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Introduction to the Standard Model

Lecture 8: Quantisation and Feynman Rules

Quantisation of Gauge Fields

problem with gauge fields:Given the field equation: M we see that becauseMμν∂ν= 0,Mis not invertible. The problem can be solved by using the fact that not all degrees of freedom forAμare physical (observable). This can be seen by applying a gauge transformation toAμ: A We can now choose Λ such that∂μAμ= 0 which removes one degree of freedom from the vector fieldAμ. The gauge function Λ is not completely determined; there isanother gauge freedom such that?Λ?= 0. Then we have A μ→A??μ=Aμ+∂μΛ +∂μΛ?

where∂μΛ can be used to remove∂μAμand the term∂μΛ?can be used to remove another

degree of freedom,e.g.A0= 0 . Thus,Aμnow only has two degrees of freedom; the other two can be "gauged away." The mode expansion for the 4-component gauge field is A

μ(x) =?d3k

(2π)33 r=0? Notice thatAμ(x) =A?μ(x); the gauge fields are real-valued.

Appyling the gauge condition:

μAμ= 0 =?kμεrμ= 0

A

0= 0 =?εμ= (0,ε)?

=?k·ε= 0 This is thetransversalitycondition; it is consistent with the observation that EM radiation is tranversly polarised. Note:kμεμr= 0 is manifestly covariant whereasA0= 0 is not. 1 By choosing a reference framekμ=ω(1,0,0,1), the polarisation vectors read:

1μ= (0,1,0,0)

2μ= (0,0,1,0)?

Linearly polarised

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

±μ=?1

⎷2(εμ1±εμ2) =?1⎷2(0,1,±i,0)

μare thehelicity eigenstatesof the photon.

Note:ε1,2,ε±μcorrespond to the two observable degrees of freedom of the gauge fieldAμ.

The quantisation of gauge fields is non-trivial because∂μAμ= 0 cannot be implemented at the operator level due to a contradiction with the canonicalcommutation relations. This issue is solved by the Gupta-Bleuler formalism (see Rel. QFTnotes for more detail): •only quantum states which correspond to transverse photons(ε±/ε1,2) are relevant for observables. •unphysical degrees of freedom do not contribute in the scattering matrix (S-matrix) elements. The Gupta-Bleuler formalism works for anyU(1) gauge theory but fails for Non-Abelian theories (this was later solved by Fedeev and Popov in 1958 -see Modern QFT).

Feynman Rules and Feynman Diagrams

The dynamics of a theory are determined by the propagation ofthe fields and the interactions 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 interaction of the particles (or field quanta). •The quantum field theory foralism leads to computational rules to evaluate theS-matrix elements:Sif=?i|?S|f?where?Sis the scattering operator (see further lectures). 2 Propagators:Thepropagatoris the Green"s function for the inhomogeneous field equation. i) Scalar propagator (?+m2)φ(x) =J(x) whereJ(x) is the source term that creates the inhomogeneity requiredfor this definition; this result follows from the inhomogeneous Lagrangian:L=LKG+J(x)φ(x).

φ(x) =φ0(x) +i?

d

4y?G(x-y)J(y)

-(?+m2)?G(x-y) =iδ(x-y)

We use a Fourier ansatz:

G(x) =?d4k

(2π)4G(k)e-ik·x

δ(x) =?d4k

(2π)4e-ik·x to find k ?G(k) =ik2-m2+i? ii) Fermion propagator (i∂/-m)?S(x-y) =iδ(x-y)

S(x) =?d4k

(2π)4S(k)e-ik·x whereS(k)≡(S(k))αβis a matrix in spinor space. k ?S(k)αβ=i? k/+mk2-m2+i?? =?ik/-m? iii) Gauge boson propagator -gμν+?1-1 λ?∂μ∂nu??Dνρ(x) =-gρνδ(x) k

μν?Dμν(k) =ik2+i??

-gμν+kμkνk2-i??1-λ?? 3 whereλis a gauge fixing term:

λ= 1 Feynman Gauge

λ= 0 Landau Gauge

In the Landau gauge,Dμν(k) obeys transversality condition,kμDμν(k) = 0.

Interaction Vertices

Derivation of Feynman RulesEach term in a Lagrangian that contains products of fields,?1,...,?N;?j??φ,ψ,Aμ?, leads to ann-point vertex:

φ4(x4)

n(xn)φ

2(x2)φ

1(x1)φ

3(x3) ≂Vφ1...φn(x1,...,xn) =δδφ1(x1)···δδφn(xn)? i? d 4xL? We desire a momentum space representation→Fourier Transform

Noether"s theorem?

translational invariance?

Poincar´e invariance?

energy and momentum are conserved:

δ(p1+···+pn) overall

A propagator (up to a minus sign) is the inverse two-vertex: -1?????k2→k2-i?=

Pφφ†=-?

V -1???k2→k2-i? Note:Thei?term is called the 'Feynman prescription" (or simply the 'i?prescription") and it ensures causality. 4 Recipe for deriving Feynman rules:Rather than actually performing the Fourier trans- form and the functional derivative, the following rules canbe used:

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

2.)Replace all derivatives by (-i) times the incoming momenta of the respective fields

(Fourier Transform):

3.)Symmetrize indices of all identical bosonic fields:

igq and omit external fields.

The Feynman rule for the example vertex is then:

AμA

B νq q? pig(qμgνρ+q?ρgμν)? 5quotesdbs_dbs14.pdfusesText_20