Elementary Particle Physics Quantum Electro Dynamics QED Mark
- Slides: 130
Elementary Particle Physics Quantum Electro. Dynamics (QED) Mark Thomson: Chapters 4, 5 & 6 Lectures 4, 5, 6, 7, 8 Halzen&Martin: Frank Linde, Nikhef, H 044, f. linde@nikhef. nl, 020 -5925140 Chapter 4
QED: Klein-Gordon equation Dirac equation: free particles Dirac equation: interactions + ee + cross section
Klein Gordon equation
Lorentz invariant Schrödinger eqn. ? With the quantum mechanical energy & momentum operators: You simply ‘derive’ the Schrödinger equation from classical mechanics: Schrödinger equation With the relativistic relation between E, p & m you get: Klein-Gordon equation
Probability & current densities Use 4 -derivatives to make Klein-Gordon equation Lorentz invariant: =0 ‘Simple’ plane-wave solutions for : Analogous to classical QM, you can derive probability & current densities corresponding to the Klein-Gordon equation: Gives you for KG probability & current densities:
KG-equation , problem? ‘Surprise’ of the plane-wave if you plug them in the KG equation you find: solutions, solutions with E<0 solutions with <0 One way out: drop KG equation! That is what Dirac successfully did! Other way out: re-interpret in terms of charge density & charge flow: q>0 particles q<0 particles In reality (electrons negatively charged) just the opposite way …: particles with anti-particles with First hint of anti-particles! & Multi-particle theory
Particles & anti-particles e ( E, p) time absorption +e emission (+E, +p) As far as this system is concerned: emission: equivalent to: absorption: e with p =( E, p) e+ with p =(+E, +p) Expressed in terms of the charge/current densities: e+ ( E, p) e 2 2 (+E, p) Interpretation of Pauli & Weisskopf / Stückelberg & Feynman
One vertex two vertices e 1 vertex: e e+ negative energy e backwards in time e e e e+ 2 vertices: e Intermediate state: e e e Intermediate state: e e+e
Electrons & photons recap of the essentials of electro-magnetism
Maxwell eqn. & Lorentz invariance Inhomogeneous (sources!) homogeneous Maxwell eqn. : (HL units h=c=1) Maxwell eqn. with 4 -potential A =(V, A) & 4 -current j =( , j) potentials: And ‘nicer’ with F A A : Inhomogeneous terms: in terms of A :
Maxwell eqn. & gauge invariance 4 -potential A is not unique (E & B determine the physics!) Because E & B remain the same: Use gauge invariance to simplify A e. g. by imposing: , Lorentz gauge, co-variant A is still not unique: Could e. g. require: , Coulomb gauge, not co-variant
Free photon wave functions j =0 in vacuum & solving becomes easy: Plane waves do the job: photons ( ) are massless! with And in the Lorentz gauge: Coulomb gauge: Hence only 2 of the original 4 degrees of freedom in survive! transversal (assuming p // z-axis) circular
Vertices & 4 -momentum conservation time pi pf k T e e p pi T p+ k p+ p T e e T p p+ e+ e p+ pf pf k e e+ i k e+
Interactions of photons with KG field Klein-Gordon equation without interactions: + = = NOT EXPLAINED IN THIS COURSE Klein-Gordon equation with interaction via ‘minimal substitution’ (electron charge: -e) Yields: higher orders, ignore for now (later relevant in view of gauge invariance)
Interactions of photons with KG field Plug V in the expression for Tfi: partial integration Define ‘transition current’ jfi’: (recall: = Ne ip x) To obtain:
Example: K K scattering K Assume scatters from an A caused by K How do you find this A ? Ansatz: find solution of Maxwell equation with as current source the transition current belonging to K (A) (C) Solve for A : Simple!: (D) K (B) How to handle? 1 And the transition matrix element Tfi: note that Tfi is symmetric in j. AC & j. BD 1 p. C p. A 1
K scattering: amplitude (D) K (B) And using the earlier introduced calculation of the transition probability Wfi in terms of Tfi: K (A) You get for the amplitude (q p. D-p. B=p. A-p. C): 1 (C)
K scattering: Standard expression for d A+B C+D d cross section (D) K (B) in c. m. frame: K A B (A) (C) In the c. m. frame: & Assume particle masses to be << particle energies: And hence:
K scattering: Feynman rules (D) K (B) K The matrix element contains several factors which each can be associated with parts of the diagram to give so-called Feynman rules: (A) (C) K - vertex - propagator - vertex
Feynman rules S=0 electrodynamics 20 p 1 in: out: External lines k p Vertices: p p’ k k Propagators: q p’ k’ q
+ + KK scattering Just follow the rules: p. B Feynman diagram(s): p. A anti-particles time reversed particles amplitude: K+ + q=p. A+p. B =p. C+p. D p. B + p. A K K+ q=p. A+p. B =p. C+p. D K p. D p. C
+ + KK scattering Just follow the rules: p. B + Feynman diagram(s): anti-particles time reversed particles amplitude: p. A A D relativistic limit (or set particle masses to zero): q=p. A+p. B =p. C+p. D K p. C C pick frame (c. m. ) & make 4 -vectors explicit: standard d /d expression: K+ B
+ p. B + p. D q=p. A-p. C =p. D-p. B (a) p. A p. B + p. A Feynman diagram(s): p. D p. C p. B + p. A q=p. A+p. B p. C =p. C+p. D (b) p. C q=p. A-p. C =p. D-p. B (a) scattering + + p. D anti-particles time reversed particles p. B (b) p. A + + p. D q=p. A+p. B p. C =p. C+p. D
+ p. D + p. B q=p. A-p. C =p. D-p. B (a) p. A amplitude: etcetera! p. C scattering + p. B (b) p. A + + p. D q=p. A+p. B p. C =p. C+p. D
* Feynman diagram(s): p k q=p+k =p’+k’ (a) direct interaction amplitude (a): scattering p k’ p’ q=p-k’ =p’-k k k’ p (c) p’ (b) p’ exchange interaction k k’ contact interaction k = k’ ’ = 0
* Feynman diagram(s): p k q=p+k =p’+k’ (a) direct interaction amplitude (b): scattering p k’ p’ q=p-k’ =p’-k k k’ p (c) p’ (b) p’ exchange interaction k k’ contact interaction k = k’ ’ = 0
* Feynman diagram(s): p k q=p+k =p’+k’ (a) direct interaction amplitude (c): scattering p k’ p’ q=p-k’ =p’-k k k’ p (c) p’ (b) p’ exchange interaction k k’ contact interaction
* + k, p q=p-k =k’-p’ scattering (a) k’, ’ direct interaction amplitude (a): p k, q=p-k’ =k-p’ , p’ -p’ (b) k’, ’ exchange interaction , p’ anti-particle k, p -p’ particle (c) k’, ’ contact interaction
* + k, p q=p-k =k’-p’ scattering (a) k’, ’ direct interaction amplitude (b): p k, q=p-k’ =k-p’ , p’ -p’ (b) k’, ’ exchange interaction , p’ anti-particle k, p -p’ particle (c) k’, ’ contact interaction
* + k, p q=p-k =k’-p’ scattering (a) k’, ’ direct interaction amplitude (c): p k, q=p-k’ =k-p’ , p’ -p’ (b) k’, ’ exchange interaction , p’ anti-particle k, p -p’ particle (c) k’, ’ contact interaction
Recap Klein Gordon equation
Free Klein-Gordon particle wave functions With the quantum mechanical energy & momentum operators: non-relativistic yields Schrödinger equation: or With free particle plane-wave solutions and conserved current:
Free photon wave functions Maxwell equations for A in vacuum (Lorentz gauge, no sources): With plane-wave solutions: photons ( ) are massless! with particles & photons (do not mind Lorentz-index )
Interaction Klein-Gordon particles & photons e (& , K ) interactions with electromagnetic field (=photons) ‘turned on’ by ‘minimal substitution’ (technically: local gauge invariance): Note: electron (& , K ) charge: e Plugging this into Klein-Gordon equation, you find the perturbation V:
Interaction Klein-Gordon particles & photons For now: just look at one ‘leg’ K K K p. D p. B q K p. A define ‘transition current’: p. C Note: Tfi looks awful but with plane-waves for ’s it really is simple!
Interaction Klein-Gordon particles & photons K K K p. D p. B q K p. A Last step: must find expression for A Idea: assume the ’s to scatter from an A caused by the K’s transition current p. C Easy! Note: Tfi still looks awful but only plane-waves for ’s & photon = A !
Interaction Klein-Gordon particles & photons Tfi can now be calculated! K K K p. D p. B q K p. A p. C With this Tfi you calculate Wfi as usual and hence the cross section
Interaction Klein-Gordon particles & photons And finally the cross section K K K p. D p. B q K p. A p. C With matrix element or Feynman amplitude M: cross section CM frame: A C B D Some simple – Feyman – rules allow you to ‘easily’ calculate M
Feynman rules S=0 electrodynamics Feynman rules K K K p. D p. B q K p. A p. C
Feynman rules S=0 electrodynamics with anti-particles Feynman rules +K K + K p. D p. B q p. A +K p. C +
Feynman rules S=0 electrodynamics with anti-particles Feynman rules +K K K p. D p. B q p. A +K p. C Note: do not add particle/antiparticle labels since they are already implicit in the direction of the arrows!
Dirac equation: free particles
Schrödinger Klein-Gordon Dirac Quantum mechanical E & p operators: You simply ‘derive’ the Schrödinger equation from classical mechanics: Schrödinger equation With the relativistic relation between E, p & m you get: Klein-Gordon equation The negative energy solutions led Dirac to try an equation with first order derivatives in time (like Schrödinger) as well as in space Dirac equation
Does it make sense? Also Dirac equation should reflect: Basically ‘squaring’: Tells you: 2=1 i + i = 0 i j: i j + j i = 0
Properties of i and can not be simple commuting numbers, but must be matrices Because 2= i 2=1, both and must have eigenvalues 1 Since the eigenvalues are real ( 1), both and must be Hermitean Both and must be traceless matrices: Tr(ABC) = Tr(CAB) = Tr(BCA) 2=1 cyclic anti commutation 2=1 Tr( i ) = Tr( i ) and hence Tr( i )=0 You can easily show the dimension d of the matrices , to be even: d odd d even or: with eigenvalues 1, matrices are only traceless in even dimensions either:
Explicit expressions for i and In 2 dimensions, you find at most 3 anti-commuting matrices, Pauli spin matrices: In 4 dimensions, you can find 4 anti-commuting matrices, numerous possibilities, Dirac-Pauli representation: Any other set of 4 anti-commutating matrices will give same physics (if the Dirac equation is to make any sense at all of course …. . and … if it would not: we would not be discussing it here!)
Co-variant form: Dirac -matrices does not look that Lorentz invariant Multiplying on the left with and collecting all the derivatives gives: Hereby, the Dirac -matrices are defined as: And you can verify that: As well as: and:
Co-variant form: Dirac -matrices with the Dirac -matrices defined as:
Warning! This notation is misleading, is not a 4 -vector! The are just a set of four 4 4 matrices, which do no not transform at all i. e. in every frame they are the same, despite the index. The Dirac wave-functions ( or ), so-called ´spinors´ have interesting Lorentz transformation properties which we will discuss shortly. After that it will become clear why the notation with To make things even worse, we define: is useful! & beautiful!
Spinors & (Dirac) matrices this one we will encounter later …
Dirac current & probability densities Proceed analogously to Schrödinger & Klein-Gordon equations, but with Hermitean instead of complex conjugate wave-functions: Dirac equations for & : Add these two equations to get: Conserved 4 -current: (exactly what Dirac aimed to achieve …)
Solve by splitting 4 -component in two 2 -components: with 0 (1/c) t follows: solutions: e + e
Anticipate plane-waves: And again anticipate two 2 -components: Plugging this in gives: 0 k
Solutions: pick u. A(p) & calculate u. B(p): e E>0 Similarly: pick u. B(p) & calculate u. A(p): + e E<0
Elementary Particle Physics Quantum Electro. Dynamics (QED) Mark Thomson: Chapters 4, 5 & 6 Lectures 4, 5, 6, 7, 8 Halzen&Martin: Frank Linde, Nikhef, H 044, f. linde@nikhef. nl, 020 -5925140 Chapter 4
Recap introduction Dirac equation
Dirac equation From: & classical QM ‘transcription’: We found: With , 1, 2 & 3 (4 4) matrices, satisfying: ? 2=1 i 2=1 + = 0 i j: i j + j i = 0
Co-variant form: Dirac -matrices Dirac’s original form does not look covariant: Multiplying on the left with and collecting all the derivatives gives covariant form: With Dirac -matrices defined as: From the properties of , 1, 2 & 3 follows:
Anticipate plane-waves: And again anticipate two 2 -components: Plugging this in gives: 0 k
Solutions: pick u. A(p) & calculate u. B(p): e E>0 Similarly: pick u. B(p) & calculate u. A(p): + e E<0
Dirac particle solutions: spinors Dirac eqn. : e spin ½ electrons e+ E>0 spin ½ positrons E<0
Dirac equation: more on free particles normalisation 4 -vector current anti-particles
sorry for the c’s One more look at The conditions: Imply: 1 i. e. energy-momentum relation, as expected Check this:
Normalisation of the Dirac spinors Just calculate it!: Spinors 1 & 2, E>0: To normalize @ 2 E particles/unit volume Spinors 3 & 4, E<0: To normalize @ 2 E particles/unit volume
Current & probability densities Again, just plug it in! always using particle @ rest N moving particle N not that easy, next slide!
Current & probability densities Explicit verification of jx for moving particle solution (1): And jx for moving anti-particle solution (3):
Antiparticles
Surprising applications PET – Positron Emission Tomography
Particles & Anti-particles 4 -component Dirac spinors 4 -solutions. These represent: 2 spin states of the electron 2 spin states of the anti-electron i. e. the positron Different ways how to proceed: • Use E>0 & E<0 solutions of the electron Dirac eqn. • Use E>0 & E<0 solutions of the positron Dirac eqn. • Use E>0 solutions for the ‘particle’ i. e. electron & Use E<0 solutions for the ‘anti-particle’ i. e. positron Will opt for the last option: i. e. using the physical E & to characterize states
And now: E<0 antiparticles ‘Dirac sea’: fill all E<0 states (thanks to Pauli exclusion principle) single e But: e+e e +e does not work for bosons an infinite energy sea not a nice concept …
And now: E<0 antiparticles ‘Feynman-Stückelberg’: E<0 particle solutions propagating backwards in time E>0 anti-particle solutions propagating forwards in time e (E>0) e (E<0) ‘Up-shot’: Sequel: e (E>0) E =2 E time e+ (E>0) E =2 E Dirac equation accommodates both particle & antiparticles! will use particle & anti-particle spinors labelled with their physical, E>0 & real , kinematics. (exponents remain opposite)
We had: Dirac ‘u’-spinors e E>0 + e E<0
From now on use: Dirac ‘u’- & ‘v’-spinors u-spinors: for electrons, labeled with physical E>0 & e E>0 v-spinors: for positrons, labeled with physical E>0 & + e E>0
Dirac equation: more on free particles Spin Helicity Chirality
Dirac particles & spin As you might guess, the two-fold degeneracy is because of the spin=½ nature of the particles the Dirac equation describes! How do you see this? Use commutator with Hamiltonian to find conserved quantities First attempt: orbital angular momentum tells you: Second attempt: internal angular momentum tells you: + + total spin is conserved!
Dirac particles & spin Do we indeed describe particles with spin =½? Yes! For particles with p=0: can use ( 2 , 3) to classify states For particles with p 0 we can not use 3 , but we can use spin // p: Are you sure? Check it yourself! is called helicity with eigenvalues: ½ helicity + ½ helicity ½
* Helicity states right-handed helicity + ½ RH helicity ½ LH is called helicity with eigenvalues: ½ left-handed Instead of u 1 & u 2 spinors, we could use helicity ½ : u & u spinors (& similarly for v-spinors) You ‘simply’ solve the eigenvalue equation: Eigenvalues, use : With u. A, you get u. B using the Dirac eqn. as we did before (easier now = ½ as it should ):
* Helicity states right-handed helicity + ½ RH helicity ½ LH is called helicity with eigenvalues: ½ left-handed solving easiest using spherical coordinates: yields: with follows: or: For = +½:
* Helicity states is called helicity with eigenvalues: ½ Particles right-handed helicity + ½ RH helicity ½ LH left-handed Anti-particles Remark: we have used physical E & p for the v-spinors. Nevertheless: exponents still reflect negative energy (& momentum)! This means that the physical E, p and even helicity of v-spinors are obtained using the opposite of the operators used for u-spinors! Afteral: we are re-interpreting the unwanted negative energy solutions of the Dirac eqn. !
Chirality For massless & extremely relativistic particles, helicity states become simple: Particles Anti-particles These four states are also eigenstates of: Simple check: and: Eigenstates of 5 called: Left-handed (L) Right handed (R) chiral states. Weak interactions!
Elementary Particle Physics Quantum Electro. Dynamics (QED) Mark Thomson: Chapters 4, 5 & 6 Lectures 4, 5, 6, 7, 8 Halzen&Martin: Frank Linde, Nikhef, H 044, f. linde@nikhef. nl, 020 -5925140 Chapter 4
Recap Dirac equation spinors particles & anti-particles spin, helicity & chirality
Dirac equation
Spinors Dirac eqn. : e spin ½ electrons e+ E>0 spin ½ positrons E<0
Particles & anti-particles u-spinors: for electrons, labeled with physical E>0 & e E>0 v-spinors: for positrons, labeled with physical E>0 & + e E>0
Spin, helicity & chirality Spinors represent spin ½ states! Spin operator: Particles - helicity u h=+½ h= ½ u Anti-particles - helicity v v h= ½ h=+½
Spin, helicity & chirality Spinors represent spin ½ states! Spin operator: also commutes with Dirac Hamiltonian eigenvalues 1 chirality For massless & extremely relativistic particles, chirality & helicity eigenstates are: h=+½ h= ½ Particles - chirality 5: +1 5 : 1 u. R u. L Anti-particles - chirality v. R v. L 5 : 1 h= ½ 5: +1 h=+½
Dirac equation: more on free particles * transformation properties ! ! ! normalisation orthogonality completeness
* Transformation properties Of course, all observers use the same Dirac equation Explicitly for observers S & S’ and the transformation x’ = a x Free particle solution has Lorentz invariant part & 4 -component spinor part reasonable to assume that S(a) acts on. To ease notation: S(a) SL Tedious work: with finally follows:
* Space inversion parity Coordinate transformation for space inversion is: Hence for space inversion or parity operation SP: Meaning: does the job (verify!). And therefore: SP=SP†=SP 1 For the free Dirac spinor solutions you can verify: And hence particle & anti-particle solutions have opposite intrinsic parity
* Infinitesimal transformation Introducing Lorentz transformations, we saw: with = ‘Simply’ by checking, you can show that: not that simple at all! with: obeys: With this you get for infinitesimal transformations: And since: Hence: the same holds for finite transformations!
* By the way: these 16 combinations (1(S)+4(V)+6(T)+1(P)+4(A)) exhaust the 16 number of possibilities! Finally we can justify the ‘ 4 -vector’ Using: and easy to show that: notation! With: Hence: it follows that: &
* Boosts Recall the ‘simple’ Lorentz transformation for boost along Z-axis: 03 = 30 = /N and & Use = 03 = ( 0 3 3 0)= i 3 gives spinor transformation: Infinitesimal: finite: please: you check! You find: With some goniometry:
Useful relations: normalisation Explicitly for u 1: Explicitly for v 1:
Useful relations: orthogonality Easy, just do it!
Useful relations: completeness realize: 4 4 matrices! with: you get: Key! using: done: Check order of the terms! (I cheated a bit, but OK)
Useful relations: completeness realize: 4 4 matrices! remark: completeness relations will turn out to be very useful! why? because in real experiments the spin states of incoming & outgoing particles remain unknown hence: - we sum over all the spin states of outgoing particles - we average over all the spin states of incoming particles ‘easy’ using the completeness relations!
Elementary Particle Physics Quantum Electro. Dynamics (QED) Mark Thomson: Chapters 4, 5 & 6 Lectures 4, 5, 6, 7, 8 Halzen&Martin: Frank Linde, Nikhef, H 044, f. linde@nikhef. nl, 020 -5925140 Chapter 4
Dirac equation: free particles
covered and are you able to boost & rotate spinors? Spin – Helicity – Chirality – Completeness +½ ½ Normalisation – Orthogonality
covered and are you able to boost & rotate spinors? Spin – Helicity – Chirality – Completeness +½ ½ Normalisation – Orthogonality
Dirac equation: interactions
Physicist’s prime process Future: ILC, CLIC, … (250 -500 -3000 Ge. V) e +e +
International Linear Collider (Japan? )
Interactions of photons with Dirac field Dirac eqn. without interactions (E>0): As for KG eqn. , put photons (A ) in via minimal substitution (qe=-e): p p + e. A compensates 0 in front of / t term with As for KG eqn. , you can find the transition amplitude Tfi in terms of a ‘transition current’: V 1 uf KG, S=0 ui Dirac, S=1/2 With: time 1 f i Compare:
e B p. A A D muon electron scattering Just like for K K scattering for S=0 p. D q e AC transition current: BD transition current: p. C Scatter electron (muon) from A calculated from 1 C ( (q=p. A p. C=p. D p. B) Transition amplitude Tfi becomes: 1 1 1 time )
Feynman rules S=½ electrodynamics External lines: Vertex: ie E. g. for e e you find: Propagators: q q
Key processes in QED time Möller scattering Compton scattering Bhabha scattering pair annihilation pair creation
Real life examples: LEP + ee
Real life examples: LEP + ee
Real life examples: LEP + ee
Real life examples: LEP detector + ee particle identification
Real life examples: LEP + ee
Normalization process For any measured process = Nevents/Normalisation = Accelerator: ‘integrated luminosity’ Experiment: ‘acceptance’ Luminosity monitor measures ‘simple’ well-known process elastic e+e scattering e+ e Characteristics: Ø ‘back-to-back’ Ø energy = Ebeam Ø mostly @ small angles e+ e+ e e e+ e+ e+ e
Measured distributions Max(E+ , E ) Min(E+ , E ) + For any other process: Determined by e e = Constant Nevents/Nluminosity 1. cross section (physics) & 2. acceptance (experiment)
QED @ work: higher order processes
Now towards real calculations For most experiments: spins of incoming & outgoing particles remain unobserved Note: squares because spin states can be measured What does this mean? 1. We should sum over outgoing spins 2. We should average over incoming spins For A+B C+D this means: Consider e. g. : e e the ‘ ’: interchange 2 fermions! ie ie B D B A C A D C ie ie
Non-relativistic: tedious Non relativistic limit, |p| 0, gives: incoming outgoing With this current terms reduce to: I. e. spins do not flip & only matrix elements 0: A B C D
Non-relativistic: tedious Collecting terms 0 & squaring: The leading factor 2/4 comes from: … give or take factors 4 … Hence: And with master formula for d /d : 2: sum over outgoing spins 4=2 2: averaging incoming spins
Relativistic: same with helicity states See paragraph 6. 2 in Mark Thomson’s book: A+B C+D i. e. 4 spin configuration in initial & final state 16 terms! Very instructive to read through, but also very (very!) tedious ….
Relativistic: work once joy forever Take: e e (1 diagram) instead of e e (2 diagrams) B q = p. A p. C = p D p. B |M|2 becomes: D A e C Spin summation (final states) & averaging (initial states) yields:
Lepton tensor Casimir’s trick! lepton tensor:
Products & Traces of -matrices Products of any odd number of matrices = 0 Traces weak interaction
And the pay-off: e+e + cross section Use the Feyman rules: p k q p’ k’ With Casimir’s trick & formulae for the Traces: relativistic limit: m=M 0
And the pay-off: e+e + cross section q Relativistic limit: And hence the differential cross section: Using: You find: and: =e 2/(4 )
Measured + ee cross section + Z e e
Measured angular distributions e e
And … e+e + gives you e+e qq with three colours u, c, t d, s, b udsc Rudsc = 3 [4/9+1/9+4/9]=3. 3 udsc no color Rudsc = 4/9+1/9+4/9]=1. 1 s
And … e+e + gives you e+e + Price to pay: can not neglect the -mass near threshold …. m =1. 7768 Ge. V -mass determined via very accurate threshold scan @ BES/Bejing!
* Other processes p Compton scattering: e e k k e+ Pair creation: e+e p’ k’ q=p+k k k’ p’ p e+ e k’ p q=p-k’ k p’ k’ p’
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