Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 48 / 2

Use a real scalar field and Minkowski action $S[\varphi ]$, with the vacuum selected by a Feynman propagator pole prescription or adiabatic vacuum boundary conditions. The normalized correlation functions areHere $|\Omega ⟩$ is the interacting vacuum and $\mathcal{T}$ denotes time ordering. These are the canonical vacuum expectation values, rather than unordered products of noncommuting fields. The functional integral formula assumes a regulated measure and the same vacuum prescription on both sides.

Couple an external source $J$ linearly to the scalar field and choose the normalized vacuum generating functionalThen functional derivatives insert the fields:In the convention used below, the connected generating functional isThe logarithm removes products of disconnected components. For example the connected correlation function $G_2^c=G_2-G_1{G}_1$; at three points one subtracts all three $G_2{G}_1$ products and adds $2G_1{G}_1{G}_1$. Normalization has already removed vacuum bubbles.

Keep $Z=e^{iW}$, define the mean scalar field $\phi (x)=\delta W/\delta J(x)$, and locally invert the source-field relation. The quantum effective action is the Legendre transformIts value at zero source is expanded around ${\phi }_0=⟨\varphi ⟩$, which need not vanish. The kernels ${\Gamma }^{(n)}={\delta }^n\Gamma /\delta {\phi }^n{|}_{{\phi }_0}$ are the one-particle-irreducible vertices, with the Minkowski diagram vertex factor $i{\Gamma }^{(n)}$. A one-particle-irreducible Feynman diagram remains connected after any one internal line is cut. Amputation of external propagators removes the full two-point functions on its external legs. For higher-point connected graphs amputation alone need not remove internal one-particle-reducible Feynman diagrams; the derivatives of $\Gamma$ select precisely the irreducible kernels.

To derive the three-point relation, use condensed indices that include spacetime integration. Put $D_{ij}=W_{,ij}=i{G}_{2,ij}^c$ and $K={\Gamma }^{(2)}$. Differentiating ${\Gamma }_{,i}=-J_i$ gives $K_{ia}{D}_{aj}=-{\delta }_{ij}$, so $K=-D^{-1}$. A further derivative, using $\delta D_{ab}/\delta {\phi }_k=W_{,ab\ell }(D^{-1}{)}_{\ell k}$ and the derivative of an inverse, givesSince $W^{(3)}=-G_3^c$ and $D^{-1}=-i(G_2^c{)}^{-1}$, the three-point amputation in the connected-action convention isEquivalently, attach a full two-point correlation function to each leg of $i{\Gamma }^{(3)}$ to reconstruct $G_3^c$. The connected functions are evaluated at the same zero-source background. The displayed factors of $i$ follow from the declared Minkowski convention; Euclidean generating functionals redistribute these factors.

For a bosonic quadratic action $S_2=\frac{1}{2}\int \varphi K\varphi$, completing the square in the generating functional gives $Z_0[J]/Z_0[0]=\exp (-iJ{K}^{-1}J/2)$. Two functional derivatives therefore give the Feynman propagator $\Delta =i{K}^{-1}$, with boundary conditions and a pole prescription included in the inverse. Zero modes must be removed or fixed before an inverse exists.

Take Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the Yang-Mills field strength $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$. The interaction terms do not contribute to the quadratic action. Integration by parts in $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-(\partial \cdot A^a{)}^2/(2\alpha )$ yieldsThe color kernel is diagonal. For $k^2\ne 0$, use the projectors $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$ and $P_T^{\mu \nu }={\eta }^{\mu \nu }-P_L^{\mu \nu }$. They obey $P_L^2=P_L$, $P_T^2=P_T$, $P_L{P}_T=0$, soMultiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the Feynman propagator prescription gives the gluon propagatorThe second term has the causal double-pole prescription. Away from the pole this is the usual $-i{\delta }^{ab}[{\eta }_{\mu \nu }-(1-\alpha )k_{\mu }{k}_{\nu }/k^2]/(k^2+i0)$. For $\alpha =1$ it becomes the Feynman-gauge adjoint propagator; $\alpha \to 0$ gives a transverse covariant Landau gauge propagator.

The limit $\alpha \to \infty$ removes gauge fixing and has no finite inverse on the full field space. The longitudinal eigenvalue $-k^2/\alpha$ tends to zero and its inverse diverges. A contraction with conserved external currents kills the longitudinal term, but that does not define a gauge-unfixed Gaussian integral. This massless limit is not the massive-vector unitary-gauge limit: here it exposes the unfixed gauge orbit degeneracy.