Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 (Advanced Quantum Field Theory)

Decompose the original scalar field as $\Phi =\varphi +\hat{\varphi }$, where $\tilde{\varphi }(p)$ has support in $|p|\le {\Lambda }^{\prime }$ and $\tilde{\hat{\varphi }}(p)$ has support in ${\Lambda }^{\prime }<|p|\le \Lambda$. This is a momentum-shell decomposition of a scalar field. The two collections of integration variables are disjoint. Define the lower-scale Wilsonian effective action by integrating out the second collection:A field-independent normalization may be retained as a vacuum term or absorbed into the measure. Integrating this identity over the low modes recovers the original Euclidean path integral, so it preserves all observables depending only on those modes.

Put $\Delta S=S_{\Lambda }^{\mathrm{eff}}[\varphi +\hat{\varphi }]-S_{\Lambda }^{\mathrm{eff}}[\varphi ]$. The quadratic cross terms integrate to zero: in Fourier transform variables each pairs a low momentum with its negative, which cannot be a shell momentum. Expanding the interaction therefore givesFactoring $e^{-S_{\Lambda }^{\mathrm{eff}}[\varphi ]}$ out of the shell integral and taking minus its logarithm yieldsThis definition is a Wilsonian effective action, rather than a Legendre transform generating only one-particle-irreducible Feynman diagrams.

Let the shell-restricted scalar propagator beWrite $\Delta S=S_0[\hat{\varphi }]+V[\varphi ,\hat{\varphi }]$ and average with the normalized Gaussian shell measure. The cumulant expansion givesThe subtraction removes disconnected Feynman diagrams; the logarithm retains connected Feynman diagrams made from shell contractions. This is the connected shell-contraction expansion. An external line denotes the background low field, not an additional low-momentum propagator.

At order $g$, panel A is the existing four-field interaction vertex. The new shell contractions are B, a tadpole diagram contributing a two-field vertex, and C, a Vacuum Feynman diagram contributing a constant. The Gaussian functional determinant in panel L is another field-independent term, of order $g^0$; the original quadratic kinetic term and mass term are retained as well.

For completeness, all two-vertex topologies from Wick contractions at order $g^2$ are shown. Let $r$ be the number of shell lines joining the vertices and $t_1,t_2$ their numbers of self-contractions. The external-field counts areInterchanging the two vertices identifies the same topology. Enumerating these conditions gives exactly the following eight possibilities:

The support qualification is important. In E or F, a vertex with one external low field and one bridge also has a tadpole whose two momenta cancel. Conservation forces the bridge momentum to equal that single low momentum, outside the shell. Equivalently, convolution by $C_{>}$ annihilates $\varphi$. This is momentum-support exclusion for Wilsonian bridge diagrams. D is different: its bridge carries the sum of three low momenta, which can lie in the shell. Its kernel is proportional to ${\varphi }^3(x)C_{>}(x-y){\varphi }^3(y)$ and need not vanish. Thus one-particle-reducible Feynman diagrams must not be excluded merely because the object being computed is an effective action.

These kernels need not be local before a derivative expansion. Their momentum dependence generates derivative interactions where such an expansion is valid. If all external momenta are sufficiently far below ${\Lambda }^{\prime }$, D vanishes too; in particular it does not produce a zero-momentum local ${\varphi }^6$ coupling at order $g^2$. A local six-field term is allowed and is generated at higher orders, for example by a three-vertex shell triangle.

To all orders, expect every scalar interaction allowed by the original symmetries: arbitrary even powers of the field and their allowed derivative couplings, together with vacuum terms. The Z2 symmetry $\varphi \mapsto -\varphi$ excludes odd-field vertices. The original quartic form is therefore not closed under exact Wilsonian integration.

After integrating out the shell, let a chosen operator's coefficient be $\kappa +\Delta \kappa$ and write the two-derivative quadratic term as $\frac{1}{2}(1+\Delta Z)(\partial \varphi {)}^2$. Here $\Delta \kappa$ is the shell-induced change before rescaling, and $\Delta Z$ is the shell correction to the wavefunction renormalization. Work in an operator basis in which that quadratic term has this form; the other generated operators remain in the Wilsonian effective action.

Under $x^{\prime }=bx$, the volume element changes by $d^{4}x=b^{-4}{d}^4{x}^{\prime }$ and each derivative by ${\partial }_x=b{\partial }_{x^{\prime }}$. Canonical field normalization is restored by definingIndeed, the volume factor, two derivatives and two field factors cancel in the kinetic term. A term with $n$ fields and $m$ derivatives consequently obtains the factor $b^{-4}{b}^m{b}^n(1+\Delta Z{)}^{-n/2}$. HenceThis is Wilsonian rescaling of a scalar coupling. The exponent is minus the coupling's engineering mass dimension, $[\kappa ]=4-n-m$. The rescaling also returns the low-momentum cutoff to $\Lambda$, because $p^{\prime }=p/b$.

The cubic scalar field theory has a one-particle-irreducible Feynman diagram with two cubic vertices, one external leg at each vertex, and two internal lines joining them:

The Euclidean Feynman rule for each cubic interaction vertex is $-g{\mu }^{(6-d)/2}$: the minus sign comes from expanding $e^{-S_{\mathrm{int}}}$, and $1/3!$ in the action cancels the permutations of its three fields. The mass dimension of the cubic coupling is $(6-d)/2$, so ${\mu }^{(6-d)/2}$ permits $g$ to remain dimensionless. Two vertices supply $(-g{)}^2{\mu }^{6-d}$.

Each internal scalar propagator contributes $1/(p^2+m^2)$ with its own momentum. Conservation leaves one independent loop momentum; choose the two propagator momenta to be $p$ and $p+k$. The remaining Fourier integration measure is $d^{d}p/(2\pi {)}^d$. Finally, the factor $1/2$ is the Feynman-diagram symmetry factor for exchanging the two identical internal lines. Direct Wick contraction counting gives the same factor: two choices for which vertex receives a labeled external leg, $3^2$ choices of the incident fields and $2!$ pairings of the remaining fields, divided by $2!(3!{)}^2$, yield $1/2$.

The displayed loop integral is the amputated two-point insertion. For the correction to the full propagator, multiply it by the external scalar propagators, giving $D_0(k{)}^{2}I(k)$ with $D_0(k)=1/(k^2+m^2)$.

Apply a Feynman parameter and shift the loop momentum to $\ell =p+xk$. The common denominator becomes $[{\ell }^2+M_x^2{]}^2$, whereFor positive $M_x^2$, the Gamma-integral representation $a^{-2}={\int }_0^{\infty }dss{e}^{-sa}$ and a Gaussian integral giveThis formula initially converges for $\mathrm{Re}d<4$ and defines the Euclidean massive loop integral at other dimensions by analytic continuation. Consequently,With $d=6-ϵ$, the Gamma function factor is $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. Only its pole matters: the other factors can be evaluated at $ϵ=0$ when extracting that pole. Since ${\int }_0^{1}x(1-x)dx=1/6$,This is the one-loop two-point divergence in six-dimensional cubic scalar theory. Its polynomial momentum dependence is precisely what permits subtraction by local counterterms.

Use a mass counterterm and a wavefunction renormalization counterterm, defining their additive coefficients byThe sign follows from how the Euclidean path integral expands. A quadratic counterterm inserts $-\delta K$ into the propagator, whereas the loop defined in the question inserts $+I$. To this order,Thus cancellation requires $\delta K=I_{\mathrm{div}}$, rather than its negative. In the minimal subtraction scheme, with no finite parts added,These are the minimal-subtraction two-point counterterms in cubic scalar theory. They absorb respectively the $k^2$ and constant terms in the two-point pole.

The additive mass coefficient is distinct from the shift of a bare mass when the bare field also includes wavefunction renormalization. If ${\varphi }_B=\sqrt{1+\delta Z}\varphi$ and $(1+\delta Z)m_B^2=m^2+\delta m^2$, then to one-loop orderThis last relation specifies the convention; the boxed coefficients are those multiplying the local counterterms displayed above.

A phase-space path integral is defined as a limit of finite-dimensional integrals, not by assigning a classical derivative to every path. Choose $t_1=0<t_2<\cdots <t_{N+1}=T$, let $\Delta t_i=t_{i+1}-t_i$, and set $q_{N+1}=q_f$. On slice $i$ the precise prescription isThe remaining Hamiltonian term must have a compatible operator-ordering prescription. For example, evaluate $H$ at $(p_i,(q_{i+1}+q_i)/2)$ for midpoint/Weyl ordering. A prepoint prescription $H(p_i,q_i)$ defines a corresponding ordering instead. This choice matters for a general mixed $H(p,q)$; the separable kinetic-plus-potential Hamiltonian in the next part admits the usual Trotter prescription.

This is time slicing of a phase-space path integral. Integrate the intermediate $q_i$ and the slice momenta and only then take ${\max }_i\Delta t_i\to 0$. Typical paths of the Euclidean path integral need not be differentiable; the finite difference is the meaning of the printed $\dot{q}$. For a fixed-endpoint kernel the initial coordinate is fixed as well, whereas propagation of a wavefunction includes an integral over that initial coordinate.

Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent isEach momentum integration is an ordinary Gaussian integral:Therefore the measure written in the question gives exactlyThe resulting configuration integral is $\int \mathcal{D}{q}_{\mathrm{raw},N}{e}^{-S_{E,N}}$. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses $d{p}_i/(2\pi )$ rather than $d{p}_i$. With that normalization the configuration measure isThis differs from the raw measure by $(2\pi {)}^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\dots ,q_N$ are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit $d{q}_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.

We now address the unheaded real-time continuation. Set $m=\hslash =1$. The normalized short-time Schrödinger kernel gives the final-slice recurrenceThe square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's $d{q}^{\prime }/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i{)}^{-1/2}$ fixes the identity limit and is included at every step.

Put $\eta =q^{\prime }-q$. The normalized oscillatory Gaussian has momentsTaylor-expand the smooth wavefunction and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order $(\Delta t{)}^2$. ThusSubtract the initial value, divide by $\Delta t$ and take the limit:This proves the Time-dependent Schrödinger equation by the Schrödinger equation from a short-time path integral argument. Nonuniform partitions give the same limit when their largest time step tends to zero.

Choose Lie-algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$ and an invariant orthonormal color metric. Since the coupling is placed outside the gauge kinetic term, it is absorbed into $A_{\mu }$ in the connection convention. The gauge field strength and adjoint covariant derivative areThe structure constants are antisymmetric in $b,c$ and obey the Jacobi identity. These definitions give $[D_{\mu },D_{\nu }]v=[F_{\mu \nu },v]$ on an adjoint-valued field. The field $c$ is a Grassmann-odd Faddeev-Popov ghost field; the Lie bracket in $D_{\mu }c$ contracts its color components in the same way as for any adjoint field.

Put the anticommuting parameter on the left and write $\delta \Phi =ϵs\Phi$. Then $s$ is a left-acting BRST differential: it is odd and satisfies the graded product rule $s(UV)=(sU)V+(-1{)}^{|U|}U(sV)$. The Faddeev-Popov ghost field $c$ and antighost field $b$ are odd, whereas $A_{\mu }$ and the Nakanishi-Lautrup field $h$ are even. Use $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $sb=h$, $sh=0$.

A total derivative in the Lagrangian variation must be included in the Noether current. For constant $ϵ$, the ghost covariant derivative has $s(D_{\mu }c)=0$, by the Jacobi identity and the odd statistics of $c$. The Yang-Mills term is invariant, and the remaining variation isThe term proportional to $\xi$ does not vary because $sh=0$.

To obtain the signs without an ambiguity about fermionic canonical momenta, now allow $ϵ=ϵ(x)$. In particular,The coefficients of ${\partial }_{\mu }ϵ$ in the Yang-Mills, ghost, and multiplier terms are respectivelyThe positive sign of the last ghost expression results from moving ${\partial }_{\mu }ϵ$ past the odd ${\partial }^{\mu }{b}^a$. Subtracting the total-derivative term $K^{\mu }$ therefore gives the Yang-Mills BRST Noether currentIndeed, the full localized variation is $\delta \mathcal{L}=({\partial }_{\mu }ϵ)j^{\mu }+{\partial }_{\mu }(ϵK^{\mu })$. The Noether theorem then gives ${\partial }_{\mu }{j}^{\mu }=0$ on the field equations. With spatial boundary terms vanishing, the corresponding BRST charge in four spacetime dimensions isIt is odd and has ghost number one. Overall generator phases depend on the convention relating this Noether charge to quantum commutators; one may use $s\mathcal{O}=i[Q_{\mathrm{BRST}},\mathcal{O}\}$. The displayed current fixes the classical Noether normalization for the left-parameter convention.

Let $G_1,\dots ,G_r$ be the separated gauge-invariant insertions. Their BRST symmetry variations vanish: the gauge field strength and its gauge covariant derivatives transform by adjoint commutators, and invariant color contractions remove these commutators. Thus $s{G}_i=0$.

Make the infinitesimal change of integration variables $\Phi \mapsto \Phi +ϵs\Phi$ in the normalized expectation of $\mathcal{O}(x){\prod }_i{G}_i$. The action is invariant up to its boundary term and, by assumption, the measure has no Jacobian anomaly. ThereforeSince $s\mathcal{O}$ is generated by the BRST charge, this is the graded BRST Ward identityHere $[Q,\mathcal{O}\}=Q\mathcal{O}-(-1{)}^{|\mathcal{O}|}\mathcal{O}Q$ is the graded commutator. For even $\mathcal{O}$ it is the ordinary commutator printed in the question; for odd $\mathcal{O}$ it is an anticommutator. Separation of the other insertions from $x$ avoids the additional coincident-point contact terms. An operator need not itself be BRST-closed for this identity to hold.

A BRST-exact insertion has zero correlation with physical, gauge-invariant insertions. This is the decoupling of BRST-exact insertions in physical correlation functions. In the associated BRST cohomology, physical information is represented by closed states or operators modulo exact ones, schematically $\ker Q/\mathrm{im}Q$, using the nilpotence of the BRST charge in the anomaly-free theory. Gauge-fixing fields thereby do not supply additional physical observables.

For example, the gauge parameter dependence is exact:Differentiating a normalized physical correlation function with respect to $\xi$ inserts this expression; the BRST Ward identity makes that derivative zero under the same invariant-measure and boundary assumptions. This illustrates gauge-fixing parameter independence from BRST symmetry.