Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 2

Use the source action $S_{\mathrm{src}}=\int d^{4}x[J\varphi +{\bar{\eta }}_{\alpha }{\psi }_{\alpha }+{\bar{\psi }}_{\alpha }{\eta }_{\alpha }]$, with independent Grassmann variables used as sources $\eta ,\bar{\eta }$ and left functional derivatives. A scalar insertion is $(1/i)\delta /\delta J$. For the odd fields the ordered rules areThe middle sign is not arbitrary: differentiating the rightmost odd source in $\bar{\psi }\eta$ from the left gives ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. Derivative ordering must consequently match the displayed order of the fields; the rightmost derivative acts first.

Expand $\exp [ig\int \varphi \bar{\psi }\psi ]$ in its power series. Replace each field in every term by its source insertion, preserving the fermion order. These derivatives can be pulled outside the free path integral. Re-exponentiation gives the unnormalized generating functionalSummation over the spinor index is understood. Normalized correlation functions require division by the same interacting $Z[0,0,0]$, which cancels vacuum factors; that division is separate from this operator identity.

First fix the propagator kernels in a positive-sign Gaussian generating functional. From its scalar exponent, $i^{-2}{\delta }^2{Z}_0/\delta J\delta J$ gives the contraction $D(x-y)=-i\Delta (x-y)$. The fermionic insertion rules similarly give $F(x-y)=⟨T\psi (x)\bar{\psi }(y){⟩}_0=-iS(x-y)$, writing $S$ for the kernel labelled $S_F$ in that exponent. Thus the kernels in the question do not already include the factors of $i$ in the actual time-ordered contractions.

At order $g^2$, expand the interaction operator to its second term, or equivalently apply Wick's theorem toFor a connected contribution, the external scalar insertions attach to different interaction points. The two assignments are equal after exchanging $x,y$ and cancel $1/2$. The connected fermionic contraction isThe minus sign is the closed fermion loop sign. The alternative pairing gives two separate fermion tadpoles and belongs to disconnected external components, not the requested connected function. A free external scalar line multiplying a vacuum graph also cancels in the normalized result.

Combining $(ig{)}^2=-g^2$ with the loop minus sign givesThere are four contractions, so replacing $D=-i\Delta$ and $F=-iS$ introduces $(-i{)}^4=1$. ThereforeThis is the fermion-loop diagram in the last panel of i. The trace contracts spinor indices around the loop; it is not an additional degeneracy factor. A convention using physical contractions directly obtains the preceding formula in $D,F$ with the same coefficient. Mixing one convention's Gaussian kernels with another's insertion factors would change its apparent signs.

For two distinct commuting real scalar fields ${\chi }_1,{\chi }_2$ in place of the labelled fermion pair, with interaction $g\varphi {\chi }_1{\chi }_2$, there is no anticommutation sign on closing the loop and no spinor trace. The two propagators are simply multiplied. In terms of physical time-ordered scalar contractions $C_1,C_2$, the connected contribution isThe statistics and symmetry factors of a two-propagator loop determine the replacement. Relative to the fermionic formula in ii, the loop-statistics sign reverses. Scalar source kernels normalized like the stated Gaussian give the same overall minus sign after the four contraction factors are converted. There is no spin multiplicity.

If both replacement fields instead mean the same real field $\chi$, the coupling normalization and the identical-line symmetry factor must be specified. With interaction $(g/2)\varphi {\chi }^2$, the cubic vertex is $ig$ and its two-identical-line loop carries $1/2$, giving coefficient $-g^2/2$. With the literal coefficient $g\varphi {\chi }^2$, its vertex is $2ig$ and the coefficient is $-2g^2$. These cases are distinct from two separately labelled real fields; suppressing their combinatorics would hide a factor-of-two ambiguity in the replacement.

Choose an infinitesimal gauge transformation $\delta A_{\mu }^a=D_{\mu }^{ab}{\omega }^b$, where $D_{\mu }^{ab}={\delta }^{ab}{\partial }_{\mu }+g{f}^{acb}{A}_{\mu }^c$. Varying the axial gauge function at fixed $\sigma$ givesThis is the Faddeev-Popov operator. The Faddeev-Popov determinant has a Grassmann representation $\det M\propto \int \mathcal{D}\bar{c}\mathcal{D}c{e}^{i\int \bar{c}Mc}$. Choosing this normalization for the antighost givesA simultaneous antighost rephasing changes the overall ghost-action sign and vertex conventions, but not the determinant's field-dependent content. The essential facts are the kinetic operator $n\cdot \partial$ and a ghost-gluon vertex proportional to $n^{\mu }$.

Now average the fixed-$\sigma$ constraint with the stated Gaussian:up to a field-independent normalization. ConsequentlyThe Faddeev-Popov ghost determinant is not generally field independent before imposing the strict axial condition: its interaction term contains $n\cdot A$. In the $\xi \to 0$ gauge this quantity vanishes on the integration slice, giving the decoupling established next. Defining the inverse of $n\cdot \partial$ still requires boundary conditions for residual gauge transformations.

Contract the numerator of the quoted light cone gauge propagator with $n_{\mu }$:Thus every internal gluon emerging from a ghost-gluon vertex contributes a factor $n_{\mu }{\tilde{\Delta }}^{\mu \nu }=0$. Every ghost-dependent contribution to a physical gluon or matter amplitude vanishes. Equivalently, the strict condition $n\cdot A=0$ removes the Faddeev-Popov ghost interaction in i, leaving the field-independent determinant $\det (n\cdot \partial )$, which cancels from normalized correlators.

This is ghost decoupling in strict axial gauge, not a statement that the free Faddeev-Popov ghost propagator itself is zero. The algebra assumes a consistent axial-gauge pole prescription for $n\cdot p=0$ and treatment of residual gauge zero modes. The overall sign assigned to the displayed vector propagator does not enter the contraction proof. With the mostly-minus convention and the usual $-F^2/4$ action, its conventional prefactor is $-i/(p^2+i0)$; the source prints $+i$. Either prefactor gives the same zero contraction, so no sign correction is needed to obtain the requested ghost-decoupling result.