Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 46 (Advanced Quantum Field Theory)

For the quantum rotor, the transition kernel is the path integralThe integral includes all paths on the circle, not just paths whose chosen real lifts have the same endpoint difference. Operator evolution and the stated energy-state normalization giveThe factor $1/(2\pi )$ comes from the completeness relation; it makes the zero-time limit the periodic Dirac delta distribution. For real time the oscillatory sum is understood through $T\to T-i0$, or as a distributional limit.

The Euler-Lagrange equation gives $\ddot{\theta }=0$. Lift the path to the real line beginning at ${\theta }_i$. Its endpoint can be any ${\theta }_f+2\pi m$, where $m\in \mathbb{Z}$ labels the relative winding number. With $\Delta ={\theta }_f-{\theta }_i$, every classical path and its action areThese exhaust the constant-velocity paths in the different winding sectors.

Apply the Poisson summation formula in the convention supplied, takingCompleting the square in its Fourier transform and evaluating the analytically continued Gaussian integral givesThus the spectral sum becomesThe square-root branch is fixed by continuation from positive imaginary-time damping. Each term is the classical contribution of part (b), multiplied by the same fluctuation prefactor. Since the action is quadratic, writing $\theta ={\theta }_m+\eta$ with $\eta (0)=\eta (T)=0$ removes the cross term and leaves a free Gaussian fluctuation integral independent of $m$. This real-time winding representation of a quantum rotor kernel is exact, rather than only a leading semiclassical approximation.

Fix the orientation of the Berezin integral by defining its measure as the linear functional that extracts the coefficient of ${\prod }_{i=1}^N({\theta }_i{\bar{\theta }}_i)$; all monomials of lower degree integrate to zero. In particular $\int \mathcal{D}(\bar{\theta },\theta ){\prod }_i({\theta }_i{\bar{\theta }}_i)=1$. This specifies the sign convention in the notation $d^N\bar{\theta }{d}^N\theta$.

The Grassmann change-of-variables formula givesFor a unitary matrix, $\det U\det U^{\ast }=1$, so the combined measure is invariant. Diagonalize the Hermitian matrix $B$ by a unitary change of variables. If its eigenvalues are $b_i$, the integrand factors as ${\prod }_i(1-b_i{\bar{\theta }}_i{\theta }_i)={\prod }_i(1+b_i{\theta }_i{\bar{\theta }}_i)$. Extracting its top coefficient proves the Grassmann Gaussian integralThe determinant identity also holds without Hermiticity by direct expansion; Hermiticity makes the diagonalization proof immediate.

Let $C(x)={\sum }_k{x}_k{C}^{(k)}$. Integrating out the Grassmann variables first gives the exact polynomial expressionThe requested factorization requires $B$ to be invertible, an assumption omitted from the question. Under it, use the formal perturbative matrix logarithm near $g=0$:ConsequentlyExpanding $C(x)$ gives exactly the displayed coefficients $V_{k_1\dots k_m}^{(m)}=(m-1)!\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$, and hence $I(g)=\det B\int d^{N}x{e}^{-x^{T}Ax/2-V_{\mathrm{eff}}(x)}$. The logarithmic representation is a formal series, not a globally convergent logarithm for every $x$: the determinant can have zeros. The original determinant integral remains well-defined for singular $B$, but $I(0)=0$ and the requested normalized expansion then cannot be formed.

There is a second convention to specify. The $x_i$ are real, so a complex Hermitian $A$ gives $x^{T}Ax=x^T{A}_{R}x$, where $A_R=\mathrm{Re}A=(A+A^T)/2$ is real symmetric positive definite. Its Gaussian covariance matrix is $A_R^{-1}$. It equals $A^{-1}$ only if $A$ is real symmetric. ThusThe effective Feynman rules are: a bosonic contraction joins indices $k,\ell$ with $(A_R^{-1}{)}_{k\ell }$; an $m$-leg vertex carries $-g^m\mathrm{Sym}{V}_{k_1\dots k_m}^{(m)}$; sum all closed vacuum diagrams with their symmetry factors, including disconnected diagrams. The symmetrization is necessary because the $x_k$ commute: the trace tensor is cyclically invariant but need not be fully symmetric. The minus sign comes from expanding $e^{-V_{\mathrm{eff}}}$. Taking the logarithm of the normalized integral instead keeps only connected vacuum diagrams.

Equivalently, retain the fermions: an oriented contraction is $(B^{-1}{)}_{ij}=⟨{\theta }_i{\bar{\theta }}_j⟩$, the mixed vertex has weight $+g{C}_{ij}^{(k)}$, and every closed fermion loop gives a minus sign. The same bosonic covariance joins its $x$ legs. Expanding these contractions recovers the trace vertices above. This is bosonic trace vertices from a finite fermionic determinant.

Use $\hslash =c=1$ and normalize the interaction of quartic scalar field theory as $\lambda {\varphi }^4/4!$. For a renormalized $n$-point connected correlation function, adopt ${\varphi }_B=Z_{\varphi }^{1/2}{\varphi }_R$ and ${\gamma }_{\varphi }=\frac{1}{2}\mu d\log Z_{\varphi }/d\mu$. Independence of bare quantities from the renormalization scale gives the Callan-Symanzik equationwhere $\beta =\mu d\lambda /d\mu$ and ${\gamma }_{m^2}=\mu d\log m^2/d\mu$, with bare parameters fixed. A convention using proper vertices reverses the field-normalization term, so the choice of Green's function must be stated.

Solve it by characteristics: set $\mu (s)={\mu }_0{e}^s$, $d\lambda (s)/ds=\beta (\lambda (s))$, and $d\log m^2(s)/ds={\gamma }_{m^2}(\lambda (s))$. At fixed external momenta,This is the characteristic solution of the multiplicative Callan-Symanzik equation. If the field anomalous dimension vanishes, the function is constant along the running-parameter characteristics. At a massless fixed point with constant anomalous dimension, the normalization factor is a power of $\mu /{\mu }_0$. Choosing the scale close to a characteristic external momentum avoids large perturbative logarithms.

For the supplied minimal subtraction scheme relation in dimensional regularization with $d=4-ϵ$, let $a=3/(16{\pi }^2)$ and write ${\lambda }_B={\mu }^{ϵ}[\lambda +a{\lambda }^2/ϵ+O({\lambda }^3)]$. Differentiating at fixed bare coupling givesA consistent expansion yields $\beta =-ϵ\lambda +a{\lambda }^2+O({\lambda }^3)$. The $-ϵ\lambda$ term must be retained while it multiplies the pole; setting $ϵ=0$ prematurely loses the finite result. In four dimensions the one-loop quartic scalar beta function isThe one-loop two-point diagram is a momentum-independent tadpole diagram, so there is no field wavefunction renormalization at this order and ${\gamma }_{\varphi }=O({\lambda }^2)$. Differentiating the supplied bare-mass relation in the same way gives ${\gamma }_{m^2}=\lambda /(16{\pi }^2)+O({\lambda }^2)$.

Integrating $d\lambda /d\log \mu =a{\lambda }^2$ gives the running couplingFor positive coupling it grows toward the ultraviolet and decreases toward the infrared: the theory is not asymptotically free. Its one-loop Landau pole lies at ${\mu }_L={\mu }_0\exp [16{\pi }^2/(3\lambda ({\mu }_0))]$. Perturbation theory fails before reaching that pole, so the pole in this approximation alone does not prove the existence of an exact physical divergence. The corresponding one-loop running mass satisfiesIt follows from $d\log m^2/d\lambda =1/(3\lambda )$. The mass-independent subtraction scheme keeps these beta functions independent of the mass; physical low-energy matching can still require threshold treatment. Running parameters resum the leading logarithms within the regime of small coupling.

Take metric signature $(+---)$ and the Fourier transform convention $e^{-ipx}$. Introduce an even source $J$ for the gauge field and odd sources $\eta ,\bar{\eta }$ for the Faddeev-Popov ghost field:At tree level the two-point functions come from the quadratic action. Integrating the gauge kinetic term by parts gives its momentum-space kernelIts inverse follows from transverse and longitudinal projectors: $K=-p^2(P_T+{\zeta }^{-1}{P}_L)$, so $K^{-1}=-p^{-2}(P_T+\zeta P_L)$. Completing the bosonic Gaussian square gives $Z_0[J]=\exp [-iJ{K}_F^{-1}J/2]$, with the Feynman i-epsilon prescription in the inverse. Two source derivatives and their insertion factors give the gauge-boson propagator $D_F=i{K}_F^{-1}$. HenceFor $\zeta =1$ this reduces to the Feynman-gauge adjoint propagator. The longitudinal double pole is understood with the compatible Feynman boundary prescription.

The ghost kinetic term as printed is $-{\partial }^{\mu }{\bar{c}}^a{\partial }_{\mu }{c}^a$. Integration by parts gives ${\bar{c}}^a\square c^a$, so its Fourier-space quadratic kernel is $K_c^{ab}(p)=-p^2{\delta }^{ab}$. Completing the Grassmann square in the same generating functional givesKeep the source order fixed, using a left derivative for $\bar{\eta }$ and a right derivative for $\eta$. Including the two insertion factors produces $⟨Tc\bar{c}⟩=i{K}_{c,F}^{-1}$, just as in the Gaussian generating functional for a Dirac field. ThereforeThe minus sign follows from the particular ghost Lagrangian in this paper. A convention with the opposite ghost kinetic sign has the opposite $c\bar{c}$ propagator; redefining the Faddeev-Popov antighost field changes its interaction and source signs at the same time. Mixing propagator and vertex conventions would give an erroneous loop sign. This is the ghost propagator for a negative derivative kinetic term.

Expanding the covariant derivative isolates the ghost-gluon vertex interactionThe derivative acts on the antighost. With all momenta incoming, let its momentum be $k$, the gluon's Lorentz index be $\mu$, and their color indices be $(a,b,c)$ in the order $(\bar{c},A,c)$. The convention $e^{-ikx}$ replaces ${\partial }^{\mu }$ by $-i{k}^{\mu }$, and multiplication by $i$ from $e^{iS}$ givesMomentum conservation is implicit. The structure constants reflect the Adjoint representation, and the antighost momentum follows directly from its differentiated field. Reversing the momentum convention changes the displayed vertex sign consistently; exchanging the ghost and antighost source order also carries a fermionic sign.