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A background gauge field is a nondynamical source for a conserved current, rather than a field integrated over in the path integral. For a Dirac current with source term $+B_{\mu }{j}^{\mu }$, the combined transformations $\psi \mapsto e^{i\alpha }\psi$ and $B_{\mu }\mapsto B_{\mu }+{\partial }_{\mu }\alpha$ preserve the action. Whether this also preserves the generating functional depends on the absence of the corresponding gauge anomaly.

A perturbative correlator factors into an exponential of connected Vacuum Feynman diagrams times a sum of Feynman diagrams attached to external insertions. Dividing its path integral by the vacuum generating functional at zero source cancels that exponential, including all disconnected products of vacuum components. The corresponding operator formalism uses the same denominator in the vacuum expectation of the Dyson series. This leaves externally disconnected correlators intact; the connected generating functional removes their products.

The logarithm of a generating functional generates connected correlation functions because its cumulant expansion removes products of disconnected vacuum components.

The quantum effective action is the Legendre transform of the generating functional of connected correlation functions. Its functional derivatives are the one-particle-irreducible vertex functions.

A generating functional is a path integral with a source $J$ coupled to a field. Its functional derivatives with respect to $J$ generate time-ordered correlation functions.

A Grassmann derivative is an odd differentiation operation on a Grassmann algebra. A left Grassmann derivative obeys ${\partial }_{\theta }^L(ab)=({\partial }_{\theta }^{L}a)b+(-1{)}^{|a|}a({\partial }_{\theta }^{L}b)$ for homogeneous $a$. Right derivatives use the corresponding right-sided product rule. Their placement must be fixed when differentiating fermionic sources in a generating functional, since commuting ordinary derivatives would lose fermionic signs.

The linked-cluster theorem states that the logarithm of a vacuum generating functional is the sum of connected vacuum diagrams. Exponentiating that sum reconstructs arbitrary products of disconnected connected components with their correct factorials.

Use the Minkowski generating functional convention $Z=e^W$, source insertion $iJ\varphi$, classical field $\phi =-i\delta W/\delta J$, and $i\Gamma =W-i\int J\phi$. With $G_r^c=i^{-r}{\delta }^{r}W/\delta J^r$, the response kernel is $\delta \phi /\delta J=i{G}_2^c$. Differentiating $\delta \Gamma /\delta \phi =-J$ once gives the displayed two-point vertex, where the vertex means a derivative of $i\Gamma$. Differentiating the inverse kernel gives the three-point vertex: the connected three-point function is amputated by one inverse two-point kernel on each leg. In the alternative convention $W_c=-i\log Z$, the factors of $i$ are redistributed.

For a vacuum generating functional $Z[J]$, normalization gives $\hat{Z}[J]=Z[J]/Z[0]$ and $\hat{Z}[0]=1$. Functional derivatives of $\hat{Z}$ give normalized vacuum expectation values. The linked-cluster theorem shows that this ratio cancels Vacuum Feynman diagrams, while $-i\log \hat{Z}[J]$ gives connected correlators. Components attached to external insertions are not removed merely by the normalization.

At first order the Dyson expansion inserts $i{S}_{\mathrm{int}}=\lambda \int d^{4}x{\bar{\psi }}_{\alpha }(x)({\gamma }_5{)}_{\alpha \beta }{\psi }_{\beta }(x)\varphi (x)$. Contract the scalar field at the vertex with the external scalar. The four remaining fermion fields have two admissible pairings: both external fermions attach to the vertex, or the two vertex fermions contract with each other while the external pair contracts freely. Wick's theorem givesThe minus sign in the second term is the fermion sign of the closed contraction: $⟨T{\bar{\psi }}_{\alpha }(x){\psi }_{\beta }(x)⟩=-i{S}_F(0{)}_{\beta \alpha }$. The first term is connected; the second is a scalar-attached fermion tadpole multiplied by a disconnected free external fermion propagator. The full time-ordered expectation includes both topologies, whereas the connected generating functional retains only the first. There is no order-${\lambda }^2$ contribution in this pure Yukawa expansion because the total number of scalar fields would be odd.

For a parity-preserving free propagator, the tadpole coefficient is $\mathrm{tr}[{\gamma }_5{S}_F(0)]$, whose integrand contains only $\mathrm{tr}{\gamma }_5$ and $\mathrm{tr}({\gamma }_5{\gamma }^{\mu })$. Both vanish. Thus the second term is a legitimate formal Wick contraction but is zero with a compatible regulator; normal ordering the interaction removes it from the outset.

Use the source term $\int d^{4}x[J\varphi +\bar{\varsigma }\psi +\bar{\psi }\varsigma ]$ and left Grassmann functional derivatives. Differentiating its exponential inserts fields according toThe sign for $\bar{\psi }$ follows because a left derivative with respect to $\varsigma$ must cross $\bar{\psi }$. Keep the order $\bar{\psi }{\gamma }_5\psi \varphi$ when replacing these fields. Expand $e^{i{S}_{\mathrm{int}}}$ in the original path integral; each field monomial can now be generated by these ordered derivatives acting on the free functional. Resumming givesThe arguments refer respectively to $\varphi ,\bar{\psi },\psi$ at $x$. The free functional arises by performing the free Gaussian field integrals, and $\mathcal{N}$ may be chosen so that $Z[0]=1$.

No additional field-integration measure remains in this differential-operator formula. Such a measure is printed outside the operator in the source even though $Z_0$ has already integrated out those fields. At $\lambda =0$, that literal expression would contain $\int \mathcal{D}\varphi \mathcal{D}\bar{\psi }\mathcal{D}\psi 1$ rather than simply $Z_0$, which is not the free generating functional. Also, when ${\mathcal{L}}_{\mathrm{int}}$ denotes a density, the spacetime integral in the exponent is required.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1

The dated cover establishes Trinity Term 2018, despite the containing bundle's 2017 filename. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. In question 1(a), $D=4-ϵ$ and $e_D={\mu }^{ϵ/2}e$. The supplied transverse photon propagator selects Landau gauge within the Lorenz-gauge family. In question 2(a), distinguish the kernels in the stated generating functional from time-ordered Wick contractions, as explained below.

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.

After integration by parts, the quadratic Proca action with an external source isSince $M>0$, this kernel is invertible with the Feynman i-epsilon prescription; the massless gauge redundancy is absent in this unitary gauge quadratic action. Complete the square by $A\mapsto A+K_F^{-1}J$ in the Gaussian functional integral. Its source-independent determinant cancels in the normalized generating functional:In Fourier space, the kernel isFor $k^2\ne 0$, use the transverse projector of a vector field $P_T^{\mu \nu }=g^{\mu \nu }-k^{\mu }{k}^{\nu }/k^2$ and the longitudinal projector of a vector field $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$. Their mixed-index products obey $P_T{P}_L=0$, soThe last expression also extends to $k^2=0$; that point only obstructs the intermediate projector decomposition. Two source functional derivatives yield the gauge-boson propagator $D_F=i{K}_F^{-1}$:The pole residue is the physical massive-vector massive vector polarization sum. The longitudinal term is essential for inverting the quadratic kernel, and makes the naive $M\to 0$ limit of this unitary gauge Proca propagator singular.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2020.pdf

The annual bundle's filename says 2019, but this paper's cover says Trinity Term 2020 and identifies A15272W1. These are solutions to that 2020 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $e_D={\mu }^{ϵ/2}e$, and $\mu$ is the renormalization scale; keeping the dimensionful coupling $e_D$ explicit prevents a factor-of-two ambiguity in the ultraviolet pole. In question 2(a), the factors of $i$ in the stated generating functional fix the normalization of its kernels, as explained in the solution.

First fix a convention that is implicit in the source. For the stated free generating functionalthe time-ordered Wick contractions are $S_F=⟨T\psi \bar{\psi }{⟩}_0=-iS$ and ${\Delta }_F=⟨TAA{⟩}_0=-i\Delta$. These follow directly by the source differentiation rules in part (i). Thus the kernels called propagators in the question are distinguished here from the contractions, which already include their factors of $i$.

At first order in the QED interaction, write $a$ for its integration point. The normalized three-point correlation function isThe zeroth-order term vanishes because the free photon one-point correlation function is zero. There are two nonzero Wick contractions of the four fields:The first term pairs the two external fermions directly and closes the two fields at the interaction into a fermion loop; its minus sign is the closed fermion loop sign. The second term joins the external fermions through the interaction, giving the connected contribution. This is also obtained by differentiating the quartic term $(i\bar{\eta }S\eta {)}^2/2$ in the free generating functional; the two choices of bilinear cancel its factor $1/2$.

Insert $S_F=-iS$ and ${\Delta }_F=-i\Delta$. Since $(-ie)(-i{)}^3=e$, the result is exactlyThe trace term is a disconnected product of a photon tadpole diagram with the free external fermion propagator; the other term is the connected three-point correlation function. In this QED calculation the disconnected term vanishes, as established next. With a convention using $S_F,{\Delta }_F$ themselves as the symbols for the propagators, the equivalent formula retains the prefactor $-ie$ and the connected-minus-trace ordering above; mixing these two conventions would give an erroneous overall phase.

Use the path integral with source termThe sources $\eta ,\bar{\eta }$ are Grassmann variables; all derivatives below are left functional derivatives. Because $\eta$ stands to the right of $\bar{\psi }$, differentiating with respect to it introduces a minus sign. The insertion rules are thereforeFor example ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. The order of the odd functional derivatives matters: the rightmost derivative acts first, and the displayed product inserts ${\bar{\psi }}_{\alpha }{\psi }_{\beta }$ in that order.

Expand $\exp [-ie\int \bar{\psi }{\gamma }^{\mu }\psi A_{\mu }]$ in its power series, replace each field by its source derivative, and pull the derivatives outside the free path integral. Re-exponentiating givesThis identity concerns the unnormalized generating functional. Normalized correlation functions divide by its zero-source value so that vacuum contributions cancel.

Normalize the generating functional by its value with all sources zero, and take the connected part of the time-ordered correlation function. Source differentiation of the interaction exponential is equivalent to the Dyson series followed by Wick's theorem. Write $D(x-y)=⟨T\varphi (x)\varphi (y){⟩}_0$ and $F_{ij}(x-y)=⟨T{\psi }_i(x){\bar{\psi }}_j(y){⟩}_0$ temporarily, so that every contraction has an unambiguous meaning.

The second-order Dyson series contributesThere are two connected external scalar field attachments:They become equal after exchanging the integration variables $x,y$, cancelling $2!$. The connected fermion contractions have the closed fermion loop sign:The minus sign can also be checked directly: the pairing of the first with the fourth field and the second with the third has positive pairing parity, but $⟨T{\bar{\psi }}_i(x){\psi }_j(y)⟩=-F_{ji}(y-x)$. Multiplying it by $(ig{)}^2=-g^2$ leaves a positive result:A contraction of the two external fields with each other times a vacuum bubble cancels against the normalization. Separate fermion tadpole integrals at $x$ and $y$ belong to a product of one-point functions, removed by taking the connected part.

To match the source's generating functional convention, distinguish its displayed kernels from contractions. The positive $iJ\Delta J/2$ exponent gives $D=-i\Delta$. With the source insertion order in equation (5), the positive $i\bar{\eta }{S}_F\eta$ exponent similarly gives $F=-i{S}_F$. The two scalar propagators and two Dirac propagators supply a total phase $(-i{)}^4=1$. Thus, writing $S=S_F$ as in the requested formula,The matrix trace closes the spinor indices; no extra factor of two or additional symmetry factor remains for a Dirac field with this Yukawa interaction.

For a Dirac field, the path integral is a Berezin integral over independent Grassmann fields $\psi$ and $\bar{\psi }$:The vacuum boundary conditions and the Feynman i-epsilon prescription are understood; $\mathcal{N}$ fixes the vacuum normalization. This is the vacuum generating functional without sources, rather than a finite-temperature partition function.

The Dirac adjoint transforms as $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\alpha }$. For constant $\alpha$, ${\partial }_{\mu }(e^{i\alpha }\psi )=e^{i\alpha }{\partial }_{\mu }\psi$. Both the kinetic term and the Dirac mass term therefore retain their values. This proves the global phase symmetry, whose group is the unitary group $U(1)$.

The scalar field path integral integrates over real field histories with vacuum boundary conditions. With $S_0[\varphi ]=\int d^{4}x{\mathcal{L}}_0$, the partition function isThe constant $\mathcal{N}$ fixes the desired vacuum normalization. A generating functional adds $\int d^{4}xJ\varphi$ to the action; the required partition function is its zero-source value. Equivalently, a Wick rotation gives $Z_E={\mathcal{N}}_E\int \mathcal{D}\varphi e^{-S_E[\varphi ]}$, with the positive-sign kinetic and potential terms in the Euclidean action when $m^2\ge 0$ and $\lambda >0$.