Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 48 (Advanced Quantum Field Theory)

Write the Hamiltonian operator as $\hat{H}={\hat{p}}^2/(2m)+V(\hat{q})$, with $[\hat{q},\hat{p}]=i$, and define the quantum-mechanical propagator by $U(\beta ,\alpha ;T)=⟨\beta |e^{-iT\hat{H}}|\alpha ⟩$. Use $N$ intervals of length $\epsilon =T/N$, endpoint coordinates $q_0=\alpha ,q_N=\beta$, and the Trotter product formula. Inserting the completeness relations of the position eigenstates between the short-time factors givesNormalize momentum eigenstates by $⟨q|p⟩=e^{ipq}/\sqrt{2\pi }$. The Gaussian integral over the intermediate momentum isThe square-root branch is fixed by continuation from $\mathrm{Im}\epsilon <0$. Consequently the precise time-sliced configuration-space path integral isThis formula defines the notation $\int \mathcal{D}q{e}^{iS[q]}$; it is not an unnormalized infinite product of ordinary integrals. For real time take the boundary value from $T-i\delta$, $\delta >0$, or the corresponding oscillatory-distribution limit. For a self-adjoint, bounded-below Hamiltonian operator under the usual product-formula hypotheses, the operator limit supplies the definition even where the coordinate kernel is a distribution. The discrete exponent approaches the action with kinetic term $m{\dot{q}}^2/2$, and the prefactor is part of the functional measure. A different consistent slicing gives the same kernel for this separated kinetic-plus-potential Hamiltonian.

For the quantum harmonic oscillator, expand the action about a classical path $x$, writing $q=x+\xi$ with homogeneous Dirichlet boundary conditions on $\xi$. The only mixed term isThe boundary term vanishes and the Euler-Lagrange equation sets the remaining integrand to zero. Thus $S[x+\xi ]=S[x]+S[\xi ]$ exactly, rather than only to quadratic approximation. Translation of each integrated coordinate has unit Jacobian determinant; in the continuum measure convention of the question this is $\mathcal{D}q=\mathcal{D}\xi$. The classical-path factorization of a quadratic path integral follows:The fluctuation action, measure and boundary values depend on $T,m,\omega$ but not on $\alpha ,\beta$. For real $T$ with $\sin \omega T\ne 0$ the boundary-value classical path exists uniquely. At the conjugate times $\omega T=k\pi$, generic endpoints admit no such path and the kernel is a delta distribution; the formula is interpreted by the same regulated boundary limit, not as an ordinary finite prefactor times a nonexistent classical solution.

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.

Take $M^2>0$ initially to avoid an infrared divergence. A Schwinger parameterization and the Euclidean Gaussian integral give, in the convergent range $0<\mathrm{Re}d<4$,Integration over $t$ supplies the Gamma function. Its analytic continuation defines the answer in dimensional regularization:Use $d=4-ϵ$ near four dimensions and $d=6-ϵ$ near six dimensions. The Gamma function recurrence implies $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. ThereforeBoth are simple poles. Equivalently their residues as functions of $d$ are $-2/(4\pi {)}^2$ at $d=4$ and $2M^2/(4\pi {)}^3$ at $d=6$. The negative pole near six dimensions is a feature of analytic continuation; it is not a negative value of the original convergent positive integral. A massless zero-momentum integral is scaleless and cannot be used indiscriminately to extract these ultraviolet poles, because dimensional regularization then combines ultraviolet and infrared contributions.

For the following Feynman diagrams, specify the Euclidean action and additive counterterms explicitly so their signs are unambiguous. The four-dimensional interaction is ${\mu }^{ϵ}\lambda {\varphi }^4/4!$, the six-dimensional interaction is ${\mu }^{ϵ/2}g{\varphi }^3/3!$, and the common free inverse scalar propagator is $p^2+m^2$. Here $ϵ=4-d$ or $6-d$, respectively. The factors of $\mu$ give dimensionless renormalized couplings and change no leading pole residues. The cubic Euclidean theory is treated as a formal perturbation expansion about its massive quadratic scalar field theory; its real cubic potential is not bounded below.

Let $p$ be the total external Euclidean momentum crossing one four-point channel. The two internal scalar propagators produce the scalar bubble integralUsing the Feynman parameter identity $1/(ab)={\int }_0^{1}dx[xa+(1-x)b{]}^{-2}$ and shifting the loop momentum yieldsIn phi-fourth theory the bubble diagram has two quartic vertices and a symmetry factor $1/2$ from exchanging its two identical internal lines. There are three partitions of four labelled external legs into two pairs, the $s,t,u$ channels. The connected amputated loop diagram has positive coefficient ${\lambda }^{2}B/2$, whereas the Euclidean one-particle-irreducible vertex in the effective action has the opposite sign. ThusOne can check this sign without external-line conventions by expanding the one-loop quantum effective action $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ}\lambda {\varphi }^2/2)$. Its second logarithmic term is $-{\mu }^{2ϵ}{\lambda }^2\mathrm{Tr}(K^{-1}{\varphi }^2{K}^{-1}{\varphi }^2)/16$; four field derivatives give precisely the three negative bubble terms above.

The four-dimensional pole of $I_d$ is independent of its mass argument, so each channel contributes $2/[(4\pi {)}^2ϵ]$. The divergence of the proper four-point vertex is $-3{\lambda }^2/[(4\pi {)}^2ϵ]$. The minimal subtraction scheme therefore requiresFor a Minkowski interaction $-{\mu }^{ϵ}\lambda {\varphi }^4/4!$, the corresponding local interaction counterterm is $-{\mu }^{ϵ}\delta \lambda {\varphi }^4/4!$. No momentum-dependent four-point counterterm is needed for these logarithmic poles. This calculation concerns the requested four-point bubble contributions; a massive two-point tadpole and vacuum diagrams require their own mass and vacuum-energy subtractions if those functions are also renormalized.

In phi cubed theory, a two-point bubble diagram has two cubic vertices, two internal scalar propagators and symmetry factor $1/2$. Its connected amputated insertion is $+{\mu }^{ϵ}{g}^2{B}_d(p)/2$. The inverse scalar propagator, or quadratic one-particle-irreducible vertex, therefore receives the negative insertion:Equivalently, the connected two-point correction is $G_0(p{)}^2{\mu }^{ϵ}{g}^2{B}_d(p)/2$, and expansion of its inverse changes the sign. The same sign follows from the second term of $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ/2}g\varphi )$.

Using the six-dimensional residue and ${\int }_0^{1}x(1-x)dx=1/6$, the divergent inverse-propagator correction isIts momentum term requires wave-function renormalization, unlike the massive quartic tadpole. The minimal-subtraction two-point counterterms in cubic scalar theory areThese are additive coefficients in a Lagrangian expressed in terms of the renormalized field. If ${\varphi }_0=Z^{1/2}\varphi$, $Z=1+\delta Z$, and the bare mass is separately written $m_0^2=m^2+\Delta m^2$, then $\delta m^2=m^2\delta Z+\Delta m^2$ at this order, so $\Delta m^2=-5g^2{m}^2/[6(4\pi {)}^3ϵ]$. Distinguishing these conventions prevents a spurious disagreement between mass counterterms. The local Minkowski counterterms are $\delta Z(\partial \varphi {)}^2/2-\delta m^2{\varphi }^2/2$. A complete cubic theory also has one-point and three-point subtractions; they are not the specified two-point bubble poles.

A Berezin integral over independent Grassmann variables selects the coefficient containing every variable. Fix the real orientation by $\int d{\omega }_1\cdots d{\omega }_{2m}{\omega }_{2m}\cdots {\omega }_1=1$. For an antisymmetric matrix $A$, only the order-$m$ term in the exponential survives:Anticommutation collects its top coefficient into the signed sum over pairings that defines the Pfaffian. Reversing the order of $2m$ variables gives $(-1{)}^{m(2m-1)}=(-1{)}^m$, cancelling the prefactor sign. The real Grassmann Gaussian integral is thereforewith this orientation, and a fixed sign times this answer with other orientations. In dimension two, write $A_{12}=a$; the exponential is $1-a{\omega }_1{\omega }_2=1+a{\omega }_2{\omega }_1$, and integration gives $a$. Although $\mathrm{Pf}(A{)}^2=\det A$, writing a positive square root $\sqrt{\det A}$ would give $|a|$ in this test and would lose the correct sign.

For the complex Grassmann Gaussian integral, treat ${\theta }_a$ and ${\bar{\theta }}_a$ as independent integration variables. Every surviving top term contains each $\theta$ and each $\bar{\theta }$ once. Expanding the exponential assigns one matrix entry from every row and column, and anticommutation supplies the permutation sign. Hencewhere $C_n$ is a nonzero constant fixed by the order and normalization of the differentials, independent of $M$. This argument works for singular matrices as well and does not assume that $M$ is Hermitian.

For the explicit $n=2$ check, put $Q={\bar{\theta }}_a{M}_{ab}{\theta }_b$. Moving all the $\theta$ variables before the barred ones givesThe linear and constant terms integrate to zero. Thus the integral is a fixed orientation sign times $M_{11}{M}_{22}-M_{12}{M}_{21}$, exactly the determinant. Choosing the top-monomial integral to be $-1$ gives $C_2=1$.

For the required real-complex compatibility of Grassmann Gaussian integrals, let $u_a={\omega }_a$ and $v_a={\omega }_{n+a}$. Substitution givesIndeed $v^{T}Mu=-u^T{M}^{T}v=u^{T}Mv$ because the variables anticommute and $M^T=-M$. The real quadratic matrix is consequently $A=\mathrm{diag}(M,M)$, of size $2n$. The linear change of Grassmann variables has a constant nonzero Berezin integration Jacobian, so it changes only the allowed normalization factor. For even $n$, the two blocks giveFor odd $n$, $\det M=\det (-M)=(-1{)}^n\det M$ implies $\det M=0$. Each of the separate $u$ and $v$ quadratic exponentials has only even degree and cannot saturate an odd number of variables, so the real integral vanishes too. This covers the odd case without attempting to define a Pfaffian of an odd-dimensional matrix.

Let a gauge transformation act infinitesimally as $\delta A_{\mu }^a=(D_{\mu }\omega {)}^a$, and define the Faddeev-Popov operator by variation of the chosen gauge functional:Work in a perturbative regular gauge slice, with boundary conditions that remove residual zero modes and with one local representative of each gauge orbit. The Faddeev-Popov gauge-orbit identity is $1={\Delta }_{\mathrm{FP}}[A]\int \mathcal{D}\omega \delta [\mathcal{F}[A^{\omega }]]$. Inserting it and factoring out the formal gauge-group volume givesThe functional delta imposes the gauge condition at every point and for every color. The determinant compensates the gauge-orbit Jacobian. For ordinary unoriented real gauge-orbit integration the local Jacobian is $|\det \mathcal{M}|$; perturbatively its sign is fixed and absorbed into $\mathcal{N}$, leaving the displayed determinant. Multiple global intersections, or a Gribov ambiguity, require more care; the local perturbative formula is not a global uniqueness assertion.

The complex Grassmann Gaussian integral now generalizes to spacetime and color indices. Introduce independent anticommuting Faddeev-Popov ghost fields $c^a(x),{\bar{c}}^a(x)$ and writeThe factors $i$ in the functional determinant differ from a real exponential only by a regulated, field-independent constant. These Grassmann fields are Lorentz scalars in the adjoint color representation; the antighost is independent of the ghost rather than an ordinary complex-conjugate commuting variable. They are unphysical fields, not external observable particles. Representing the functional determinant by a local ghost action when $\mathcal{F}$ is local makes ordinary Feynman rules available. Anticommutation gives the minus sign of a closed ghost loop, enabling cancellation of unphysical gauge contributions and maintaining the gauge identities in perturbation theory.

For example, the linear covariant choice ${\mathcal{F}}^a={\partial }^{\mu }{A}_{\mu }^a$ gives ${\mathcal{M}}^{ab}={\partial }^{\mu }{D}_{\mu }^{ab}$, so non-abelian ghost-gauge interactions remain. In an axial gauge ${\mathcal{F}}^a=n^{\mu }{A}_{\mu }^a$ with constant $n$, however,on the exact gauge slice. Its determinant is field-independent, so axial-gauge ghosts decouple and contribute only a normalization factor. This statement uses the strict delta-functional constraint; a finite-width Gaussian gauge weight is not the same exact slice.

The disadvantage is that choosing $n$ obscures manifest Lorentz covariance and the axial-gauge propagator contains spurious $1/(n\cdot k)$ poles. An axial-gauge pole prescription and control of residual transformations satisfying $n\cdot \partial \omega =0$ are necessary; formal decoupling alone does not fix those problems.

Abelian ghosts do not always decouple. In a field-independent linear Lorenz gauge, $\delta A_{\mu }={\partial }_{\mu }\omega$ gives $\mathcal{M}=\square$ and they are free. But the quadratic Abelian gauge fixing $\mathcal{F}[A]=\partial \cdot A+\kappa A_{\mu }{A}^{\mu }$ givesIts Faddeev-Popov operator depends on $A$, and the ghost action contains $2\kappa \bar{c}{A}^{\mu }{\partial }_{\mu }c$. This is an explicit abelian ghost-gauge interaction. Abelianity removes the commutator term in the gauge transformation, not the field dependence of an arbitrary nonlinear gauge condition.