Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2016 / 3 / c / Solution

For $M>0$, integrate the vector action by parts, dropping the boundary term, and write it as a quadratic functional:The explicit mass makes this Proca kernel invertible: the displayed action alone has no gauge redundancy left to fix. It is interpreted as unitary gauge when obtained from a Higgs or Stückelberg completion with its would-be Goldstone field removed.

Introduce a source $J_{\mu }$ and complete the square, $A=A^{\prime }-K^{-1}J$. The normalized Gaussian path integral isThe source-independent functional determinant cancels in $Z[J]/Z[0]$. For Fourier momentum $k$, the mixed-index kernel is ${\tilde{K}}^{\mu }{}_{\nu }=-(k^2+M^2){\delta }^{\mu }{}_{\nu }+k^{\mu }{k}_{\nu }$. For $k^2\ne 0$, use the projectorsTheir orthogonality givesCombining the projectors removes their apparent massless denominator. Adding the Feynman prescription gives the unitary-gauge propagatorMultiplication by $\tilde{K}$ verifies that it inverts the quadratic operator. On the physical pole $k^2=-M^2$, the numerator is transverse to $k$, as required for the three physical massive-vector polarizations.

The drawback is its off-shell ultraviolet behaviour: at large spacelike momentum the transverse term falls as $1/k^2$, but the longitudinal tensor term remains of order $1/M^2$. Naive loop power counting therefore looks worse and renormalizability is not manifest. In a finite $R_{\xi }$ gauge the corresponding expression iswhich falls as $1/k^2$ at fixed finite $\xi$. Its $\xi \to \infty$ limit gives the unitary-gauge longitudinal term, so taking that limit before the ultraviolet limit changes power counting.

This is not a claim that a renormalizable Higgs theory ceases to be physically renormalizable in this gauge. Nor does it mean that the free Proca Gaussian integral is inconsistent. For conserved external sources $k\cdot J=0$, the longitudinal numerator drops out of $J^{\mu }{D}_{\mu \nu }{J}^{\nu }$ and the tree exchange again falls as $1/k^2$. The difficulty concerns off-shell Green functions and manifest control of loop divergences, not the number of physical polarizations.