Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 304 (Advanced Quantum Field Theory)

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.

Use Lie algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$, and define the adjoint gauge covariant derivative by $(D_{\mu }X{)}^a={\partial }_{\mu }{X}^a+f^a{}_{bc}{A}_{\mu }^b{X}^c$. Varying the gauge field strength gives $\delta F_{\mu \nu }=D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$. Antisymmetry followed by integration by parts therefore givesFor variations vanishing at the boundary, the Yang-Mills equations areThe Jacobi identity for the commutators of gauge covariant derivatives, using $[D_{\mu },D_{\nu }]X=[F_{\mu \nu },X]$, gives the gauge-theory Bianchi identity:In the second form of the gauge-theory Bianchi identity ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. The identity is a geometric consequence of the definition of curvature, not a second dynamical field equation.

For the usual non-Abelian Yang-Mills theory, the quantum running coupling becomes strong at low energies. Dimensional transmutation generates a scale absent from the classically scale-invariant equations. The expected confining dynamics and massive colour-singlet spectrum involve confinement and a mass gap, rather than freely propagating weakly coupled coloured waves. These nonperturbative effects are not captured by simply solving the classical equations; this is not a claim of a mathematical proof of the Yang-Mills mass gap.

Write the gauge condition as ${\chi }^a[A]=f^a[A]$ to distinguish it from the structure constants. Under an infinitesimal gauge transformation, ${\delta }_{\omega }{A}_{\mu }=D_{\mu }\omega$. Define the Faddeev-Popov operator by ${\mathcal{M}}^{ab}=\delta {\chi }^a[A+D\omega ]/\delta {\omega }^b{|}_{\omega =0}$. In a Euclidean convention, the gauge-fixing and ghost additions can be chosen asA nonlocal kernel would require a double integral; a local functional $\chi$ instead makes $\mathcal{M}$ a local differential operator. Choosing a local gauge functional thus retains a local gauge-fixed action and the ordinary local interaction structure of perturbative quantum field theory.

The fields $c^a$ and ${\overline{c}}^a$ are independent Grassmann-valued fields in the Adjoint representation, Lorentz scalars with ghost numbers $+1$ and $-1$. They are the Faddeev-Popov ghost fields: their functional integral produces $\det \mathcal{M}$, the Faddeev-Popov determinant correcting the volume element along a gauge orbit. They are internal fields, not physical asymptotic particles. Their statistics supply a minus sign for every closed ghost loop. No auxiliary field is needed in the displayed gauge-fixing representation.

In axial gauge, ${\chi }^a=n^{\mu }{A}_{\mu }^a$, soWith the Fourier transform convention $e^{ik\cdot x}$, the ghost Feynman rules in the original normalization of $A$ areMomentum conservation accompanies the vertex; each ghost loop has the extra minus sign. With a canonically normalized field $A=g{A}_{\mathrm{can}}$, the vertex is instead $-g{f}^a{}_{bc}{n}_{\mu }$. The gauge fixing term is quadratic and introduces no further interaction vertex.

The Euclidean quadratic kernel of $A$ is $g^{-2}[k^2{\delta }_{\mu \nu }-k_{\mu }{k}_{\nu }+{\xi }^{-1}{n}_{\mu }{n}_{\nu }]$. For $n\cdot k\ne 0$, its inverse isThis follows by multiplication with the kernel: the $k^2\delta -kk$ piece gives $I-n{k}^T/(n\cdot k)$, and the ${\xi }^{-1}nn$ piece supplies the missing $n{k}^T/(n\cdot k)$. The strict axial-gauge propagator is thereforeIt obeys $n^{\mu }{D}_{\mu \nu }=0$. Every ghost attachment to an internal gauge propagator therefore vanishes in the strict limit; equivalently, $\mathcal{M}=n\cdot \partial$ on the gauge slice. Thus the interacting Faddeev-Popov ghosts decouple. Residual gauge transformations and the poles at $n\cdot k=0$ require compatible boundary conditions and an axial-gauge pole prescription.

After Wick rotation, the corresponding Minkowski propagator has ${\delta }_{\mu \nu }$ replaced by ${\eta }_{\mu \nu }$ and the usual overall factor $-i$, with a compatible causal prescription. The fixed $n$ makes the gauge-fixed propagator nonmanifestly Lorentz covariant. Nevertheless, gauge invariance makes physical observables independent of this gauge-choice vector, so their perturbative predictions are Lorentz invariant. Already at tree level, contraction with conserved external currents removes all terms containing $k_{\mu }$ or $k_{\nu }$, leaving $g^{2}j\cdot j^{\prime }/k^2$. The quantum Ward identities give the analogous cancellation in complete physical amplitudes; arbitrary gauge-dependent Green functions need not be independent of $n$.

Set $P(z)=W^{\prime }(z)$ and $\overline{P}(\bar{z})=\overline{W^{\prime }(z)}$, treating $z,\bar{z}$ as independent coordinates for Wirtinger derivatives. Use left Grassmann derivatives, so ${\partial }_{{\theta }_2}({\theta }_1{\theta }_2)=-{\theta }_1$. The given odd vector field acts by $V_{1}z={\theta }_1$, $V_1{\theta }_2=-\overline{P}$, and zero on ${\theta }_1,\bar{z},{\overline{\theta }}_i$. Consequently its square vanishes on every coordinate. The square of an odd graded derivation is an even graded derivation, so the nilpotent operator property holds on every function:

For the action in this zero-dimensional supersymmetric field theory, ${\partial }_{z}S=P^{\prime }\overline{P}-P^{\prime \prime }{\theta }_1{\theta }_2$ and ${\partial }_{{\theta }_2}S=P^{\prime }{\theta }_1$. The cubic term vanishes after multiplication by ${\theta }_1$, and the other two terms cancel:Three further independent odd symmetries areFor $V_2$, use ${\partial }_{{\theta }_1}S=-P^{\prime }{\theta }_2$; the same cancellation proves invariance. The barred calculations exchange barred and unbarred variables and coefficients. Each graded derivation is also a nilpotent operator, by its coordinate action. Their different odd coefficients and coordinate derivatives make them linearly independent. These are odd symmetries of a zero-dimensional polynomial model.

Fix the measure conventions before evaluating this zero-dimensional supersymmetric field theory. Take $dzd\bar{z}$ to mean ordinary positive area measure $d^{2}z=dxdy$, and choose Berezin integration orientations with $\int d^2\theta {\theta }_1{\theta }_2=\int d^2\overline{\theta }{\overline{\theta }}_1{\overline{\theta }}_2=1$. A different overall measure normalization multiplies all the unnormalized answers by the same constant. Assume $r=\mathrm{deg}P=\mathrm{deg}W-1\ge 1$, so $|P(z)|\to \infty$ at infinity and all polynomial insertions converge. An affine or constant $W$ does not give this confining integral and is outside this convergence hypothesis.

The Grassmann variables truncate the fermionic exponential. The coefficient of ${\theta }_1{\theta }_2{\overline{\theta }}_1{\overline{\theta }}_2$ is $P^{\prime }(z)\overline{P^{\prime }(z)}$, so the partition function becomesThe Jacobian determinant of the holomorphic map $w=P(z)$ is $|P^{\prime }(z){|}^2$. Away from its finitely many critical values, a polynomial of polynomial degree $r$ has $r$ preimages counted with multiplicity. The change of variables formula therefore givesCritical values form a set of area zero and do not affect the integral. This partition function measures the covering degree, even when some zeros of $P$ are repeated.

The odd symmetries of a zero-dimensional polynomial model preserve both the action and the flat integration measure: their coefficients have zero superdivergence. The confining polynomial weight removes boundary terms at infinity. Thus the integral of an odd-symmetry derivative is zero, a supersymmetric Ward identity for this finite-dimensional integral in this zero-dimensional supersymmetric field theory.

Since $g(z)$ is a holomorphic function, the barred symmetry obeys ${\overline{V}}_1({\overline{\theta }}_{2}g(z))=-P(z)g(z)$. Applying the supersymmetric Ward identity gives the requested unnormalized insertion:One can also verify the result after Berezin integration, without invoking the symmetry terminology:The final integration by parts uses ${\partial }_{\bar{z}}(g{P}^{\prime })=0$ and rapid decay. The derivative on $W^{\prime }$ in the insertion is essential.

Rescale $W\mapsto tW$ with $t>0$. After Berezin integration, the unnormalized holomorphic function insertion isIts derivative is an ordinary total derivative:The last equality again follows from polynomial growth and exponential decay. Thus the insertion is independent of $t$, and supersymmetric localization allows evaluation as $t\to \infty$ at the zeros of $P$, as in localization of a zero-dimensional polynomial model.

For a simple zero $z_a$, $P(z)=P^{\prime }(z_a)(z-z_a)+O((z-z_a{)}^2)$. The local Gaussian integral cancels the factor $|P^{\prime }(z_a){|}^2$ and contributes $\pi f(z_a)$. For a zero with multiplicity of a root $m_a$, write $P(z)=a_a(z-z_a{)}^{m_a}+\cdots$. Its leading radial integral isThe insertion tends to $f(z_a)$ there; contributions away from the zeros vanish. Therefore the full answer, including degenerate critical points, isThis is an unnormalized expectation. For simple zeros all $m_a=1$; division by $Z=\pi r$ would give the normalized average, but is not part of the requested insertion. Taking $f=1$ recovers the partition function, and taking $f=Pg$ recovers its vanishing supersymmetric Ward identity.

Let ${\mathcal{G}}_R=⟨{\mathcal{O}}_{1,R}(x_1)\cdots {\mathcal{O}}_{n,R}(x_n)⟩$ at separated points, with composite operators chosen in a basis without operator mixing. Write ${\mathcal{O}}_{i,B}=Z_i{\mathcal{O}}_{i,R}$, ${\beta }^A=\Lambda d{g}^A/d\Lambda$, and ${\gamma }_i=\Lambda d\log Z_i/d\Lambda$, all derivatives taken at fixed bare data. Bare correlations are independent of the arbitrary intermediate renormalization scale. Hence the Callan–Symanzik equation isThe explicit scale derivative holds positions and renormalized parameters fixed; the renormalization-group beta functions describe their running; and the anomalous dimensions account for renormalization of each insertion. Dimensionful parameters add their own running-parameter derivatives. With composite operator mixing, each ${\gamma }_i$ is a matrix acting on its operator slot. The displayed sign follows the stated convention for $Z_i$. Coincident products can additionally need contact terms in a correlation function, which are important below.

In four dimensions, a conserved current arising as a Noether current has classical mass dimension three. Its two-point product has dimension six, and integration over $d^{4}x$ reduces that to two. The transverse tensor $k_{\mu }{k}_{\nu }-{\delta }_{\mu \nu }{k}^2$ already has dimension two. Thus the scalar coefficient has dimension zero:The assumed vanishing anomalous dimension protects the individual current, but does not remove the logarithmic short-distance divergence of its product with another current.

Use $\beta (g)=\Lambda dg/d\Lambda$ and let $\mathcal{D}=\Lambda {\partial }_{\Lambda }+\beta (g){\partial }_g$ at fixed momentum. Bare-scale independence gives $\mathcal{D}{G}_0=0$ at fixed ${\Lambda }_0$. Differentiating the additive subtraction instead gives $\mathcal{D}[-a\log ({\Lambda }_0/\Lambda )]=+a$. Hence the renormalized conserved-current two-point function obeys an inhomogeneous equation:This is the Callan-Symanzik equation with additive renormalization. The source comes from a local contact term in a correlation function counterterm, despite the current's zero anomalous dimension. A homogeneous equation would miss this contribution.

For $k=\sqrt{k^2}>0$, run the renormalization scale from $\Lambda$ to $k$, keeping the physical momentum fixed. Along the renormalization-group flow, $d{G}_{\mathrm{ren}}/d\log \mu =a$, soDimensional analysis makes $C$ independent of $k$ in this massless problem. Asymptotic freedom means $g(k)\to 0$ as $k/\Lambda \to \infty$. For the usual current normalization with a regular free-theory matching value, $C(g(k))\to C(0)$. Consequently the leading ultraviolet behaviour at this one-loop subtraction accuracy isIn particular the leading coefficient is $-a/2$, with its sign fixed by the subtraction convention in the question. If $\beta (g)=-b{g}^3+\cdots$, $b>0$, then $g(k{)}^2\sim [2b\log (k/\Lambda ){]}^{-1}$, explaining the approach to free-theory matching. Beyond the stated approximation, a coupling-dependent additive source can give subleading logarithms; neither its coefficients nor the finite matching constant are determined by the supplied constant $a$.