Lectures in Quantum Field Theory Lecture 1

Instituto Superior T´ecnico

Jorge C. Rom˜ao Instituto Superior T´ecnico, Departamento de F´ısica & CFTP A. Rovisco Pais 1, 1049-001 Lisboa, Portugal

January 2014

Jorge C. Rom˜ao IDPASC School Braga – 1 IST Summary of Lectures

Summary ❐ Part 1 – Relativistic Quantum Mechanics (Lecture 1) Introduction Klein-Gordon Eq. ◆ The basic principles of Quantum Mechanics and Special Relativity Dirac Equation Covariance Dirac Eq. ◆ Klein-Gordon and Dirac equations Free Particle Solutions ◆ Gamma matrices and spinors Minimal Coupling NR Limit Dirac Eq. ◆ Covariance E < 0 Solutions ◆ Solutions for the free particle Charge Conjugation Massless Spin 1/2 ◆ Minimal coupling ◆ Non relativistic limit and the Pauli equation ◆ Charge conjugation and antiparticles ◆ Massless spin 1/2 particles

Jorge C. Rom˜ao IDPASC School Braga – 2 IST Summary ...

Summary ❐ Part 2 – Quantum Field Theory (Two Lectures) Introduction Klein-Gordon Eq. ◆ QED as an example (Lecture 2) Dirac Equation Covariance Dirac Eq. ■ QED as a gauge theory Free Particle Solutions ■ Propagators and Green functions Minimal Coupling ■ NR Limit Dirac Eq. Feynman rules for QED E < 0 Solutions ■ Example 1: Compton scattering Charge Conjugation − + − + ■ Example 2: e + e µ + µ in QED Massless Spin 1/2 → ◆ Non Abelian Gauge Theories (NAGT) (Lecture 3)

■ Radiative corrections and renormalization ■ Non Abelian gauge theories: Classical theory ■ Non Abelian gauge theories: Quantization ■ Feynman rules for a NAGT ■ Example: Vacuum polarization in QCD

Jorge C. Rom˜ao IDPASC School Braga – 3 IST Summary ...

Summary ❐ Part 3 – The Standard Model (Lecture 4) Introduction Klein-Gordon Eq. ◆ Gauge group and particle content of the massless SM Dirac Equation Covariance Dirac Eq. ◆ Spontaneous Symmetry Breaking and the Higgs mechanism Free Particle Solutions ◆ The SM with masses after Spontaneous Symmetry Breaking Minimal Coupling NR Limit Dirac Eq. ◆ Interactions dictated by the gauge symmetry group E < 0 Solutions ◆ Example 1: The decay Z ff in the SM Charge Conjugation → + − + − Massless Spin 1/2 ◆ Example 2: e e µ µ in the SM → ◆ Example 3: Muon decay ◆ How the gauge symmetry corrects for the bad high energy behaviour in + − νeνe W W → L L ◆ The need for the Higgs to correct the bad high energy behaviour in W +W − W +W − L L → L L

Jorge C. Rom˜ao IDPASC School Braga – 4 IST Basic Ideas of Quantum Mechanics and Special Relativity

Summary ❐ In this lecture we will put together the ideas of Quantum Mechanics and Introduction • QM Special Relativity • Special Relativity Klein-Gordon Eq. ❐ This will lead to the substitution of Schr¨odinger equation by the Dirac Equation Klein-Gordon (spin 0) and Dirac equations (spin 1/2) Covariance Dirac Eq. Free Particle Solutions ❐ The idea of describing Quantum Physics by an equation will be abandoned Minimal Coupling in favour of a description in terms of a variable number of particles allowing NR Limit Dirac Eq. for the creation and annihilation of particles E < 0 Solutions Charge Conjugation ❐ However, the formalism that we will develop in this ﬁrst lecture will be very Massless Spin 1/2 useful for the Quantum Field Theory that we will address in the following lectures

Jorge C. Rom˜ao IDPASC School Braga – 5 IST Principles of Quantum Mechanics

We list the basic principles of Quantum Mechanics: Summary ❐ For a given physical state there is a state function Φ that contains all the Introduction | i • QM possible information about the system • Special Relativity Klein-Gordon Eq. ◆ In many cases we will deal with the representation of state Φ in terms | i Dirac Equation of the coordinates, the so-called wave function Ψ(qi,si,t). Covariance Dirac Eq. 2 ◆ Ψ(qi,si,t) 0 has the interpretation of a density of probability of Free Particle Solutions | | ≥ Minimal Coupling ﬁnding the system in a state with coordinates qi, internal quantum NR Limit Dirac Eq. numbers si at time t E < 0 Solutions Charge Conjugation ❐ Physical observables are represented by hermitian linear operators, as Massless Spin 1/2 ∂ ∂ pi i¯h ,E i¯h → − ∂qi → ∂t

❐ A state Φn is an eigenstate of the operator Ω if | i

Ω Φn = ωn Φn | i | i

where Φn is the eigenstate that corresponds to the eigenvalue ωn. If Ω is | i hermitian then ωn are real numbers.

Jorge C. Rom˜ao IDPASC School Braga – 6 IST Principles of Quantum Mechanics ...

❐ For a complete set of commuting operators Ω1, Ω2,... , there exists a Summary { } complete set of orthonormal eigenfunctions, Ψn. An arbitrary state (wave Introduction • QM function) can expanded in this set • Special Relativity

Klein-Gordon Eq. Ψ = anΨn Dirac Equation n Covariance Dirac Eq. X ❐ The result of the measurement of the observable Ω is any of the eigenvalues Free Particle Solutions 2 ωn with probability an . The average value of the observable is Minimal Coupling | | NR Limit Dirac Eq.

E < 0 Solutions ∗ 2 < Ω >Ψ= dq1...Ψ (qi,si,t)ΩΨ(qi,si,t) = an ωn Charge Conjugation | | s Z n Massless Spin 1/2 X X ❐ The time evolution of a system is given by ∂Ψ i¯h = HΨ (The Hamiltonian H is a linear and hermitian operator) ∂t ❐ Linearity implies the superposition principle and hermiticity leads to the conservation of probability d i dq Ψ∗Ψ = dq [(HΨ)∗Ψ Ψ∗(HΨ)] = 0 dt 1 ¯h 1 s ··· s ··· − X Z X Z Jorge C. Rom˜ao IDPASC School Braga – 7 IST Principles of Special Relativity

Special relativity is based on the principles of relativity and of the constancy of the Summary speed of light in all reference frames. For our purposes it is enough to recall: Introduction • QM ❐ The coordinates of two reference frames are related by • Special Relativity Klein-Gordon Eq. ′µ µ ν Dirac Equation x = a ν x , µ,ν = 0, 1, 2, 3

Covariance Dirac Eq.

Free Particle Solutions Minimal Coupling ❐ The invariance of the interval NR Limit Dirac Eq. 0 2 µ ν µ E < Solutions ds = gµνdx dx = dx dxµ Charge Conjugation

Massless Spin 1/2 where the metric is diagonal and given by gµν = diag(+ ), restricts the µ − −− coeﬃcients a ν to obey

µ ν α β α β gµνa αa βdx dx = gαβdx dx

which implies

µ ν T a αgµν a β = gαβ or in matrix form a ga = g

Jorge C. Rom˜ao IDPASC School Braga – 8 IST Principles of Special Relativity ...

❐ The matrices that obey aT ga = g constitute the Lorentz group, designated Summary

Introduction by O(3, 1) • QM • Special Relativity ❐ We can easily verify that Klein-Gordon Eq. Dirac Equation det a = 1 Covariance Dirac Eq. ± Free Particle Solutions ❐ Transformations with det a = +1 constitute the proper Lorentz group a Minimal Coupling subgroup of the Lorentz group. They can be built from inﬁnitesimal NR Limit Dirac Eq.

E < 0 Solutions transformations. Examples are rotations and Lorentz transformations

Charge Conjugation

Massless Spin 1/2 1 0 00 γ γβ 0 0 0 cos θ sin θ 0 γβ− γ 0 0 a = , a = 0 sin θ cos θ 0 −0 0 10 − 0 0 01  0 0 01         where 1 V γ = ; β = 1 β2 c − and V is thep relative velocity of reference frame S′ with respect to S

Jorge C. Rom˜ao IDPASC School Braga – 9 IST Principles of Special Relativity ...

Summary ❐ Examples of transformations with det a = 1 are the space (Parity) and Introduction − • QM time (Time Reversal) inversions. For instance • Special Relativity

Klein-Gordon Eq.

Dirac Equation 10 0 0 1 0 0 0 0 1 0 0 −0 100 Covariance Dirac Eq. a =  −  a =   Free Particle Solutions 0 0 1 0 0 010 − Minimal Coupling 0 0 0 1  0 001  −    NR Limit Dirac Eq.     0 E < Solutions Parity Time Reversal Charge Conjugation

Massless Spin 1/2 ❐ The transformations with det a = 1 do not form a subgroup of the full Lorentz group (they do not contain− the identity)

Jorge C. Rom˜ao IDPASC School Braga – 10 IST The Klein-Gordon Equation

We start with the free particle: Summary Introduction ❐ In non-relativistic quantum mechanics we get Schr¨odinger equation from the Klein-Gordon Eq. fundamental equation Dirac Equation Covariance Dirac Eq. ∂ i¯h ψ = Hψ Free Particle Solutions ∂t Minimal Coupling NR Limit Dirac Eq. using the non-relativistic free particle Hamiltonian E < 0 Solutions Charge Conjugation p2 H = Massless Spin 1/2 2m

and making the substitution ~p i¯h~ . We get → − ∇ ∂ψ ¯h2 i¯h = 2ψ ∂t −2m∇

❐ The idea is to use the relativistic form of H = E, the energy of the free particle.

Jorge C. Rom˜ao IDPASC School Braga – 11 IST The Klein-Gordon Equation ...

❐ In special relativity the energy and momentum are related through Summary

Introduction µ 2 2 µ E Klein-Gordon Eq. pµp = m c , p , ~p ≡ c Dirac Equation Covariance Dirac Eq. giving Free Particle Solutions

Minimal Coupling E2 = p2c2 + m2c4 E = p2c2 + m2c4 NR Limit Dirac Eq. → ± 0 E < Solutions p Charge Conjugation ❐ Classically we require energies to be positive, so we could choose Massless Spin 1/2 H = p2c2 + m2c4 p ❐ We are lead to interpret the square root of an operator. To avoid this problem we can ﬁnd an equation for H2. Iterating the original equation and ∂ noticing that i¯h ∂t ,H = 0, we get

2 2 2 ∂ 2 2 ~2 2 4 mc µ ¯h ψ =( ¯h c + m c )ψ + ψ = 0, = ∂µ∂ − ∂t2 − ∇ → ⊔⊓ ¯h ⊔⊓

Jorge C. Rom˜ao IDPASC School Braga – 12 IST The Klein-Gordon Equation ...

❐ Now we have no problem with the operators but we reintroduce the negative Summary energy solutions. We will see that the negative energy solutions can not be Introduction

Klein-Gordon Eq. avoided and its interpretation will be related to the antiparticles. Dirac Equation ❐ It was not because of the negative energy solutions that the Klein-Gordon Covariance Dirac Eq.

Free Particle Solutions equation was abandoned, but because of the density of probability. Using the

Minimal Coupling Klein-Gordon equation and its hermitian conjugate we get

NR Limit Dirac Eq. 2 2 0 mc mc E < Solutions ψ∗ + ψ ψ + ψ∗ = 0 Charge Conjugation ⊔⊓ ¯h − ⊔⊓ ¯h Massless Spin 1/2 or ↔µ ↔µ ∗ ∗ ∗ µ µ ∗ 0 = ψ ψ ψ ψ = ∂µ(ψ ∂ ψ) ∂µJ =0 ; J = ψ ∂ ψ ⊔⊓ − ⊔⊓ →

❐ In the usual identiﬁcation J µ =(ρc, J~) gives

1 ∂ψ ∂ψ∗ ρ = ψ∗ ψ Not positive deﬁned c2 ∂t − ∂t → ❐ The Klein-Gordon equation was abandoned by the wrong reasons: Spin 0

Jorge C. Rom˜ao IDPASC School Braga – 13 IST Dirac Equation

❐ Dirac approach was to treat time and space on the same footing, in the Summary spirit of relativity. As in the fundamental equation the time derivative Introduction

Klein-Gordon Eq. appears linearly, Dirac postulated the same behaviour for the space

Dirac Equation derivatives, writing,

Covariance Dirac Eq. ∂ψ Free Particle Solutions i¯h = i¯hc~α ~ + βmc2 ψ Hψ Minimal Coupling ∂t − · ∇ ≡ NR Limit Dirac Eq. ❐ The dimensionless constants αi and β can not be numbers as we will see in 0 E < Solutions a moment. So, Dirac postulated that ~α and β are N N hermitian matrices Charge Conjugation × Massless Spin 1/2 (for H to be hermitian) acting on a column vector with N entries,

ψ1 . ψ =  .  ψ  N    ❐ This matrix equation must obey the conditions: ◆ Give the correct relation E2 = p2c2 + m2c4 for the free particle ◆ Give a probability that is positive deﬁned ◆ Must be covariant under Lorentz transformations

Jorge C. Rom˜ao IDPASC School Braga – 14 IST Dirac Equation ...

❐ For the energy-momentum relation, it is enough that each component Summary satisﬁes the Klein-Gordon equation. We iterate the equation to get Introduction Klein-Gordon Eq. ∂2ψ ∂ψ Dirac Equation 2 i 2 ¯h 2 = i¯hcα i + βmc i¯h Covariance Dirac Eq. − ∂t − ∇ ∂t Free Particle Solutions i j j i 2 2 α α + α α 2 i i 2 2 4 Minimal Coupling = ¯h c i j i¯hmc (α β + βα ) i + β m c ψ − 2 ∇ ∇ − ∇ NR Limit Dirac Eq. E < 0 Solutions ❐ This gives the relations Charge Conjugation Massless Spin 1/2 αiαj + αjαi = 2δij

 i i  α β + βα = 0   (αi)2 = β2 = 1 Eigenvalues = 1  → ±  ❐ We have to construct 4 anti-commuting matrices with square equal to unity. This is not possible for N = 2 (only 3 Pauli matrices). It is not possible for N odd, because αi = βαiβ Tr(αi) = Tr( βαiβ) = Tr(αi) Tr(αi) = 0 − → − − → Jorge C. Rom˜ao IDPASC School Braga – 15 IST Dirac Equation ...

❐ The smallest is N = 4. An explicit representation (not unique) is Summary

Introduction Klein-Gordon Eq. 0 σ 1 0 αi = i ; β = Blocks 2 2 Dirac Equation σi 0 0 1 × Covariance Dirac Eq. −

Free Particle Solutions where σi are the Pauli matrices: Minimal Coupling

NR Limit Dirac Eq.

E < 0 Solutions 0 1 0 i 1 0 σ1 = ; σ2 = − ; σ3 = Charge Conjugation 1 0 i 0 0 1 − Massless Spin 1/2 ❐ Let us now look at the probability current. As αi and β are hermitian we get

† ← ∂ψ † i 2 i¯h = ψ (i¯hcα ∂ i + βmc ) − ∂t

❐ Doing the usual trick of multiplying and subtracting we get

∂ † † i ∂ † † i¯h (ψ ψ) = i¯hc i(ψ α ψ) (ψ ψ) + ~ (ψ c~αψ) = 0 ∂t − ∇ → ∂t ∇·

Jorge C. Rom˜ao IDPASC School Braga – 16 IST Dirac Equation ...

❐ Using Summary

Introduction † † ∂ρ Klein-Gordon Eq. ρ = ψ ψ, ~j = ψ c~αψ + ~ ~j = 0 ∂t Dirac Equation → ∇·

Covariance Dirac Eq. Free Particle Solutions ❐ Integrating in all the space we get Minimal Coupling d NR Limit Dirac Eq. d3x ψ†ψ = 0 E < 0 Solutions dt Charge Conjugation Z † Massless Spin 1/2 which allow us to identify ψ ψ with a positive density of probability ❐ The notation anticipates the fact that ~j is a vector in 3-dimensional space. Actually, we will show in the following that jµ =(cρ,~j) is a conserved µ 4-vector, ∂µj = 0, and that Dirac equation is covariant, that is it, maintains its form in all inertial reference frames

Jorge C. Rom˜ao IDPASC School Braga – 17 IST Dirac gamma matrices

❐ We start by introduction a convenient notation due to Dirac. We multiply Summary the Dirac equation on the left by 1 β and introduce the matrices, Introduction c

Klein-Gordon Eq. γ0 β ; γi βαi i = 1, 2, 3 Dirac Equation ≡ ≡ Covariance Dirac Eq. ❐ • γ matrices The Dirac equation reads • Equivalence Transf. • Covariance Proof µ • Parity (i¯hγ ∂µ mc)ψ = 0 or (i¯h∂/ mc)ψ = 0 • Bilinears − − Free Particle Solutions µ where we introduced Feynman slash notation, ∂/ γ ∂µ Minimal Coupling ≡ NR Limit Dirac Eq. ❐ In the Dirac representation we have E < 0 Solutions

Charge Conjugation 0 1 0 i 0 σi Massless Spin 1/2 γ = ; γ = 0 1 σi 0 − − ❐ The normalization and anti-commuting relations can be written as

γµγν + γν γµ = 2gµν

❐ We should note that, despite the suggestive form of the above equation, we have not yet proved its covariance Jorge C. Rom˜ao IDPASC School Braga – 18 IST Diﬀerent γ representations: Equivalence transformations

❐ Let us consider two representations, γµ and γ˜µ. Both satisfy Dirac equation, Summary Introduction µ µ (i¯hγ ∂µ mc)ψ = 0, (i¯hγ˜ ∂µ mc)ψ˜ = 0 Klein-Gordon Eq. − − Dirac Equation Covariance Dirac Eq. ❐ Both describe the same Physics, so must exist a relation between ψ and ψ˜. • γ matrices • Equivalence Transf. Let it be • Covariance Proof • Parity • Bilinears ψ = Uψ˜ Free Particle Solutions Minimal Coupling where U is a constant matrix that admits inverse. Substituting in the above NR Limit Dirac Eq. equation and multiplying on the left by U −1 we get E < 0 Solutions

Charge Conjugation µ −1 µ Massless Spin 1/2 γ˜ = U γ U

❐ This kind of transformations are called equivalence transformations and they should not change the observable physical quantities despite the fact that the wave function is changed. For future use, we also note that

Tr γ˜µγ˜ν γ˜σ = Tr γµγν γσ ··· ··· Jorge C. Rom˜ao IDPASC School Braga – 19 IST Proof of the Covariance

❐ Let us consider the Dirac equation in two inertial frames O and O′ Summary Introduction µ ′µ ′ ′ ′ (i¯hγ ∂µ mc)ψ(x) = 0, (i¯hγ ∂ mc)ψ (x ) = 0 Klein-Gordon Eq. − µ − Dirac Equation ′µ µ Covariance Dirac Eq. ❐ γ satisﬁes the same anti-commuting relations as γ , as well as the • γ matrices ′0† ′0 ′i† ′i ′µ µ • Equivalence Transf. relations, γ = γ and γ = γ . We can then show that γ and γ • Covariance Proof − • Parity are related by an equivalence transformation, • Bilinears Free Particle Solutions γ′µ = U −1γµU Minimal Coupling NR Limit Dirac Eq. where U is an unitary matrix. E < 0 Solutions Charge Conjugation ❐ We can therefore transfer all the changes into the wave function and keep Massless Spin 1/2 the same representation in all inertial frames. The wave functions ψ′(x′) and ψ(x) must then be related through

′ ′ ′ −1 ′ ′µ µ ν ψ (x ) = ψ (ax) = S(a)ψ(x) = S(a)ψ(a x ), x = a νx

and the matrix S(a) must depend on the relative velocity and/or the rotation between the two frames (for proper Lorentz transformations)

Jorge C. Rom˜ao IDPASC School Braga – 20 IST Proof of the Covariance ...

❐ Substituting ψ′(x′) = S(a)ψ(x) we get Summary

Introduction µ ∂ Klein-Gordon Eq. (i¯hγ mc)S(a)ψ(x) = 0 ∂x′µ − Dirac Equation

Covariance Dirac Eq. • γ matrices ❐ Now we use • Equivalence Transf. • Covariance Proof ν • ∂ ∂x ∂ Parity = =(a−1)ν ∂ • Bilinears ∂x′µ ∂x′µ ∂xν µ ν Free Particle Solutions Minimal Coupling to get NR Limit Dirac Eq.

E < 0 −1 µ −1 ν Solutions i¯hS (a)γ S(a)(a ) µ ∂ν mc ψ(x) = 0 Charge Conjugation − Massless Spin 1/2 ❐ Comparing with the equation in frame O, we have the same equation if

−1 µ −1 ν ν µ −1 ν ν S (a)γ S(a)(a ) µ = γ S(a)γ S (a)a µ = γ → These are the fundamental relations that will allow to ﬁnd S(a). If we succeed we have proved the covariance

Jorge C. Rom˜ao IDPASC School Braga – 21 IST Proof of the Covariance ...

❐ To obtain S(a) we start by considering infinitesimal transformations, Summary Introduction ν ν ν µν νµ a µ = g µ + ω µ + with ω = ω Klein-Gordon Eq. ··· − Dirac Equation Covariance Dirac Eq. ❐ The last equation means that there are only six independent parameters. We • γ matrices • Equivalence Transf. will see that they will be identified with the three degrees of freedom of a • Covariance Proof • Parity rotation plus the three degrees of freedom of a Lorentz transformation • Bilinears (boost) in an arbitrary direction Free Particle Solutions Minimal Coupling ❐ For infinitesimal transformations we define then NR Limit Dirac Eq. E < 0 Solutions i µν −1 i µν S = 1 σµν ω + , S = 1 + σµνω + Charge Conjugation − 4 ··· 4 ··· Massless Spin 1/2 where the matrices σµν = σνµ are anti-symmetric. − ❐ Substituting in the original equation and keeping terms only linear in ωµν we get

µ µ µ [γ ,σαβ] = 2i(g αγβ g βγα) −

Jorge C. Rom˜ao IDPASC School Braga – 22 IST Proof of the Covariance ...

❐ Using the anti-commutation relations for the γ’s we can verify that the Summary solution is Introduction

Klein-Gordon Eq. i σµν = [γµ,γν ] Dirac Equation 2 Covariance Dirac Eq. • γ matrices ❐ This determines S and S−1 for inﬁnitesimal transformations. However, for • Equivalence Transf. • Covariance Proof these continuous transformations (Lie groups) the solution exponentiates, • Parity • Bilinears giving the ﬁnal result

Free Particle Solutions i µν − 4 σµν ω Minimal Coupling S = e NR Limit Dirac Eq. E < 0 Solutions ❐ To find an explicit form we distinguish the case of Lorentz boosts from the Charge Conjugation rotations. For rotations we define the vectors, Massless Spin 1/2 (θ1,θ2,θ3) (ω2 , ω3 , ω1 ), (Σ1, Σ2, Σ3) (σ23,σ31,σ12) ≡ 3 1 2 ≡ ❐ Then i ~ ~ 2 θ·Σ ~σ 0 SR = e , Σ~ (Dirac representation) ≡ 0 ~σ a generalization of the way 2-spinors transform under rotations. Jorge C. Romão IDPASC School Braga – 23 IST Proof of the Covariance ...

❐ Using Summary

Introduction (θ~ Σ)(~ θ~ Σ)~ = θ~ θ~ Klein-Gordon Eq. · · · Dirac Equation

Covariance Dirac Eq. we can get another useful form • γ matrices • Equivalence Transf. • Covariance Proof θ ˆ ~ θ • Parity SR = cos + iθ Σ sin • Bilinears 2 · 2

Free Particle Solutions Minimal Coupling where θˆ is an unit vector in the direction of the rotation NR Limit Dirac Eq. i 0i E < 0 Solutions ❐ For the Lorentz boosts we deﬁne the 3-vector ~ω such that (ω ω ) Charge Conjugation ≡ Massless Spin 1/2 ωˆ Vˆ ≡  V  tanh ω = c

❐ Using now i σ0i = γ0,γi = iγ0γi = iαi 2 Jorge C. Rom˜ao IDPASC School Braga – 24 IST Proof of the Covariance ...

❐ We obtain for the case of the Lorentz transformations (boosts) Summary

Introduction 1 ω ω − 2 ~ω·~α Klein-Gordon Eq. SL = e , SL = cosh ωˆ ~α sinh Dirac Equation 2 − · 2

Covariance Dirac Eq. • γ matrices where we have used (~ω ~α)2 = ~ω ~ω • Equivalence Transf. · · • Covariance Proof • Parity ❐ We can easily convince ourselves that while SR is an unitary matrix but the • Bilinears 0 same is not true for SL. However, we can show (using [γ , Σ]=0~ and Free Particle Solutions γ0, ~α = 0) that Minimal Coupling { } NR Limit Dirac Eq. −1 0 † 0 E < 0 Solutions S = γ S γ

Charge Conjugation Massless Spin 1/2 both for SR and SL ❐ This is important to show that the current,

jµ(x) = cψ†(x)γ0γµψ(x)

is a 4-vector as we will prove now

Jorge C. Rom˜ao IDPASC School Braga – 25 IST Proof of the Covariance ...

❐ The current in frame O′ is Summary Introduction ′µ ′† ′ 0 µ ′ ′ Klein-Gordon Eq. j =cψ (x )γ γ ψ (x ) Dirac Equation =cψ†(x)S†γ0γµSψ(x) Covariance Dirac Eq. • γ matrices † 0 0 † 0 µ • Equivalence Transf. =cψ (x)γ γ S γ γ Sψ(x) • Covariance Proof • Parity † 0 −1 µ • Bilinears =cψ (x)γ S γ Sψ(x)

Free Particle Solutions −1 µ µ ν Minimal Coupling and using S γ S = a ν γ we get

NR Limit Dirac Eq. ′µ µ ν E < 0 Solutions j = a ν j Charge Conjugation Massless Spin 1/2 as it should for a 4-vector ❐ The combination ψ†γ0 appears so frequently that it is convenient to deﬁne a symbol for it,

ψ ψ†γ0 ≡ the so-called Dirac adjoint Jorge C. Rom˜ao IDPASC School Braga – 26 IST Parity

Summary Let us consider one case of discrete symmetries, the space inversion or parity, P . Introduction Klein-Gordon Eq. ❐ Parity corresponds to the Lorentz transformation with det a = 1, Dirac Equation − Covariance Dirac Eq. • γ matrices 1 • Equivalence Transf. µ 1 • Covariance Proof a ν  −  • Parity ≡ 1 • Bilinears −  1 Free Particle Solutions  −  Minimal Coupling   ❐ NR Limit Dirac Eq. We want to ﬁnd a matrix SP that satisﬁes

E < 0 Solutions

Charge Conjugation −1 µ µ ν 0 i SP γ SP = a ν γ γ ,SP = 0, γ ,SP = 0 Massless Spin 1/2 → ❐ We can easily verify that the relation is satisﬁed by,

iϕ 0 SP = e γ P ≡ where eiϕ is an arbitrary phase

Jorge C. Rom˜ao IDPASC School Braga – 27 IST Covariant Bilinears

❐ Any 4 4 matrix can be expanded in terms of a basis of 16 linearly Summary independent,× 4 4 matrices. Introduction × Klein-Gordon Eq. ❐ To deﬁne this basis it is convenient to introduce the matrix Dirac Equation

Covariance Dirac Eq. • γ matrices 0 0 1 0 • Equivalence Transf. 0 1 2 3 0 0 0 1 • Covariance Proof γ5 iγ γ γ γ , γ5 =   (In the Dirac representation) • Parity ≡ 1 0 0 0 • Bilinears 0 1 0 0 Free Particle Solutions     Minimal Coupling ❐ NR Limit Dirac Eq. From the deﬁnition we have the properties,

E < 0 Solutions µ 2 Charge Conjugation γ5,γ = 0, (γ5) = 1 Massless Spin 1/2 { } ❐ Now we deﬁne the 16 matrices of the basis

S P Γ =1 Γ = γ5 V A Γ = γµ Γ γ γµ µ µ ≡ 5 i ΓT = σ = [γ ,γ ] µν µν 2 µ ν Jorge C. Rom˜ao IDPASC School Braga – 28 IST Covariant Bilinears ...

❐ The labels S, V , T , A and P have the meaning of Scalar, Vector, Tensor, Summary Axial-Vector and Pseudo-Scalar and have to do with the transformations Introduction

Klein-Gordon Eq. properties, under Lorentz transformations, of the bilinears

Dirac Equation a Covariance Dirac Eq. ψ Γ ψ a = S,V,T,A and P • γ matrices • Equivalence Transf. • Covariance Proof ❐ As an example • Parity • Bilinears ′ ′ A ′ ′ ′ ′ µ ′ ′ Free Particle Solutions ψ (x ) Γ ψ (x ) =ψ (x )γ5γ ψ (x ) Minimal Coupling −1 µ NR Limit Dirac Eq. =ψ(x)S γ5γ Sψ(x) E < 0 Solutions µ ν Charge Conjugation =det a a νψ(x)γ5γ ψ(x) Massless Spin 1/2 where we have used [S,γ5] = 0 for proper Lorentz transformations (rotations and boosts) and ,γ = 0 for the space inversion (parity). This shows {P 5} that ψ(x)γ5γµψ(x) is an axial-vector

Jorge C. Rom˜ao IDPASC School Braga – 29 IST Plane Wave Solutions

❐ From now on we take a system of units where ¯h = c = 1. The Dirac Summary equation then reads Introduction

Klein-Gordon Eq. (i∂/ m)ψ(x) = 0 Dirac Equation − Covariance Dirac Eq. Free Particle Solutions ❐ This equation has plane wave solutions • Plane Waves • Rest Frame • µ Arbitrary Frame −ipµx • Spin ψ(x) = w(~p)e • Wave Packets µ 2 Minimal Coupling if pµp = m . NR Limit Dirac Eq. 0 2 2 2 E < 0 Solutions ❐ This implies (p ) = E = ~p ~p + m showing that we have negative energy · Charge Conjugation solutions Massless Spin 1/2 ❐ We always take p0 = E = ~p 2 + m2 > 0 | | p µ ψr(x) = wr(~p)e−iεrpµx

where εr = 1 for the positive and negative energy solutions, respectively ±

Jorge C. Rom˜ao IDPASC School Braga – 30 IST Plane Waves ...

❐ We start by ﬁnding wr(~p) in the rest frame of the particle. Dirac equation Summary reduces then to Introduction

Klein-Gordon Eq. 0 ∂ Dirac Equation iγ m ψ = 0 ∂t − Covariance Dirac Eq. Free Particle Solutions • Plane Waves ❐ • Rest Frame Using • Arbitrary Frame • Spin r r −iεrmt • Wave Packets ψ = w (0)e Minimal Coupling NR Limit Dirac Eq. we have E < 0 Solutions Charge Conjugation 0 r +1 r = 1, 2 0 +1 0 m εrγ 1 ψ = 0, εr = , γ = Massless Spin 1/2 − 1 r = 3, 4 0 1 − − ❐ With the solutions (N = √2m)

1 0 0 0 0 1 0 0 w(1)(0) = N ,w2(0) = N ,w3(0) = N ,w4(0) = N  0   0   1   0  0 0 0 1         Jorge C. Rom˜ao      IDPASC School Braga – 31 IST Plane Waves ...

❐ To get wr(~p) we perform a Lorentz boost to the instantaneous frame that Summary moves with velocity ~v, that is Introduction − Klein-Gordon Eq. tanh ω = ~v = β cosh ω = γ, sinh ω = γβ Dirac Equation | | → Covariance Dirac Eq. Free Particle Solutions ❐ We get then • Plane Waves • Rest Frame • 1 Arbitrary Frame r − 2 ~ω·~α r • Spin w (~p) =e w (0) • Wave Packets ω ω r Minimal Coupling = cosh 1 ωˆ ~α sinh w (0) NR Limit Dirac Eq. 2 − · 2 0 h i E < Solutions 1 r = (εrp/ + m)w (0) Charge Conjugation √2m√E + m Massless Spin 1/2 and 1 wr†(~p) =wr†(0)(p/γ0 + m) √2m√E + m

r r 1 w (~p) =w (0)(εrp/ + m) √2m√E + m

Jorge C. Rom˜ao IDPASC School Braga – 32 IST Plane Waves ...

Summary Using these expressions we can derive the following relations: Introduction

Klein-Gordon Eq.

Dirac Equation r (p/ εrm)w (~p) = 0 Covariance Dirac Eq. − Free Particle Solutions r w (~p)(p/ εrm) = 0 • Plane Waves − • Rest Frame ′ • Arbitrary Frame r r ′ • Spin w (~p)w (~p) = 2m δrr εr • Wave Packets

Minimal Coupling 4 r r NR Limit Dirac Eq. εrwα(~p)wβ(~p) = 2m δαβ E < 0 Solutions r=1 Charge Conjugation X ′ r† r Massless Spin 1/2 w (εr~p)w (εr′ ~p) = 2E δrr′

❐ wr(~p)wr(~p) and ψψ are scalars

r† r † ❐ w (εr~p)w (εr~p) and ψ ψ are the time component of a 4-vector.

Jorge C. Rom˜ao IDPASC School Braga – 33 IST Spin of the Dirac Equation

❐ The Poincar´egroup, translations, P µ, plus Lorentz transformations, J µν, Summary 2 µ 2 µ has two invariants P = PµP and W = WµW where Wµ is the Introduction

Klein-Gordon Eq. Pauli-Lubanski 4-vector,

Dirac Equation 1 νρ σ Covariance Dirac Eq. Wµ = εµνρσJ P Free Particle Solutions −2 • Plane Waves • Rest Frame • Arbitrary Frame ❐ One can show that their eigenvalues are • Spin • Wave Packets 2 2 2 2 Minimal Coupling P = m , W = m s(s + 1) NR Limit Dirac Eq. − E < 0 Solutions where m is the mass and s the spin (integer or half-integer) Charge Conjugation Massless Spin 1/2 ❐ For the Dirac equation and for inﬁnitesimal transformations

′ i µν ′ ′ i µν ψ (x) 1 Jµνω ψ(x), ψ (x ) = 1 σµνω ψ(x) ≡ − 2 − 4 leading to 1 Jµν = σµν + i(xµ∂ν xν ∂µ) 2 − Jorge C. Rom˜ao IDPASC School Braga – 34 IST Spin of the Dirac Equation ...

❐ Using this Jµν into Wµ we get Summary

Introduction i νρ σ Klein-Gordon Eq. Wµ = εµνρσσ ∂ −4 Dirac Equation

Covariance Dirac Eq. ❐ We can easily show that for the Dirac equation we have, Free Particle Solutions • Plane Waves • Rest Frame 2 3 2 • Arbitrary Frame W = m • Spin −4 • Wave Packets

Minimal Coupling 1 which conﬁrms that the Dirac equation describes spin s = 2 . NR Limit Dirac Eq. E < 0 Solutions ❐ For the plane wave solutions we obtain Charge Conjugation Massless Spin 1/2 1 νρ σ 1 Wµ = εr εµνρσσ p = γ [γµ, p/] εr −4 −4 5

❐ In the rest frame

W~ 1 ~σ 0 W 0 = 0 , = Σ~ ε with Σ~ = m 2 r 0 ~σ

Jorge C. Rom˜ao IDPASC School Braga – 35 IST Spin of the Dirac Equation ...

❐ The spin operator in an arbitrary direction is Summary

Introduction W s 1 µ µ Klein-Gordon Eq. · = γ s/p/εr, s sµ = 1, pµs = 0 − m 2m 5 − Dirac Equation

Covariance Dirac Eq. ❐ µ Free Particle Solutions Using this operator and choosing s = (0, 0, 0, 1) in the rest frame we get • Plane Waves • Rest Frame • Arbitrary Frame • Spin 0 1 0 σ3 • Wave Packets 1 1 3 1 γ5s/p/εr = γ5γ = Minimal Coupling 2m − 2 −2 " 1 0 #" σ3 0 # NR Limit Dirac Eq. − E < 0 Solutions 1 Charge Conjugation 2 1 1 Massless Spin 1/2 σ3 0   = 2 = − 2 1 1 " 0 σ3 #   − 2  − 2     1   2    ❐ This shows that wr(0) are the eigenstates with eigenvalues 1/2 along the z axis: +1/2 for r = 1, 4 and 1/2 for r = 2, 3 ± −

Jorge C. Rom˜ao IDPASC School Braga – 36 IST Spinors u and v

❐ Instead of using the spinors wr(p) it is conventional to use different names Summary for the positive energy solutions, u(p,s) and for the negative energy, v(p,s) Introduction Klein-Gordon Eq. ❐ The u spinors satisfy Dirac Equation Covariance Dirac Eq. ~ Free Particle Solutions (p/ m)u(p,s) = 0 and in rest frame Σ ~su(p,s) = u(p,s) • Plane Waves − · • Rest Frame • Arbitrary Frame ❐ • Spin The v spinors satisfy • Wave Packets Minimal Coupling (p/ + m)v(p,s) = 0 and in rest frame Σ~ ~sv(p,s) = v(p,s) NR Limit Dirac Eq. · − 0 E < Solutions showing that it has spin ~s in the rest frame. Charge Conjugation − Massless Spin 1/2 ❐ With these definitions we have the identification

1 w (~p) =u(p,sz) 2 w (~p) =u(p, sz) − 3 w (~p) =v(p, sz) − 4 w (~p) =v(p,sz)

Jorge C. Rom˜ao IDPASC School Braga – 37 IST Spinors u and v . . .

❐ The explicit expressions for u and v are (Dirac representation) Summary

Introduction ~σ·~p Klein-Gordon Eq. χ(s) χ( s) E+m − Dirac Equation u(p,s) = √E + m v(p,s) = √E + m  ~σ·~p χ(s)  χ( s)  Covariance Dirac Eq. E+m − Free Particle Solutions • Plane Waves     • Rest Frame where χ(s) is a Pauli two component spinor. • Arbitrary Frame • Spin • Wave Packets ❐ As an example Minimal Coupling p− NR Limit Dirac Eq. E+m E < 0 Solutions ~σ·~ρ pz Charge Conjugation E+m χ( ) E+m ↓ −  4 Massless Spin 1/2 v(p, ) = √E + m = √E + m = w (~p) ↑  χ( )   0  ↓        1      where p− = px ipy. −

Jorge C. Rom˜ao IDPASC School Braga – 38 IST Wave Packets

❐ Dirac equation is linear so we can form superpositions of plane wave Summary solutions, the so-called wave-packets Introduction Klein-Gordon Eq. ❐ Consider one formed only by positive energy solutions Dirac Equation Covariance Dirac Eq. 3 (+) d p 1 −ip·x Free Particle Solutions ψ (x) = 3 b(p,s)u(p,s)e • Plane Waves (2π) 2E ±s • Rest Frame Z • Arbitrary Frame X • Spin • Wave Packets ❐ We can show that Minimal Coupling 3 NR Limit Dirac Eq. d p 1 d3xψ†(x)ψ(x) = b(p,s) 2 = 1 E < 0 Solutions (2π)3 2E s | | Charge Conjugation Z Z X Massless Spin 1/2 and

d3p 1 ~p ~p J~(+) = b(p,s) 2 =< > (2π)3 2E E E s | | Z X

❐ ~p < E > is the group velocity, so we recover a familiar result from wave packets in non-relativistic quantum mechanics

Jorge C. Rom˜ao IDPASC School Braga – 39 IST Wave Packets ...

❐ However the complete set of solutions must include also the negative energy Summary solutions. Therefore Introduction Klein-Gordon Eq. 3 d p 1 −ip·x ∗ ipx Dirac Equation ψ(x) = b(p,s)u(p,s)e + d (p,s)v(p,s)e (2π)3 2E Covariance Dirac Eq. Z s Free Particle Solutions X h i • Plane Waves • Rest Frame ❐ The probability gives • Arbitrary Frame • Spin 3 • d p 1 Wave Packets d3xψ†ψ = b(p,s) 2 + d(p,s) 2 = 1 Minimal Coupling (2π)3 2E | | | | Z Z s NR Limit Dirac Eq. X h i E < 0 Solutions and the current, (p˜ (p0, ~p)) Charge Conjugation ≡ − Massless Spin 1/2 d3p 1 pk J k = d3xψγkψ = b(p,s) 2 + d(p,s) 2 (2π)3 2E | | | | E Z Z s The cross terms oscillate very rapidly n X h i with frequencies ω = 2E > 2m + i b∗(˜p, s′)d∗(p,s)e2iEtu(˜p, s′)σk0v(p,s) 1 MeV > 1.5 1021 s−1. They be-≃ s,s′ come important× if we try to localize X the electron at distances of order of its ′ −2iEt k0 λ = 1 4 10−11 cm i b(˜p,s )d(p,s)e v(p,s)σ u(˜p, s) c m − ≃ × s,s′ X o Jorge C. Rom˜ao IDPASC School Braga – 40 IST Minimal Coupling

❐ The interaction of charged particles with the electromagnetic ﬁeld is given by Summary the minimal coupling prescription, Introduction Klein-Gordon Eq. µ µ µ p p eA , or ∂µ ∂µ + ieAµ, e = e Dirac Equation −→ − −→ −| | Covariance Dirac Eq. Free Particle Solutions ❐ Therefore we get for the Dirac equation Minimal Coupling • Minimal Coupling µ µ (iγ ∂µ eγ Aµ m) ψ(x) = 0 NR Limit Dirac Eq. − − E < 0 Solutions Charge Conjugation ❐ In the original form Massless Spin 1/2 ∂ψ i = i~α (~ ieA~) + βm + eA0 ψ (H + H′)ψ ∂t − · ∇ − ≡ 0 h i where H = i~α ~ + βm, H′ = e~α A~ + eA0 0 − · ∇ − · ❐ Notice the analogy ′ ~v 0 Hclassic = e A~ + eA ~vop = c~α − c · → Jorge C. Rom˜ao IDPASC School Braga – 41 IST Non-relativistic Limit of Dirac Equation: Free Particle

❐ We start by the free particle, deﬁning Summary Introduction ϕˆ Klein-Gordon Eq. ψ = χˆ Dirac Equation Covariance Dirac Eq. where χˆ and ϕˆ are two component Pauli spinors. Free Particle Solutions Minimal Coupling ❐ Using ~α and β in the Dirac representation we get NR Limit Dirac Eq. • Free Particle • Pauli Equation ∂ϕˆ ~ • − 2 i = i~σ χˆ + mϕˆ g ∂t − · ∇ E < 0 Solutions  ∂χˆ Charge Conjugation  i = i~σ ~ ϕˆ mχˆ Massless Spin 1/2 ∂t − · ∇ −  ❐ In the non-relativistic limit E m m and therefore we make the redeﬁnition − ≪

ϕˆ ϕ = e−imt χˆ χ

Jorge C. Rom˜ao IDPASC School Braga – 42 IST Non-relativistic Limit of Dirac Equation: Free Particle ...

❐ Substituting back we get for the redeﬁned spinors Summary Introduction ∂ϕ i = i~σ ~ χ Klein-Gordon Eq. ∂t − · ∇ Dirac Equation  i∂χ = i~σ ~ ϕ 2mχ Covariance Dirac Eq.  ∂t − · ∇ − Free Particle Solutions  Minimal Coupling ❐ As χ changes slowly we solve approximately the second equation, getting NR Limit Dirac Eq. • Free Particle • Pauli Equation ~σ ~ ~σ ~p • g − 2 χ i · ∇ϕ = · ϕ ϕ E < 0 Solutions ≃ − 2 m 2 m ≪

Charge Conjugation ❐ Massless Spin 1/2 Substituting in the first ∂ϕ 2 i = ∇ ϕ ∂t −2 m which is the Schrödinger equation for the free particle. Therefore in the non-relativistic limit the big components, ϕ, obey the Schrödinger equation and the small components, χ, are neglected. So the negative energy solutions, the small components, disappear in the non-relativistic limit.

Jorge C. Rom˜ao IDPASC School Braga – 43 IST Non-relativistic Limit of Dirac Equation: Pauli Equation

❐ We start now from Summary

Introduction ∂ψ 0 Klein-Gordon Eq. i = i~α ~π + βm + eA ψ, ~π ~ ieA~ ∂t − · ≡ ∇ − Dirac Equation Covariance Dirac Eq. ❐ Free Particle Solutions With the previous decomposition we get

Minimal Coupling ∂ϕ 0 NR Limit Dirac Eq. i = ~σ ~πχ + eA ϕ • Free Particle ∂t · • Pauli Equation  ∂χ 0 • g − 2 i = ~σ ~πϕ + eA χ 2 mχ  ∂t · − E < 0 Solutions Charge Conjugation  ❐ Assuming eA0 2 m, and slow time variation Massless Spin 1/2 ≪ ~σ ~π χ = · ϕ 2 m

❐ And for the big components

∂ϕ (~σ ~π)(~σ ~π) i = · · + eA0 ϕ ∂t 2 m

Jorge C. Rom˜ao IDPASC School Braga – 44 IST Non-relativistic Limit of Dirac Equation: Pauli Equation ...

❐ To understand the meaning of this equation we notice that Summary

Introduction (~σ ~π)(~σ ~π) = ~π ~π e~σ B~ Klein-Gordon Eq. · · · − · Dirac Equation Covariance Dirac Eq. ❐ Then we get Pauli equation for the electron Free Particle Solutions

Minimal Coupling ∂ϕ (~p eA~)2 e NR Limit Dirac Eq. i = − ~σ B~ + eA0 ϕ • Free Particle ∂t 2 m − 2 m · • Pauli Equation " # • g − 2

E < 0 Solutions Charge Conjugation ❐ Putting back ¯h and c we get Massless Spin 1/2 e¯h e¯h e ¯h~σ H = ~σ B~ ~µ B,~ with ~µ = ~σ = 2 mag −2mc · ≡ − · 2mc 2mc 2 which shows that gyromagnetic ratio is g = 2 ❐ Being able to predict the correct value for g was one of the biggest successes of the Dirac equation ❐ Higher order corrections (in QED and in the SM) deviate g from 2.

Jorge C. Rom˜ao IDPASC School Braga – 45 IST Non-relativistic Limit of Dirac Equation: Pauli Equation ...

❐ We deﬁne Summary Introduction g 2 a − Klein-Gordon Eq. ≡ 2 Dirac Equation

Covariance Dirac Eq. ❐ Free Particle Solutions The present situation for the electron is very precise (QED only)

Minimal Coupling th exp −14 NR Limit Dirac Eq. ae = ae = (115965218073 28) 10 • Free Particle ± × • Pauli Equation • g − 2 and we use it to determine the ﬁne structure constant. E < 0 Solutions ❐ Charge Conjugation For the muon there is a 2σ diﬀerence

Massless Spin 1/2 ath =(116591841 81) 10−11 µ ± × aexp =(116592080 58) 10−11 µ ± ×

Jorge C. Rom˜ao IDPASC School Braga – 46 IST Dirac Hole Theory

❐ Despite all the successes of the Dirac equation the interpretation of the Summary negative energy solutions still remains to be addressed Introduction Klein-Gordon Eq. ❐ This is not an academic problem. We have to explain why the electrons do Dirac Equation not make transitions to the negative energy states. A simple calculation Covariance Dirac Eq. gives a rate of 108 s−1 to decay into the energy interval [ mc2, 2mc2] Free Particle Solutions − − Minimal Coupling ❐ Dirac hole theory makes use of the Pauli exclusion principle and deﬁnes the NR Limit Dirac Eq.

E < 0 Solutions vacuum as the state with all the negative energy states filled. So, no • Dirac Hole Theory transitions are allowed from states with E > 0 and atoms are stable • Anti-particles Charge Conjugation ❐ Of course the vacuum has infinite energy and charge but as we only measure Massless Spin 1/2 differences with respect to the vacuum this is not a problem ❐ The second quantization allows for a better understanding of this question and allows for the extension also to bosons ❐ The main consequence is the prediction of anti-particles. Consider that the vacuum has a hole. This means the absence of an electron with negative energy E and charge e . But this can be interpreted as the presence of a particle− with positive energy−| | +E and charge + e , the positron | |

Jorge C. Rom˜ao IDPASC School Braga – 47 IST Dirac Hole Theory: pair creation and annihilation

E e− E e− Summary mc2 mc2 Introduction

Klein-Gordon Eq. 0 γ 0 γ mc2 mc2 Dirac Equation − − Covariance Dirac Eq.

Free Particle Solutions hole hole

Minimal Coupling

NR Limit Dirac Eq. ❐ E < 0 Solutions Pair creation: An electron is excited from a negative energy state into a • Dirac Hole Theory positive energy state. It leaves behind a hole interpreted as the positron. • Anti-particles Therefore it was created a pair e+e− Charge Conjugation Massless Spin 1/2 ❐ Pair annihilation: An electron with E > 0 makes a transition to a non-occupied E < 0 state, the hole corresponding to the positron. In the end both electron and positron disappear with the emission of radiation ❐ With Dirac hole theory we abandon the wave function interpretation as we can vary the number of particles. Only the second quantized creation and annihilation operators can provide a consistent description ❐ However, Dirac interpretation was a great success conﬁrmed by the discovery of the anti-particles in the following years Jorge C. Rom˜ao IDPASC School Braga – 48 IST Charge Conjugation

❐ From Dirac hole theory emerges a new symmetry of Nature: for each particle Summary there exists an anti-particle. We call this symmetry Charge Conjugation. We Introduction

Klein-Gordon Eq. now will deﬁne it more precisely. Dirac Equation ❐ We should have a correspondence between the negative energy solutions of Covariance Dirac Eq.

Free Particle Solutions Dirac equation for electrons,

Minimal Coupling NR Limit Dirac Eq. (i∂/ eA/ m)ψ = 0 E < 0 Solutions − − Charge Conjugation and the positive energy solutions of Dirac equation for positrons Massless Spin 1/2

(i∂/ + eA/ m)ψc = 0 −

where ψc is the wave function for the positron. ❐ We take the complex conjugate of the ﬁrst equation

∗ ∗ µ µ ∗ ( iγ ∂µ eγ Aµ m)ψ = 0 − − −

Jorge C. Rom˜ao IDPASC School Braga – 49 IST Charge Conjugation ...

T ∗ ❐ 0T ∗ 0T µ 0T µT Summary Using the relations γ ψ = ψ and γ γ γ = γ we get

Introduction µT T Klein-Gordon Eq. γ (+i∂µ + eAµ) m ψ = 0 Dirac Equation − − Covariance Dirac Eq. ❐ If we ﬁnd a matrix C such that Free Particle Solutions

Minimal Coupling µT −1 µ NR Limit Dirac Eq. Cγ C = γ E < 0 Solutions − Charge Conjugation we can identify Massless Spin 1/2 T ψc Cψ ≡

❐ This can be shown by explicit construction. In the Dirac representation

C = iγ2γ0 = C−1 = C† = CT − − − or 0 0 0 1 0 iσ2 0 0 1− 0 C = = iσ2 −0 0 1 0 0  − − 1 0 0 0    Jorge C. Rom˜ao   IDPASC School Braga – 50 IST Charge Conjugation ...

❐ As an example, let us consider a negative energy electron with spin down Summary Introduction 0 Klein-Gordon Eq. 0 imt Dirac Equation ψ = N   e Covariance Dirac Eq. 0

Free Particle Solutions  1    Minimal Coupling   NR Limit Dirac Eq. ❐ Then ψc corresponds to a positive energy positron with spin up E < 0 Solutions Charge Conjugation 00 01 0 1 Massless Spin 1/2 T 0 0 1 0 0 0 ψ = Cψ = Cγ0ψ∗ = N e−imt = N e−imt c 0 1− 0 0  0   0  − 10 00  1   0              ❐ For an arbitrary wave function we can show

εp/ + m 1 + γ5s/ εp/ + m 1 + γ5s/ ψ = ψ ψc = − ψc 2m 2 → 2m 2 that is, inverts the sign of energy and the direction of the spin

Jorge C. Rom˜ao IDPASC School Braga – 51 IST Massless Spin 1/2 Particles

Summary ❐ We always considered the Dirac equation for a massive electron. However, Introduction we could put the question, is there in Nature any massless spin 1/2 particle? Klein-Gordon Eq. Dirac Equation ❐ For many years the Standard Model of particle physics assumed that this Covariance Dirac Eq. was the case for neutrinos. Today we know that they have a very small mass Free Particle Solutions (less than an eV), but they are not massless Minimal Coupling NR Limit Dirac Eq. ❐ Despite this, massless spin 1/2 particles can be very useful. For instance, in E < 0 Solutions many situations we can neglect the electron mass compared with the Charge Conjugation

Massless Spin 1/2 energies in the center of mass of the process. For more reason the same is • Weyl Equation true for neutrinos • Massless 4-Spinors ❐ Also, gauge theories are formulated in terms of massless particles, before the spontaneous breaking of the gauge theory ❐ All these are good reasons to look at massless spin 1/2 particles and we will see that some surprises appear

Jorge C. Rom˜ao IDPASC School Braga – 52 IST Description in terms of 2-spinors: Weyl Equation

❐ For the massless case Dirac equation reads Summary Introduction ∂ψ Klein-Gordon Eq. i = i~α ~ ψ ∂t − · ∇ Dirac Equation Covariance Dirac Eq. ❐ Free Particle Solutions We see that the β matrix disappears. This has the consequence that the

Minimal Coupling relations

NR Limit Dirac Eq. i j j i ij E < 0 Solutions α α + α α = 2δ Charge Conjugation Massless Spin 1/2 can be veriﬁed for 2 2 matrices, for instance the Pauli matrices. There are • Weyl Equation × • Massless 4-Spinors two possible choices

~α = ~σ ± ❐ Let us consider plane wave solutions

ψ = χ(p,s)e−ip·x ~σ ~pχ(p,s) = Eχ(p,s) → ± ·

Jorge C. Rom˜ao IDPASC School Braga – 53 IST Description in terms of 2-spinors: Weyl Equation ...

Summary ❐ Consider ﬁrst the case α = +~σ. We take ~p along the positive z direction. Introduction

Klein-Gordon Eq. Dirac Equation 1 ~σ.~p χ(p, +) = χ(p, +) = +χ(p, +) Covariance Dirac Eq. 0 → ~p Free Particle Solutions | | Minimal Coupling where we used ( ~p = E). This solution corresponds to positive helicity NR Limit Dirac Eq. | | E < 0 Solutions ❐ If we consider the case ~α = ~σ we get Charge Conjugation −

Massless Spin 1/2 • Weyl Equation 0 ~σ.~p • Massless 4-Spinors χ(p, ) = χ(p, ) = χ(p, ) − 1 → ~p − − − | | corresponding to negative helicity

Jorge C. Rom˜ao IDPASC School Braga – 54 IST Description in terms of 4-Spinors

❐ Although it is enough to use 2-component spinors for massless spin 1/2 Summary particles in many applications it is convenient to use 4-component spinors Introduction Klein-Gordon Eq. ❐ For this it is convenient to choose the Weyl representation for the γ matrices Dirac Equation

Covariance Dirac Eq. ~σ 0 0 0 1 1 0 Free Particle Solutions ~α = ; β = γ = ; γ = 0 ~σ 1− 0 5 0 1 Minimal Coupling − − − NR Limit Dirac Eq. E < 0 Solutions ❐ If we write Charge Conjugation Massless Spin 1/2 χ(+) • Weyl Equation ψ = • χ( ) Massless 4-Spinors − we get ∂ i χ(+) = i~σ ~ χ(+) mχ( ) ∂t − · ∇ − − ∂ i χ( ) =i~σ ~ χ( ) mχ(+) ∂t − · ∇ − − showing that the equations are coupled by the mass term

Jorge C. Rom˜ao IDPASC School Braga – 55 IST Description in terms of 4-Spinors

Summary ❐ In the limit m 0 the two equations become independent and they are just Introduction the two Weyl equations→ for the cases ~α = ~σ Klein-Gordon Eq. ± Dirac Equation ❐ We notice also that Covariance Dirac Eq.

Free Particle Solutions

Minimal Coupling

NR Limit Dirac Eq. χ(+) 0 γ5ψ( ) = ψ( ) ψ(+) = ; ψ( ) = E < 0 Solutions ± ± ± 0 − χ( ) − Charge Conjugation Massless Spin 1/2 and therefore • Weyl Equation • Massless 4-Spinors

1 + γ χ(+) 1 γ 0 5 ψ = , − 5 ψ = 2 0 2 χ( ) − ❐ This shows that chirality and helicity are the same thing in the limit of massless particles

Jorge C. Rom˜ao IDPASC School Braga – 56