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Particle Physics · AI note · graduate · #particle

Complete Particle Physics

A concise guide to particles, relativistic kinematics, symmetries, quarks, Feynman amplitudes, QED, weak interactions, QCD, the Standard Model, and physics beyond it.

· 12 min
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Historical Introduction to the Elementary Particles

Core ideas

Particle physics grew from cosmic rays, nuclear physics, accelerators, and the discovery of hadrons, leptons, quarks, neutrinos, and gauge bosons. The modern picture is the Standard Model: matter fermions interacting through gauge fields, with masses from electroweak symmetry breaking.

For review, be able to name the Standard Model particles, distinguish leptons, quarks, gauge bosons, and hadrons, and explain why accelerators probe short distances. 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

Δx∼ℏ/Δp,SM gauge group=SU(3)C×SU(2)L×U(1)Y\Delta x\sim\hbar/\Delta p,\qquad \mathrm{SM\ gauge\ group}=SU(3)_C\times SU(2)_L\times U(1)_Y

Worked example

The LHC accelerates protons to a momentum of roughly p=7 TeV/cp=7\,\mathrm{TeV}/c. Using the de Broglie relation we can estimate the shortest distance scale that such a beam can resolve:

λ=hp=4.136×10−15 eV s×2.998×108 m/s7×1012 eV/c×c=1.2398×10−6 eV m7×1012 eV≈1.8×10−19 m=0.18 fm.\lambda = \frac{h}{p} = \frac{4.136\times10^{-15}\,\mathrm{eV\,s}\times 2.998\times10^{8}\,\mathrm{m/s}}{7\times10^{12}\,\mathrm{eV}/c \times c} = \frac{1.2398\times10^{-6}\,\mathrm{eV\,m}}{7\times10^{12}\,\mathrm{eV}} \approx 1.8\times10^{-19}\,\mathrm{m}=0.18\,\mathrm{fm}.

This is well below the charge radius of the proton (≈0.84 fm\approx0.84\,\mathrm{fm}), which is why the LHC can probe the internal structure of the proton and produce heavy particles such as the Higgs boson (mH≈125 GeV/c2m_H\approx125\,\mathrm{GeV}/c^2). The energy available in the proton—proton center-of-mass frame at the LHC is s=13 TeV\sqrt{s}=13\,\mathrm{TeV}, enough to create on-shell pairs of top quarks (2mt≈346 GeV2m_t\approx346\,\mathrm{GeV}) with abundant phase space.

Problems with Solutions

Problem 1. How many elementary fermions (quarks + leptons, including neutrinos) does the Standard Model contain? Do not count antiparticles separately. Solution. There are 6 quark flavors (u,d,c,s,t,bu,d,c,s,t,b), each coming in 3 colors, giving 6×3=186\times3=18 quarks. There are 6 leptons (e,μ,τ,νe,νμ,ντe,\mu,\tau,\nu_e,\nu_\mu,\nu_\tau). Each fermion has a left-handed and right-handed chirality state, but counting distinct particle species once, the total is 18+6=2418+6=24 fermions.

Problem 2. Estimate the de Broglie wavelength of an electron with kinetic energy K=1.0 GeVK=1.0\,\mathrm{GeV}. Use ℏc≈197.3 MeV fm\hbar c\approx197.3\,\mathrm{MeV\,fm} and mec2=0.511 MeVm_ec^2=0.511\,\mathrm{MeV}. Solution. Since K≫mec2K\gg m_ec^2, the electron is ultra-relativistic and p≈E/c≈K/cp\approx E/c\approx K/c. Thus

λ=hp≈2πℏcK=2π(197.3 MeV fm)1000 MeV≈1.24 fm.\lambda = \frac{h}{p} \approx \frac{2\pi\hbar c}{K} = \frac{2\pi(197.3\,\mathrm{MeV\,fm})}{1000\,\mathrm{MeV}} \approx 1.24\,\mathrm{fm}.

This wavelength is comparable to the size of a nucleus, which is why GeV-scale electron scattering (as at JLab) resolves nuclear substructure.

Problem 3. The Higgs boson has mass mH=125.1 GeV/c2m_H=125.1\,\mathrm{GeV}/c^2. What is the minimum photon energy required to produce an on-shell Higgs boson in the process γ+γ→H\gamma+\gamma\to H in the center-of-mass frame? What is the corresponding photon wavelength? Solution. In the center-of-mass frame, each photon must carry half the total invariant mass energy:

Eγ=mHc22=125.12 GeV≈62.6 GeV.E_\gamma = \frac{m_Hc^2}{2} = \frac{125.1}{2}\,\mathrm{GeV} \approx 62.6\,\mathrm{GeV}.

The wavelength is

λ=hcEγ=4.136×10−15 eV s×2.998×108 m/s62.6×109 eV≈1.98×10−17 m=1.98×10−8 fm.\lambda = \frac{hc}{E_\gamma} = \frac{4.136\times10^{-15}\,\mathrm{eV\,s}\times 2.998\times10^{8}\,\mathrm{m/s}}{62.6\times10^{9}\,\mathrm{eV}} \approx 1.98\times10^{-17}\,\mathrm{m}=1.98\times10^{-8}\,\mathrm{fm}.

Section summary. The Standard Model organizes the particle zoo into fields and symmetries.

Elementary Particle Dynamics

Core ideas

Scattering and decay experiments measure probabilities from amplitudes. Cross sections, decay rates, luminosity, phase space, and matrix elements connect theory to event counts. Relativistic normalization and conservation laws constrain every process.

For review, be able to compute event-rate relations, distinguish cross section and decay width, and use phase space qualitatively. 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

dN=Lσ dt,dΓ=12M∣M∣2dΦndN=\mathcal L\sigma\,dt,\qquad d\Gamma=\frac{1}{2M}|\mathcal M|^2d\Phi_n

Worked example

In a typical LHC run the ATLAS detector records data with an instantaneous luminosity L=2×1034 cm−2 s−1\mathcal L=2\times10^{34}\,\mathrm{cm^{-2}\,s^{-1}}. The cross section for producing a Higgs boson via gluon fusion is approximately σgg→H≈50 pb\sigma_{gg\to H}\approx50\,\mathrm{pb} at s=13 TeV\sqrt{s}=13\,\mathrm{TeV}. The event rate is

dNdt=Lσ=(2×1034 cm−2 s−1)(50×10−12 cm2)=1.0×103 s−1.\frac{dN}{dt}=\mathcal L\sigma =(2\times10^{34}\,\mathrm{cm^{-2}\,s^{-1}})(50\times10^{-12}\,\mathrm{cm^2}) =1.0\times10^{3}\,\mathrm{s^{-1}}.

Over a data-taking period of T=107 sT=10^7\,\mathrm{s} (roughly one LHC year), the expected number of Higgs events is

N=Lintσ=(2×1034×107 cm−2)(50×10−12 cm2)=107 pb−1×50 pb=5×105.N = \mathcal L_{\rm int}\sigma = (2\times10^{34}\times10^7\,\mathrm{cm^{-2}})(50\times10^{-12}\,\mathrm{cm^2}) = 10^{7}\,\mathrm{pb}^{-1}\times50\,\mathrm{pb}=5\times10^{5}.

Only a fraction of these are cleanly reconstructed; the rest are buried in background.

Problems with Solutions

Problem 1. The total cross section for e+e−→μ+μ−e^+e^-\to\mu^+\mu^- near s=10 GeV\sqrt{s}=10\,\mathrm{GeV} is about σ≈1.4 nb\sigma\approx1.4\,\mathrm{nb}. If the BaBar experiment had an integrated luminosity Lint=500 fb−1\mathcal L_{\rm int}=500\,\mathrm{fb^{-1}}, how many μ+μ−\mu^+\mu^- events were produced? Solution. Convert units: 1 fb−1=10−39 cm21\,\mathrm{fb^{-1}}=10^{-39}\,\mathrm{cm^2}, and 1 nb=10−33 cm2=106 fb1\,\mathrm{nb}=10^{-33}\,\mathrm{cm^2}=10^{6}\,\mathrm{fb}. Then

N=σLint=(1.4×106 fb)(500 fb−1)=7×108 events.N = \sigma\mathcal L_{\rm int} = (1.4\times10^{6}\,\mathrm{fb})(500\,\mathrm{fb^{-1}})=7\times10^{8}\,\text{events}.

Problem 2. A particle has a proper lifetime τ=1.5×10−23 s\tau=1.5\times10^{-23}\,\mathrm{s}. Compute its decay width Γ\Gamma in MeV. Solution. Using Γ=ℏ/τ\Gamma=\hbar/\tau with ℏ=6.582×10−22 MeV s\hbar=6.582\times10^{-22}\,\mathrm{MeV\,s},

Γ=6.582×10−22 MeV s1.5×10−23 s≈43.9 MeV.\Gamma = \frac{6.582\times10^{-22}\,\mathrm{MeV\,s}}{1.5\times10^{-23}\,\mathrm{s}}\approx 43.9\,\mathrm{MeV}.

Problem 3. A beam of 10610^{6} neutrons per second passes through a thin target of thickness d=1 cmd=1\,\mathrm{cm} containing n=5×1022 cm−3n=5\times10^{22}\,\mathrm{cm^{-3}} hydrogen nuclei. The total neutron-proton scattering cross section is σtot≈20 b\sigma_{\rm tot}\approx20\,\mathrm{b}. How many scattering events occur per second? Solution. The interaction probability for a thin target is P=nσdP=n\sigma d. The number of events per second is

dNdt=Nbeamnσd=(106)(5×1022)(20×10−24)(1)=1.0×106 s−1.\frac{dN}{dt}=N_{\rm beam}n\sigma d=(10^{6})(5\times10^{22})(20\times10^{-24})(1)=1.0\times10^{6}\,\mathrm{s^{-1}}.

The beam is completely attenuated because nσd=1n\sigma d=1; this is the definition of one mean free path.

Section summary. Dynamics is inferred from scattering rates and decay probabilities.

Relativistic Kinematics

Core ideas

Particle reactions obey four-momentum conservation. Mandelstam variables, invariant masses, thresholds, rapidity, and center-of-mass frames make high-energy reactions frame independent.

For review, be able to use invariant mass, threshold conditions, Mandelstam variables, and two-body kinematics. 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=(p1+p2)2,t=(p1−p3)2,u=(p1−p4)2s=(p_1+p_2)^2,\qquad t=(p_1-p_3)^2,\qquad u=(p_1-p_4)^2

Worked example

Consider the decay π0→γγ\pi^0\to\gamma\gamma with the pion at rest (mπ0=135 MeV/c2m_{\pi^0}=135\,\text{MeV}/c^2). Energy conservation gives each photon energy Eγ=mπ0c2/2=67.5 MeVE_\gamma=m_{\pi^0}c^2/2=67.5\,\text{MeV}. In the lab frame where the pion has momentum pπ=200 MeV/cp_\pi=200\,\text{MeV}/c, the invariant mass is s=(Eπ)2−(pπc)2=((135)2+(200)2)2−(200)2=(240.5)2−(200)2=1.82×104 MeV2s=(E_\pi)^2-(p_\pi c)^2=(\sqrt{(135)^2+(200)^2})^2-(200)^2=(240.5)^2-(200)^2=1.82\times10^4\,\text{MeV}^2, and s=240.5 MeV/c2\sqrt{s}=240.5\,\text{MeV}/c^2, which is the total center-of-mass energy.

Problems with Solutions

Problem 1. Two protons each with kinetic energy K=50 GeVK=50\,\text{GeV} collide head-on. What is s\sqrt{s}? Solution. For identical particles colliding head-on, s=2Ebeam=2(50+0.938)≈102 GeV\sqrt{s}=2E_{\text{beam}}=2(50+0.938)\approx102\,\text{GeV}.

Problem 2. In fixed-target scattering of a 10 GeV10\,\text{GeV} electron on a proton at rest, what is s\sqrt{s}? Solution. s=me2+mp2+2mpElab≈(0.511)2+(938)2+2(938)(10000)≈1.876×107 MeV2s=m_e^2+m_p^2+2m_pE_{\text{lab}}\approx(0.511)^2+(938)^2+2(938)(10000)\approx1.876\times10^7\,\text{MeV}^2, so s≈4.3 GeV\sqrt{s}\approx4.3\,\text{GeV}.

Problem 3. For e+e−→μ+μ−e^+e^-\to\mu^+\mu^-, what is the threshold center-of-mass energy? Solution. The minimum s\sqrt{s} is 2mμ=2×105.7=211.4 MeV2m_\mu=2\times105.7=211.4\,\text{MeV}. At threshold the muons are produced at rest in the CM frame.

Section summary. Relativistic invariants are the language of collisions.

Symmetries

Core ideas

Symmetries classify particles and constrain interactions. Continuous symmetries give conserved charges; discrete symmetries include parity, charge conjugation, and time reversal. Gauge symmetry is not optional decoration: it determines interactions.

For review, be able to apply Noether reasoning, identify internal quantum numbers, and distinguish global from gauge symmetries. 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

∂μjμ=0,Q=∫j0 d3x\partial_\mu j^\mu=0,\qquad Q=\int j^0\,d^3x

Worked example

The decay π0→γγ\pi^0\to\gamma\gamma is allowed because the pion has JPC=0−+J^{PC}=0^{-+} and two photons can combine to JP=0−J^P=0^- (with the two photons in a state antisymmetric under exchange, consistent with C-parity C=+1C=+1 for the two-photon system). In contrast, π0→γγγ\pi^0\to\gamma\gamma\gamma is strongly suppressed because C-parity would be C=−1C=-1 for three photons, violating C-conservation in electromagnetic interactions.

Problems with Solutions

Problem 1. A particle has spin J=1J=1 and decays into two identical spin-0 particles. What are the allowed values of the orbital angular momentum ℓ\ell between them? Solution. J=ℓJ=\ell for two spin-0 particles, so ℓ=1\ell=1. Since the particles are identical bosons, the spatial wavefunction must be symmetric under exchange, requiring even ℓ\ell. But ℓ=1\ell=1 is odd---thus the decay is forbidden by Bose symmetry (this is why the ZZ boson does not decay to two identical neutral pions).

Problem 2. Show that the electromagnetic current jμ=ψˉγμψj^\mu=\bar\psi\gamma^\mu\psi is conserved for the Dirac equation. Solution. Using the Dirac equation (iγμ∂μ−m)ψ=0(i\gamma^\mu\partial_\mu-m)\psi=0 and its adjoint, one finds ∂μjμ=0\partial_\mu j^\mu=0, confirming charge conservation.

Problem 3. In strong interactions, isospin is conserved. A Δ++\Delta^{++} (I=3/2I=3/2, I3=+3/2I_3=+3/2) decays to pp (I=1/2I=1/2, I3=+1/2I_3=+1/2) and π+\pi^+ (I=1I=1, I3=+1I_3=+1). Verify I3I_3 conservation. Solution. Initial I3=+3/2I_3=+3/2. Final I3=+1/2+1=+3/2I_3=+1/2+1=+3/2. Conserved.

Section summary. Symmetries determine conservation laws and allowed processes.

Bound States and the Quark Model

Core ideas

Hadrons are bound states of quarks and gluons. Mesons are quark-antiquark states, baryons contain three valence quarks, and flavor symmetry organizes multiplets. Color confinement prevents isolated quarks.

For review, be able to construct basic meson and baryon quantum numbers, use flavor multiplets, and explain color singlets. 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

qqˉ meson,qqq baryon,physical states are color singletsq\bar q\ \mathrm{meson},\qquad qqq\ \mathrm{baryon},\qquad \mathrm{physical\ states\ are\ color\ singlets}

Worked example

The Δ++\Delta^{++} baryon is composed of three up quarks (uuuuuu). Each quark has spin 1/21/2, charge +2/3+2/3, and color can be red, green, or blue. To make a color singlet, the three quarks must carry different colors. The total spin is S=3/2S=3/2 (all spins aligned), giving the observed JP=3/2+J^P=3/2^+. The charge is Q=3×(+2/3)=+2Q=3\times(+2/3)=+2, matching experiment.

Problems with Solutions

Problem 1. Construct the quark content of the K+K^+ meson and determine its strangeness, isospin, and charge. Solution. K+=usˉK^+=u\bar s. Strangeness S=+1S=+1 (since sˉ\bar s has S=+1S=+1), isospin I=1/2I=1/2, I3=+1/2I_3=+1/2, charge Q=+2/3−(−1/3)=+1Q=+2/3-(-1/3)=+1.

Problem 2. The Ω−\Omega^- baryon has S=−3S=-3 and Q=−1Q=-1. What is its quark content? Solution. Three strange quarks: ssssss. Each ss has S=−1S=-1 and Q=−1/3Q=-1/3, so total S=−3S=-3 and Q=−1Q=-1.

Problem 3. Why cannot a hadron with quark content uuuu exist as a free particle? Solution. A uuuu diquark is not a color singlet. In QCD, only color-singlet states have finite energy at large distances; colored states experience a confining potential that grows with separation. Also, baryons need three quarks (or antibaryons three antiquarks) to form a color singlet with the antisymmetric color wavefunction.

Section summary. The quark model organizes hadron spectroscopy.

Feynman Calculus and QED Processes

Core ideas

Feynman diagrams encode perturbative amplitudes. In QED, vertices couple charged fermions to photons; propagators describe virtual particles; spin sums and phase space produce cross sections.

For review, be able to translate simple diagrams to amplitudes, identify propagators and vertices, and know standard QED processes like annihilation and Compton scattering. 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(iγμDμ−m)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 e−μ−→e−μ−e^-\mu^-\to e^-\mu^-, the tree-level amplitude is M=(−ie)2(uˉeγμue)(−gμν/q2)(uˉμγνuμ)\mathcal M=(-ie)^2(\bar u_e\gamma^\mu u_e)(-g_{\mu\nu}/q^2)(\bar u_\mu\gamma^\nu u_\mu). Squaring and summing over spins gives the Mott cross section. At low energies this reduces to Rutherford scattering, but with relativistic corrections.

Problems with Solutions

Problem 1. Write down the Feynman amplitude for e+e−→μ+μ−e^+e^-\to\mu^+\mu^- at tree level. Solution. M=e2s(vˉe−γμue+)(uˉμ−γμvμ+)\mathcal M=\frac{e^2}{s}(\bar v_{e^-}\gamma^\mu u_{e^+})(\bar u_{\mu^-}\gamma_\mu v_{\mu^+}).

Problem 2. In Compton scattering, what are the two tree-level diagrams? Solution. (1) Electron absorbs incoming photon, emits outgoing photon. (2) Electron emits outgoing photon first, then absorbs incoming photon. The sum gives the Klein—Nishina formula.

Problem 3. The running coupling in QED is α(Q2)=α(0)/(1−Π(Q2))\alpha(Q^2)=\alpha(0)/(1-\Pi(Q^2)). At Q=mZQ=m_Z, α≈1/128\alpha\approx1/128. Why is it larger than α(0)=1/137\alpha(0)=1/137? Solution. The vacuum polarization Π(Q2)\Pi(Q^2) is negative at Q2>0Q^2>0, so the denominator is <1<1. Physically, the charge is screened at long distances by virtual e+e−e^+e^- pairs; at short distances (high Q2Q^2) one penetrates the screening cloud and sees more of the bare charge.

Section summary. QED is the prototype precision gauge theory.

Weak Interactions

Core ideas

Weak interactions change flavor and violate parity. Charged currents involve W±W^\pm, neutral currents involve ZZ, and low-energy weak processes are described by Fermi theory. Neutrino mixing shows that flavor and mass eigenstates differ.

For review, be able to describe beta decay, charged and neutral currents, parity violation, CKM/PMNS mixing, and the Fermi limit. 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

LFermi∼−GF2Jμ†Jμ,GF/2=g28MW2\mathcal L_{\rm Fermi}\sim-\frac{G_F}{\sqrt2}J_\mu^\dagger J^\mu,\qquad G_F/\sqrt2=\frac{g^2}{8M_W^2}

Worked example

The muon lifetime is τμ=2.2 \mus\tau_\mu=2.2\,\text{\mus}. In Fermi theory, the decay width is

Γμ=GF2mμ5192π3.\Gamma_\mu=\frac{G_F^2m_\mu^5}{192\pi^3}.

Using GF=1.166×10−5 GeV−2G_F=1.166\times10^{-5}\,\text{GeV}^{-2} and mμ=105.7 MeVm_\mu=105.7\,\text{MeV}:

Γμ=(1.166×10−5)2(0.1057)5192π3≈3.0×10−19 GeV,\Gamma_\mu=\frac{(1.166\times10^{-5})^2(0.1057)^5}{192\pi^3}\approx3.0\times10^{-19}\,\text{GeV},

giving τμ=ℏ/Γ≈6.6×10−25 GeV s/3.0×10−19 GeV≈2.2×10−6 s\tau_\mu=\hbar/\Gamma\approx6.6\times10^{-25}\,\text{GeV\,s}/3.0\times10^{-19}\,\text{GeV}\approx2.2\times10^{-6}\,\text{s}, matching experiment.

Problems with Solutions

Problem 1. Estimate the decay width of the WW boson (MW=80.4 GeVM_W=80.4\,\text{GeV}) using ΓW≈GFMW3/(62π)\Gamma_W\approx G_FM_W^3/(6\sqrt{2}\pi). Solution. ΓW=(1.166×10−5)2(80.4)3/(62π)\Gamma_W=(1.166\times10^{-5})^2(80.4)^3/(6\sqrt{2}\pi)… Actually, a better estimate is ΓW≈2.1 GeV\Gamma_W\approx2.1\,\text{GeV}, which corresponds to τW≈3×10−25 s\tau_W\approx3\times10^{-25}\,\text{s}.

Problem 2. In beta decay, the maximum electron kinetic energy is Kmax⁡=0.782 MeVK_{\max}=0.782\,\text{MeV} for n→p+e−+νˉen\to p+e^-+\bar\nu_e. What is the minimum neutrino energy in the neutron rest frame? Solution. The neutrino energy is minimized when the electron and proton recoil together opposite to the neutrino. Using energy-momentum conservation, Eνmin⁡≈Kmax⁡−QE_{\nu}^{\min}\approx K_{\max}-Q… In the limit, Eνmin⁡≈0E_\nu^{\min}\approx0 (when the electron takes all the energy), while Eνmax⁡=Q=0.782 MeVE_\nu^{\max}=Q=0.782\,\text{MeV} (when the electron and proton are at rest relative to each other).

Problem 3. The CKM matrix element ∣Vus∣≈0.22|V_{us}|\approx0.22. By what factor is s→us\to u suppressed relative to d→ud\to u transitions? Solution. The amplitude is proportional to ∣Vus∣2≈0.048|V_{us}|^2\approx0.048, so s→us\to u transitions are suppressed by roughly a factor of 2020 compared to d→ud\to u.

Section summary. Weak interactions are short-range, chiral, and flavor-changing.

Gauge Theories and the Standard Model

Core ideas

The Standard Model is a renormalizable gauge theory with spontaneous electroweak symmetry breaking. Gauge symmetry fixes the interactions; the Higgs field gives masses while preserving consistency at high energy.

For review, be able to state the gauge group, identify representations, explain Higgs mechanism, and connect masses to 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

SU(3)C×SU(2)L×U(1)Y→SU(3)C×U(1)emSU(3)_C\times SU(2)_L\times U(1)_Y\to SU(3)_C\times U(1)_{\rm em}

Worked example

The Higgs mechanism gives the WW boson mass via MW=gv2M_W=\frac{gv}{2} where v≈246 GeVv\approx246\,\text{GeV} is the Higgs vacuum expectation value and g≈0.65g\approx0.65 is the SU(2)LSU(2)_L gauge coupling. Thus MW≈0.65×246/2≈80 GeVM_W\approx0.65\times246/2\approx80\,\text{GeV}, close to the measured 80.4 GeV80.4\,\text{GeV}. The Higgs mass itself is MH=2λvM_H=\sqrt{2\lambda}v where λ\lambda is the Higgs self-coupling; experimentally MH≈125 GeVM_H\approx125\,\text{GeV}, giving λ≈0.13\lambda\approx0.13.

Problems with Solutions

Problem 1. The weak mixing angle is sin⁡2θW≈0.23\sin^2\theta_W\approx0.23. Compute the ZZ boson mass from MZ=MW/cos⁡θWM_Z=M_W/\cos\theta_W. Solution. cos⁡θW=1−0.23=0.877\cos\theta_W=\sqrt{1-0.23}=0.877, so MZ=80.4/0.877=91.7 GeVM_Z=80.4/0.877=91.7\,\text{GeV}, matching the measured 91.2 GeV91.2\,\text{GeV}.

Problem 2. How many physical Higgs bosons does the Standard Model contain? How many would a supersymmetric extension contain? Solution. The SM has one neutral scalar Higgs (h0h^0). The Minimal Supersymmetric Standard Model (MSSM) has five: two CP-even neutral scalars (h0h^0, H0H^0), one CP-odd pseudoscalar (A0A^0), and two charged scalars (H±H^\pm).

Problem 3. A fermion acquires mass via the Yukawa coupling yfy_f as mf=yfv/2m_f=y_fv/\sqrt{2}. For the top quark (mt≈173 GeVm_t\approx173\,\text{GeV}), what is yty_t? Solution. yt=2mt/v=2(173)/246=0.996≈1y_t=\sqrt{2}m_t/v=\sqrt{2}(173)/246=0.996\approx1. The top quark has a Yukawa coupling of order unity, the largest in the SM.

Section summary. The Standard Model is a symmetry-based quantum field theory.

QCD and Hadron Spectroscopy

Core ideas

QCD is the non-Abelian gauge theory of color. Asymptotic freedom makes quarks weakly coupled at high momentum; confinement and chiral symmetry breaking dominate low-energy hadrons. Jets reveal quarks and gluons experimentally.

For review, be able to explain color, running coupling, confinement, asymptotic freedom, jets, and hadron multiplets. 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

αs(Q2)∼1ln⁡(Q2/ΛQCD2),LQCD=−14GμνaGaμν+qˉ(iγμDμ−m)q\alpha_s(Q^2)\sim\frac{1}{\ln(Q^2/\Lambda_{\rm QCD}^2)},\qquad \mathcal L_{\rm QCD}=-\frac14G^a_{\mu\nu}G^{a\mu\nu}+\bar q(i\gamma^\mu D_\mu-m)q

Worked example

The running strong coupling at Q=MZQ=M_Z is αs(MZ2)≈0.118\alpha_s(M_Z^2)\approx0.118. Using the one-loop renormalization-group equation with ΛQCD≈200 MeV\Lambda_{\text{QCD}}\approx200\,\text{MeV}:

αs(Q2)=4π(11−23nf)ln⁡(Q2/Λ2).\alpha_s(Q^2)=\frac{4\pi}{(11-\frac23n_f)\ln(Q^2/\Lambda^2)}.

At Q=2 GeVQ=2\,\text{GeV} with nf=4n_f=4, αs≈4π/(9.33ln⁡(4/0.04))=4π/(9.33×4.6)≈0.29\alpha_s\approx4\pi/(9.33\ln(4/0.04))=4\pi/(9.33\times4.6)\approx0.29, showing the growth of the coupling at lower energies.

Problems with Solutions

Problem 1. A proton has radius R≈0.84 fmR\approx0.84\,\text{fm}. Estimate the characteristic QCD energy scale inside it. Solution. E∼ℏc/R=(197 MeV fm)/0.84 fm≈235 MeVE\sim\hbar c/R=(197\,\text{MeV\,fm})/0.84\,\text{fm}\approx235\,\text{MeV}, comparable to ΛQCD\Lambda_{\text{QCD}} and the confinement scale.

Problem 2. In deep inelastic scattering, Bjorken scaling says F2(x,Q2)F_2(x,Q^2) becomes independent of Q2Q^2 at high Q2Q^2. Why is this evidence for pointlike quarks? Solution. At high Q2Q^2 (short wavelength), the probe resolves distances much smaller than the hadron size. If the constituents were extended, the cross section would fall with Q2Q^2. The observed scaling implies the scattering is from structureless, pointlike objects---quarks.

Problem 3. Why are free quarks not observed? Solution. Color confinement: the potential between a quark and antiquark grows linearly with separation, V(r)≈σrV(r)\approx\sigma r with string tension σ∼1 GeV/fm\sigma\sim1\,\text{GeV/fm}. Pulling quarks apart requires enough energy to create a new qqˉq\bar q pair, producing two mesons rather than isolated quarks.

Section summary. QCD explains strong interactions through color gauge fields.

Beyond the Standard Model

Core ideas

The Standard Model is incomplete: it omits gravity, dark matter, baryon asymmetry, and a full explanation of neutrino masses and hierarchy. Effective field theory parameterizes possible heavy physics without knowing its details.

For review, be able to state major open problems, use EFT logic, and distinguish direct, indirect, and cosmological probes. 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

LEFT=LSM+∑iciΛdi−4Oi\mathcal L_{\rm EFT}=\mathcal L_{\rm SM}+\sum_i\frac{c_i}{\Lambda^{d_i-4}}\mathcal O_i

Worked example

The dimension-6 operator O6=(ℓˉγμℓ)(qˉγμq)\mathcal O_6=(\bar\ell\gamma^\mu\ell)(\bar q\gamma_\mu q) could be generated by a heavy Z′Z' boson. In the EFT framework,

LEFT=LSM+c6Λ2O6.\mathcal L_{\text{EFT}}=\mathcal L_{\text{SM}}+\frac{c_6}{\Lambda^2}\mathcal O_6.

If LEP measured the effective four-fermion coupling to be ∣Geff∣<10−6 GeV−2|G_{\text{eff}}|<10^{-6}\,\text{GeV}^{-2}, then for c6∼1c_6\sim1, Λ>1 TeV\Lambda>1\,\text{TeV}. This means any Z′Z' must be heavier than about 1 TeV1\,\text{TeV} unless its couplings are suppressed.

Problems with Solutions

Problem 1. The Higgs mass receives quadratically divergent corrections ΔmH2∼3g28π2Λ2\Delta m_H^2\sim\frac{3g^2}{8\pi^2}\Lambda^2. For Λ=1 TeV\Lambda=1\,\text{TeV}, estimate ΔmH2\Delta m_H^2 and compare to mH2=(125 GeV)2m_H^2=(125\,\text{GeV})^2. Solution. With g≈0.65g\approx0.65, ΔmH2∼3(0.65)2(1000)2/(8π2)≈1.6×104 GeV2\Delta m_H^2\sim3(0.65)^2(1000)^2/(8\pi^2)\approx1.6\times10^4\,\text{GeV}^2, which is much larger than mH2=1.56×104 GeV2m_H^2=1.56\times10^4\,\text{GeV}^2… Actually both are ∼104\sim10^4, so the correction is of order the mass itself, requiring fine-tuning of 10−210^{-2} for Λ=10 TeV\Lambda=10\,\text{TeV}.

Problem 2. Neutrino oscillations imply Δm212≈7.5×10−5 eV2\Delta m^2_{21}\approx7.5\times10^{-5}\,\text{eV}^2. If the seesaw mechanism gives mν∼mD2/MRm_\nu\sim m_D^2/M_R, with mD∼100 GeVm_D\sim100\,\text{GeV} (like the top quark), what is the right-handed neutrino mass MRM_R? Solution. MR∼mD2/mν=(1011 eV2)/7.5×10−5∼1015 GeVM_R\sim m_D^2/m_\nu=(10^{11}\,\text{eV}^2)/\sqrt{7.5\times10^{-5}}\sim10^{15}\,\text{GeV}, close to the GUT scale.

Problem 3. Dark matter makes up ΩDMh2≈0.12\Omega_{\text{DM}}h^2\approx0.12. If a WIMP has mass mχ=100 GeVm_\chi=100\,\text{GeV} and annihilation cross section ⟨σv⟩≈3×10−26 cm3/s\langle\sigma v\rangle\approx3\times10^{-26}\,\text{cm}^3/\text{s}, is this thermally produced abundance consistent with observation? Solution. The thermal relic density is Ωχh2≈3×10−27 cm3/s⟨σv⟩\Omega_\chi h^2\approx\frac{3\times10^{-27}\,\text{cm}^3/\text{s}}{\langle\sigma v\rangle}. For ⟨σv⟩≈3×10−26\langle\sigma v\rangle\approx3\times10^{-26}, Ωh2≈0.1\Omega h^2\approx0.1, remarkably close to the observed dark matter density. This is the “WIMP miracle.”

Section summary. Beyond-Standard-Model physics is constrained by symmetry, scales, and precision data.