Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2016/3/c/solution

For , 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 and complete the square, . The normalized Gaussian path integral is
The source-independent functional determinant cancels in . For Fourier momentum , the mixed-index kernel is . For , use the projectors
Their orthogonality gives
Combining the projectors removes their apparent massless denominator. Adding the Feynman prescription gives the unitary-gauge propagator
Multiplication by verifies that it inverts the quadratic operator. On the physical pole , the numerator is transverse to , 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 , but the longitudinal tensor term remains of order . Naive loop power counting therefore looks worse and renormalizability is not manifest. In a finite gauge the corresponding expression is
which falls as at fixed finite . Its 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 , the longitudinal numerator drops out of and the tree exchange again falls as . The difficulty concerns off-shell Green functions and manifest control of loop divergences, not the number of physical polarizations.

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