functional , the following rules can be used:

1.) Search for all terms in which contain a certain selection of the fields, e.g.: L

g(∂ A )AµBν = g(∂ Aρ)g AµBν − µ ν − µ νρ

2.) Replace all by ( i) times the incoming momenta of the respective fields (Fourier Transform): −

g(∂ Aρ(a))g Aµ(a′)Bν(b) igq g Aρ(q)Aµ(q′)Bν (p) − µ νρ → µ νρ

3.) Symmetrize indices of all identical bosonic fields:

igq g AρAµBν ig(q g + q′ g )Aρ(q)Aµ(q′)Bν(p) µ νρ → µ νρ ρ µν and omit external fields. The Feynman rule for the example vertex is then: Aρ q

p ′ ig(q g + q g ) ν µ νρ ρ µν ⇔ B q ′

Aµ

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