Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 44 (Advanced Quantum Field Theory)

For a unitary matrix, $U{U}^{†}=1$, so $(U^{†}MU{)}^r=U^{†}{M}^{r}U$. Cyclicity of the trace then gives $\mathrm{tr}(U^{†}{M}^{r}U)=\mathrm{tr}(M^r)$ for both powers entering the action. Thus the action is invariant under unitary conjugation.

The spectral theorem for normal operators diagonalizes a finite-dimensional Hermitian matrix by a unitary matrix. Applying the invariance to that diagonal form givesOnly the eigenvalues enter, not the choice of eigenvectors.

Use the normalized Gaussian integral with action $\mathrm{tr}(M^2)/2$. Its Wick contraction isThere is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized Hermitian matrix model integral to first order givesThe subtracted term removes Vacuum Feynman diagrams disconnected from the external pair. Since the one-point function vanishes by $M\mapsto -M$, the resulting two-point function is a connected correlation function.

Write the vertex as $M^a{}_b{M}^b{}_c{M}^c{}_d{M}^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a tadpole diagram. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N{\delta }^i{}_l{\delta }^k{}_j$ in each case. The other four attach them at opposite positions and give ${\delta }^i{}_j{\delta }^k{}_l$, with no free index loop. ThereforeBoth index structures are required at finite $N$. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal perturbative quantum field theory expansion; the real integral is convergent for $g\ge 0$.

The off-diagonal auxiliary integrations impose $M^i{}_j=0$ for $i\ne j$, by the Fourier representation of a Dirac delta function. After this constraint, $M=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ andThe Grassmann Gaussian integral over each off-diagonal pair gives its coefficient. Consequently the ghost determinant isThe prime omits the diagonal directions, which were excluded from the outset. The constant sign depends on the Berezin integral ordering and can be absorbed in normalization. Thus the Vandermonde determinant squared is the eigenvalue measure factor in this matrix diagonalization ghost determinant.

Up to an eigenvalue-independent constant, the remaining integral is $\int {\prod }_{i}d{\lambda }_i{e}^{-S_{\mathrm{eig}}}$, whereThe determinant supplies logarithmic eigenvalue repulsion. An ordering restriction on the eigenvalues changes only a constant factorial; no remaining eigenvalue integral needs to be performed. The off-diagonal bosonic contours and the displayed normalization are understood in the usual Fourier-delta prescription.

In the Wilsonian effective action picture, integrate out field modes in a high-momentum shell and encode their effects in the action for the retained modes. Rescale lengths and fields to compare the resulting theory at the original cutoff. The couplings then follow a renormalization-group flow, while low-energy predictions are preserved. In general all local interactions allowed by the symmetries are generated, even if only a few appear in the initial action.

Near a renormalization-group fixed point, a perturbation $u\int d^{d}x\mathcal{O}(x)$ with scaling dimension $\Delta$ has linearized eigenvalue $y=d-\Delta$. Under coarse graining by $b>1$, its dimensionless coupling scales as $u^{\prime }\simeq b^{y}u$. A relevant operator has $y>0$, so its perturbation grows toward long distances; an irrelevant operator has $y<0$ and decreases; a marginal operator has $y=0$ and needs nonlinear flow to determine its behavior. Marginality at linear order need not mean exact scale independence. Interactions can be marginally relevant or marginally irrelevant operators.

Choose $\varphi (x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ip\cdot x}\tilde{\varphi }(p)$ in Euclidean signature. The inverse quadratic kernel is the smooth-cutoff scalar propagatorThe subscript records the explicit cutoff dependence implied by the condition $f_{\Lambda }(z)={\Lambda }^{2}z$ at small $z$. For $p^2+m^2\le {\Lambda }^2$, this reduces to $1/(p^2+m^2)$, the usual Euclidean scalar propagator. For $p^2+m^2\gg {\Lambda }^2$, the kernel $f_{\Lambda }$ grows rapidly and its inverse tends to zero.

High-momentum modes are strongly suppressed. For a finite smooth regulator they are not literally identically zero; that statement would require a sharp cutoff. The field variance carried by those Fourier transform modes is correspondingly negligible.

It is useful to regulate the number of modes first, so the functional integral identities reduce to ordinary integration by parts. Let $C$ be the Gaussian covariance, $F=C^{-1}$, $W=e^{-S_1}$ and put a dot for $\Lambda {\partial }_{\Lambda }$. The matrix identity $\dot{F}=-F\dot{C}F$ givesThe second field derivative of the Gaussian isHence, after two integrations by parts,The imposed flow makes the bracket vanish. The remaining trace is independent of the fields and only changes the Gaussian normalization. Since the free Gaussian normalization is $\mathcal{N}=(\det (2\pi C){)}^{1/2}$, $\dot{\log \mathcal{N}}=\mathrm{Tr}(F\dot{C})/2$. ThereforeEquivalently, $Z$ is cutoff independent after discarding the stated overall rescaling. This is the Gaussian covariance differentiation identity behind the Polchinski equation.

With the Fourier convention above and functional derivatives satisfying $\delta \tilde{\varphi }(p)/\delta \tilde{\varphi }(q)={\delta }^{(4)}(p-q)$, contraction with $\dot{C}$ becomes $\int d^{4}p(2\pi {)}^4{\dot{C}}_{\Lambda }(p){\delta }^2/[\delta \tilde{\varphi }(p)\delta \tilde{\varphi }(-p)]$. Thus the numerator $(2\pi {)}^4$ in the printed flow is consistent with this derivative convention; it must not be changed independently of the convention.

Using functional derivatives of the interaction functional,Dividing the flow by $e^{-S_1}$ therefore givesIn the first term, remove one leg from each of two interaction vertices and join them with a line weighted by ${\dot{C}}_{\Lambda }(p)$. This produces the tree joining of two vertices. In the second, remove two legs from one vertex and contract them with that line, raising the loop order by one, with tadpole diagrams as the simplest example. The factor one half accounts for interchanging the contracted ends; the displayed signs are the signs in the interaction-action flow.

The varying cutoff replaces an internal propagator by its cutoff derivative. Repeated tree joins and loop closures express how eliminated high-momentum fluctuations generate the vertices of the Wilsonian effective action. The action is not restricted to one-particle-irreducible Feynman diagrams: connected tree joins also occur. Field-independent vacuum contributions can again be absorbed into normalization.

Fix the sign convention by taking the proper two-point insertion to be $i\Sigma (/p)$. This convention matches the loop expression and mass conversion printed later. Dyson resummation of successive fermion self-energy insertions givesThe subscript denotes the renormalized self-energy. If the insertion is instead named $-i{\Sigma }_{\mathrm{usual}}$, then ${\Sigma }_{\mathrm{usual}}=-{\Sigma }_R$ and the denominator is written $/p-m-{\Sigma }_{\mathrm{usual}}$. These are the same physical convention.

Lorentz covariance allows ${\Sigma }_R=\mathcal{A}(p^2)/p+\mathcal{B}(p^2)m$. The physical mass-shell condition isIt locates the mass-shell singularity of the Dirac propagator. In infrared-regulated perturbation theory this is the pole mass; near a simple pole the quantum field theory propagator has the form $i{Z}_{\mathrm{pole}}(/p+m_{\mathrm{phys}})/(p^2-m_{\mathrm{phys}}^2+i0)$. The mass and pole residue are different quantities: the former fixes the singularity's location, the latter the field normalization. The calculation below uses this standard perturbative pole-mass definition.

An on-shell renormalization scheme fixes the mass parameter at the physical pole mass and fixes the renormalized field by the chosen pole residue. Its counterterms include finite contributions needed to enforce those conditions. A coupling is likewise defined by a specified physical amplitude or mass-shell condition.

The minimal subtraction scheme removes only poles in the dimensional regulator. The modified minimal subtraction scheme, which is the overbarred scheme in the PDF, removes the accompanying universal combination $\log 4\pi -{\gamma }_E$ as well. In $D=4-ϵ$, the one-loop subtraction is proportional toA subtraction-scheme mass is a running parameter and generally differs from the physical mass. Finite redefinitions relate the schemes while leaving physical quantities unchanged. The overbar is essential for the stated finite formula in part (d).

The divergent part follows from the elementary integrals ${\int }_0^{1}xdx=1/2$ and ${\int }_0^{1}dx=1$:Thus a wave-function counterterm and a mass counterterm are required. Write their contribution asIn the insertion convention of part (a), the inverse Dirac propagator receives $\delta Z_2/p-\delta m_{\mathrm{coeff}}$. With $\kappa =e^2/(8{\pi }^2ϵ)$, cancellation requiresTo distinguish the coefficient counterterm from the multiplicative mass renormalization, write $m_0=Z_{m}m$ and ${\psi }_0=Z_2^{1/2}\psi$. Then $\delta m_{\mathrm{coeff}}=m(\delta Z_2+\delta Z_m)$ at this order, giving $\delta Z_m=-3\kappa$. In modified minimal subtraction, replace $\kappa$ by $e^2{\Delta }_{\overline{\mathrm{MS}}}/(16{\pi }^2)$. No new derivative structure is needed for this two-point divergence; charge and photon counterterms are determined from other functions.

After modified minimal subtraction, the surviving logarithm in the given fermion self-energy is $\log [{\mu }^2/((1-x)(m^2-p^{2}x))]$. For a one-loop mass shift, set $p^2=m^2$ and let $/p$ act as $m$ on an on-shell spinor inside that correction; changing these arguments by the mass shift contributes only at order $e^4$. ThusThe needed integrals areFor the second, put $u=1-x$ and use ${\int }_0^1\log udu=-1$ and ${\int }_0^{1}u\log udu=-1/4$. ConsequentlyThe pole mass condition $m_{\mathrm{phys}}-m+{\Sigma }_R=0$ gives $m_{\mathrm{phys}}=m[1+e^2(5+3\log ({\mu }^2/m^2))/(16{\pi }^2)]+O(e^4)$. Invert this relation and replace $m$ by $m_{\mathrm{phys}}$ inside the already one-loop term to obtainThe negative sign in the running-mass conversion follows from the explicitly chosen self-energy convention. Subtracting poles alone, instead of the overbarred combination, would leave additional $\log 4\pi -{\gamma }_E$ terms.

Use Hermitian generators $T^a$ of the special unitary group, with $[T^a,T^b]=i{f}^{ab}{}_c{T}^c$. An adjoint field is the Lie-algebra-valued matrix $\psi ={\psi }^a{T}^a$, transforming as $\psi \mapsto U\psi U^{-1}$. The adjoint covariant derivative isIn components, $(D_{\mu }\psi {)}^a={\partial }_{\mu }{\psi }^a+g{f}^{abc}{A}_{\mu }^b{\psi }^c$. With $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$, direct substitution gives $D_{\mu }\psi \mapsto U(D_{\mu }\psi )U^{-1}$. This is covariance in the Adjoint representation of a Lie group. In the following BRST symmetry formulas, absorb the coupling into the connection, so $D_{\mu }={\partial }_{\mu }-i[A_{\mu },\cdot ]$.

Factor the odd parameter on the left, $\delta =ϵs$. The resulting left-acting BRST differential obeys the graded Leibniz ruleFor the odd Grassmann field $c$, the bracket in the transformation is a graded commutator: $[c,c{]}_{\mathrm{gr}}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=i{c}^2$, while $s{A}_{\mu }=D_{\mu }c$, $s\bar{c}=h$ and $sh=0$.

On the ghost,On the gauge field, variation of the connection and the adjoint covariant derivative givesHere $D_{\mu }$ is even and therefore obeys the ordinary product rule. Also $s^2\bar{c}=sh=0$ and $s^{2}h=0$, without using any field equation; this is off-shell nilpotence supplied by the Nakanishi-Lautrup field.

Applying the graded Leibniz rule twice cancels the two cross terms:The square is consequently an even graded derivation. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence $s^2\mathcal{O}=0$ for every such operator. This genuine result is stronger than the automatic vanishing obtained by merely setting ${ϵ}^2=0$; two independent transformation parameters also give a vanishing commutator.

Use the Minkowski metric, path-integral weight $e^{iS}$ and the Abelian gauge theory transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$. The gauge functional $F[A]=\partial \cdot A+A^2$ varies asThus the Faddeev-Popov operator is ${\mathcal{M}}_A=\square +2A\cdot \partial$. It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.

Choose the gauge-fixing fermion $\Psi =\int d^{4}x\bar{c}(F[A]+\xi h/2)$. The gauge-fixed action $S=S_{\mathrm{Maxwell}}+s\Psi$ isHere the tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is distinct from the scalar gauge functional $F[A]$. With $\xi =0$, integrating over $h$ imposes the exact printed constraint. For nonzero $\xi$, eliminating $h=-F[A]/\xi$ instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its $\xi \to 0$ limit.

For the Fourier transform convention $c(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}c(p)$, the quadratic ghost kernel is $p^2$. With the ordering $⟨c(p)\bar{c}(q)⟩$,The term $-2\bar{c}{A}^{\mu }{\partial }_{\mu }c$ has Fourier coefficient $2i{p}_{\mu }$, where $p$ is the incoming ghost momentum. Multiplication by $i$ in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.

A gauge-invariant observable is BRST-closed: replacing its infinitesimal gauge parameter by the ghost gives $s\mathcal{O}=0$. Change the gauge functional continuously, or interpolate between two admissible choices, by changing the gauge-fixing fermion to ${\Psi }_t$. The action changes by the BRST-exact operator ${\partial }_{t}S=s({\partial }_t{\Psi }_t)$.

For a normalized correlator of $\mathcal{O}={\prod }_i{\mathcal{O}}_i$ with each insertion BRST-closed, differentiation of the functional integral givesThe BRST Ward identity says $⟨sX⟩=0$ for an invariant measure and action, with appropriate boundary conditions. Since $s\mathcal{O}=0$, the graded Leibniz rule makes the first insertion an exact variation of $\mathcal{O}{\partial }_t{\Psi }_t$ up to its harmless parity sign, and both terms vanish. ThusPhysical gauge-invariant correlation functions are independent of this gauge condition, even though individual gauge-field and ghost Feynman diagrams change. This argument assumes an admissible perturbative gauge fixing, a BRST symmetry-preserving regulator/measure and no uncanceled boundary contribution. A global failure of those assumptions is not settled by the formal local calculation.