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Astronomy

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Astronomical Distance Units .
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Celestial Coordinates .
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Celestial Navigation .
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Location of North and South Celestial Poles .

Chemistry

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Avogadro's Number
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Balancing Chemical Equations
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Stochiometry
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The Periodic Table .

Classical Physics

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Archimedes Principle
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Bernoulli Principle
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Blackbody (Cavity) Radiation and Planck's Hypothesis
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Center of Mass Frame
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Comparison Between Gravitation and Electrostatics
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Compton Effect .
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Coriolis Effect
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Cyclotron Resonance
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Dispersion
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Doppler Effect
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Double Slit Experiment
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Elastic and Inelastic Collisions .
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Electric Fields
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Error Analysis
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Fick's Law
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Fluid Pressure
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Gauss's Law of Universal Gravity .
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Gravity - Force and Acceleration
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Hooke's law
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Ideal and Non-Ideal Gas Laws (van der Waal)
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Impulse Force
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Inclined Plane
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Inertia
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Kepler's Laws
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Kinematics
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Kinetic Theory of Gases .
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Kirchoff's Laws
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Laplace's and Poisson's Equations
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Lorentz Force Law
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Maxwell's Equations
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Moments and Torque
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Nuclear Spin
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One Dimensional Wave Equation .
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Pascal's Principle
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Phase and Group Velocity
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Planck Radiation Law .
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Poiseuille's Law
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Radioactive Decay
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Refractive Index
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Rotational Dynamics
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Simple Harmonic Motion
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Specific Heat, Latent Heat and Calorimetry
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Stefan-Boltzmann Law
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The Gas Laws
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The Laws of Thermodynamics
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The Zeeman Effect .
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Wien's Displacement Law
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Young's Modulus

Climate Change

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Keeling Curve .

Cosmology

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Baryogenesis
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Cosmic Background Radiation and Decoupling
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CPT Symmetries
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Dark Matter
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Friedmann-Robertson-Walker Equations
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Geometries of the Universe
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Hubble's Law
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Inflation Theory
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Introduction to Black Holes .
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Olbers' Paradox
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Penrose Diagrams
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Planck Units
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Stephen Hawking's Last Paper .
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Stephen Hawking's PhD Thesis .
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The Big Bang Model

Finance and Accounting

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Amortization
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Annuities
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Brownian Model of Financial Markets
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Capital Structure
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Dividend Discount Formula
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Lecture Notes on International Financial Management
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NPV and IRR
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Periodically and Continuously Compounded Interest
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Repurchase versus Dividend Analysis

Game Theory

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The Truel .

General Relativity

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Accelerated Reference Frames - Rindler Coordinates
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Catalog of Spacetimes .
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Curvature and Parallel Transport
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Dirac Equation in Curved Spacetime
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Einstein's Field Equations
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Geodesics
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Gravitational Time Dilation
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Gravitational Waves
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One-forms
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Quantum Gravity
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Relativistic, Cosmological and Gravitational Redshift
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Ricci Decomposition
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Ricci Flow
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Stress-Energy Tensor
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Stress-Energy-Momentum Tensor
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Tensors
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The Area Metric
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The Equivalence Principal
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The Essential Mathematics of General Relativity
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The Induced Metric
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The Metric Tensor
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Vierbein (Frame) Fields
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World Lines Refresher

Lagrangian and Hamiltonian Mechanics

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Classical Field Theory .
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Euler-Lagrange Equation
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Ex: Newtonian, Lagrangian and Hamiltonian Mechanics
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Hamiltonian Formulation .
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Liouville's Theorem
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Symmetry and Conservation Laws - Noether's Theorem

Macroeconomics

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Lecture Notes on International Economics
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Lecture Notes on Macroeconomics
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Macroeconomic Policy

Mathematics

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Amplitude, Period and Phase
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Arithmetic and Geometric Sequences and Series
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Asymptotes
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Augmented Matrices and Cramer's Rule
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Basic Group Theory
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Basic Representation Theory
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Binomial Theorem (Pascal's Triangle)
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Building Groups From Other Groups
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Completing the Square
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Complex Numbers
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Composite Functions
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Conformal Transformations .
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Conjugate Pair Theorem
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Contravariant and Covariant Components of a Vector
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Derivatives of Inverse Functions
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Double Angle Formulas
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Eigenvectors and Eigenvalues
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Euler Formula for Polyhedrons
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Factoring of a3 +/- b3
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Fourier Series and Transforms .
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Fractals
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Gauss's Divergence Theorem
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Grassmann and Clifford Algebras .
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Heron's Formula
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Index Notation (Tensors and Matrices)
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Inequalities
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Integration By Parts
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Introduction to Conformal Field Theory .
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Inverse of a Function
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Law of Sines and Cosines
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Line Integrals, ∮
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Logarithms and Logarithmic Equations
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Matrices and Determinants
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Matrix Exponential
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Mean Value and Rolle's Theorem
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Modulus Equations
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Orthogonal Curvilinear Coordinates .
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Parabolas, Ellipses and Hyperbolas
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Piecewise Functions
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Polar Coordinates
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Polynomial Division
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Quaternions 1
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Quaternions 2
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Regular Polygons
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Related Rates
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Sets, Groups, Modules, Rings and Vector Spaces
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Similar Matrices and Diagonalization .
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Spherical Trigonometry
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Stirling's Approximation
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Sum and Differences of Squares and Cubes
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Symbolic Logic
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Symmetric Groups
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Tangent and Normal Line
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Taylor and Maclaurin Series .
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The Essential Mathematics of Lie Groups
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The Integers Modulo n Under + and x
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The Limit Definition of the Exponential Function
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Tic-Tac-Toe Factoring
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Trapezoidal Rule
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Unit Vectors
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Vector Calculus
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Volume Integrals

Microeconomics

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Marginal Revenue and Cost

Particle Physics

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Feynman Diagrams and Loops
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Field Dimensions
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Helicity and Chirality
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Klein-Gordon and Dirac Equations
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Regularization and Renormalization
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Scattering - Mandelstam Variables
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Spin 1 Eigenvectors .
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The Vacuum Catastrophe

Probability and Statistics

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Box and Whisker Plots
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Categorical Data - Crosstabs
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Chebyshev's Theorem
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Chi Squared Goodness of Fit
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Conditional Probability
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Confidence Intervals
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Data Types
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Expected Value
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Factor Analysis
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Hypothesis Testing
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Linear Regression
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Monte Carlo Methods
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Non Parametric Tests
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One-Way ANOVA
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Pearson Correlation
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Permutations and Combinations
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Pooled Variance and Standard Error
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Probability Distributions
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Probability Rules
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Sample Size Determination
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Sampling Distributions
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Set Theory - Venn Diagrams
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Stacked and Unstacked Data
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Stem Plots, Histograms and Ogives
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Survey Data - Likert Item and Scale
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Tukey's Test
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Two-Way ANOVA

Programming and Computer Science

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Hashing
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How this site works ...
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More Programming Topics
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MVC Architecture
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Open Systems Interconnection (OSI) Standard - TCP/IP Protocol
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Public Key Encryption

Quantum Computing

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The Qubit .

Quantum Field Theory

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Creation and Annihilation Operators
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Field Operators for Bosons and Fermions
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Lagrangians in Quantum Field Theory
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Path Integral Formulation
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Relativistic Quantum Field Theory

Quantum Mechanics

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Basic Relationships
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Bell's Theorem
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Bohr Atom
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Clebsch-Gordan Coefficients .
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Commutators
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Dyson Series
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Electron Orbital Angular Momentum and Spin
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Entangled States
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Heisenberg Uncertainty Principle
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Ladder Operators .
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Multi Electron Wavefunctions
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Pauli Exclusion Principle
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Pauli Spin Matrices
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Photoelectric Effect
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Position and Momentum States
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Probability Current
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Schrodinger Equation for Hydrogen Atom
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Schrodinger Wave Equation
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Schrodinger Wave Equation (continued)
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Spin 1/2 Eigenvectors
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The Differential Operator
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The Essential Mathematics of Quantum Mechanics
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The Observer Effect
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The Quantum Harmonic Oscillator .
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The Schrodinger, Heisenberg and Dirac Pictures
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The WKB Approximation
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Time Dependent Perturbation Theory
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Time Evolution and Symmetry Operations
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Time Independent Perturbation Theory
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Wavepackets

Semiconductor Reliability

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The Weibull Distribution

Solid State Electronics

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Band Theory of Solids .
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Fermi-Dirac Statistics .
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Intrinsic and Extrinsic Semiconductors
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The MOSFET
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The P-N Junction

Special Relativity

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4-vectors .
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Electromagnetic 4 - Potential
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Energy and Momentum, E = mc2
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Lorentz Invariance
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Lorentz Transform
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Lorentz Transformation of the EM Field
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Newton versus Einstein
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Spinors - Part 1 .
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Spinors - Part 2 .
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The Lorentz Group
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Velocity Addition

Statistical Mechanics

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Black Body Radiation
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Entropy and the Partition Function
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The Harmonic Oscillator
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The Ideal Gas

String Theory

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Bosonic Strings
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Extra Dimensions
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Introduction to String Theory
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Kaluza-Klein Compactification of Closed Strings
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Strings in Curved Spacetime
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Toroidal Compactification

Superconductivity

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BCS Theory
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Introduction to Superconductors
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Superconductivity (Lectures 1 - 10)
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Superconductivity (Lectures 11 - 20)

Supersymmetry (SUSY) and Grand Unified Theory (GUT)

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Chiral Superfields
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Generators of a Supergroup
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Grassmann Numbers
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Introduction to Supersymmetry
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The Gauge Hierarchy Problem

The Standard Model

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Electroweak Unification (Glashow-Weinberg-Salam)
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Gauge Theories (Yang-Mills)
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Gravitational Force and the Planck Scale
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Introduction to the Standard Model
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Isospin, Hypercharge, Weak Isospin and Weak Hypercharge
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Quantum Flavordynamics and Quantum Chromodynamics
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Special Unitary Groups and the Standard Model - Part 1 .
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Special Unitary Groups and the Standard Model - Part 2
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Special Unitary Groups and the Standard Model - Part 3 .
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Standard Model Lagrangian
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The Higgs Mechanism
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The Nature of the Weak Interaction

Topology

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Units, Constants and Useful Formulas

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Constants
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Formulas
Last modified: November 18, 2021 ✓

Special Unitary Groups and the Standard Model - Part 1 Special Unitary Groups and the Standard Model - Part 2 Special Unitary Groups and the Standard Model - Part 3 ------------------------------------------------------ Multi-Particle States --------------------- _ In constructing the 3 and the 3 we made use of the raising and lowering operators I±, U± and V± which acted on single particles. We now need to consider how we handle multi-particle states. Tensor Methods -------------- Let particle 1 live in a vector space, V, with an operator T. Let particle 2 live in a vector space, W, with an operator S. The 2 particles form a system. We represent the state of the 2 particle system as v ⊗ w in a new vector space V ⊗ W. Will will use the angular momentum operators, J, as our example. Our claim is that the ‘sum’ of angular momenta is an angular momentum in the tensor product: Ji ≡ Ji(1) ⊗ 1 + 1 ⊗ Ji(2) This acts on v ⊗ w and satisfies [Ji,Jj] = ihεijkJk Certainly the sum operator, as defined above, is an operator on v ⊗ w. We just need to check that the commutator holds. [Ji,Jj] = [Ji(1) ⊗ 1 + 1 ⊗ Ji(2),Jj(1) ⊗ 1 + 1 ⊗ Jj(2)] = [Ji(1) ⊗ 1,Jj(1) ⊗ 1] + [1 ⊗ Ji(2),1 ⊗ Jj(2)] = [Ji(1),Jj(1)] ⊗ 1 + 1 ⊗ [Ji(2),Jj(2)] = ihεijk(Jk(1) ⊗ 1 + 1 ⊗ Jk(2)) = ihεijkJk Therefore, we can write: Ji|v ⊗ w> = (Ji(1) ⊗ 1 + 1 ⊗ Ji(2))|(v ⊗ w)> Lets look at the individual terms with the understanding that each operator on the right hand side acts on the appropriate factor in the tensor product.: (Ji(1) ⊗ 1)(v ⊗ w) = Ji(1)v ⊗ w and, (1 ⊗ Ji(2))(v ⊗ w) = v ⊗ Ji(2)w In ket form we get: Ji|v ⊗ w> = Ji(1)|v> ⊗ w + v ⊗ Ji(2)|w> The eigenvalue for this equation is: λi|v ⊗ w> = λi(1)|v> ⊗ w + v ⊗ λi(2)|w> Where we have replaced Ji|v ⊗ w> = λi|v ⊗ w>, Ji(1)|v> = λi(1)|v> and Ji(2)|w> = λi(2)|w>. This becomes: λi|v ⊗ w> = (λi(1) + λi(2))|v ⊗ w> and, (Ji(1) ⊗ Ji(2))|(v ⊗ w)> = Ji(1)|v> ⊗ Ji(2)|w> = λi(1)|v> ⊗ λi(2)|w> = λi(1)λi(2)|(v ⊗ w)> Therefore, the eigenvalues ADD in the tensor product representation. Let us now apply this to the Cartan generators. We can write: H1'R1 ⊗ ψR2> = H1R1> ⊗ |ψR2> + |ψR1> ⊗ H1R2> and, H2'R1 ⊗ ψR2> = H2R1> ⊗ |ψR2> + |ψR1> ⊗ H2R2> So the weight of |ψR1 ⊗ ψR2> is found by adding the weights of the individual representations, ψR1 and ψR2. 2 ⊗ 2 Decomposition -------------------- The weights for the 2 are: (1/2,1/2√3) ... u (-1/2,1/2√3) ... d 2 ⊗ 2 is spanned by: u ⊗ u = (1,1/√3) d ⊗ d = (-1,1/√3) u ⊗ d = (0,1/√3) d ⊗ d = (0,1/√3) This looks like: _ 2 ⊗ 2 Decomposition -------------------- _ The weights for the 2 and 2 are: (1/2,1/2√3) ... u (-1/2,1/2√3) ... d _ (-1/2,-1/2√3) ... u _ (1/2,-1/2√3) ... d _ 2 ⊗ 2 is spanned by: _ u ⊗ u = (0,0) _ d ⊗ d = (0,0) _ d ⊗ u = (1,0) _ u ⊗ d = (-1,0) This looks like: _ 3 ⊗ 3 Decomposition -------------------- _ The weights for the 3 are: _ (-1/2,-1/2√3) ... u _ (1/2,-1/2√3) ... d _ (0,1/√3) ... s _ 3 ⊗ 3 is spanned by: _ u ⊗ s = (1/2,1/2√3) + (0,1/√3) = (1/2,√3/2) _ u ⊗ d = (1/2,1/2√3) + (1/2,-1/2√3) = (1,0) _ d ⊗ s = (-1/2,1/2√3) + (0,1/√3) = (-1/2,√3/2) _ u ⊗ u = (1/2,1/2√3) + (-1/2,-1/2√3) = (0,0) _ d ⊗ d = (1/2,1/2√3) + (-1/2,-1/2√3) = (0,0) _ s ⊗ s = (1/2,1/2√3) + (-1/2,-1/2√3) = (0,0) _ d ⊗ u = (-1/2,1/2√3) + (-1/2,-1/2√3) = (-1,0) _ s ⊗ u = (0,-1/√3) + (-1/2,-1/2√3) = (-1/2,-√3/2) _ s ⊗ d = (0,-1/√3) + (1/2,-1/2√3) = (1/2,-√3/2) This looks like: 3 ⊗ 3 Decomposition ------------------- The weights for the 3 are: (1/2,1/2√3) ... u (-1/2,1/2√3) ... d (0,-1/√3) ... s 3 ⊗ 3 is spanned by: u ⊗ u = (1,1/√3) d ⊗ d = (-1,1/√3) s ⊗ s = (0,-2/√3) u ⊗ d = (0,1/√3) d ⊗ u = (0,1/√3) u ⊗ s = (1/2,-1/2√3) s ⊗ u = (1/2,-1/2√3) d ⊗ s = (-1/2,-1/2√3) s ⊗ d = (-1/2,-1/2√3) This looks like: 6 ⊗ 3 Decomposition ------------------- Now that we know the '6' we can easily compute the 6 ⊗ 3 as: d ⊗ d ⊗ u = (-1,1/√3) + (1/2,1/2√3) = (-1/2,√3/2) u ⊗ d ⊗ u = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2) d ⊗ u ⊗ u = (0,1/√3) + (1/2,1/2√3) = (1/2,√3/2) u ⊗ u ⊗ u = (1,1/√3) + (1/2,1/2√3) = (3/2,√3/2) d ⊗ s ⊗ u = (-1/2,-1/2√3) + (1/2,1/2√3) = (0,0) s ⊗ d ⊗ u = (-1/2,-1/2√3) + (1/2,1/2√3) = (0,0) u ⊗ s ⊗ u = (1/2,-1/2√3) + (1/2,1/2√3) = (1,0) s ⊗ u ⊗ u = (1/2,-1/2√3) + (1/2,1/2√3) = (1,0) s ⊗ s ⊗ u = (0,-2/√3) + (1/2,1/2√3) = (1/2,-√3/2) d ⊗ d ⊗ d = (-1,1/√3) + (-1/2,1/2√3) = (-3/2,√3/2) u ⊗ d ⊗ d = (0,1/√3) + (-1/2,1/2√3) = (-1/2,√3/2) d ⊗ u ⊗ d = (0,1/√3) + (-1/2,1/2√3) = (-1/2,√3/2) u ⊗ u ⊗ d = (1,1/√3) + (-1/2,1/2√3) = (1/2,√3/2) d ⊗ s ⊗ d = (-1/2,-1/2√3) + (-1/2,1/2√3) = (-1,0) s ⊗ d ⊗ d = (-1/2,-1/2√3) + (-1/2,1/2√3) = (-1,0) u ⊗ s ⊗ d = (1/2,-1/2√3) + (-1/2,1/2√3) = (0,0) s ⊗ u ⊗ d = (1/2,-1/2√3) + (-1/2,1/2√3) = (0,0) s ⊗ s ⊗ d = (0,-2/√3) + (-1/2,1/2√3) = (-1/2,-√3/2) d ⊗ d ⊗ s = (-1,1/√3) + (0,-1/√3) = (-1,0) u ⊗ d ⊗ s = (0,1/√3) + (0,-1/√3) = (0,0) d ⊗ u ⊗ s = (0,1/√3) + (0,-1/√3) = (0,0) u ⊗ u ⊗ s = (1,1/√3) + (0,-1/√3) = (1,0) d ⊗ s ⊗ s = (-1/2,-1/2√3) + (0,-1/√3) = (-1/2,-√3/2) s ⊗ d ⊗ s = (-1/2,-1/2√3) + (0,-1/√3) = (-1/2,-√3/2) u ⊗ s ⊗ s = (1/2,-1/2√3) + (0,-1/√3) = (1/2,-√3/2) s ⊗ u ⊗ s = (1/2,-1/2√3) + (0,-1/√3) = (1/2,-√3/2) s ⊗ s ⊗ s = (0,-2/√3) + (0,-1/√3) = (0,-√3) This looks like: Invariant Tensors ----------------- We want Qi*Qi to be invariant under a transformation, U. Qi -> Q'i = UQi Qi* -> Q'i* = U*Qi* Qi†Qi -> Q'i†Q'i = Qi†UUQi Therefore, the requirement is: UU = I We write (Qi)* as Qi and get: Qi -> Q'i = UijQj Qi -> Q'i = Qj(U)ji QiQi -> Q'iQ'i = Qj(U)jiUikQk = QjδjkQk = Q Now consider the following mixed tensor: Qijk -> Q'ijk = UilUjm(U)nkQlmn So the upper indeces transform with U and the lower indeces transform with U. Multiply both sides by δkj to get: δkjQijk -> δkjQ'ijk = δkjUilUjm(U)njQlmn = Qi Therefore, δkjQijk represents the trace of Qijk. δkj is invariant because: δkj = δilUli(U)kj Such that, δkjQijk = δilUli(U)kjQijk = Qi as before. Note that δ is the Kroneker Delta which is nothing more than the unit matrix in N dimensions. The Levi-Civita Tensor ---------------------- There 2 other invariant tensors in SU(3) - the completely antisymmetric tensors εijk and εijk. These are the Levi-Civita symbols in tensor form. The L-C symbol is not a tensor and does not change under coordinate transformations. In order to make it a tensor we need to look at Leibniz formula for determinants. Leibnitz Formula ---------------- The Leibnitz formula expresses the determinant of a square matrix in terms of permutations of the matrix elements. The formula is: det M = ΣMM ... M n! Proof by example: | a b c | aei - afh | d e f | = + bfg - bdi | g h i | + cdh - ceg Term Row Column #Perm Sign ---- --- ------ ----- ---- aei 123 123 0 + afh 123 132 1 - bfg 123 231 2 + bdi 123 213 1 - cdh 123 312 2 + ceg 123 321 1 - #Perm is the number of permutations (flips) it takes to get back to the row index with even = + and odd = -. Therefore, det M = M11M22M33 - M11M23M32 + M12M23M31 - M12M21M33 + M13M21M32 - M13M22M31 For SU(3): det M = ΣMμ1μ1'Mμ2μ2'Mμ3μ3' n! Multiplying both sides by εijk gives: εμ1'μ2'μ3'det(M) = εμ1μ2μ3(∂xμ1/∂xμ1')(∂xμ2/∂xμ2')(∂xμ3/∂xμ3') M is the Jacobean matrix. Therefore, we can write det(M) as det(∂xμ/∂xμ'). This gives: εμ1'μ2'μ3'det(∂xμ/∂xμ') = εμ1μ2μ3(∂xμ1/∂xμ1')(∂xμ2/∂xμ2')(∂xμ3/∂xμ3') Or, εμ1'μ2'μ3' = det(∂xμ'/∂xμμ1μ2μ3(∂xμ1/∂xμ1')(∂xμ2/∂xμ2')(∂xμ3/∂xμ3') This is close to the tensor transformation law, except for the determinant out front. Objects which transform in this way are known as TENSOR DENSITIES. It is a quantity whose transformation law under change of basis involves the determinant of the transformation matrix (as opposed to a tensor, whose transformation law does not involve the determinant). Now consider, gμ'ν' = (∂xμ1/∂xμ1')(∂xμ2/∂xμ2')gμν The RHS can be viewed as the product of 3 matrices. Taking the determinant of both sides of the equation, and using the fact that the determinant of a matrix product is the product of the determinants, this becomes: det(gμ'ν') = det(∂xμ1/∂xμ1')det(∂xμ2/∂xμ2')det(gμν) = (det(∂xμ/∂xμ'))2det(gμν) Therefore, √{det(gμ'ν')/det(gμν)} = det(∂xμ/∂xμ') Or, √{det(gμν)/det(gμ'ν')} = det(∂xμ'/∂xμ) εμ1'μ2'μ3' = √{det(gμν)/det(gμ'ν')}εμ1μ2μ3(∂xμ1/∂xμ1')(∂xμ2/∂xμ2')(∂xμ3/∂xμ3') In Euclidean space det(gμ'ν') = 1 leaving: εμ1'μ2'μ3' = √{det(gμν)}εμ1μ2μ3(∂xμ1/∂xμ1')(∂xμ2/∂xμ2')(∂xμ3/∂xμ3') Switching notation we get: εijk = √|g|εijk ^ ^ | | tensor symbol Likewise, εijk = (1/√|g|)εijk Where g = det(|gμν|) = 1 for Euclidean space. The symbol now transforms according to the tensor transformation law. Since this is a real tensor, we can use εij and εij to raise and lower indices in SU(2) and εijk and εijk to raise and lower indices in SU(3). We will now prove the invariance of ε. εμ1'μ2'μ3'det(U) = (Uμ1μ1'Uμ2μ2'Uμ3μ3'μ1μ2μ3 Bet det(U) = 1. Therefore, εμ1'μ2'μ3' = (Uμ1μ1'Uμ2μ2'Uμ3μ3'μ1μ2μ3 Therefore, the antisymmetric tensor, εijk, is invariant when transformed by the U's. This is also true for εijk. Levi-Civita Symbol and Kronecker Delta Relationship --------------------------------------------------- In 3D, the Levi-Civita symbol is related to the Kronecker delta in the following manner:        | δik δim δin | εijlεkmn = | δjk δjm δjn |        | δlk δlm δln | or, εijlεkmn = δikjmδln - δjnδlm) - δimjkδln - δjnδlk) + δinjkδlm - δjmδlk) For the special case: εijlεlmn = δimδjn - δjnδim Symmetric and Antisymmetric Tensors ----------------------------------- Just as we do with wavefunctions in QM, we can decompose any tensor, Qij, into symmetric and antisymmetric parts as follows: Q{ij} = (1/2)(Qij + Qji) ∴ Qij = Qji Q[ij] = (1/2)(Qij - Qji) ∴ Qij = -Qji Where we have used the notation {} to indicate that the tensor is symmetric in i and j and [] to indicate that the tensor is antisymmetric in i and j. (note the analogy with the anti-commutator and the commutator). Therefore, Qij = Q{ij} + Q[ij] = (1/2)(Qij + Qji) + (1/2)(Qij - Qji) = Qij Example: Qij = Q{ij} + Q[ij] - - - - - - | 7 10 3 | | 7 7 6 | | 0 3 -3 | | 4 -1 -2 | = | 7 -1 1 | + | -3 0 -3 | | 9 4 5 | | 6 1 5 | | 3 3 0 | - - - - - - Also, note that: Q[ij]Q{ij} = 0 Proof: Q[ij]Q{ij} = -QjiQij = -QjiQji = -QijQij (after swapping i <-> j) Therefore, Q[ij]Q{ij} = 0 Notice that a completely antisymmetric tensor cannot have diagonal elements, because we have the equation Qii... = -Qii... ∴ Qii... = 0 Note that in the case of mixed tensors, the upper indices and lower indices are to be permuted separately. It doesn't make sense to interchange an upper index and a lower index since they are are different entities. A tensor with one upper and one lower index is considered to be symmetric. Trivially, all scalars and vectors are totally antisymmetric (as well as being totally symmetric). Separating the Trace -------------------- When a tensor has both upper and lower indeces it can be contracted to obtain the trace which can then be subtracted out. The process is illustrated in the following examples. Example: ~ Qij = Qij - (1/3)δijQkk - - - - - - | 0 10 3 | | 7 10 3 | | 1 0 0 | | 4 0 -2 | = | 4 -1 -2 | - (11/3)| 0 1 0 | | 9 4 0 | | 9 4 5 | | 0 0 1 | - - - - - - Where ~ means traceless (i.e. Tr = 0). This was found in the following way: ~ Qij = Qij - αδijQkk If we contract i with j we can write this as: ~ Qii = Tr(Qij) = 0 = Qii - αδiiQkk But δii = 3. Therefore, 0 = Qii - 3αQkk Therefore, α = 1/3 and Qkk = 11. We can generalize this result to the following: ~ Qijkl = Qijkl - β(δikQjl + δjkQil + δilQjk + δjiQik) where Qjl = QjlQnn etc. Contract k with i: ~ Qijil = 0 = Qijil - β(δiiQjl + δjiQil + δilQji + δjiQii) = Qijil - β(δiiQjl + δjiQil + δilQji) = Qijil - β(δii + δji + δil)Qjl = Qijil - β(3 + 1 + 1)Qjl = Qijil - β5Qjl ∴ β = 1/5 Likewise, ~ Sijk = Qijk - β(δikQj + δjkQi) Contract k with i: Siji = 0 = Qiji - β(δikQj + δjkQi) = Qiji - β(δiiQj + δjkQi) = Qiji - β(3 + 1)Qj ∴ β = 1/4 Note, that number of traces that can be subtracted equals the number of possible contractions. For example: Qij has 1 trace. Qijk has 2 traces (k with i, k with j) Qijkl has 3 traces (l with i, l with j, l with k) Qijkl has 4 traces (l with i, l with j, k with i and k with j). etc. etc. ----------------------------------------------------- Digression: We stated that contraction on a pair of indices that are either both contravariant or both covariant is not possible. However, in differential geometry the metric tensor can be used used to raise or lower one of the indices, as needed, i.e. Tkl = gjkTij Such that contraction is possible. This operation is known as METRIC CONTRACTION. ----------------------------------------------------- Dimensions of Symmetric Treceless Tensors ----------------------------------------- The dimensions of a completely symmetric tensor is given by: D(n,m) = (1/2)(n + 1)(m + 1)(n + m + 2) The formula comprehends the traceless condition ΣQijklmk = 0 for tensors with upper and lower indeces that can be contracted. Proof: Assume the tensor is completely symmetric with n upper indeces and m lower indeces. For the upper indeces: = (d + n - 1)!/n!(d - 1)! = (2 + n)!/2!n! = Γ(n + 3)/2Γ(n + 1) = (1/2)(n + 1)(n + 2) And for the lower indeces: = (d + m - 1)!/m!(d - 1)! = (2 + m)!/2!m! = Γ(m + 3)/2Γ(m + 1) = (1/2)(m + 1)(m + 2) The total is then: = (1/4)(n + 1)(n + 2)(m + 1)(m + 2) Contraction of 1 upper and 1 lower index reduces the number of independent components by: = (1/4)nm(n + 1)(m + 1) = # of components in EACH trace. Examples: Q{ij} (not contractable): - - | 11 | | 21 22 | = 6 | 31 32 33 | - - Qij (contractable): - - | 11 12 13 | | 21 22 23 | | 31 32 33 | - - 3 Tr = ΣSkk = Q11 + Q22 + Q33 = 0 k=1 Now since the RHS are just scalars we can write: Q33 = -(Q11 + Q22) So Q33 is not independent and we can ignore it. As a consequence D(1,1) = 9 - 1 = 8 - - | 11 12 13 | | 21 22 23 | | 31 32 33 | - - Q{ij}k (contractable): - - - - - - | 111 | | 112 | | 113 | | 211 221 | | 212 222 | | 213 223 | | 311 321 331 | | 312 322 332 | | 313 323 333 | - - - - - - Contract i and j with k and sum to get the 2 traces: 3 Tr = ΣSikk = Q111 + Q222 + Q333 k=1 = Q1 + Q2 + Q3 and, 3 Tr = ΣSkjk = Q111 + Q222 + Q333 k=1 = Q1 + Q2 + Q3 - - - - - - | 111 | | 112 | | 113 | | 211 221 | | 212 222 | | 213 223 | | 311 321 331 | | 312 322 332 | | 313 323 333 | - - - - - - So the 2 traces are the same and removing them only reduces the number of independent components by 3. Note that we are now dealing with vectors so we cannot say. Q3 = -(Q1 + Q2) As this requires Q1 + Q2 + Q3 = 0 which can only be satisfied if Q1 = Q2 = Q3 = 0. Summarizing: (0,0) = 1 (1,0) = (0,1) = 3 (2,0) = (0,2) = 6 (1,1) = 8 (2,1) = (1,2) = 15 (2,2) = 27 (3,0) = 10 (4,0) = 15 Dimensions of Antisymmetric Tensors ----------------------------------- The dimensions of a completely antisymmetric tensor is given by: D[] = d!/r!(d - r)! Where d is the dimension of the space and r is the tensor rank. Therefore, for d = 3 we get: Q: 1 Qi = Qi: 3 Q[ij] = Q[ij]: 3 Q[ijk] = Q[ijk]: 1 Using the example from before we can see that: - - - - - - | 7 10 3 | | 7 7 6 | | 0 3 -3 | | 4 -1 -2 | = | 7 -1 1 | + | -3 0 -3 | | 9 4 5 | | 6 1 5 | | 3 3 0 | - - - - - - Qij = Q{ij} + Q[ij] Dim(Qij) = 9 Dim(Q{ij}) = (1/2)(n + 1)(m + 1)(n + n + 2) = 6 Dim(Q[ij]) = d!/r!(d - r)! = 3 Alternative Formulas for Dimensions ----------------------------------- By plugging in numbers it is easy to show that for the following specific cases of SU(N) we can also write: Qi = N Antisymmetric: Q[ij] = d!/r!(d - r)! ≡ N(N - 1)/2 Symmetric: Q{ij} = (n + 1)(m + 1)(n + m + 2)/2 ≡ N(N + 1)/2 Q{i}{j} = (n + 1)(m + 1)(n + m + 2)/2 ≡ N2 - 1 Mixed: Q[ij]k = N(N - 2)(N + 1)/2 Tensors Components as States ---------------------------- It should be easy to see that there is a correspondence between the states of the 3, 6, 8, 10 etc. and the number of independent components of traceless symmetric tensors. Qi has 3 independent elements. If we make the substitution 1 = u, 2 = d, 3 = s we get: 2 1 d u 3 s Q{ij} has 6 independent elements. 22 12 11 dd ud uu du 23 13 ds us sd su 33 ss Q{ijk} has 10 independent elements. - - - - - - | 111 | | | | | | 211 221 | | 222 | | | | 311 321 331 | | 322 332 | | 333 | - - - - - - 222 221 211 111 (I3 = 3/2) ddd ddu duu uuu 322 321 311 (I3 = 1) sdd sud suu 332 331 (I3 = 1/2) ssd ssu 333 (I3 = 0) sss Clebsch-Gordan Decomposition for SU(3) -------------------------------------- The Clebsch-Gordan decomposition for SU(3) is the decomposition of the tensor product of 2 representations into a direct sum of irreducible representations. The basic process is as follows: 1). Split the tensor into the sum of symmetric and antisymmetric parts. 2). When contraction is possible, subtract out all traces and then add them back as separate terms. Removing all the traces is necessary otherwise the tansor is reducible and can be contracted into smaller dimensional representation. IN OTHER WORDS, THE TENSOR IS IRREDUCIBLE IF A LOWER RANK TENSOR CANNOT BE FOUND. 3). Use the Levi-Civita tensor to symmetrize the antisymmetric parts, i.e. Qk = εijkQij This now transforms like a tensor with a single index which is symmetric. Note: Symmetrization does not alter the underlying antisymmetry and is performed only for the purpose of determining the C-G decomposition. Clebsch-Gordan Series --------------------- 3 ⊗ 3 ------ ui ⊗ vj = Q{ij} + Q[ij] Q{ij} = (1/2)(uivj + ujvi) Q[ij] = (1/2)(uivj - ujvi) We can write the 2nd terms as (1/2)εijllmnumvn) Proof: (uivj - ujvi) = (δimδjn - δjnδim)umvn = εijlεlmnumvn The antisymmetric term is automatically traceless since uivi = -uivi. Therefore, Qij = (1/2((uivj + ujvi) + (1/2)εijllmnumvn)    = (1/2)(uivj + ujvi) + (1/2)εijl(uvl) The first term is the '6' (2,0). The second _ term is the '3' (0,1). _ 3 ⊗ 3 ------ ui ⊗ vj = Qij ~ Qij = uivj - βδijukvk Contract i with j and demand that Tr(Sij) = 0: ~ Qii = 0 = uivi - βδiiukvk = uivi - β3ukvv ∴ β = 1/3 Therefore, uivj = (uivj - (1/3)δijukvk) + (1/3)δijukvk The term inside the () is a traceless symmetric tensor and the second term is antisymmetric. Note the scalar terms can be regarded as being totally symmetric or totally antisymmetric. The first term is the '8' (1,1). The second term is the '1' (0,0). ----------------------------------------------------- Digression: The product of the fundamental representation with its complex conjugate is always reducible into a sum that contains at least the singlet state. We can see this in the following manner by considering the Kronecker δ in tensor form: _ δij -> (R ⊗ R)δij It is convenient to introduce upper and lower indices to write: (Ta(C))ij = -(Ta(R))ij So that complex conjugation is achieved by changing the lower indices to upper ones, and vice versa. Also, transposition is achieved by exchanging rows and columns. Thus, (Ta(R))ji is the transpose of (Ta(C))ij. We can now write the infinitessimal generators as: δij -> (1 + iθTa(R))ik(1 + iθTa(C))jlδkl    -> (1 + iθTa(R))ik(1 - iθTa(R))jlδkl    -> (1 + iθTa(R))ik(1 - iθTa(R))ljδkl    -> δij + O(θ2) _ This means that R ⊗ R must contain the singlet representation. Now, the dimension of the tensor product is the product of the dimensions of R and its conjugate. Therefore, the non-singlet states must number N2 - 1. Thus, _ R ⊗ R = 1 ⊕ (N2 - 1) For SU(3) N2 - 1 = 8 which is the dimension of _ the adjoint representation. So the 3 ⊗ 3 is reducible into a sum that contains at least the singlet and the adjoint representations. i.e. _ - - 3 ⊗ 3 = 1 ⊕ 8 = | [1x1] 0 |   | 0 [8x8] |   - - ----------------------------------------------------- 3 ⊗ 8 ------ ui ⊗ vjk = Q{ij}k + Q[ij]k ~ Q{ij}k = (1/2)(uivjk + ujvik - α(δikulvjl + δjkulvil)) ~ Q[ij]k = (1/2)(uivjk - ujvik - β(δikulvjl - δjkulvil)) ~ Consider Q{ij}k. Contract i with k to get: ~ Q{ij}i = 0 = (1/2)(uivji + ujvii - α(δiiulvjl + δjkulvil)) = (1/2)(uivji - α(δii + δjk)ulvjl) = (1/2)(uivji - α(3 + 1)ulvjl) since δii = 3 Therefore, α = 1/4. Rewriting we get: ~ Q{ij}k = (1/2)(uivjk + ujvik) - (1/8)δikulvjl - (1/8)δjkulvil ~ Now consider Q[ij]k. Contract i with k to get: Q[ij]i = 0 = (1/2)(uivji - β(δiiulvjl - δjkulvil)) = (1/2)(uivji - β(δii - δjk)ulvjl) = (1/2)(uivji - β(3 - 1)ulvjl) Therefore, β = 1/2. Rewriting we get: ~ Q[ij]k = (1/2)(uivjk - ujvik) - (1/4)δikulvjl + (1/4)δjkulvil) So that, ~ ~ Q{ij}k + Q[ij]k = (1/2)(uivjk + ujvik) - (1/8)δikulvjl - (1/8)δjkulvil + (1/2)(uivjk - ujvik) - (1/4)δikulvjl + (1/4)δjkulvil) = uivjk - (3/8))δikulvjl + (1/8)δjkulvil Need to add (3/8)δikulvjl - (1/8)δjkulvil to make both sides balance. Therefore: ~ Qijk = (1/2)(uivjk + ujvik) - (1/8)δikulvjl - (1/8)δjkulvil + (1/2)(uivjk - ujvik) - (1/4)δikulvjl + (1/4)δjkulvil) + (3/8)δikulvjl - (1/8)δjkulvil The blue term is: (1/2)εijlεlmnumvnk The red term is: (1/4)δikulvjl + (1/4)δjkulvil) = (1/4)uivjk + (1/4)ujvik Which can be written as: (1/4)εijlεkmnumvnl Proof: We use the identity: | δik δim δin | | δjk δjm δjn | = εijlεkmn | δlk δlm δln | (1/4)εijlεkmnumvnl = (1/4)(δikjmδln - δjnδlm)umvnl - δimjkδln - δjnδlk)umvnl = δim(umvjk - umvjk) + δinjkδlm - δjmδlk)umvnl = δin(umvnk - ujvnk)) The 2nd and 3rd terms are 0 so we get: (1/4)εijlεkmnumvnl = (1/4)(δikjmδln - δjnδlm)umvnl = (1/4)δik(ujvll - ulvjl) = (1/4)(ujvik - uivjk) = -(1/4)uivjk + (1/4)ujvik Finally, we get: ~ Qijk = (1/2)(uivjk + ujvik) - (1/8)δikulvjl - (1/8)δjkulvil + (1/4)(εijllmnumvnk + εkmnumvnl)) + (1/8)(3δikulvjl - δjkulvil) The first term is the '15' (2,1). The second term _ is the '6' (0,2) and the third terms is the '3' (1,0). 6 ⊗ 3 ------ ui ⊗ vjk = Q{ijk} + Q[ij]k + Q[ik]j Q{ijk} = (uivjk + ujvki + ukvij) Q[ij]k = (uivjk - ujvik) Q[ik]j = (uivkj - ukvij) ui ⊗ vjk = Q{ijk} + (uivjk - ujvik) + (uivkj - ukvij) As before, we can write: (uivjk - ujvik) = εijllmnumvnk) and, (uivkj - ukvij) = εikllmnumvnj) ui ⊗ vjk = (uivjk + ujvki + ukvij) + εijllmnumvnk) + εikllmnumvnj) = (uivjk + ujvki + ukvij) + εijl(uvkl) + εikl(uvjl) The first term is the '10' (3,0). The second term is the '8' (1,1). Note that the second term is traceless because it is antisymmetric (Aijk = -Ajik and Aijk = -Akij). Note that the non-separability of the trace in the 6 ⊗ 3 prevents the decomposition into 9 ⊕ 8 ⊕ 1. Young's Tableaux ---------------- We can also represent tensor products and irreps using YOUNG'S TABLEAUX. This an alternative (and simpler) way of looking at tensor decompositions. A Young tableau is a diagram of left-justified rows of boxes where any row is not longer than the row on top of it and each column is no longer than the one to the left of it. There is 1 box for each particle. The rows represent symmetric tensors, and the columns represent antisymmetric tensors. The remaining combinations represent mixed states. A general tableau will be of mixed symmetry. A tableau can consist of any number of boxes provided that no more than N boxes occur in any column (for SU(N)) We will restrict the discussion to SU(2) and SU(3). In these cases the maximum column height is 2 and 3 respectively. Let: n = # of boxes beyond the 2nd row. m = # of boxes beyond the 3rd row. We can then describe the box combinations in tensor notation (n,m) as follows (tensors are denoted by Q to differentiate them from group generators. SU(2): Qi = (1,0): --- | | --- Q{ij) = (2,0): ------- | | | ------- Q = (0,0): --- | | |---| | | --- Note: This is not (0,1) because SU(2) can only have a maximum of 2 rows. SU(3): Qi = (1,0): --- | | --- Q{ij} = (2,0): ------- | | | ------- Q{ijk} = (3,0): ----------- | | | | ----------- Q = (0,0): --- | | |---| | | |---| | | --- Qij = (1,1): ------- | | | |---|--- | | --- Q{ij}k = (2,1): ----------- | | | | |---|------- | | --- Relation to Dynkin Labels ------------------------- Note that the numbers in the brackets are identical to the Dynkin labels for the highest weights described earlier, i.e. for SU(3), Rep. Dynkin Q Highest weight (n,m) (ν) ------------------------------------ 3 (1,0) Qi (1/2,1/2√3) - 3 (0,1) Qi (1/2,-1/2√3) 8 (1,1) Qij (1,0) Dimensions ---------- 1. Write down the quotient of 2 copies of the tableau. 2. In the numerator, fill the boxes as follows: ----------- | N |N+1|N+2| |---|---|---| |N-1| N |N+1| |---|---|--- |N-2|N-1| ------- 3. In the denominator, the number to be entered in a given box is called the HOOK LENGTH. It is the number of boxes to the right of the given box PLUS the number of boxes below the given box PLUS 1 for itself. Therefore, the abpve would look like: ----------- | 5 | 4 | 2 | |---|---|---| | 4 | 3 | 1 | |---|---|--- | 2 | 1 | ------- 4. The dimension can be found by evaluating the quotient by calculating the product of their respective entries and dividing. For example, for SU(3) we get: 3.4.5.2.3.4.1.2/5.4.2.4.3.1.2.1 = 3 For SU(2) we have: --- --- | 2 | ÷ | 1 | --- --- D = 2/1 = 2 ------- ------- | 2 | 3 | ÷ | 2 | 1 | ------- ------- D = 2.3/2.1 = 3 --- --- | 2 | | 2 | |---| ÷ |---| | 1 | | 1 | --- --- D = 2.2/2.1 = 1 For SU(3) we have: --- --- | 3 | ÷ | 1 | --- --- D = 3/1 = 3 ------- ------- | 3 | 4 | ÷ | 2 | 1 | ------- ------- D = 3.4/2.1 = 6 ----------- ----------- | 3 | 4 | 5 | ÷ | 3 | 2 | 1 | ----------- ----------- D = 3.4.5/3.2.1 = 10 --- --- | 3 | | 3 | |---| |---| | 2 | ÷ | 2 | |---| |---| | 1 | | 1 | --- ... D = 3.2.1/3.2.1 = 1 ------- ------- | 3 | 4 | ÷ | 3 | 1 | |---|--- |---|--- | 2 | | 1 | --- --- D = 3.4.2/3.1.1 = 8 ----------- ----------- | 3 | 4 | 5 | | 4 | 2 | 1 | |---|------- ÷ |---|------- | 2 | | 1 | --- --- D = 3.4.5.2/4.2.1.1 = 15 Number of States ---------------- In terms of multiplets, the integers n and m also represent the number of steps across the top and bottom levels of the multiplet diagram. This is illustated below, The number of states associated with each combination is given by: SU(2): (n + 1) (top diagram) SU(3): (n + 1)(m + 1)(n + m + 2)/2 (bottom 2 diagrams) For example, (3,0) represents (4)(1)(3 + 2)/2 = 10 states. Conjugate Representation ------------------------ The complex conjugate of a given irrep is represented by a tableaux obtained by switching any column of k boxes with a column of (N - k) boxes. For example, for SU(3) N = 3. ----------- ----------- | | | | -> | | | | ----------- |---|---|---| | | | | ----------- SU(2): Any irrep is self conjugate. (0,1) = '2': --- | | --- (0,2) = '3': ------- | | | ------- (0,0) = '1': --- | | |---| | | --- SU(3): _ (0,1) = '3': --- | | |---| | | --- With dimension: --- --- | 3 | | 2 | |---| ÷ |---| | 2 | | 1 | --- --- D = 3.2/2/1 = 3 _ (0,2) = '6': ------- | | | |---|---| | | | ------- With dimension: ------- ------- | 3 | 4 | | 3 | 2 | |---|---| ÷ |---|---| | 2 | 3 | | 2 | 1 | ------- ------- D = 3.4.2.3/3.2.2.1 = 6 Tensor Product of Multiplets ---------------------------- The following recipe tells us how to find the multiplets that occur in coupling 2 multiplets together. To couple together more than 2 multiplets, first couple 2, then couple a third with each of the multiplets obtained from the first 2, etc. Draw the Young diagrams for the 2 multiplets, but in one of the diagrams replace the boxes in the first row with a's and the boxes in the second row with b’s, etc. The unlettered diagram forms the upper left hand corner of all the enlarged diagrams constructed below. Add the a’s from the lettered diagram to the right hand ends of the rows of the unlettered diagram to form all possible legitimate Young diagrams that have no more than one a per column. In general, there will be several distinct diagrams, and all the a’s appear in each diagram Use the b’s to further enlarge the diagrams already obtained, subject to the same rules. Then throw away any diagram in which the full sequence of letters formed by reading right to left in the first row, then the second row, etc., is not admissible. Note that the height of the left most column cannot exceed N (3 in this case). As an example we consider the somewhat complex 8 ⊗ 8 product to illustrate the process. This gives: __ 8 ⊗ 8 = 27 ⊕ 10 ⊕ 10 ⊕ 8 ⊕ 8 ⊕ 1 2 ⊗ 2 Decomposition -------------------- (1,0) ⊗ (1,0) = (2,0) ⊕ (0,0) 2 ⊗ 2 = 3 ⊕ 1 3 ⊗ 3 Decomposition -------------------- Ti ⊗ Tj = Tij (1,0) ⊗ (1,0) = (2,0) ⊕ (0,1) _ 3 ⊗ 3 = 6 ⊕ 3 _ 3 ⊗ 3 Decomposition -------------------- Ti ⊗ Tj = Tij (1,0) ⊗ (0,1) = (1,1) ⊕ (0,0) _ 3 ⊗ 3 = 8 ⊕ 1 3 ⊗ 3 ⊗ 3 Decomposition ------------------------ Ti ⊗ Tj ⊗ Tk = Tijk (1,0) ⊗ (1,0) ⊗ (1,0) = (0,0) ⊕ (3,0) ⊕ (1,1) ⊕ (1,1) 3 ⊗ 3 ⊗ 3 = 1 ⊕ 10 ⊕ 8 ⊕ 8 We saw from before that this the decomposition: 3 ⊗ 3 ⊗ 3 _ = (3 ⊕ 6) ⊗ 3 _ = (3 ⊗ 3) ⊕ (6 ⊗ 3) = 1 ⊕ 8 ⊕ 8 ⊕ 10 6 ⊗ 3 = 10 ⊕ 8: This gives the baryon octet and the baryon decuplet. _ 3 ⊗ 3 = 1 ⊕ 8: This gives the meson octet and the anti-symmetric singlet. Boxes as Particles ------------------ In the sense of u -> d -> s as increasing values: - Rows cannot decrease. - Columns must increase. The various admissible combinations of u, d and s _ for the 3 ⊗ 3 composition for mesons. ------- ------- | u | u | = p | u | u | = Σ+ ---|--- |---|--- | d | | s | --- --- ------- ------- | u | d | = n | u | d | = Γ0 ---|--- |---|--- | d | | s | --- --- ------- ------- | u | s | = Σ0 | u | s | = Ξ0 ---|--- |---|--- | d | | s | --- --- ------- ------- | d | d | = Σ- | d | s | = Ξ- ---|--- |---|--- | s | | s | --- --- --- | u | |---| | d | = singlet |---| | s | --- Flavor Symmetry --------------- It was shown before that the values of (n,m) (the Dynkin labels) correspond to the highest weight state in the representation. We can therefore construct all states in (n,m) by acting on the tensor with lowering operators. Orthonormality of States ------------------------ Two states are orthogonal if there is 0 probability to measure one of the states when the system is prepared in the other one. For states to be orthogonal their dot product has to be 0. Therefore, ψ1 and ψ2 are orthogonal if: <ψ12> = 0 The rules are: <u|d> = <d|u> = 0 etc. _ _ <uu|uu> = 1 etc. _ _ <uu|dd> = 0 etc. <ud|du> = 0 etc. States also need to be normalized such that: <ψ11> = 1 Deriving the Isospin Wavefunctions ---------------------------------- Of particular interest is the isospin operator, I±. When we consider multiple particle states we need to take into account how their individual isospins combine. Now isospin combines in the same way as angular momentum, therefore we need to include the CLEBSCH-GORDAN coefficients in the calculations. I± is then written as: I±|I,I3> = √[I(I + 1) - I3(I3 ± 1)]|I,I3 ± 1> = C±|I,I3 ± 1> Then, C±|I,I3 ± 1> = I±|ψ> = ψ± Where ψ is the combination of quarks being raised or lowered. |I,I3> is a notation that labels states in terms of total isospin and the third component of isospin. This is directly analagous to the way states of angular momentum are specified, i.e. |eigenvalue of total isospin|eigenvalue of z component> For example |1,0> corresponds to the combination of 2 isospins |ud + du>. The Clebsch-Gordan calculation automatically ensures the orthonormality of the states. Therefore, I±|ψ> = Cψ± Such that, <Cψ±|Cψ±> = 1 For example, _ _ _ _ <C(uu - dd)|C(uu - dd)> = C2(1 - 0 - 0 + 1) = 2C2 ∴ C = √(1/2) Therefore, the correct state is: _ _ √(1/2)(uu - dd) In essence the C-G are just the normalization coefficients for composite states. Using the raising and lowering operators to find the coefficients can be very tedious. Fortunately, these coefficients can also be found directly from published tables (see the separate note entitled Clebsch-Gordan coefficients). From before we had: T±|v ⊗ w> = T±|v>w + vT±|w> Let's return to the reaising and lowering operators again. They operate on quarks and anti-quarks as follows: I+|d> = |u> _ _ I+|u> = -|d> I-|u> = |d> _ _ I-|d> = -|u> U+|s> = |d> _ _ U+|d> = -|s> U-|d> = |s> _ _ U-|s> = -|d> V+|s> = |u> _ _ V+|u> = -|s> V-|u> = |s> _ _ V-|s> = -|u> All other combinations give 0. _ The 2 ⊗ 2 ---------- _ The ordering and the minus sign in the 2 ensures that anti-quarks and quarks transform in the same way. To see this we start with defining the anti-quark doublet as: - - q* = | -d* | | u* | - - This transforms as: - - - - | -d'* | = U*| -d* | | u'* | | u* | - - - - - - Now mutiply both sides by the matrix | 0 -1 |. | 1 0 | - - - - - - - - - - | 0 -1 || -d*' | = U*| 0 -1 || -d* | | 1 0 || u*' | | 1 0 || u* | - - - - - - - - Which gives: - - - - | -u'* | = U*| -u* | | -d'* | | -d* | - - - - This compares to the quark doublet which transforms as: - - - - | u' | = U| u | | d' | | d | - - - - So far so good. However, let's stick with the previous form. Now multiply both sides by the - - inverse of | 0 -1 | to get: | 1 0 | - - - - - - - - - - | -d'* | = | 0 1 |U*| 0 -1 || -d* | | u'* | | -1 0 | | 1 0 || u* | - - - - - - - - In pecial Unitary Groups and the Standard Model - Part 1 we stated that a 2 x 2 unitary matrix can be written as: - - U = | a b | | -b* a* | - - Therefore, - - U* = | a* b* | | -b a | - - Substituting this into the previous equation gives: - - - - - - - - - - | -d'* | = | 0 1 || a* b* || 0 -1 || -d* | | u'* | | -1 0 || -b a || 1 0 || u* | - - - - - - - - - - Which gives: - - - - - - | -d*' | = | a b || -d* | | u*' | | -b* a* || u* | - - - - - - Which is just: - - - - | -d*' | = U| -d* | | u*' | | u* | - - - - So the quark and anti-quark doublets as defined transform in the same way. Note: This is a special property of SU(2). There is no analogous representation for SU(3)! The quark/anti-quark I3's are: _ u = d = 1/2 _ d = u = -1/2 Therefore, _ |1,1> = -ud _ |1,-1> = du _ _ _ I-|-ud> = -I-|u>d - uI-|d> _ _ = uu - dd and, I-|1,1> = √[1(1 + 1) - 1(1 - 1)]|1,0> = √2|1,0> Therefore, _ _ |1,0> = √(1/2)(uu - dd) The second state needs to be orthogonal to this state. This state is: _ _ ψ2 = √(1/2)(uu + dd) Check: _ _ _ _ <ψ12> = <uu - dd|uu + dd> _ _ _ _ _ _ _ _ = <uu|uu> + <uu|dd> - <dd|uu> - <dd|dd> = 1 - 0 - 0 - 1 = 0 The singlet state is ALWAYS a dead-end from the point of view of the raising and lowering operators. In other words for the state, ψ to be a singlet it must satisfy: I±|ψ> = 0 The only (0,0) state that satifies this is ψ2. _ The Mesons, 3 ⊗ 3 ------------------ Quark isospins can combine to give a total of 3/2 or 1/2 This can be achieved as follows: _ I3 for s, s = 0 _ |1,1> = -ud _ |1,-1> = du _ _ |1,1/2> = us, ds _ _ |1,-1/2> = sd, su Starting from the outer states we can reach the center in 6 ways. _ _ _ I+|du> = √(1/2)(|uu> - |dd>) _ _ _ I-|ud> = √(1/2)(|dd> - |uu>) _ _ _ V+|su> = √(1/2)(|uu> - |ss>) _ _ _ V-|us> = √(1/2)(|ss> - |uu>) _ _ _ U+|sd> = √(1/2)(|dd> - |ss>) _ _ _ U-|ds> = √(1/2)(|ss> - |dd>) _ _ We have already found the state ψ1 = √(1/2)(|uu> - |dd>) _ at (0,0) from the 2 ⊗ 2 analysis. However, there are 2 more states at (0,0) that we need to find. Whatever these states are, they must be a linear combination of: _ _ _ _ _ |uu> - |dd>, |uu> - |ss> and |dd> - |ss> The second state can be obtained by taking the linear combination of the other two states which is orthogonal to ψ1. _ _ _ _ ψ2 = |uu> - |ss> + |dd> - |ss> _ _ _ = |uu> + |dd> - 2|ss> Orthogonality check: _ _ _ _ _ <ψ21> = <uu + dd - 2ss|uu - dd> _ _ _ _ _ _ _ _ = <uu|uu> - <uu|dd> + <dd|uu> - <dd|dd> _ _ _ _ - 2<ss|uu> + 2<ss|dd> = 1 - 0 + 0 - 1 - 0 + 0 = 0 Therefore, ψ1 and ψ2 are orthogonal. _ _ _ _ _ _ <uu + dd - 2ss|uu + dd - 2ss> = 1 + 0 - 0 + 0 + 1 + 0 - 0 - 0 + 4 = 6 The normalized state is: _ _ _ _ _ _ <C(uu + dd - 2ss)|C(uu + dd - 2ss)> = 6C2 Therefore, _ _ _ ψ2 = √(1/6)(uu + dd - 2ss) Likewise, the final state can be obtained by requiring it to be orthonormal to both ψ1 and ψ2. _ _ _ ψ3 = √(1/3)(uu + dd + ss) This is the singlet state. Again, the singlet state is ALWAYS a dead-end from the point of view of the raising and lowering operators. In other words for the state, ψ to be a singlet it must satisfy: I±|ψ> = 0 and U±|ψ> = 0 and V±|ψ> = 0 The only (0,0) state that satifies this is ψ3. The complete list for the mesons is: _ |1,1> = -ud _ _ |0,0> = √(1/2)(uu - dd) _ |1,-1> = du _ _ _ |0,0> = √(1/6)(uu + dd - 2ss) _ _ |0,0> = √(1/2)(uu + dd) _ _ _ |0,0> = √(1/3)(uu + dd + ss) _ |1/2,1/2> = us _ |1/2,-1/2> = ds _ |1/2,1/2> = sd _ |1/2,-1/2> = su The Baryons ----------- Quark isospins can combine to give a total of 3/2 or 1/2 This can be achieved as follows: |3/2,3/2> = |uuu> |3/2,1/2> = positive combinations |duu>, |udu> and |uud> |3/2,-1/2> = positive conbinations of |udd>, |dud> and |ddu> |3/2,-3/2> = |ddd> |1/2,1/2> = positive and negative combinations of |duu>, |udu> and |uud> |1/2,-1/2> = positive and negative combinations of |udd>, |dud> and |ddu> We start with the 2 ⊗ 2 = 3 ⊕ 1 where the '2' is: - - 2 = | u | | d | - - The LHS is the same as the top row of the '6'. We can now form a 'parial' tensor product consisting of this and the top row of the '3'. We get: (3 ⊕ 1) ⊗ 2 = (3 ⊗ 2) ⊕ (1 ⊗ 2) = 4 ⊕ 2 ⊕ 2 3 ⊗ 2 is: - - - - | uu | | u | | ud + du |   ⊗ | | | dd | | d | - - - - Which gives the combinations uuu, uud, udu, duu, ddu and ddd. We can now use I± operator to find the intermediate states as follws: I+|ddd> = I+|d>dd + dI-|dd> = udd + d[I+|d>d + dI+|d>] = udd + d[ud + du] = udd + dud + ddu and, I+|3/2,-3/2> = √[3/2(3/2 + 1) - (-3/2)(-3/2 + 1)]|3/2,-1/2> = √3|3/2,-1/2> Therefore, √3|3/2,-1/2> = udd + dud + ddu or, |3/2,-1/2> = √(1/3)(udd + dud + ddu) We now raise this state. √(1/3)I+|udd + dud + ddu> I+|udd> = I+|u>dd + uI+|dd> = 0 + u[I+|d>d + dI+|d>] = u[ud + du] = uud + udu I+|dud> = I+|d>ud + dI+|ud> = uud + d[I+|u>d + uI+|d>] = uud + d[0 + uu| = uud + duu I+|ddu> = I+|d>du + dI+|du> = udu + d[I+|d>u + dI-|u>] = udu + d[uu + 0] = udu + duu Therefore, √(1/3)I+|udd + dud + ddu> = 2√(1/3)(uud + udu + duu) and, I+|3/2,-1/2> = √[3/2(3/2 + 1) - (-1/2)(-1/2 + 1)]|3/2,1/2> = 2|3/2,1/2> Therefore, 2|3/2,1/2> = 2√(1/3)(uud + udu + duu) or, |3/2,1/2> = √(1/3)(uud + udu + duu) Finally, after repeating the procedure again we get: I+|udd + dud + ddu> = uuu The 4 states, ddd, udd + dud + ddu, uud + udu + duu and uuu found above form a symmetric quadruplet called the '4'. 1 ⊗ 2 is: - - - - | u | | ud - du |   ⊗ | | - - | d | - - Which gives the states udu, udd, -duu and -dud. These are paired as: udu - duu = (ud - du)u and udd - dud = (ud - du)d as required and form the top row of one octet. So we have found the '4' and one of the '2's. The remaining '2' can be found by making use of the orthonormality property. The states are: √(1/6)(udd + dud - 2ddu) and, √(1/6)(2uud - udu - duu) We can use the raising and lowering operators I±, U± and V± to move around the '10' and '6' because they have the same equilateral triangular structure as the '3' and '8'. The complete list for the '10S' is: |3/2,3/2> = uuu: uuu |3/2,1/2> = uud: √(1/3)(uud + udu + duu) |3/2,-1/2> = ddu: √(1/3)(ddu + dud + udd) |3/2,-3/2> = ddd: ddd uus: √(1/3)(uus + usu + suu) uds: √(1/6)(uds + usd + dsu + dus + sud + sdu) dds: √(1/3)(dds + dsd + sdd) ssu: √(1/3)(ssu + sus + uss) ssd: √(1/3)(ssd + sds + dss) sss: sss These are symmetric for all quark interchanges. Again we have used the orthonormality condition to determine the coefficients. For example, <uus|uus> = <(uus + usu + suu)|(uus + usu + suu)> = 3 Mixed Symmetry Octet, 8MS ------------------------- √(1/6)I-|2uud - udu - duu> 2I-|uud> = 2(I-|u>ud + uI-|ud>) = 2(dud + u[I-|u>d + uI-|d>]) = 2dud + 2udd I-|udu> = I-|u>du + uI-|du> = ddu + u[I-|d>u + dI-|u>] = ddu + udd I-|duu> = I-|d>uu + dI-|uu> = d[I-|u>u + uI-|u>] = d[du + ud] = ddu + dud √(1/6)I-|2uud - udu - duu> = -√(1/6)(2ddu + dud + udd) and, I-|1/2,1/2> = √[1/2(1/2 + 1) - (1/2)(1/2 = 1)]|1/2,-1/2> = 1|1/2,-1/2> Therefore, |1/2,-1/2> = -√(1/6)(2ddu + dud + udd) and so on. The complete list for the '8MS' is: |1/2,1/2> = uud: √(1/6)(2uud - udu - duu) |1/2,-1/2> = ddu: √(1/6)(-2ddu + dud + udd) uus: √(1/6)(2uus - usu - suu) 0,0: √(1/12)(2uds - usd - dsu + 2dus - sud - sdu) 0,0: (1/2)(usd + sud - sdu - dsu) dds: √(1/6)(2dds - dsd - sdd) ssu: √(1/6)(sus + uss - 2ssu) ssd: √(1/6)(sds + dss - 2ssd) These are all symmetric for q1 <-> q2. This looks like: Mixed Symmetry Octet, 8MA ------------------------- I-|udu - duu> I-|udu> = I-|u>du + uI-|du> = ddu + u[I-|d>u + d|I-|u>] = ddu + udd I-|duu> = I-|d>uu + dI-|uu> = d[I-|u>u + uI-|u>] = d[du + ud] = ddu + dud I-|udu - duu> = udd - dud and so on. The complete list for the '8MA' is: |1/2,1/2> = ddu: √(1/2)(udd - dud) |1/2,-1/2> = uud: √(1/2)(udu - duu) uus: √(1/2)(usu - suu) 0,0: (1/2)(usd + dsu - sdu - sud) 0,0: √(1/12)(2uds - dsu - sud - 2dus - sdu + usd) dds: √(1/2)(dsd - sdd) ssu: √(1/2)(uss - sus) ssd: √(1/2)(dss - sds) These are all antisymmetric for q1 <-> q2. Totally Antisymmetric Singlet, 1A --------------------------------- The singlet state is: ψ = √(1/6)(uds - usd + dsu - dus + sud - sdu) As before, this can be verified by checking that: I±|ψ> = 0 and U±|ψ> = 0 and V±|ψ> = 0 This state is antisymmetric for all interchanges: This looks like: Again, we can use the raising and lowering operators to move around the 8. For example: If we put the '10S', '8MS', '8MA' and '1A' together we get: The 3 ⊗ 3 ⊗ 3 Approach ----------------------- Constructing Baryon states is a fairly tedious process. We can also obtain the same results if we use: 3 ⊗ 3 ⊗ 3 = 3 ⊗ (3 ⊗ 3) _ Now 3 ⊗ 3 = 6 ⊕ 3. This looks like: So we can write: _ 3 ⊗ 3 ⊗ 3 = 3 ⊗ (6 ⊕ 3) _ = 3 ⊗ 6 ⊕ 3 ⊗ 3 Now 3 ⊗ 6 looks like: We can treat each side of the '6' as a '3' as we did before and use the U± and V± operators to get the different states. For example, consider: - - - - | dd | | d | | ds + sd |   ⊗ | | | ss | | s | - - - - Which gives the combinations ddd, dds, dsd, sdd, ssd and sss. Likewise, - - - - | d | | ds - sd |   ⊗ | | - - | s | - - We can then use U-|ddd> to generate the '4' and, using orthogonality, the corresponding '2' for that particular side. If we do this for all 3 sides of the '6' and then add them together we have: 3 ⊗ 2 ⊕ 3 ⊗ 2 ⊕ 3 ⊗ 2 = 3 ⊗ (2 ⊕ 2 ⊕ 2) = 3 ⊗ 6 So we have performed the tensor product giving a total of 18 states. _ 3 ⊗ 3 = 8 ⊕ 1 yields the second mixed symmetry '2' ad the singlet. This gives: dsd - ssd and so on. Combining Spin -------------- Spin is not the same as isospin. However, both share exactly the same mathematics. Therefore, we can take any of the isospin wavefunctions derived previously and create a corresponding spin wavefunction. For the mesons we have, Spin 1 (symmetric) Triplet: |1,1> = ↑↑ |1,0> = √(1/2)(↑↓ + ↓↑) |1,-1> = ↓↓ Spin 0 (antisymmetric) Singlet: |0,0> = √(1/2)(↑↓ - ↓↑) This is often written as: 1/2 ⊗ 1/2 = 1 ⊕ 0 Isospin 1 (symmetric) Triplet: _ |1,1> = -ud _ _ |1,0> = √(1/2)(uu - dd) _ |1,-1> = du Isospin 0 (antisymmetric) Singlet: _ _ |0,0> = √(1/2)(uu + dd) |-ud> ⊗ ↑↑ _ Note: The reason the signs are flipped is that d isospin transforms in the opposite way under transformations. We can now create a composite wavefunction as: _ π+ = -ud ⊗ √(1/2)(↑↓ - ↓↑) with S = 0 _ _ = -√(1/2)(u↑d↓ - u↓d↑) For the baryons we have: For the '10S' with S = 3/2 ∴ J = 3/2: |3/2,3/2> = ↑↑↑: ↑↑↑ |3/2,1/2> = ↑↑↓: √(1/3)(↑↑↓ + ↑↓↑ + ↓↑↑) |3/2,-1/2> = ↓↓↑: √(1/3)(↓↓↑ + ↓↑↓ + ↑↓↓) |3/2,-3/2> = ↓↓↓: ↓↓↓ For the '8MA' S = 1/2 ∴ J = 1/2: |1/2,1/2> = ↓↓↑: √(1/2)(↑↓↓ - ↓↑↓) |1/2,-1/2> = ↑↑↓: √(1/2)(↑↓↑ - ↓↑↑) For the '8MS'S = 1/2 ∴ J = 1/2: |1/2,1/2> = ↑↑↓: √(1/6)(2↑↑↓ - ↑↓↑ - ↓↑↑) |1/2,-1/2> = ↓↓↑: √(1/6)(-2↓↓↑ + ↓↑↓ + ↑↓↓) We can now create a composite wavefunction as: ψFlavor ⊗ ψSpin. For example, ψFlavor ⊗ ψSpin = √(1/6)(2uud - udu - duu)√(1/6)(2↑↑↓ - ↑↓↑ - ↓↑↑) Color Symmetry -------------- The shrewd observer will note that there is a problem when we consider flavor states such as |uuu> and |ddd> in the Baron decuplet because they appear to violate the Pauli Exclusion Principle that 2 or more identical particles cannot occupy the same quantum state. In other words, it is impossible to construct a totally antisymmetric wavefunction function for the state. This led researchers to believe that an additional quantum number had to exist. Enter color! Exactly analagous to labelling u, d, s flavor states by I3 and Y we can assign color quantum numbers I3C and YC to color states. - - - - - - | 1 | | 0 | | 0 | r = | 0 | g = | 1 | b = | 0 | | 0 | | 0 | | 1 | - - - - - - _ The 3 and the 3 then look like: Flavor symmetry is not exact because the quark masses are not identical. On the other hand color symmetry is is exact. Only combinations of particles that have a COLOR SINGLET state can be observed in nature. Color singlets are colourless combinations that obey: 1. Their quantum numbers, I3C = YC = 0 2. They are invariant under SU(3) colour transformations. 3. The raising and lowering operatos I±, U± and V± all yield 0. The implication of this is that we can never observe a free quark (which would carry a color charge). Consider: _ 3 ⊗ 3 has the completely symmetric singlet: _ _ _ √(1/3)(rr + gg + bb). Proof: _ _ _ _ _ I+|rr + gg + ss> = -rg + rg = 0 _ Therefore, qq objects are colorless and can exist as free particles (mesons). Likewise, 3 ⊗ 3 ⊗ 3 has completely antisymmetric singlet: √(1/6)(rgb - grb + gbr - bgr + brg - rbg). Proof: I+|rgb - grb + gbr - bgr + brg - rbg> = rrb - rbr + rbr - rrb + brr - brr = 0 Therefore, qqq objects are colorless and can exist as free particles (baryons). However, the 3 ⊗ 3 = 6 ⊗ 3 has no color singlet therefore the qq combination does not appear in nature. Quark Properties ---------------- _ _ _ u d s u d s --- --- --- --- --- --- Baryon Number, B 1/3 1/3 1/3 -1/3 -1/3 -1/3 Strangeness, S 0 0 -1 0 0 1 Hypercharge, Y 1/3 1/3 -2/3 -1/3 -1/3 2/3 Charge, Q 2/3 -1/3 -1/3 -2/3 1/3 1/3 I3 1/2 -1/2 0 -1/2 1/2 0 Spin, J ±1/2 ±1/2 ±1/2 ±1/2 ±1/2 ±1/2 Intrinsic Parity, P +1 +1 +1 -1 -1 -1 These relationships are encapsulated on the following formulas: Y = S + B = 2(Q - I3) The Total Wavefunction ---------------------- ψTotal = ψSpin ⊗ ψSpacial ⊗ ψFlavor ⊗ ψColorSpacial aka ψOrbital) There are 2 types of mesons, PSEUDOSCALAR and VECTOR. Pseudoscalar mesons have antialigned spins resulting in spin 0. Vector mesons have aligned spins resulting in spin 1. The list is: State Pseudoscalar Vector ------- ------------ ------ |1,1> π+ ρ+ |1,0> π0 ρ0 |1,-1> π- ρ- The Mesons are bosons so it is required that the total wavefunction is symmetric. For the ground state L = 0. _ P(ud) = PuPanti-d(-1)0 = -1 (see note on Parity) Therefore, the spatial wavefunction is antisymmetric. Now, _ _ _ ψColor = √(1/3)(rr + gg + bb) is totally symmetric. Therefore, ψSpacial ⊗ ψColor is totally antisymmetric. This means that the combination ψFlavor ⊗ ψSpin needs to be be totally antisymmetric for the particles to be bosons. Consider: ψFlavor ⊗ ψSpin = -ud ⊗ √(1/2)(↑↓ - ↓↑) = -√(1/2)(u↑d↓ - u↓d↑) = √(1/2)(u↓d↑ - u↑d↓) This is antisymmetric under the interchange of any 2 quarks. ψTotal is now symmetric. This is, in fact, the spin 0 wavefunction for the π+ pseudoscalar meson. The ρ+ meson is considered to be an excited state of the π+. For the excited state L = 1 we get. _ P(ud) = PuPanti-d(-1)1 = 1 _ ρ+ = -ud ⊗ ↑↓ with S = 1 = -u↑d↓ So now ψSpatial, ψFlavor, ψSpin and ψColor are all symmetric so ψTotal is symmetric; The Baryons are fermions so it is required that the total wave-function is anti-symmetric (Pauli Exclusion Principle). P(qqq) = Pq Pq Pq(-1)L = 1 for L = 0 Therefore, the spatial wavefunction is symmetric. Now, ψColor = √(1/6)(rgb - grb + gbr - bgr + brg - rbg) is totally antisymmetric. Therefore, ψSpacial ⊗ ψColor is totally antisymmetric. This means that the combination ψFlavor ⊗ ψSpin needs to be be totally symmetric for the particles to be fermions. This can be achieved by constructing the normalized linear combination: ψFlavor ⊗ ψSpin = √(1/2)ψMSχMS + √(1/2)ψMAχMA Where ψMS is from the 8MS and ψMA is the corresponding state from the 8MA. Consider the state ψ = |uud> constructed as follows: ψFlavor ⊗ ψSpin = √(1/2)[√(1/6)(2uud - udu - duu) ⊗ √(1/6)(2↑↑↓ - ↑↓↑ - ↓↑↑)] + √(1/2)[√(1/2)(udu - duu) ⊗ √(1/2)(↑↓↑ - ↓↑↑)] = √(1/18)(2u↑u↑d↓ - u↑u↓d↑ - u↓u↑d↑ + 2u↑d↓u↑ - u↓d↑u↑ - u↑d↑u↓ + 2d↓u↑u↑ - d↑u↓u↑ - d↑u↑u↓) This is symmetric under the interchange of any 2 quarks. ψTotal is now antisymmetric. This is, in fact, the spin up wavefunction for the proton, |p↑>. The Particle Zoo ---------------- _ In summary, the 3 ⊗ 3 gives the following particles: The 3 ⊗ 3 ⊗ 3 gives the following particles: Triality, τ ----------- Triality is defined as: τ = (n - m) mod 3 (the remainder of (n - m)/3) D R τ ----- -- - (0,0) 0 (1,0) 3 1 _ (0,1) 3 2 (1,1) 8 0 (2,0) 6 2 _ (0,2) 6 1 (3,0) 10 0 Irreps with triality 0 correspond to observable particles. Those with triality ≠ 0 correspond to particles that have not been observed in nature.