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Particle Physics · AI note · graduate · #particle

Complete Quantum Field Theory

A concise guide to relativistic fields, canonical quantization, Feynman diagrams, QED, radiative corrections, path integrals, renormalization, gauge theory, QCD, electroweak theory, and anomalies.

· 14 min
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Part I: Feynman Diagrams and Quantum Electrodynamics

Core ideas

QFT combines quantum mechanics, special relativity, and many-particle physics. Particles are excitations of fields, interactions are local terms in a Lagrangian, and perturbation theory organizes scattering amplitudes as Feynman diagrams.

For review, be able to explain why fields replace fixed-particle wave functions, identify propagators and vertices, and connect amplitudes to observables. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

S=d4xL,fSi=δfi+i(2π)4δ4(PfPi)MS=\int d^4x\,\mathcal L,\qquad \langle f|S|i\rangle=\delta_{fi}+i(2\pi)^4\delta^4(P_f-P_i)\mathcal M

Worked example

The free real scalar field has Lagrangian L=12(μϕ)212m2ϕ2\mathcal L=\frac12(\partial_\mu\phi)^2-\frac12m^2\phi^2. The equation of motion is (+m2)ϕ=0(\Box+m^2)\phi=0. For a plane wave ϕ(x)=eikx\phi(x)=e^{-ik\cdot x}, this gives k2=m2k^2=m^2, i.e., E2=p2+m2E^2=\bm p^2+m^2. The on-shell condition enforces the relativistic energy-momentum relation.

Problems with Solutions

Problem 1. Show that the free scalar Lagrangian is invariant under the global phase transformation ϕeiαϕ\phi\to e^{i\alpha}\phi. Solution. For a real scalar field, ϕ\phi does not transform by a phase. For a complex scalar, ϕϕ\phi^*\phi is invariant. Since L\mathcal L depends only on ϕ2\phi^2 and (μϕ)2(\partial_\mu\phi)^2, and both are invariant under ϕϕ\phi\to-\phi, the Lagrangian has a discrete Z2\mathbb Z_2 symmetry, not a continuous U(1).

Problem 2. What are the dimensions of the scalar field ϕ\phi in d=4d=4 spacetime dimensions? Solution. The kinetic term (ϕ)2(\partial\phi)^2 has dimension 4, so []+[ϕ]=2[\partial]+[\phi]=2, giving [ϕ]=1[\phi]=1 (mass dimension). The mass term m2ϕ2m^2\phi^2 then has dimension 2+2=42+2=4, consistent.

Section summary. QFT computes relativistic quantum processes through fields and amplitudes.

The Klein-Gordon Field

Core ideas

The Klein-Gordon field is the simplest relativistic quantum field. Quantization turns each momentum mode into a harmonic oscillator and introduces creation and annihilation operators. The propagator is the Green function for relativistic propagation.

For review, be able to derive the equation of motion, quantize modes, interpret particles and antiparticles, and write the scalar propagator. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

L=12μϕμϕ12m2ϕ2,(+m2)ϕ=0,ΔF(p)=ip2m2+iϵ\mathcal L=\frac12\partial_\mu\phi\,\partial^\mu\phi-\frac12m^2\phi^2,\qquad (\Box+m^2)\phi=0,\qquad \Delta_F(p)=\frac{i}{p^2-m^2+i\epsilon}

Worked example

For a scalar field of mass m=125GeVm=125\,\mathrm{GeV} (the Higgs boson), the Feynman propagator at momentum pμ=(E,0)p^\mu=(E,\bm 0) is ΔF(E)=i/(E2m2+iϵ)\Delta_F(E)=i/(E^2-m^2+i\epsilon). At E=200GeVE=200\,\mathrm{GeV}, ΔF1/(20021252)1/(2.34×104)4.3×105GeV2|\Delta_F|\sim1/(200^2-125^2)\approx1/(2.34\times10^4)\approx4.3\times10^{-5}\,\mathrm{GeV^{-2}}. Near resonance EmE\approx m, the propagator blows up as i/(2m(Em)+iϵ)i/(2m(E-m)+i\epsilon), producing the characteristic Breit-Wigner peak.

Problems with Solutions

Problem 1. Compute the equal-time commutator [ϕ(x,t),π(y,t)][\phi(\bm x,t),\pi(\bm y,t)] for the free Klein-Gordon field. Solution. Canonical quantization gives [ϕ(x,t),π(y,t)]=iδ(3)(xy)[\phi(\bm x,t),\pi(\bm y,t)]=i\delta^{(3)}(\bm x-\bm y), which is the field-theoretic analogue of [qi,pj]=iδij[q_i,p_j]=i\delta_{ij}.

Problem 2. A real scalar field has Fourier expansion ϕ(x)=d3p(2π)312Ep(apeipx+apeipx)\phi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\bm p}}}(a_{\bm p}e^{-ip\cdot x}+a_{\bm p}^\dagger e^{ip\cdot x}). Show that apa_{\bm p}^\dagger creates a particle of momentum p\bm p. Solution. The Hamiltonian is H=d3p(2π)3EpapapH=\int\frac{d^3p}{(2\pi)^3}E_{\bm p}a_{\bm p}^\dagger a_{\bm p}. Acting on the vacuum: Hap0=Epap0H a_{\bm p}^\dagger|0\rangle=E_{\bm p}a_{\bm p}^\dagger|0\rangle, confirming ap0a_{\bm p}^\dagger|0\rangle is a one-particle state of energy Ep=p2+m2E_{\bm p}=\sqrt{\bm p^2+m^2}.

Section summary. Scalar fields show how particles emerge from quantized modes.

The Dirac Field

Core ideas

The Dirac field describes spin-1/21/2 fermions. Gamma matrices linearize the relativistic dispersion relation, spinors carry Lorentz representation data, and anticommutation enforces Pauli exclusion and positive energy.

For review, be able to use the Dirac equation, spinor completeness, conserved current, and the fermion propagator. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

L=ψˉ(iγμμm)ψ,(iγμμm)ψ=0,SF(p)=i(+m)p2m2+iϵ\mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi,\qquad (i\gamma^\mu\partial_\mu-m)\psi=0,\qquad S_F(p)=\frac{i(\not p+m)}{p^2-m^2+i\epsilon}

Worked example

For an electron at rest (p=0\bm p=0), the Dirac equation reduces to iγ00ψ=mψi\gamma^0\partial_0\psi=m\psi. In the Dirac representation, γ0=(I00I)\gamma^0=\begin{pmatrix}I&0\\0&-I\end{pmatrix}, so the positive-energy solution has upper two components ψ+eimtu(0)\psi_+\sim e^{-imt}u(0) and the negative-energy solution has lower components ψe+imtv(0)\psi_-\sim e^{+imt}v(0). The existence of negative-energy solutions requires the interpretation of antiparticles (holes in the Dirac sea).

Problems with Solutions

Problem 1. Show that {γμ,γν}=2gμν\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu} implies ()2=p2(\not p)^2=p^2. Solution. ()2=γμγνpμpν=12{γμ,γν}pμpν=gμνpμpν=p2(\not p)^2=\gamma^\mu\gamma^\nu p_\mu p_\nu=\frac12\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu=g^{\mu\nu}p_\mu p_\nu=p^2.

Problem 2. Compute the trace Tr(γμγν)\mathrm{Tr}(\gamma^\mu\gamma^\nu). Solution. Using γμγν+γνγμ=2gμν\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu} and taking the trace: 2Tr(γμγν)=2gμνTr(I)=8gμν2\mathrm{Tr}(\gamma^\mu\gamma^\nu)=2g^{\mu\nu}\mathrm{Tr}(I)=8g^{\mu\nu} in 4D (since Tr(I)=4\mathrm{Tr}(I)=4). Thus Tr(γμγν)=4gμν\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu}.

Section summary. Dirac fields are relativistic quantum fields for fermions.

Interacting Fields and Feynman Diagrams

Core ideas

Interactions make fields scatter and decay. Perturbation theory expands time-ordered correlation functions in coupling constants; Wick’s theorem reduces products of free fields to propagators and vertices.

For review, be able to derive simple Feynman rules, count symmetry factors, and relate diagrams to powers of coupling. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

L=L0+Lint,Z[J]=DϕeiS[ϕ]+iJϕ\mathcal L=\mathcal L_0+\mathcal L_{\rm int},\qquad Z[J]=\int\mathcal D\phi\,e^{iS[\phi]+i\int J\phi}

Worked example

Consider ϕ4\phi^4 theory with Lint=λ4!ϕ4\mathcal L_{\rm int}=-\frac{\lambda}{4!}\phi^4. The first-order correction to the two-point function ϕ(x)ϕ(y)\langle\phi(x)\phi(y)\rangle comes from the diagram with one vertex connected to two external lines by propagators. The symmetry factor for this “tadpole” diagram is 4!/2=124!/2=12, but the two external lines can attach in 4×3=124\times3=12 ways, giving a factor of λ/2\lambda/2. The corrected propagator is ΔF(xy)iλ2ΔF(0)d4zΔF(xz)ΔF(zy)\Delta_F(x-y)-\frac{i\lambda}{2}\Delta_F(0)\int d^4z\,\Delta_F(x-z)\Delta_F(z-y).

Problems with Solutions

Problem 1. In ϕ4\phi^4 theory, how many distinct ways can four external lines attach to a single ϕ4\phi^4 vertex? Solution. The vertex has 4 identical legs. The number of ways to choose 4 external lines (with replacement, order matters) is 4!=244!=24, but permutations of the legs give the same diagram. The symmetry factor for the 4-point function at tree level is 4!/4!=14!/4!=1, so there is exactly one tree-level diagram.

Problem 2. Write Wick’s theorem for ϕ1ϕ2ϕ3ϕ4\langle\phi_1\phi_2\phi_3\phi_4\rangle in terms of two-point functions. Solution. ϕ1ϕ2ϕ3ϕ4=ϕ1ϕ2ϕ3ϕ4+ϕ1ϕ3ϕ2ϕ4+ϕ1ϕ4ϕ2ϕ3\langle\phi_1\phi_2\phi_3\phi_4\rangle=\langle\phi_1\phi_2\rangle\langle\phi_3\phi_4\rangle+\langle\phi_1\phi_3\rangle\langle\phi_2\phi_4\rangle+\langle\phi_1\phi_4\rangle\langle\phi_2\phi_3\rangle.

Section summary. Feynman diagrams are bookkeeping for perturbative correlation functions.

Elementary Processes of QED

Core ideas

QED couples the Dirac field to the electromagnetic gauge field. Tree-level processes such as electron-muon scattering, annihilation, pair production, and Compton scattering illustrate spinor algebra, gauge invariance, and cross-section calculations.

For review, be able to write QED Feynman rules, use Ward identities conceptually, and compute the structure of basic amplitudes. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

LQED=14FμνFμν+ψˉ(iγμDμm)ψ,Dμ=μ+ieAμ\mathcal L_{\rm QED}=-\frac14F_{\mu\nu}F^{\mu\nu}+\bar\psi(i\gamma^\mu D_\mu-m)\psi,\qquad D_\mu=\partial_\mu+ieA_\mu

Worked example

The QED vertex factor is ieγμ-ie\gamma^\mu. For electron-muon scattering eμeμe^-\mu^-\to e^-\mu^-, the tree-level amplitude is M=(ie)2(uˉeγμue)(igμν/q2)(uˉμγνuμ)\mathcal M=(-ie)^2(\bar u_{e'}\gamma^\mu u_e)(-ig_{\mu\nu}/q^2)(\bar u_{\mu'}\gamma^\nu u_\mu). In the non-relativistic limit this reduces to the Coulomb scattering amplitude, with the spinor structure giving the correct magnetic moment coupling.

Problems with Solutions

Problem 1. Write the Feynman amplitude for Compton scattering γ+eγ+e\gamma+e^-\to\gamma+e^- at tree level. Solution. There are two diagrams: (s-channel) Ms=e2uˉe\slashedϵ(++m)\slashedϵue/((p+k)2m2)\mathcal M_s=-e^2\bar u_{e'}\slashed\epsilon'(\not p+\not k+m)\slashed\epsilon u_e/((p+k)^2-m^2) and (u-channel) with kkk\leftrightarrow -k'. The sum gives the Klein-Nishina formula.

Problem 2. What is the Ward identity for the QED 3-point function? Solution. kμMμ=0k_\mu\mathcal M^\mu=0, where kμk_\mu is the photon momentum. This ensures gauge invariance and guarantees that the photon remains massless to all orders.

Section summary. QED is the cleanest example of a quantum gauge theory.

Radiative Corrections

Core ideas

Loops correct masses, charges, magnetic moments, and scattering amplitudes. They introduce divergences that must be regularized and renormalized. Physical predictions depend on measured parameters at a scale, not on bare quantities.

For review, be able to identify self-energy, vertex, and vacuum polarization corrections, explain regularization, and interpret running charge. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

α(q2)=α(μ2)1Π(q2,μ2),g2 starts at α2π\alpha(q^2)=\frac{\alpha(\mu^2)}{1-\Pi(q^2,\mu^2)},\qquad g-2\ \mathrm{starts\ at}\ \frac{\alpha}{2\pi}

Worked example

The electron anomalous magnetic moment is ae=(g2)/2a_e=(g-2)/2. Schwinger’s 1948 one-loop calculation gave ae(2)=α/(2π)0.00116a_e^{(2)}=\alpha/(2\pi)\approx0.00116. Using α1/137.036\alpha\approx1/137.036, ae(2)0.0011614a_e^{(2)}\approx0.0011614. The current best experimental value is aeexp=0.00115965218059(13)a_e^{\exp}=0.00115965218059(13), matching QED predictions to 10 significant figures after including four-loop corrections.

Problems with Solutions

Problem 1. Estimate the one-loop vacuum polarization contribution to the running of α\alpha at Q2=mZ2Q^2=m_Z^2. Solution. The one-loop result is Π(Q2)=α3πln(Q2/me2)\Pi(Q^2)=-\frac{\alpha}{3\pi}\ln(Q^2/m_e^2) for Q2me2Q^2\gg m_e^2. At Q=mZQ=m_Z, Π(1/137)/(3π)ln((91GeV)2/(0.511MeV)2)0.00077×ln(3.2×1010)0.018\Pi\approx-(1/137)/(3\pi)\ln((91\,\mathrm{GeV})^2/(0.511\,\mathrm{MeV})^2)\approx-0.00077\times\ln(3.2\times10^{10})\approx-0.018. Thus α(mZ2)α(0)/(1+0.018)1/128\alpha(m_Z^2)\approx\alpha(0)/(1+0.018)\approx1/128.

Problem 2. Why must the photon self-energy Πμν(q)\Pi^{\mu\nu}(q) be transverse, qμΠμν=0q_\mu\Pi^{\mu\nu}=0? Solution. Gauge invariance (Ward identity) requires that the photon remain massless. A massive photon would correspond to a pole at q20q^2\neq0, but the transversality condition ensures Πμν(q)=(gμνq2qμqν)Π(q2)\Pi^{\mu\nu}(q)=(g^{\mu\nu}q^2-q^\mu q^\nu)\Pi(q^2), with no mass term.

Section summary. Radiative corrections make QFT predictive beyond tree level.

Part II: Renormalization

Core ideas

Renormalization separates short-distance unknowns from long-distance predictions. Counterterms absorb divergences, while renormalized couplings depend on scale. Relevant, marginal, and irrelevant operators explain why low-energy physics can be universal.

For review, be able to classify operators by dimension, state what counterterms do, and distinguish bare from renormalized parameters. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

L=Lren+Lcounter,μdgdμ=β(g)\mathcal L=\mathcal L_{\rm ren}+\mathcal L_{\rm counter},\qquad \mu\frac{dg}{d\mu}=\beta(g)

Worked example

In ϕ4\phi^4 theory in d=4d=4, the bare coupling λ0\lambda_0 is related to the renormalized coupling λ\lambda by λ0=μϵ(λ+δλ)\lambda_0=\mu^\epsilon(\lambda+\delta_\lambda) in dimensional regularization (d=4ϵd=4-\epsilon). At one loop, δλ=3λ2/(16π2ϵ)\delta_\lambda=-3\lambda^2/(16\pi^2\epsilon). The beta function is β(λ)=3λ2/(16π2)+O(λ3)\beta(\lambda)=3\lambda^2/(16\pi^2)+O(\lambda^3), showing that the coupling grows with energy ( Landau pole).

Problems with Solutions

Problem 1. In QED, the one-loop beta function is β(e)=e3/(12π2)\beta(e)=e^3/(12\pi^2). Solve for the running coupling e(μ)e(\mu). Solution. dedlnμ=e312π2\frac{de}{d\ln\mu}=\frac{e^3}{12\pi^2}. Integrating: 12e2e0e=112π2ln(μ/μ0)-\frac{1}{2e^2}\big|_{e_0}^{e}=\frac{1}{12\pi^2}\ln(\mu/\mu_0), giving 1e2(μ)=1e0216π2ln(μ/μ0)\frac{1}{e^2(\mu)}=\frac{1}{e_0^2}-\frac{1}{6\pi^2}\ln(\mu/\mu_0). The coupling increases with energy.

Problem 2. What is the superficial degree of divergence of a QED diagram with VV vertices, IeI_e internal electron lines, and IγI_\gamma internal photon lines? Solution. D=4L2IγIeD=4L-2I_\gamma-I_e where L=Ie+IγV+1L=I_e+I_\gamma-V+1. Using 2Ie+Ee=2V2I_e+E_e=2V and 2Iγ+Eγ=V2I_\gamma+E_\gamma=V, one finds D=4Eγ32EeD=4-E_\gamma-\frac32E_e, showing that only a finite number of amplitudes diverge.

Section summary. Renormalization is scale-dependent bookkeeping of physical parameters.

Functional Methods

Core ideas

Path integrals compute generating functionals for correlation functions. Sources generate expectation values, effective actions generate one-particle-irreducible vertices, and saddle points connect quantum field theory to classical field equations.

For review, be able to use generating functionals, functional derivatives, Wick rotation, and effective action concepts. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

Z[J]=Dϕexp(iS[ϕ]+iJϕd4x),ϕ(x)ϕ(y)=1i2Zδ2ZδJ(x)δJ(y)Z[J]=\int\mathcal D\phi\,\exp\left(iS[\phi]+i\int J\phi\,d^4x\right),\qquad \langle\phi(x)\phi(y)\rangle=\frac{1}{i^2Z}\frac{\delta^2Z}{\delta J(x)\delta J(y)}

Worked example

For a free scalar field, the generating functional is Z0[J]=Z0[0]exp(12d4xd4yJ(x)ΔF(xy)J(y))Z_0[J]=Z_0[0]\exp(-\frac12\int d^4x\,d^4y\,J(x)\Delta_F(x-y)J(y)). The two-point function is ϕ(x)ϕ(y)=ΔF(xy)\langle\phi(x)\phi(y)\rangle=\Delta_F(x-y), the Feynman propagator. This follows from taking two functional derivatives of Z0[J]Z_0[J] with respect to JJ and setting J=0J=0.

Problems with Solutions

Problem 1. Compute the four-point function ϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)\langle\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle for the free scalar field from the generating functional. Solution. Taking four functional derivatives of Z0[J]Z_0[J] gives the sum over all pairings: ΔF(x1x2)ΔF(x3x4)+ΔF(x1x3)ΔF(x2x4)+ΔF(x1x4)ΔF(x2x3)\Delta_F(x_1-x_2)\Delta_F(x_3-x_4)+\Delta_F(x_1-x_3)\Delta_F(x_2-x_4)+\Delta_F(x_1-x_4)\Delta_F(x_2-x_3), which is Wick’s theorem.

Problem 2. What is the effective action Γ[ϕcl]\Gamma[\phi_{\rm cl}] and how does it relate to 1PI diagrams? Solution. Γ[ϕcl]\Gamma[\phi_{\rm cl}] is the Legendre transform of W[J]=ilnZ[J]W[J]=-i\ln Z[J]. Its expansion in ϕcl\phi_{\rm cl} generates all one-particle-irreducible (1PI) vertex functions, which are the building blocks of Feynman diagrams.

Section summary. Functional methods make symmetries and correlation functions systematic.

Systematics of Renormalization

Core ideas

Renormalization requires all counterterms allowed by symmetries. Power counting tells which diagrams diverge; schemes such as dimensional regularization and minimal subtraction define finite parts. Symmetry identities constrain counterterms.

For review, be able to perform superficial degree-of-divergence counting, explain schemes, and use symmetry to restrict terms. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

D=4L2IBIF+vdv,1ϵ poles are absorbed into countertermsD=4L-2I_B-I_F+\sum_v d_v,\qquad \frac{1}{\epsilon}\ \mathrm{poles\ are\ absorbed\ into\ counterterms}

Worked example

Consider a scalar theory with interaction Lint=gϕ3\mathcal L_{\rm int}=-g\phi^3 in d=4d=4. The superficial degree of divergence of a diagram with LL loops, II internal lines, and VV vertices is D=4L2I+3VD=4L-2I+3V. Using L=IV+1L=I-V+1 and 3V=2I+E3V=2I+E (each vertex has 3 lines, EE external), we get D=4ED=4-E. Thus only the 2-point (E=2E=2, D=2D=2) and 3-point (E=3E=3, D=1D=1) functions are superficially divergent, requiring mass, field-strength, and coupling counterterms.

Problems with Solutions

Problem 1. In ϕ4\phi^4 theory in d=4d=4, which correlation functions are superficially divergent? Solution. D=4ED=4-E, so only E=2E=2 (D=2D=2) and E=4E=4 (D=0D=0) are divergent. This means only the propagator and 4-point vertex need renormalization.

Problem 2. Why does dimensional regularization preserve gauge invariance? Solution. Because the regulator ϵ=4d\epsilon=4-d does not introduce any mass scale or violate the symmetry. The integration measure and Feynman rules are analytically continued in dd, maintaining the Ward identities at every step.

Section summary. The structure of divergences is organized by dimension and symmetry.

Renormalization Group

Core ideas

The renormalization group describes how theories change with scale. Fixed points, beta functions, anomalous dimensions, and running couplings explain asymptotic freedom, critical phenomena, and effective field theory.

For review, be able to interpret beta functions, fixed points, relevant perturbations, and running couplings. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

μdgidμ=βi({g}),O(λx)=λΔO(x)\mu\frac{dg_i}{d\mu}=\beta_i(\{g\}),\qquad \mathcal O(\lambda x)=\lambda^{-\Delta}\mathcal O(x)

Worked example

In ϕ4\phi^4 theory, β(λ)=3λ2/(16π2)\beta(\lambda)=3\lambda^2/(16\pi^2). Integrating from μ0\mu_0 to μ\mu: λ(μ)=λ(μ0)/(13λ(μ0)16π2ln(μ/μ0))\lambda(\mu)=\lambda(\mu_0)/(1-\frac{3\lambda(\mu_0)}{16\pi^2}\ln(\mu/\mu_0)). As μ\mu\to\infty, the denominator vanishes at the Landau pole μL=μ0exp(16π2/(3λ(μ0)))\mu_L=\mu_0\exp(16\pi^2/(3\lambda(\mu_0))). For λ(μ0)0.1\lambda(\mu_0)\sim0.1, μLμ0exp(526)\mu_L\sim\mu_0\exp(526), exponentially far above any physical scale.

Problems with Solutions

Problem 1. QCD has β(g)=g316π2(1123nf)\beta(g)=-\frac{g^3}{16\pi^2}(11-\frac23n_f). For nf=3n_f=3, find the running coupling at μ=10GeV\mu=10\,\mathrm{GeV} if αs(1GeV)=0.35\alpha_s(1\,\mathrm{GeV})=0.35. Solution. With b0=112=9b_0=11-2=9, 1αs(μ)=1αs(μ0)+b02πln(μ/μ0)=2.86+92πln(10)=2.86+3.29=6.15\frac{1}{\alpha_s(\mu)}=\frac{1}{\alpha_s(\mu_0)}+\frac{b_0}{2\pi}\ln(\mu/\mu_0)=2.86+\frac{9}{2\pi}\ln(10)=2.86+3.29=6.15. Thus αs(10GeV)0.163\alpha_s(10\,\mathrm{GeV})\approx0.163.

Problem 2. What is the anomalous dimension of the scalar field in ϕ4\phi^4 theory at one loop? Solution. The field renormalization gives Zϕ=1λ212(16π2)2ϵ+Z_\phi=1-\frac{\lambda^2}{12(16\pi^2)^2\epsilon}+\dots, so the anomalous dimension is γϕ=12μdlnZϕdμ=O(λ2)\gamma_\phi=\frac12\mu\frac{d\ln Z_\phi}{d\mu}=O(\lambda^2). At one loop in coupling, γϕ=0\gamma_\phi=0.

Section summary. RG flow is the map of physics across length scales.

Critical Phenomena

Core ideas

QFT methods apply to continuous phase transitions because long-wavelength fluctuations dominate near criticality. Universality classes depend on symmetry, dimension, and order-parameter components rather than microscopic details.

For review, be able to connect Landau-Ginzburg fields to critical exponents, use correlation length, and explain universality. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

L=12(ϕ)2+12rϕ2+u4!ϕ4,ξTTcν\mathcal L=\frac12(\partial\phi)^2+\frac12r\phi^2+\frac{u}{4!}\phi^4,\qquad \xi\sim |T-T_c|^{-\nu}

Worked example

For the Ising model in d=3d=3, the Landau-Ginzburg theory predicts mean-field exponents: β=1/2\beta=1/2, γ=1\gamma=1, ν=1/2\nu=1/2. However, experiments give β0.326\beta\approx0.326, γ1.24\gamma\approx1.24, ν0.63\nu\approx0.63. The discrepancy arises because mean field theory neglects fluctuations. RG calculations in 4ϵ4-\epsilon dimensions give ν1/2+ϵ/12+\nu\approx1/2+\epsilon/12+\dots, and for ϵ=1\epsilon=1 this is already closer to the observed value.

Problems with Solutions

Problem 1. In mean-field theory, the correlation length diverges as ξr1/2\xi\sim|r|^{-1/2} where rTTcr\sim T-T_c. What is the critical exponent ν\nu? Solution. Since ξrν\xi\sim|r|^{-\nu} and rTTcr\sim T-T_c, we have ν=1/2\nu=1/2 in mean-field theory.

Problem 2. Estimate the upper critical dimension for the Ising model using power counting. Solution. The Gaussian fixed point is stable when the quartic coupling uu is irrelevant. In dd dimensions, [u]=4d[u]=4-d. Thus uu is irrelevant for d>4d>4, making dc=4d_c=4 the upper critical dimension above which mean-field theory is exact.

Section summary. Critical phenomena are statistical field theories near fixed points.

Part III: Non-Abelian Gauge Theories

Core ideas

Non-Abelian gauge theories generalize electromagnetism by making gauge fields carry the charge they mediate. This self-interaction is responsible for asymptotic freedom, confinement, and the structure of the Standard Model.

For review, be able to explain local gauge symmetry, covariant derivatives, field strengths, and gauge-boson self-interactions. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

Dμ=μigAμaTa,Fμνa=μAνaνAμa+gfabcAμbAνcD_\mu=\partial_\mu-igA_\mu^aT^a,\qquad F_{\mu\nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+gf^{abc}A_\mu^bA_\nu^c

Worked example

In SU(2)SU(2) Yang-Mills theory with generators Ta=σa/2T^a=\sigma^a/2, the structure constants are fabc=ϵabcf^{abc}=\epsilon^{abc}. The field strength Fμνa=μAνaνAμa+gϵabcAμbAνcF_{\mu\nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+g\epsilon^{abc}A_\mu^bA_\nu^c contains a quadratic term in the gauge fields, unlike electromagnetism. This self-interaction is responsible for asymptotic freedom and confinement.

Problems with Solutions

Problem 1. Show that the covariant derivative transforms as DμψU(Dμψ)D_\mu\psi\to U(D_\mu\psi) under a local gauge transformation U(x)U(x). Solution. Dμψ(μ+igAμ)Uψ=Uμψ+(μU)ψ+igAμUψD_\mu\psi\to(\partial_\mu+igA'_\mu)U\psi=U\partial_\mu\psi+(\partial_\mu U)\psi+igA'_\mu U\psi. With Aμ=UAμU+ig(μU)UA'_\mu=UA_\mu U^\dagger+\frac{i}{g}(\partial_\mu U)U^\dagger, this becomes U(μ+igAμ)ψ=U(Dμψ)U(\partial_\mu+igA_\mu)\psi=U(D_\mu\psi).

Problem 2. How many physical gluon polarizations exist in SU(3)SU(3) QCD? Solution. There are 8 gauge bosons (N21=8N^2-1=8 for SU(N)SU(N) with N=3N=3). Each massless gauge boson has 2 physical polarizations (transverse). Thus there are 8×2=168\times2=16 physical polarization states.

Section summary. Non-Abelian gauge symmetry is the backbone of modern particle physics.

Yang-Mills Theory

Core ideas

Pure Yang-Mills theory contains only non-Abelian gauge fields. Gauge fixing and ghosts are needed for covariant quantization. The theory is classically simple but quantum mechanically rich, with confinement and topological sectors.

For review, be able to write the Yang-Mills action, explain gauge fixing, ghosts, and why gauge redundancy is not a physical degree of freedom. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

LYM=14FμνaFaμν,Lgf=12ξ(μAaμ)2\mathcal L_{\rm YM}=-\frac14F_{\mu\nu}^aF^{a\mu\nu},\qquad \mathcal L_{\rm gf}=-\frac{1}{2\xi}(\partial_\mu A^{a\mu})^2

Worked example

The Faddeev-Popov determinant for SU(2)SU(2) Yang-Mills theory in covariant gauge gives ghost fields cac^a with Lagrangian Lghost=cˉa(2δabgϵabcμAμc)cb\mathcal L_{\rm ghost}=\bar c^a(-\partial^2\delta^{ab}-g\epsilon^{abc}\partial^\mu A_\mu^c)c^b. The ghosts are complex scalars that couple to gauge fields but obey anticommutation relations. They cancel the unphysical longitudinal and timelike polarizations of the gauge bosons in loop diagrams.

Problems with Solutions

Problem 1. What is the gauge-fixing parameter ξ\xi and how does it affect the gluon propagator? Solution. The gluon propagator in covariant gauge is Δμνab(q)=iδabq2+iϵ(gμν(1ξ)qμqνq2)\Delta_{\mu\nu}^{ab}(q)=\frac{-i\delta^{ab}}{q^2+i\epsilon}\left(g_{\mu\nu}-(1-\xi)\frac{q_\mu q_\nu}{q^2}\right). For ξ=1\xi=1 (Feynman gauge) the propagator is simplest; for ξ=0\xi=0 (Landau gauge) it is transverse.

Problem 2. Why are Faddeev-Popov ghosts necessary in non-Abelian gauge theories but not in QED? Solution. In QED the gauge transformation is Abelian and the Faddeev-Popov determinant is field-independent, so it contributes only an overall constant. In non-Abelian theories the determinant depends on AμA_\mu, requiring dynamical ghost fields to maintain unitarity.

Section summary. Yang-Mills theory is the core non-Abelian gauge field theory.

QCD

Core ideas

QCD is Yang-Mills theory with quarks in the fundamental representation of color. It is weakly coupled at high energy and strongly coupled at low energy. Chiral symmetry, confinement, jets, and hadrons are all consequences of color dynamics.

For review, be able to state QCD Lagrangian, explain asymptotic freedom and confinement, and connect quarks/gluons to hadrons and jets. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

LQCD=14GμνaGaμν+qˉ(iγμDμm)q,β(g)<0\mathcal L_{\rm QCD}=-\frac14G_{\mu\nu}^aG^{a\mu\nu}+\bar q(i\gamma^\mu D_\mu-m)q,\qquad \beta(g)<0

Worked example

At one loop, the QCD beta function is β(g)=g316π2(1123nf)\beta(g)=-\frac{g^3}{16\pi^2}(11-\frac23n_f). For nf=3n_f=3 light quarks, the coefficient is b0=9b_0=9. The running coupling is αs(Q2)=4πb0ln(Q2/ΛQCD2)\alpha_s(Q^2)=\frac{4\pi}{b_0\ln(Q^2/\Lambda_{\rm QCD}^2)}. With ΛQCD200MeV\Lambda_{\rm QCD}\approx200\,\mathrm{MeV}, at Q=2GeVQ=2\,\mathrm{GeV} we get αs=4π/(9ln(4/0.04))=4π/(9×4.6)0.30\alpha_s=4\pi/(9\ln(4/0.04))=4\pi/(9\times4.6)\approx0.30, consistent with lattice and experimental values.

Problems with Solutions

Problem 1. At what energy scale does QCD become strongly coupled if ΛQCD=200MeV\Lambda_{\rm QCD}=200\,\mathrm{MeV}? Solution. By definition, αs\alpha_s\to\infty as QΛQCDQ\to\Lambda_{\rm QCD}. The perturbative formula breaks down near QΛQCDQ\sim\Lambda_{\rm QCD}, which sets the confinement scale. Hadrons have sizes Rc/ΛQCD1fmR\sim\hbar c/\Lambda_{\rm QCD}\sim1\,\mathrm{fm}.

Problem 2. Why does asymptotic freedom imply that quarks behave as free particles at short distances? Solution. As Q2Q^2\to\infty, αs(Q2)0\alpha_s(Q^2)\to0. In high-energy scattering, the relevant distance scale is r/Qr\sim\hbar/Q, so at short distances the coupling is weak. This justifies the parton model and perturbative calculations for jets.

Section summary. QCD is the quantum field theory of strong interactions.

Electroweak Theory

Core ideas

Electroweak theory unifies weak and electromagnetic interactions through SU(2)L×U(1)YSU(2)_L\times U(1)_Y gauge symmetry. The Higgs mechanism breaks it to electromagnetism, giving masses to WW, ZZ, and fermions while leaving the photon massless.

For review, be able to write the symmetry breaking pattern, identify charged and neutral currents, and explain masses from the Higgs vacuum expectation value. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

SU(2)L×U(1)YU(1)em,MW=12gv,MZ=12g2+g2vSU(2)_L\times U(1)_Y\to U(1)_{\rm em},\qquad M_W=\frac12gv,\qquad M_Z=\frac12\sqrt{g^2+g'^2}\,v

Worked example

With measured values MW=80.4GeVM_W=80.4\,\mathrm{GeV} and MZ=91.2GeVM_Z=91.2\,\mathrm{GeV}, the weak mixing angle is cosθW=MW/MZ=0.881\cos\theta_W=M_W/M_Z=0.881, giving sin2θW=10.777=0.223\sin^2\theta_W=1-0.777=0.223. The SU(2)LSU(2)_L coupling is g=2MW/v=2(80.4)/246=0.654g=2M_W/v=2(80.4)/246=0.654, and the U(1)YU(1)_Y coupling is g=gtanθW=0.654(0.487)=0.318g'=g\tan\theta_W=0.654(0.487)=0.318. The electric charge is e=gg/g2+g2=0.313e=gg'/\sqrt{g^2+g'^2}=0.313, or α=e2/(4π)1/128\alpha=e^2/(4\pi)\approx1/128 at the weak scale.

Problems with Solutions

Problem 1. The Higgs vacuum expectation value is v=(2GF)1/2246GeVv=(\sqrt{2}G_F)^{-1/2}\approx246\,\mathrm{GeV}. Derive this from the muon lifetime formula. Solution. Fermi theory gives GF/2=g2/(8MW2)G_F/\sqrt{2}=g^2/(8M_W^2). Using MW=gv/2M_W=gv/2, we get GF/2=1/(2v2)G_F/\sqrt{2}=1/(2v^2), so v=(2GF)1/2v=(\sqrt{2}G_F)^{-1/2}. With GF=1.166×105GeV2G_F=1.166\times10^{-5}\,\mathrm{GeV^{-2}}, v=(2×1.166×105)1/2246GeVv=(\sqrt{2}\times1.166\times10^{-5})^{-1/2}\approx246\,\mathrm{GeV}.

Problem 2. How many Goldstone bosons are eaten in electroweak symmetry breaking? Solution. The gauge group SU(2)L×U(1)YSU(2)_L\times U(1)_Y has 4 generators. The unbroken U(1)emU(1)_{\rm em} has 1 generator. Thus 41=34-1=3 Goldstone bosons are eaten by the W±W^\pm and ZZ bosons, giving them longitudinal polarizations. The photon remains massless.

Section summary. Electroweak theory is chiral gauge theory with spontaneous symmetry breaking.

Anomalies and Beyond

Core ideas

Anomalies occur when a classical symmetry fails after quantization. Gauge anomalies must cancel for consistency; global anomalies can have physical consequences. Effective field theory extends the Standard Model by higher-dimension operators.

For review, be able to distinguish global and gauge anomalies, state anomaly cancellation, and use EFT as a language for new physics. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

μj5μ=g216π2FμνF~μν,LEFT=LSM+iciΛdi4Oi\partial_\mu j_5^\mu=\frac{g^2}{16\pi^2}F_{\mu\nu}\tilde F^{\mu\nu},\qquad \mathcal L_{\rm EFT}=\mathcal L_{\rm SM}+\sum_i\frac{c_i}{\Lambda^{d_i-4}}\mathcal O_i

Worked example

The chiral anomaly in QED with one fermion gives μj5μ=e28π2EB\partial_\mu j_5^\mu=\frac{e^2}{8\pi^2}\bm E\cdot\bm B. In a constant background with E=105V/mE=10^5\,\mathrm{V/m} and B=1TB=1\,\mathrm T, the anomaly density is e28π22c3EB(1.6×1019)28π2(1.05×1034)2(3×108)3(105)(1)2.9×1033m3s1\frac{e^2}{8\pi^2\hbar^2c^3}EB\approx\frac{(1.6\times10^{-19})^2}{8\pi^2(1.05\times10^{-34})^2(3\times10^8)^3}(10^5)(1)\approx2.9\times10^{33}\,\mathrm{m^{-3}s^{-1}}. This enormous rate reflects that the anomaly is a UV effect, not a physical particle production rate in a static background.

Problems with Solutions

Problem 1. Why must gauge anomalies cancel in the Standard Model? Solution. A gauge anomaly would make the theory inconsistent: the Ward identities fail, unitarity is violated, and the theory cannot be renormalized. The SM cancels anomalies because the quark and lepton contributions add to zero: generations(2NcQq2Q+2Q2)=0\sum_{\text{generations}}(2N_cQ_q^2-Q_{\ell^+}^2-Q_{\ell^-}^2)=0.

Problem 2. Write the dimension-6 operator O6ψ=(ˉγμ)(eˉγμe)\mathcal O_{6\psi}=(\bar\ell\gamma^\mu\ell)(\bar e\gamma_\mu e) that contributes to muon decay. If the coefficient is c/Λ2c/\Lambda^2 and c1c\sim1, what limit on Λ\Lambda does the muon lifetime constraint GFG_F imply? Solution. Since GF/21/(2v2)1.17×105GeV2G_F/\sqrt{2}\sim1/(2v^2)\approx1.17\times10^{-5}\,\mathrm{GeV^{-2}}, requiring the new contribution to be smaller gives Λ1/GF300GeV\Lambda\gtrsim1/\sqrt{G_F}\approx300\,\mathrm{GeV}. More precisely, precision electroweak data push this to ΛfewTeV\Lambda\gtrsim few\,\mathrm{TeV}.

Section summary. Anomalies and EFT show how quantum consistency and high scales shape low-energy physics.