Elementary Particle Physics Quantum ElectroDynamics (QED) Lectures 4, 5, 6, 7, 8 Mark Thomson: Chapters 4, 5 & 6 Halzen&Martin: Chapter 4 Frank Linde, Nikhef, H044, f.linde@nikhef.nl, 020-5925140

QED:  Klein-Gordon equation  Dirac equation: free particles  Dirac equation: interactions e+e + cross section

Klein Gordon equation

Lorentz invariant Schrödinger eqn.? With the quantum mechanical energy & momentum operators: 𝑝 =𝑖  recall: 𝒑 𝝁 = 𝑬, 𝒑 and 𝝏 𝝁 = 𝝏 𝝏𝒕 ,− 𝝏 𝝏𝒙 ,− 𝝏 𝝏𝒚 ,− 𝝏 𝝏𝒛 = 𝝏 𝝏𝒕 ,− 𝜵 You simply ‘derive’ the Schrödinger equation from classical mechanics: E = 𝒑 2 2𝑚 𝑖  𝑡  =  1 2𝑚  2   Schrödinger equation With the relativistic relation between E, p & m you get: 𝐸 2 = 𝒑 2 + 𝑚 2  2  𝑡 2  =  2   𝑚 2   Klein-Gordon equation

Probability & current densities Use 4-derivatives to make Klein-Gordon equation Lorentz invariant:  2  𝑡 2  =  2   𝑚 2   = 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? 𝜕 𝜇 𝜕 𝜇 + 𝑚 2 =0 𝜕 𝜇 𝜕 𝜇 + 𝑚 2 =0 ‘Surprise’ of the plane-wave solutions, if you plug them in the KG equation you find: 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

Particles & anti-particles (E,p) emission +e (+E,+p) absorption time As far as this system is concerned: emission: e with p =(E, p) equivalent to: absorption: e+ with p =(+E, +p) e e+ ( E, p) Expressed in terms of the charge/current densities: 2 (+E,p) Interpretation of Pauli & Weisskopf / Stückelberg & Feynman

One vertex  two vertices   e+  negative energy e backwards in time e e time e Intermediate state: ee+e e e 2 vertices: 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: 𝑨 𝝁 ≡ 𝑽, 𝑨   And ‘nicer’ with F A  A: Inhomogeneous terms: in terms of A: 𝒋 𝝁 ≡ 𝝆, 𝒋

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: with photons () are massless! And in the Lorentz gauge: Coulomb gauge: Hence only 2 of the original 4 degrees of freedom in   survive! transversal circular (assuming p // z-axis)

Vertices & 4-momentum conservation time T  pi  pf e pi pf k e T pi pf k e+  p+ e p p+ k e e+  p+ e T p p+ k e+ e T

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 watch out: really: 𝝏 𝝁 𝑨 𝝁  𝒊 (recall:  = Neipx) Define ‘transition current’ jfi’: To obtain:

Example:  K   K scattering  (A) (B) (C) (D) How to handle? 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 Solve for A:  Simple!: 1 And the transition matrix element Tfi: note that Tfi is symmetric in jAC & jBD 1 pC  pA

 K   K scattering: amplitude  (A) (B) (C) (D) And using the earlier introduced calculation of the transition probability Wfi in terms of Tfi: 1 You get for the amplitude (qpD-pB=pA-pC): =𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪 𝝁 × −𝒊 𝒈 𝝁 𝒒 𝟐 ×𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫 

 K   K scattering: cross section  (A) (B) (C) (D) dA+BC+D d Standard expression for in c.m. frame: B A In the c.m. frame: & Assume particle masses to be << particle energies: 𝜶≡ 𝒆 𝟐 𝟒𝝅 And hence:

 K   K scattering: Feynman rules  (A) (B) (C) (D) 𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫  The matrix element contains several factors which each can be associated with parts of the diagram to give so-called Feynman rules: −𝒊 𝒈 𝝁 𝒒 𝟐 𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪 𝝁 × −𝒊 𝒈 𝝁 𝒒 𝟐 ×𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫  𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪 𝝁   - vertex 𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪 𝝁 K  - vertex 𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫   - propagator −𝒊 𝒈 𝝁 𝒒 𝟐 1

Feynman rules S=0 electrodynamics External lines p 1 k in:   out:   Vertices: p p’ k k’ Propagators: q

 +   K+K scattering Just follow the rules: pB pD pA pC    + K q=pA+pB =pC+pD pD pA pC    + K K+ Just follow the rules: Feynman diagram(s): anti-particles  time reversed particles  pB q=pA+pB =pC+pD  pD pA pC    + K K+ amplitude:

 +   K+K scattering Just follow the rules:  pB  pD pA pC   q=pA+pB =pC+pD  pD pA pC    + K K+ Just follow the rules: Feynman diagram(s): anti-particles  time reversed particles amplitude: pick frame (c.m.) & make 4-vectors explicit: A B C D  standard d/d expression: relativistic limit (or set particle masses to zero): 𝒑 𝑨 − 𝒑 𝑩 ∙ 𝒑 𝑪 − 𝒑 𝑫 →− 𝒑 ∙ 𝒑 ′− 𝒑 ∙ 𝒑 ′ − 𝒑 ∙ 𝒑 ′ − 𝒑 ∙ 𝒑 ′= −𝟒 𝑬 𝟐 cos𝜽= − scos𝜽

 +    +  scattering (a) (b) (a) (b) pB q=pA-pC =pD-pB pD pA pC q=pA+pB =pC+pD Feynman diagram(s): (a)  pB q=pA-pC =pD-pB  pD pA pC    + (b) q=pA+pB =pC+pD anti-particles  time reversed particles

 +    +  scattering (a) (b)  pB q=pA-pC =pD-pB  pD pA pC   q=pA+pB =pC+pD amplitude: etcetera!

‘Classical’ example: Coulomb scattering * ‘Classical’ example: Coulomb scattering V(x) pi pf Compare previous to ‘classical’ Coulomb  Rutherford  scattering: 1/4 missing! damned units … So: no photons, but ‘simple’ Coulomb potential scattering  Transition amplitude: (use: ) Transition probability:

‘Classical’ example: Coulomb scattering * ‘Classical’ example: Coulomb scattering V(x) pi pf Incoming flux: Final state phase space: Norm. wave functions: N= 𝟏 𝑽 Transition probability: 1  = e2 instead of  = e2/4  Rutherford Lecture # 1 Classical limit: 

Lecture 4 Study Halzen & Martin Chapter 4

*      scattering (c) (b) (a) q=p+k =p’+k’ p k p’ k’ q=p-k’ Feynman diagram(s): direct interaction q=p-k’ =p’-k p k p’ k’ (b) exchange interaction p p’ k k’ (c) contact interaction amplitude (a): k = k’’ = 0

*      scattering (c) (b) (a) q=p+k =p’+k’ p k p’ k’ q=p-k’ Feynman diagram(s): direct interaction q=p-k’ =p’-k p k p’ k’ (b) exchange interaction p p’ k k’ (c) contact interaction amplitude (b): k = k’’ = 0

*      scattering (c) (b) (a) q=p+k =p’+k’ p k p’ k’ q=p-k’ direct interaction Feynman diagram(s): q=p-k’ =p’-k p k p’ k’ (b) exchange interaction p p’ k k’ (c) contact interaction amplitude (c):

     scattering: using gauge invariance *      scattering: using gauge invariance Apology: some spurious factors because ‘Gaussian’ instead of Heaviside-Lorentz units … 𝟒𝝅 Total amplitude: Lorentz condition: (used already) Remaining freedom: with: using you show M to be invariant under: or with: gauge invariance (nice check!)      + 𝒌  ∝ 𝒆 𝒊𝒌𝒙

     scattering: using gauge invariance *      scattering: using gauge invariance Apology: some spurious factors because ‘Gaussian’ instead of Heaviside-Lorentz units … 𝟒𝝅 Checking this for  k and’ k’, you find for –iMa-iMb:  =  k use:  ’ k’  0 But by adding –iMc you do restore gauge invariance, as it should!

     scattering: using gauge invariance *      scattering: using gauge invariance Apology: some spurious factors because ‘Gaussian’ instead of Heaviside-Lorentz units … 𝟒𝝅 I can use gauge invariance to make p=0 & ’p=0  remains Take this time the lab. frame: Standard A+B C+D cross section expression yields: with 

     scattering: using gauge invariance *      scattering: using gauge invariance Apology: some spurious factors because ‘Gaussian’ instead of Heaviside-Lorentz units … 𝟒𝝅 d’ integral can be achieved using the -function, paying attention to ’ dependence: Using property of -function (see Thomson Appendix A) You get the cross section:

     scattering: using gauge invariance *      scattering: using gauge invariance Apology: some spurious factors because ‘Gaussian’ instead of Heaviside-Lorentz units … 𝟒𝝅 For the total cross section: you must average over the incoming photon polarizations & you must sum over the outgoing photon polarizations: ‘brute force’: find photon polarizations in Lab. Frame:  𝑘′ = 𝑐𝑜𝑠𝜑𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜑𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑘 = 0 0 1 Incoming : 𝜀= 1 0 0 , 0 1 0 1 2 𝜀,𝜀′ 𝜀∙ 𝜀′ ∗ 2 𝑑 = 1 2 𝑐𝑜𝑠 2 𝜃+1 𝑑= 8𝜋 3 Outgoing : 𝜀 ′ = 𝑐𝑜𝑠𝜑𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜑𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 , 𝑠𝑖𝑛𝜑 −𝑐𝑜𝑠𝜑 0  Essentials: Feynman: simple rules to calculate diagrams Gauge invariance: use all diagrams to each order n

*  +    scattering (a) (b) (c)  q=p-k =k’-p’ p k’,’ -p’ k, particle anti-particle  q=p-k =k’-p’ p k’,’ -p’ k,    q=p-k’ =k-p’ p k’,’ -p’ k,    p -p’ k, k’,’    (a) (b) (c) direct interaction exchange interaction contact interaction amplitude (a):

*  +    scattering (a) (b) (c)  q=p-k =k’-p’ p k’,’ -p’ k, particle anti-particle  q=p-k =k’-p’ p k’,’ -p’ k,    q=p-k’ =k-p’ p k’,’ -p’ k,    p -p’ k, k’,’    (a) (b) (c) direct interaction exchange interaction contact interaction amplitude (b):

*  +    scattering (a) (b) (c)  q=p-k =k’-p’ p k’,’ -p’ k, particle anti-particle  q=p-k =k’-p’ p k’,’ -p’ k,    q=p-k’ =k-p’ p k’,’ -p’ k,    p -p’ k, k’,’    (a) (b) (c) direct interaction exchange interaction contact interaction amplitude (c):

Elementary Particle Physics Quantum ElectroDynamics (QED) Lectures 4, 5, 6, 7, 8 Mark Thomson: Chapters 4, 5 & 6 Halzen&Martin: Chapter 4 Frank Linde, Nikhef, H044, f.linde@nikhef.nl, 020-5925140

Recap Klein Gordon equation

Free Klein-Gordon particle wave functions With the quantum mechanical energy & momentum operators: 𝑝 =𝑖  recall: 𝒑 𝝁 = 𝑬, 𝒑 and 𝝏 𝝁 = 𝝏 𝝏𝒕 ,− 𝝏 𝝏𝒙 ,− 𝝏 𝝏𝒚 ,− 𝝏 𝝏𝒛 = 𝝏 𝝏𝒕 ,− 𝜵 E = 𝒑 𝟐 𝟐𝒎 𝒊  𝒕  =  𝟏 𝟐𝒎  𝟐  yields Schrödinger equation: non-relativistic We ‘derived’ Klein-Gordon equation from relativistic 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐  2  𝑡 2  =  2   𝑚 2  𝜕 𝜇 𝜕 𝜇 + 𝑚 2 =0 or With free particle plane-wave solutions and conserved current:

Free photon wave functions Maxwell equations for A in vacuum (Lorentz gauge, no sources): 𝜕 𝜇 𝜕 𝜇 𝐴  =0 With plane-wave solutions: with photons () are massless! Now we have 𝒆 −𝒊𝒑∙𝒙 plane-wave solutions for particles & photons In transition amplitudes like:  𝒇 ∗ 𝑨 𝝁  𝒊 𝒅𝒙 you hereby ‘automatically’ get a 4-momentum conserving -function:  𝒇 ∗ 𝑨 𝝁  𝒊 𝒅𝒙 ∝ 𝟐𝝅 𝟒 𝜹 𝟒 𝒑 𝒊 + 𝒑 𝜸 − 𝒑 𝒇 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 𝒊𝝏 𝝁 → 𝒊𝝏 𝝁 + 𝒆 𝑨 𝝁 𝝏 𝝁 → 𝝏 𝝁 −𝒊𝒆 𝑨 𝝁  𝒑 𝝁 → 𝒑 𝝁 +𝒆 𝑨 𝝁 Plugging this into Klein-Gordon equation, you find the perturbation V: 𝟎= 𝝏 𝝁 𝝏 𝝁 + 𝒎 𝟐 → 𝟎= 𝝏 𝝁 −𝒊𝒆 𝑨 𝝁 𝝏 𝝁 −𝒊𝒆 𝑨 𝝁 + 𝒎 𝟐  𝟎= 𝝏 𝝁 𝝏 𝝁 + 𝒎 𝟐 −𝒊𝒆 𝑨 𝝁 𝝏 𝝁 −𝒊𝒆 𝝏 𝝁 𝑨 𝝁 − 𝒆 𝟐 𝑨 𝝁 𝑨 𝝁  𝑽=−𝒊𝒆 𝑨 𝝁 𝝏 𝝁 −𝒊𝒆 𝝏 𝝁 𝑨 𝝁 − 𝒆 𝟐 𝑨 𝝁 𝑨 𝝁 ~−𝒊𝒆 𝑨 𝝁 𝝏 𝝁 −𝒊𝒆 𝝏 𝝁 𝑨 𝝁

Interaction Klein-Gordon particles & photons For now: just look at one ‘leg’ 𝑻 𝒇𝒊 =−𝒊  𝒇 ∗ 𝑽  𝒊 𝒅𝒙 𝒋 𝑩𝑫 𝝁 K    pB pD pC pA q  K   K =−𝒆  𝒇 ∗ 𝑨 𝝁 𝝏 𝝁 + 𝝏 𝝁 𝑨 𝝁  𝒊 𝒅𝒙 =−𝒆  𝒇 ∗ 𝑨 𝝁 𝝏 𝝁  𝒊 − 𝝏 𝝁  𝒇 ∗ 𝑨 𝝁  𝒊 𝒅𝒙 define ‘transition current’: 𝒋 𝒇𝒊 𝝁 =−𝒊𝒆  𝒇 ∗ 𝝏 𝝁  𝒊 − 𝝏 𝝁  𝒇 ∗  𝒊 𝑻 𝒇𝒊 =−𝒊 𝒋 𝒇𝒊 𝝁 𝑨 𝝁 𝒅𝒙 𝒋 𝑨𝑪 𝝁 Note: Tfi looks awful  but with plane-waves for  ’s it really is simple!

Interaction Klein-Gordon particles & photons 𝒋 𝑩𝑫 𝝁 K    pB pD pC pA q  K   K Last step: must find expression for A Idea: assume the  ’s to scatter from an A caused by the K’s transition current Must solve Maxwell with source 𝒋 𝑩𝑫 𝝁 : 𝝏  𝝏  𝑨 𝝁 = 𝒋 𝑩𝑫 𝝁 =−𝒊𝒆  𝑫 ∗ 𝝏 𝝁  𝑩 − 𝝏 𝝁  𝑫 ∗  𝑩 =−𝒆 𝑵 𝑩 𝑵 𝑫 ∗ 𝒑 𝑩 𝝁 + 𝒑 𝑫 𝝁 𝒆 𝒊 𝒑 𝑫 − 𝒑 𝑩 ∙𝒙 𝒋 𝑨𝑪 𝝁 Easy! 𝑨 𝝁 = 𝒆 𝑵 𝑩 𝑵 𝑫 ∗ 𝒑 𝑩 𝝁 + 𝒑 𝑫 𝝁 𝒑 𝑫 − 𝒑 𝑩 𝟐 𝒆 𝒊 𝒑 𝑫 − 𝒑 𝑩 ∙𝒙 Note: Tfi still looks awful  but only plane-waves for ’s & photon = A!

Interaction Klein-Gordon particles & photons 𝑻 𝒇𝒊 =−𝒊  𝒇 ∗ 𝑽  𝒊 𝒅𝒙 Tfi can now be calculated! 𝒋 𝑩𝑫 𝝁 K    pB pD pC pA q  K   K →+𝒊 𝒋 𝑨𝑪 𝝁 𝟏 𝒑 𝑫 − 𝒑 𝑩 𝟐 𝒋 𝑩𝑫 𝝁 𝒅𝒙 = 𝒅𝒙 𝑵 𝑨 𝑵 𝑪 ∗ 𝒑 𝑨 𝝁 + 𝒑 𝑪 𝝁 𝒆 𝒊 𝒑 𝑪 − 𝒑 𝑨 ∙𝒙 × 𝒊 𝒆 𝟐 𝒑 𝑫 − 𝒑 𝑩 𝟐 × 𝑵 𝑩 𝑵 𝑫 ∗ 𝒑 𝝁 𝑩 + 𝒑 𝝁 𝑫 𝒆 𝒊 𝒑 𝑫 − 𝒑 𝑩 ∙𝒙 And with that 𝑻 𝒇𝒊 becomes: 𝑵 𝑨 𝑵 𝑪 ∗ 𝒑 𝑨 𝝁 + 𝒑 𝑪 𝝁 × 𝑵 𝑩 𝑵 𝑫 ∗ 𝒑 𝝁 𝑩 + 𝒑 𝝁 𝑫 𝒑 𝑫 − 𝒑 𝑩 𝟐 × 𝒊 𝒆 𝟐 × 𝟐𝝅 𝟒 𝜹 𝟒 𝒑 𝑨 + 𝒑 𝑩 − 𝒑 𝑪 − 𝒑 𝑫 𝒋 𝑨𝑪 𝝁 With this Tfi you calculate Wfi as usual and hence the cross section

Interaction Klein-Gordon particles & photons 𝑾 𝒇𝒊 = 𝒍𝒊𝒎 𝑻,𝑽→∞ 𝑻 𝒇𝒊 𝟐 𝑻𝑽 And finally the cross section 𝒋 𝑩𝑫 𝝁 K    pB pD pC pA q  K   K = 𝑵 𝑨 𝑵 𝑩 𝑵 𝑪 𝑵 𝑫 𝟐 × 𝑴 𝟐 × 𝟐𝝅 𝟒 𝜹 𝟒 𝒑 𝑨 + 𝒑 𝑩 − 𝒑 𝑪 − 𝒑 𝑫 −𝒊𝑴=𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪 𝝁 × −𝒊 𝒈 𝝁 𝒒 𝟐 ×𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫  With matrix element or Feynman amplitude M: with: 𝒒= 𝒑 𝑫 − 𝒑 𝑩 = 𝒑 𝑪 − 𝒑 𝑨 cross section CM frame: A B C D  𝒋 𝑨𝑪 𝝁 𝒅𝝈 𝒅 = 𝟏 𝟔𝟒 𝝅 𝟐 𝒔 𝒑 ′ 𝒑 𝑴 𝟐 Some simple – Feyman – rules allow you to ‘easily’ calculate M

Feynman rules S=0 electrodynamics 𝒋 𝑩𝑫 𝝁 K    pB pD pC pA q  K   K 𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫 𝝁 −𝒊𝑴= × −𝒊 𝒈 𝝁 𝒒 𝟐 × 𝒊𝒆 𝒑 𝑨 + 𝒑 𝑪  𝒋 𝑨𝑪 𝝁 with: 𝒒= 𝒑 𝑫 − 𝒑 𝑩 = 𝒑 𝑪 − 𝒑 𝑨

Feynman rules S=0 electrodynamics with anti-particles Feynman rules 𝒋 𝑩𝑫 𝝁 K  +  pB pD pC pA q  +K   +K Interpret anti-particles as: particles with: 𝒑→−𝒑= −𝑬,− 𝒑 running backwards in time This way one equation (Klein-Gordon) deals with both particles & anti-particles And  we solved the E<0 problem! 𝒋 𝑨𝑪 𝝁 with: 𝒒= 𝒑 𝑫 − 𝒑 𝑩 = 𝒑 𝑪 − 𝒑 𝑨

Feynman rules S=0 electrodynamics with anti-particles Feynman rules 𝒋 𝑩𝑫 𝝁 K    pB pD  pC  pA q  +K   +K −𝒊𝑴=𝒊𝒆 𝒑 𝑩 + 𝒑 𝑫 𝝁 × −𝒊 𝒈 𝝁 𝒒 𝟐 × 𝒊𝒆 𝒑 𝑨  𝒑 𝑪  𝒋 𝑨𝑪 𝝁 with: 𝒒= 𝒑 𝑫 − 𝒑 𝑩 = 𝒑 𝑨 − 𝒑 𝑪

Today 23 February problem sessions SP D1.111 & SP D1.162 15:00 – 17:00

Dirac equation: free particles

Schrödinger  Klein-Gordon  Dirac Quantum mechanical E & p operators: 𝑝 =𝑖  𝒑 𝝁 = 𝑬, 𝒑  i 𝝏 𝝁 =𝒊 𝝏 𝝏𝒕 ,− 𝜵 You simply ‘derive’ the Schrödinger equation from classical mechanics: E= 𝒑 2 2𝑚 𝑖  𝑡  =  1 2𝑚  2   Schrödinger equation With the relativistic relation between E, p & m you get: 𝐸 2 = 𝒑 2 + 𝑚 2  2  𝑡 2  =  2   𝑚 2   Klein-Gordon equation Dirac equation The negative energy solutions led Dirac to try an equation with first order derivatives in time (like Schrödinger) as well as in space 𝑖 𝜕 𝜕𝑡 =−𝑖   𝛻 +𝑚

Does it make sense? 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐 𝐄 𝟐 𝐦 𝟐 𝐩 𝟐 Also Dirac equation should reflect: 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐 Basically squaring: 𝒊 𝝏 𝝏𝒕 =−𝒊   𝜵 +𝒎=   𝒑 + 𝒎 Tells you: 𝐄 𝟐  2=1 𝐦 𝟐 i 2=1 i +i = 0 𝐩 𝟐 ij: i j +j i = 0

Properties of i and   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 𝑨 𝒊𝒋 𝑩 𝒋𝒌 𝑪 𝒌𝒊 = 𝑪 𝒌𝒊 𝑨 𝒊𝒋 𝑩 𝒋𝒌 = 𝑩 𝒋𝒌 𝑪 𝒌𝒊 𝑨 𝒊𝒋 Both  and  must be traceless matrices: Tr(ABC) = Tr(CAB) = Tr(BCA) anti commutation  2=1 cyclic  2=1 Tr(i ) = Tr(i) = Tr(i) = 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: 𝜷= 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜶 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 𝟏 𝟎 𝟎 −𝟏 𝟎 𝟎 𝟎 𝟎 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 𝑖 𝜕 𝜕𝑡 =−𝑖   𝛻 +𝑚 does not look that Lorentz invariant 𝑚=𝑖 𝜕 𝜕𝑡 +𝑖 𝛽   𝛻 ≡𝑖   𝜕   Multiplying on the left with  and collecting all the derivatives gives: note: 𝝏 𝝁 = 𝝏 𝒕 ,+ 𝜵 Hereby, the Dirac -matrices are defined as: And you can verify that: As well as: and: → 𝜸 𝝁+ = 𝜸 𝟎 𝜸 𝝁 𝜸 𝟎

Co-variant form: Dirac -matrices 𝜷= 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜶 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 𝟏 𝟎 𝟎 −𝟏 𝟎 𝟎 𝟎 𝟎 with the Dirac -matrices defined as: 𝒎=𝒊   𝝏   𝜸 𝟎 = 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜸 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝒊 −𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎

Warning! 𝑚=𝑖   𝜕       𝑚=𝑖   𝜕   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 is useful!   & beautiful! To make things even worse, we define:

Spinors & (Dirac) matrices 𝝓= ∗ ∗ ∗ ∗ 𝝓 + = ∗ ∗ ∗ ∗ 𝜸 𝝁 = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 𝜸 𝝁 𝝓= ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ = ∗ ∗ ∗ ∗ 𝜸 𝝁 𝝓 + = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ =  𝝓 + 𝜸 𝝁 = ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ = ∗ ∗ ∗ ∗ 𝝓 𝜸 𝝁 = ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ =  𝝓 + 𝝓= ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ = ∗ 𝝓 𝝓 + = ∗ ∗ ∗ ∗ × ∗ ∗ ∗ ∗ = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 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: ×𝟎 Dirac equations for : & × ×  Add these two equations to get: Conserved 4-current: (exactly what Dirac aimed to achieve …)

Solutions: particles @ rest 𝒑 = 𝟎 Dirac equation for 𝒑 = 𝟎 is simple: Solve by splitting 4-component in two 2-components: with  0(1/c) t follows: solutions: e e+

Solutions: moving particles 𝒑 ≠ 𝟎 Dirac equation for 𝒑 ≠ 𝟎 less simple: Anticipate plane-waves: And again anticipate two 2-components: Plugging this in gives: k 

Solutions: moving particles 𝒑 ≠ 𝟎 Solutions: pick uA(p) & calculate uB(p): 𝒖 𝑨 = 𝟏 𝟎 → 𝒖 𝑨 = 𝟎 𝟏 → e E>0 In limit 𝒑 → 𝟎 you retrieve the E>0 solutions, hence these are 𝒑 ≠ 𝟎 electron solutions Similarly: pick uB(p) & calculate uA(p): 𝒖 𝑩 = 𝟏 𝟎 → 𝒖 𝑩 = 𝟎 𝟏 → e+  E<0 In limit 𝒑 → 𝟎 you retrieve the E<0 solutions, hence these are 𝒑 ≠ 𝟎 positron solutions

Lecture 5 Study Thomson 4

Elementary Particle Physics Quantum ElectroDynamics (QED) Lectures 4, 5, 6, 7, 8 Mark Thomson: Chapters 4, 5 & 6 Halzen&Martin: Chapter 4 Frank Linde, Nikhef, H044, f.linde@nikhef.nl, 020-5925140

Recap introduction Dirac equation

Dirac equation 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐 ? 𝐄 𝟐 𝐦 𝟐 𝐩 𝟐 𝝏 𝝁 = 𝝏 𝒕 ,− 𝜵 From: 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐 E=𝑖  𝑡 𝑝 =𝑖  & classical  QM ‘transcription’: 𝒊 𝝏 𝝏𝒕 =−𝒊   𝜵 +𝒎=   𝒑 +𝒎 We found: 𝑬 𝟐 = 𝒑 𝟐 + 𝒎 𝟐 ? With  , 1, 2 & 3 (44) matrices, satisfying:  2=1 𝐄 𝟐 𝐦 𝟐 i 2=1  + = 0 𝐩 𝟐 ij: i j +j i = 0 𝜷= 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜶 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜶 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 𝟏 𝟎 𝟎 −𝟏 𝟎 𝟎 𝟎 𝟎

Co-variant form: Dirac -matrices Dirac’s original form does not look covariant: 𝑖 𝜕 𝜕𝑡 =−𝑖   𝛻 +𝑚 Multiplying on the left with  and collecting all the derivatives gives covariant form: note: 𝝏 𝝁 = 𝝏 𝒕 ,+ 𝜵 With Dirac -matrices defined as: 𝛾 0 =𝛽= 1 0 0 −1 𝛾 𝑘 =𝛽 𝛼 𝑘 = 0 𝜎 𝑘 − 𝜎 𝑘 0 𝜸 𝟎 = 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜸 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝒊 −𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 From the properties of  , 1, 2 & 3 follows: → 𝜸 𝝁+ = 𝜸 𝟎 𝜸 𝝁 𝜸 𝟎

Dirac particle solutions: spinors 𝒖 𝑨 𝒑 = 𝒑 ∙ 𝝈 𝑬−𝒎 𝒖 𝑩 𝒑 𝒖 𝑩 𝒑 = 𝒑 ∙ 𝝈 𝑬+𝒎 𝒖 𝑨 𝒑 Ansatz solution: = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 Dirac eqn.: 𝒑 = 𝟎 solutions: e 𝒖 𝑨 = 𝟎 𝟏 e+ 𝒖 𝑩 = 𝟎 𝟏 𝒖 𝑨 = 𝟏 𝟎 𝒖 𝑩 = 𝟏 𝟎 E>0 E<0 spin ½ electrons spin ½ positrons 𝒑 ≠ 𝟎 solutions:  

Dirac equation: more on free particles normalisation 4-vector current anti-particles

One more look at 𝑝   1 The conditions: Imply:   sorry for the c’s 𝑝   The conditions: Imply:  1  𝒑 𝟐 𝒄 𝟐 = 𝑬 𝟐 − 𝒎 𝟐 𝒄 𝟒 i.e. energy-momentum relation, as expected Check this: 

Normalisation of the Dirac spinors Just calculate it!: Spinors 1 & 2, E>0: To normalize @ 2E particles/unit volume Spinors 3 & 4, E<0: To normalize @ 2E particles/unit volume 

Current & probability densities always using Again, just plug it in! 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): 𝒋 𝒙 = 𝑵 𝟐 𝟐𝒄 𝒑 𝒙 𝑬−𝒎 𝒄 𝟐 →−𝟐 𝒑 𝒙

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 𝒑

And now: E<0  antiparticles ‘Dirac sea’: fill all E<0 states (thanks to Pauli exclusion principle) single e  e+e e+e   But:  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 −𝒊(−𝑬)(−𝒕) = 𝒆 −𝒊𝑬𝒕 e (E>0) e (E<0) E =2E  e (E>0) e+ (E>0) E =2E  time ‘Up-shot’: Dirac equation accommodates both particle & antiparticles! Sequel: will use particle & anti-particle spinors labelled with their physical, E>0 & real , kinematics. (exponents remain opposite) 𝒑

We had: Dirac ‘u’-spinors  1 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 ≡ 𝑢 1 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥  2 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 0 1 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 − 𝑝 𝑧 𝐸+𝑚 ≡ 𝑢 2 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥 e E>0  3 = 𝐸 +𝑚 𝑒 −𝑖𝑝∙𝑥 𝑝 𝑧 𝐸−𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸−𝑚 1 0 ≡ 𝑢 3 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥  4 = 𝐸 +𝑚 𝑒 −𝑖𝑝∙𝑥 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸−𝑚 −𝑝 𝑧 𝐸−𝑚 0 1 ≡ 𝑢 4 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥 e+ E<0

From now on use: Dirac ‘u’- & ‘v’-spinors u-spinors: for electrons, labeled with physical E>0 & 𝒑  1 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 ≡ 𝒖 𝟏 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙  2 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 0 1 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 − 𝑝 𝑧 𝐸+𝑚 ≡ 𝒖 𝟐 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 e E>0 v-spinors: for positrons, labeled with physical E>0 & 𝒑 e+ 𝑢 4 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 ≡ 𝒗 𝟏 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 E>0 𝑢 3 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 ≡ 𝒗 𝟐 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙

Lecture 6 Study Thomson 4

28 February 2017

Today 28 February problem sessions SP B0.207 & SP B0.209 13:00 – 15:00

Dirac equation 𝒊 𝝏 𝝏𝒕 𝝍=−𝒊   𝜵 𝝍+𝒎𝝍 𝒊 𝝏 𝝏𝒕 𝝍=−𝒊   𝜵 𝝍+𝒎𝝍 Dirac equation in original form with matrices 𝜶 & 𝜷: With plane-wave solutions: =𝒖(𝒑) 𝒆 −𝒊𝒑∙𝒙 = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 you find for spinor u(p): Eu(𝒑)=   𝒑 𝒖(𝒑)+𝒎𝒖(𝒑) This algabraic equation for u(p) you can solve for particles with 𝒑 𝝁 = 𝑬, 𝒑 Co-variant form of Dirac equation with matrices 𝜸 𝝁 : 𝒎𝝍=𝒊   𝝏  𝝍   𝒑  −𝒎 𝒖 𝒑 =𝟎 with =𝒖(𝒑) 𝒆 −𝒊𝒑∙𝒙 = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 you get 𝜸 𝟎 = 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜸 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝒊 −𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 Explicit expressions for the 𝜸 𝝁 matrices: And the algebra for the 𝜸 𝝁 matrices: 𝜸 𝝁 𝜸 𝝂 + 𝜸 𝝂 𝜸 𝝁 =𝟐 𝒈 𝝁𝝂

Spinors E>0 E<0 𝒑 = 𝟎 solutions: e e+ 𝒑 ≠ 𝟎 solutions: 𝒖 𝑨 𝒑 = 𝒑 ∙ 𝝈 𝑬−𝒎 𝒖 𝑩 𝒑 𝒖 𝑩 𝒑 = 𝒑 ∙ 𝝈 𝑬+𝒎 𝒖 𝑨 𝒑 Ansatz solution: = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 Dirac eqn.: 𝒑 = 𝟎 solutions: e 𝒖 𝑨 = 𝟎 𝟏 e+ 𝒖 𝑩 = 𝟎 𝟏 𝒖 𝑨 = 𝟏 𝟎 𝒖 𝑩 = 𝟏 𝟎 E>0 E<0 spin ½ electrons spin ½ positrons 𝒑 ≠ 𝟎 solutions:  

Particles & anti-particles u-spinors: for electrons, labeled with physical E>0 & 𝒑  1 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 = 𝒖 𝟏 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙  2 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 0 1 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 − 𝑝 𝑧 𝐸+𝑚 = 𝒖 𝟐 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 e E>0 v-spinors: for positrons, labeled with physical E>0 & 𝒑 e+  1 = 𝑢 4 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 = 𝒗 𝟏 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 E>0  2 = 𝑢 3 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 = 𝒗 𝟐 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙

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    =  ½ as it should  helicity + ½ RH right-handed is called helicity with eigenvalues:  ½ helicity + ½ RH helicity  ½ LH left-handed Instead of u1 & u2 spinors, we could use helicity ½ : u & u spinors (& similarly for v-spinors) You ‘simply’ solve the eigenvalue equation:  Eigenvalues, use :   =  ½ as it should With uA, you get uB using the Dirac eqn. as we did before (easier now ): 𝝈∙ 𝒑 𝒖 𝑨 =𝟐𝒑 𝒖 𝑨 

* Helicity states   helicity + ½ RH right-handed is called helicity with eigenvalues:  ½ helicity + ½ RH helicity  ½ LH left-handed solving easiest using spherical coordinates: yields: follows: with 𝒂 𝐜𝐨𝐬 𝜽 +𝒃 𝐬𝐢𝐧 𝜽 𝒆 −𝒊𝝋 =𝟐𝝀𝒂 𝒂 𝐬𝐢𝐧 𝜽 𝒆 +𝒊𝝋 −𝒃 𝐜𝐨𝐬 𝜽 =𝟐𝝀𝒃 or:   For  = +½:

* Helicity states Particles Anti-particles 𝒄≡𝐜𝐨𝐬 𝜽/𝟐 𝒔≡𝐬𝐢𝐧 𝜽/𝟐 right-handed is called helicity with eigenvalues:  ½ helicity + ½ RH helicity  ½ LH left-handed Particles 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 Particles Anti-particles Weak interactions! For massless & extremely relativistic particles, helicity states become simple: 𝑢  = 𝐸 𝑐 𝑠 𝑒 𝑖𝜑 𝑐 𝑠 𝑒 𝑖𝜑 𝑢  = 𝐸 −𝑠 𝑐 𝑒 𝑖𝜑 𝑠 −𝑐 𝑒 𝑖𝜑 Particles  𝒖 𝑹  𝒖 𝑳  𝒗 𝑹  𝒗 𝑳 Eigenstates of  5 called: Left-handed (L) Right handed (R) chiral states. Weak interactions! 𝑣  = 𝐸 𝑠 −𝑐 𝑒 𝑖𝜑 −𝑠 𝑐 𝑒 𝑖𝜑 𝑣  = 𝐸 𝑐 𝑠 𝑒 𝑖𝜑 𝑐 𝑠 𝑒 𝑖𝜑 Anti-particles These four states are also eigenstates of: 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 ≡ 𝛾 5 𝜸 𝟓 =𝒊 𝜸 𝟎 𝜸 𝟏 𝜸 𝟐 𝜸 𝟑 Simple check: 𝛾 5 𝑢  =+ 𝑢  𝛾 5 𝑢  =− 𝑢  𝛾 5 𝑣  =− 𝑣  𝛾 5 𝑣  =+ 𝑣  and:

Elementary Particle Physics Quantum ElectroDynamics (QED) Lectures 4, 5, 6, 7, 8 Mark Thomson: Chapters 4, 5 & 6 Halzen&Martin: Chapter 4 Frank Linde, Nikhef, H044, f.linde@nikhef.nl, 020-5925140

Recap Dirac equation spinors particles & anti-particles spin, helicity & chirality

Dirac equation 𝒊 𝝏 𝝏𝒕 𝝍=−𝒊   𝜵 𝝍+𝒎𝝍 𝒊 𝝏 𝝏𝒕 𝝍=−𝒊   𝜵 𝝍+𝒎𝝍 Dirac equation in original form with matrices 𝜶 & 𝜷: With plane-wave solutions: =𝒖(𝒑) 𝒆 −𝒊𝒑∙𝒙 = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 you find for spinor u(p): Eu(𝒑)=   𝒑 𝒖(𝒑)+𝒎𝒖(𝒑) This algabraic equation for u(p) you can solve for particles with 𝒑 𝝁 = 𝑬, 𝒑 Co-variant form of Dirac equation with matrices 𝜸 𝝁 : 𝒎𝝍=𝒊   𝝏  𝝍   𝒑  −𝒎 𝒖 𝒑 =𝟎 with =𝒖(𝒑) 𝒆 −𝒊𝒑∙𝒙 = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 you get 𝜸 𝟎 = 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 𝟎 −𝟏 𝟎 𝟎 −𝟏 𝜸 𝟏 = 𝟎 𝟎 𝟎 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟐 = 𝟎 𝟎 𝟎 𝟎 𝟎 −𝒊 𝒊 𝟎 𝟎 𝒊 −𝒊 𝟎 𝟎 𝟎 𝟎 𝟎 𝜸 𝟑 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 −𝟏 −𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 Explicit expressions for the 𝜸 𝝁 matrices: And the algebra for the 𝜸 𝝁 matrices: 𝜸 𝝁 𝜸 𝝂 + 𝜸 𝝂 𝜸 𝝁 =𝟐 𝒈 𝝁𝝂

Spinors E>0 E<0 𝒑 = 𝟎 solutions: e e+ 𝒑 ≠ 𝟎 solutions: 𝒖 𝑨 𝒑 = 𝒑 ∙ 𝝈 𝑬−𝒎 𝒖 𝑩 𝒑 𝒖 𝑩 𝒑 = 𝒑 ∙ 𝝈 𝑬+𝒎 𝒖 𝑨 𝒑 Ansatz solution: = 𝒖 𝑨 𝒑 𝒖 𝑩 𝒑 𝒆 −𝒊𝒑∙𝒙 Dirac eqn.: 𝒑 = 𝟎 solutions: e 𝒖 𝑨 = 𝟎 𝟏 e+ 𝒖 𝑩 = 𝟎 𝟏 𝒖 𝑨 = 𝟏 𝟎 𝒖 𝑩 = 𝟏 𝟎 E>0 E<0 spin ½ electrons spin ½ positrons 𝒑 ≠ 𝟎 solutions:  

Particles & anti-particles u-spinors: for electrons, labeled with physical E>0 & 𝒑  1 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 = 𝑢 1 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥  2 = 𝐸+𝑚 𝑒 −𝑖𝑝∙𝑥 0 1 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 − 𝑝 𝑧 𝐸+𝑚 = 𝑢 2 𝐸, 𝑝 𝑒 −𝑖𝑝∙𝑥 e E>0 v-spinors: for positrons, labeled with physical E>0 & 𝒑 e+  1 = 𝑢 4 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 = 𝑣 1 𝐸, 𝑝 𝑒 +𝑖𝑝∙𝑥 E>0  2 = 𝑢 3 −𝐸,− 𝑝 𝑒 +𝑖𝑝∙𝑥 = 𝑣 2 𝐸, 𝑝 𝑒 +𝑖𝑝∙𝑥

Spin, helicity & chirality Spinors represent spin ½ states! Spin operator: ½ 𝜮 =½ 𝝈 𝟎 𝟎 𝝈 ½ 𝜮 𝟐 =½ ½+𝟏 ½ 𝜮 does not commute with the Dirac Hamiltonian 𝜮 ∙ 𝒑 𝟐 𝒑 does commute with Dirac Hamiltonian, eigenvalues  ½  helicity 𝒑 = 𝒑𝐬𝐢𝐧𝜽𝐜𝐨𝐬𝝋,𝒑𝐬𝐢𝐧𝜽𝐬𝐢𝐧𝝋,𝒑𝐜𝐨𝐬𝜽 s=𝐬𝐢𝐧 𝟏 𝟐 𝜽 c=𝐜𝐨𝐬 𝟏 𝟐 𝜽 Particles - helicity 𝒖 ↑ = 𝑬+𝒎 𝒄 𝒔 𝒆 𝒊𝝋 𝒑 𝑬+𝒎 𝒄 𝒑 𝑬+𝒎 𝒔 𝒆 𝒊𝝋 𝒖 ↓ = 𝑬+𝒎 −𝒔 𝒄 𝒆 𝒊𝝋 𝒑 𝑬+𝒎 𝒔 − 𝒑 𝑬+𝒎 𝒄 𝒆 𝒊𝝋 u u h=+½ h=½ Anti-particles - helicity 𝒗 ↑ = 𝑬+𝒎 𝒑 𝑬+𝒎 𝒔 − 𝒑 𝑬+𝒎 𝒄 𝒆 𝒊𝝋 −𝒔 𝒄 𝒆 𝒊𝝋 𝒗 ↓ = 𝑬+𝒎 𝒑 𝑬+𝒎 𝒄 𝒑 𝑬+𝒎 𝒔 𝒆 𝒊𝝋 𝒄 𝒔 𝒆 𝒊𝝋 v v h= ½ h=+½

Spin, helicity & chirality Spinors represent spin ½ states! Spin operator: ½ 𝜮 =½ 𝝈 𝟎 𝟎 𝝈 ½ 𝜮 𝟐 =½ ½+𝟏 also commutes with Dirac Hamiltonian 𝜸 𝟓 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 ≡𝒊 𝜸 𝟎 𝜸 𝟏 𝜸 𝟐 𝜸 𝟑 eigenvalues  1  chirality For massless & extremely relativistic particles, chirality & helicity eigenstates are: h=+½ h=½ 𝑢 𝑅 = 𝐸 𝑐 𝑠 𝑒 𝑖𝜑 𝑐 𝑠 𝑒 𝑖𝜑 𝑢 𝐿 = 𝐸 −𝑠 𝑐 𝑒 𝑖𝜑 𝑠 −𝑐 𝑒 𝑖𝜑 Particles - chirality uR uL  5: +1  5: 1 c=𝐜𝐨𝐬 𝟏 𝟐 𝜽 s=𝐬𝐢𝐧 𝟏 𝟐 𝜽 𝑣 𝑅 = 𝐸 𝑠 −𝑐 𝑒 𝑖𝜑 −𝑠 𝑐 𝑒 𝑖𝜑 𝑣 𝐿 = 𝐸 𝑐 𝑠 𝑒 𝑖𝜑 𝑐 𝑠 𝑒 𝑖𝜑 Anti-particles - chirality vR vL  5: 1  5: +1 h= ½ h=+½

* Dirac equation: more on free particles transformation properties normalisation orthogonality completeness ! ! !

Transformation properties * 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: use 𝝏′ 𝝁 = 𝒂 𝝁 𝝂 𝝏 𝝂 × 𝑺 𝑳 −𝟏  with finally follows:

Space inversion  parity * 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 * Infinitesimal transformation Introducing Lorentz transformations, we saw: with  =  ‘Simply’ by checking, you can show that: with: obeys: not that simple at all! 𝝈 𝝁𝝂 + = 𝝁 & 𝝂≠𝟎:+ 𝝈 𝝁𝝂  or  = 0: − 𝝈 𝝁𝝂 With this you get for infinitesimal transformations: → 𝜸 𝟎 𝑺 𝑳 −𝟏 = 𝑺 𝑳 + 𝜸 𝟎 And since: the same holds for finite transformations! Hence:

Finally we can justify the   ‘4-vector’ * Finally we can justify the   ‘4-vector’ By the way: these 16 combinations (1(S)+4(V)+6(T)+1(P)+4(A)) exhaust the 16 number of possibilities! Using: and easy to show that:   notation! With: it follows that: & use: 𝑺 𝑳 = 𝟏− 𝒊 𝟒 𝝈 𝝁𝝂 ∆ 𝝁𝝂 use: 𝑺 𝑷 = 𝜸 𝟎 Hence:

* Boosts Recall the ‘simple’ Lorentz transformation for boost  along Z-axis: 03 = 30 =  /N and = Use & gives spinor transformation:  03 = ( 0 3  3 0)= i3 𝒊 𝟐 𝑺 𝑳 =𝟏− 𝒊 𝟒     →𝟏+ 𝝎 𝟐𝑵  𝟑 =1+ 𝝎 𝟐𝑵 𝟎 𝝈 𝟑 𝝈 𝟑 𝟎 Infinitesimal: finite: cosh⁡(𝜔/2) 𝜎 3 sinh⁡(𝜔/2) 𝜎 3 sinh⁡(𝜔/2) cosh⁡(𝜔/2) please: you check! With some goniometry: You find:

Transformation, charge conjugation, from  to C & vice-versa? * Charge conjugation Dirac eqn. for electron (charge –|e|) in e.m. field Dirac eqn. for positron (charge +|e|) in e.m. field Transformation, charge conjugation, from  to C & vice-versa? Manipulate electron Dirac eqn. to resemble positron Dirac eqn.  Bit tricky, but a C satisfying does the job: simply verify: − 𝒊 𝜸 𝟐 𝜸 𝝁∗ = 𝜸 𝝁 𝒊 𝜸 𝟐 Because multiplication with C 0 yields: Explicitly: does the job Which gives the positron Dirac eqn. with:

Useful relations: normalisation note: this is not 𝒋 𝟎 = 𝝍 𝜸 𝟎 𝝍=𝟐 𝑬 Explicitly for u1: 𝑢 𝑢= 𝐸+𝑚 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 1 0 0 1 0 0 0 0 0 0 0 0 −1 0 0 −1 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 =𝐸+𝑚− 𝑝 2 𝐸+𝑚 =𝐸+𝑚 − 𝐸−𝑚 =2m Explicitly for v1: 𝑣 𝑣= 𝐸+𝑚 −𝑝 𝑥 −𝑖 𝑝 𝑦 −𝐸−𝑚 𝑝 𝑧 −𝐸−𝑚 0 1 1 0 0 1 0 0 0 0 0 0 0 0 −1 0 0 −1 −𝑝 𝑥 +𝑖 𝑝 𝑦 −𝐸−𝑚 𝑝 𝑧 −𝐸−𝑚 0 1 =−𝐸−𝑚+ 𝑝 2 𝐸+𝑚 =−𝐸−𝑚+ 𝐸−𝑚 =−2m

Useful relations: orthogonality 𝑢 (1) 𝑢 (2) = 𝐸+𝑚 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 1 0 0 1 0 0 0 0 0 0 0 0 −1 0 0 −1 0 1 𝑝 𝑥 −𝑖 𝑝 𝑦 𝐸+𝑚 −𝑝 𝑧 𝐸+𝑚 = ∙∙∙ Easy, just do it! 𝑣 (1) 𝑣 (2) = 𝐸+𝑚 −𝑝 𝑥 −𝑖 𝑝 𝑦 −𝐸−𝑚 𝑝 𝑧 −𝐸−𝑚 0 1 1 0 0 1 0 0 0 0 0 0 0 0 −1 0 0 −1 −𝑝 𝑧 −𝐸−𝑚 −𝑝 𝑥 −𝑖 𝑝 𝑦 −𝐸−𝑚 1 0 = ∙∙∙ Easy, just do it!

Useful relations: completeness realize: 44 matrices! with: you get: using: done:

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!

Lecture 7 Study Thomson 4 & 5

Today 2 March problem sessions SP A1.30 & SP D1.162 15:00 – 17:00

Elementary Particle Physics Quantum ElectroDynamics (QED) Lectures 4, 5, 6, 7, 8 Mark Thomson: Chapters 4, 5 & 6 Halzen&Martin: Chapter 4 Frank Linde, Nikhef, H044, f.linde@nikhef.nl, 020-5925140

Dirac equation: free particles

covered Dirac equation: 𝒎𝝍=𝒊   𝝏  𝝍 𝒖 𝟏 𝑬, 𝒑 = 𝐸+𝑚 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 Free particle solutions: 𝒖 𝟏 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 𝒖 𝟐 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 𝒗 𝟏 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 𝒗 𝟐 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 and are you able to boost & rotate spinors? Spin – Helicity – Chirality ½ 𝜮 =½ 𝝈 𝟎 𝟎 𝝈 + ½  ½ 𝜸 𝟓 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 ½ 𝜮 ∙ 𝒑 Normalisation – Orthogonality – Completeness 𝒖 𝒖=+𝟐𝒎 𝒗 𝒗=−𝟐𝒎 𝒖 𝟏 𝒖 𝟐 =𝟎 𝒗 𝟏 𝒗 𝟐 =𝟎 𝒔=𝟏,𝟐 𝒖 𝒔 𝒖 𝒔 =𝒑+𝒎

covered Dirac equation: 𝒎𝝍=𝒊   𝝏  𝝍 𝒖 𝟏 𝑬, 𝒑 = 𝐸+𝑚 1 0 𝑝 𝑧 𝐸+𝑚 𝑝 𝑥 +𝑖 𝑝 𝑦 𝐸+𝑚 Free particle solutions: 𝒖 𝟏 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 𝒖 𝟐 𝑬, 𝒑 𝒆 −𝒊𝒑∙𝒙 𝒗 𝟏 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 𝒗 𝟐 𝑬, 𝒑 𝒆 +𝒊𝒑∙𝒙 and are you able to boost & rotate spinors? Spin – Helicity – Chirality ½ 𝜮 ∙ 𝒑 𝜸 𝟓 = 𝟎 𝟎 𝟎 𝟎 𝟏 𝟎 𝟎 𝟏 𝟏 𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 ½ 𝜮 =½ 𝝈 𝟎 𝟎 𝝈 + ½  ½ Normalisation – Orthogonality – Completeness 𝒔=𝟏,𝟐 𝒗 𝒔 𝒗 𝒔 =𝒑−𝒎 𝒖 𝒖=+𝟐𝒎 𝒖 𝟏 𝒖 𝟐 =𝟎 𝒗 𝒗=−𝟐𝒎 𝒗 𝟏 𝒗 𝟐 =𝟎

Dirac equation: interactions

Physicist’s prime process Future: ILC, CLIC, … (250-500-3000 GeV) 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 + eA with compensates 0 in front of / t term f i V As for KG eqn., you can find the transition amplitude Tfi in terms of a ‘transition current’: KG, S=0 1 Dirac, S=1/2 ui uf Compare: 𝒊𝒆 𝒑 𝒇 + 𝒑 𝒊 𝝁 𝒊𝒆 𝜸 𝝁 time With:

e e scattering B A pA pB C D pC pD muon electron C D pC pD Just like for  K  K scattering for S=0 AC transition current: BD transition current: q Scatter electron (muon) from A calculated from ( ) (q=pA pC=pD pB) 1 Transition amplitude Tfi becomes: 1 𝒖 𝑩 𝒖 𝑫 𝒊𝒆 𝜸 𝝁 1 −𝒊 𝒈 𝝁 𝒒 𝟐 𝒊𝒆 𝜸  1 𝒖 𝑨 𝒖 𝑪 time

Feynman rules S=½ electrodynamics External lines: Vertex: ie  −𝒊𝑴= 𝒊𝒆 𝒖 𝑪 𝜸 𝝁 𝒖 𝑨 × −𝒊 𝒈 𝝁𝝂 𝒒 𝟐 × 𝒊𝒆 𝒖 𝑫 𝜸 𝝂 𝒖 𝑩 E.g. for e   e   you find: Propagators: q

Key processes in QED time Möller scattering Compton scattering Bhabha scattering pair annihilation pair creation

Real life examples: LEP e+e

Real life examples: LEP e+e

Real life examples: LEP e+e

Real life examples: LEP e+e particle identification detector

Real life examples: LEP e+e

Normalization process For any measured process  = Nevents/Normalisation Characteristics: ‘back-to-back’ energy = Ebeam mostly @ small angles Luminosity monitor measures ‘simple’ well-known process e+ e elastic e+e scattering e e+

Measured distributions  Max(E+ ,E ) Min(E+ ,E )  + For any other process:  = Constant  Nevents/Nluminosity

QED @ work: higher order processes

Now towards real calculations For most experiments: spins of incoming & outgoing particles remain unobserved What does this mean? Note: squares because spin states can be measured We should sum over outgoing spins We should average over incoming spins For A+B  C+D this means: A B C D Consider e.g.: ee ee  ie the ‘’: interchange 2 fermions! 

Non-relativistic: tedious 𝑴=− 𝒆 𝟐 𝒖 𝑪 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑫 𝜸 𝝁 𝒖 𝑩 𝒑 𝑨 − 𝒑 𝑪 𝟐 + 𝒆 𝟐 𝒖 𝑫 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑪 𝜸 𝝁 𝒖 𝑩 𝒑 𝑨 − 𝒑 𝑫 𝟐 =− 𝒆 𝟐 𝒖 𝑪 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑫 𝜸 𝝁 𝒖 𝑩 𝒕 + 𝒆 𝟐 𝒖 𝑫 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑪 𝜸 𝝁 𝒖 𝑩 𝒖 Non relativistic limit, |p| 0, gives: incoming outgoing With this current terms reduce to: I.e. spins do not flip & only matrix elements  0:

Non-relativistic: tedious 𝑴=− 𝒆 𝟐 𝒖 𝑪 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑫 𝜸 𝝁 𝒖 𝑩 𝒕 + 𝒆 𝟐 𝒖 𝑫 𝜸 𝝁 𝒖 𝑨 × 𝒖 𝑪 𝜸 𝝁 𝒖 𝑩 𝒖 Collecting terms  0 & squaring: 𝑴 𝟐 = 𝟐 𝟒 𝒆 𝟐 𝟒 𝒎 𝟐 𝟐 𝟏 𝒕 − 𝟏 𝒖 𝟐 + 𝟏 𝒕 𝟐 + 𝟏 𝒖 𝟐 The leading factor 2/4 comes from: 2: sum over outgoing spins 4=22: averaging incoming spins Relating t & u in c.m. frame to 𝒑 and  : 𝒕=−𝟐 𝒑 𝟐 𝟏−𝐜𝐨𝐬𝜽 =−𝟒 𝒑 𝟐 𝐬𝐢𝐧 𝟐 𝜽 𝟐 𝒖=−𝟐 𝒑 𝟐 𝟏+𝐜𝐨𝐬𝜽 =−𝟒 𝒑 𝟐 𝐜𝐨𝐬 𝟐 𝜽 𝟐 Hence: And with master formula for d/d :

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 ….

Today 7 March problem sessions SP B0.207 & SP B0.209 13:00 – 15:00

Feynman rules S=½ electrodynamics External lines: Vertex: ie  −𝒊𝑴= 𝒊𝒆 𝒖 𝑪 𝜸 𝝁 𝒖 𝑨 × −𝒊 𝒈 𝝁𝝂 𝒒 𝟐 × 𝒊𝒆 𝒖 𝑫 𝜸 𝝂 𝒖 𝑩 E.g. for e   e   you find: Propagators: q

Relativistic: work once  joy forever Take: e e (1 diagram) instead of ee ee (2 diagrams) A B C D e  q = pApC = pDpB  |M|2 becomes: Spin summation (final states) & averaging (initial states) yields:

Lepton tensor  Casimir’s trick!

Products & Traces of -matrices weak interaction Traces of any odd number of  matrices = 0

And the pay-off: e+e + cross section Use the Feyman rules: q 𝒊𝒆 𝜸 𝝁 𝒊𝒆 𝜸  −𝒊 𝒈 𝝁 𝒒 𝟐 𝒗 𝑩 𝒖 𝑨 𝒗 𝑫 𝒖 𝑪 p p’ −𝒊𝑴= 𝒊𝒆 𝒗 𝑩 𝜸 𝝁 𝒖 𝑨 × −𝒊 𝒈 𝝁𝝂 𝒑 𝑨 + 𝒑 𝑩 𝟐 × 𝒊𝒆 𝒖 𝑪 𝜸 𝝂 𝒗 𝑫 k k’ →𝑴=− 𝒆 𝟐 𝒗 𝑩 𝜸 𝝁 𝒖 𝑨 × 𝟏 𝒑 𝑨 + 𝒑 𝑩 𝟐 × 𝒖 𝑪 𝜸 𝝁 𝒗 𝑫 With Casimir’s trick & formulae for the Traces: 𝒆 𝟐 𝒌+𝒑 𝟐 𝟐 × 𝟏 𝟒 𝒗 𝒑 𝜸 𝝁 𝒖(𝒌) × 𝒖 𝒌 ′ 𝜸 𝝁 𝒗(𝒑′) × 𝒗 𝒑 ′ 𝜸 𝝂 𝒖(𝒌′) × 𝒖 𝒌 𝜸 𝝂 𝒗(𝒑) 𝑀 2 = = 𝒆 𝟐 𝒌+𝒑 𝟐 𝟐 × 𝟏 𝟒 ×𝐓𝐫𝐚𝐜𝐞 𝒑−𝒎 𝜸 𝝁 𝒌+𝒎 𝜸 𝝂 ×𝐓𝐫𝐚𝐜𝐞 𝒑′−𝑴 𝜸 𝝂 𝒌′+𝑴 𝜸 𝝁 = 𝒆 𝟐 𝒌+𝒑 𝟐 𝟐 × 𝟏 𝟒 ×𝐓𝐫𝐚𝐜𝐞 𝒑 𝜸 𝝁 𝒌 𝜸 𝝂 − 𝒎 𝟐 𝜸 𝝁 𝜸 𝝂 ×𝐓𝐫𝐚𝐜𝐞 𝒑′ 𝜸 𝝂 𝒌′ 𝜸 𝝁 −𝑴 𝟐 𝜸 𝝂 𝜸 𝝁 = 𝒆 𝟐 𝒌+𝒑 𝟐 𝟐 × 𝟏 𝟒 ×𝟒 𝒑 𝝁 𝒌 𝝂 + 𝒑 𝝂 𝒌 𝝁 −𝒑∙𝒌 𝒈 𝝁𝝂 − 𝒎 𝟐 𝒈 𝝁𝝂 ×𝟒 𝒑′ 𝝁 𝒌′ 𝝂 + 𝒑′ 𝝂 𝒌′ 𝝁 −𝒑′∙𝒌′ 𝒈 𝝁𝝂 −𝑴 𝟐 𝒈 𝝁𝝂 relativistic limit: m = M  0 = 𝒆 𝟐 𝒌+𝒑 𝟐 𝟐 ×𝟒 𝟐 𝒑∙𝒑′ 𝒌∙𝒌′ +𝟐 𝒑∙𝒌′ 𝒑′∙𝒌 +𝐦𝐚𝐬𝐬 𝐭𝐞𝐫𝐦𝐬 =𝟐 𝒆 𝟒 𝒔 𝟐 𝒕 𝟐 + 𝒖 𝟐

And the pay-off: e+e + cross section q 𝒊𝒆 𝜸 𝝁 𝒊𝒆 𝜸  −𝒊 𝒈 𝝁 𝒒 𝟐 𝒗 𝑩 𝒖 𝑨 𝒗 𝑫 𝒖 𝑪 𝑴 𝟐 = 𝒆 𝟐 𝒔 𝟐 ×𝟐× 𝒕 𝟐 + 𝒖 𝟐 Relativistic limit: And hence the differential cross section: Using: ⇒ 𝑡 2 + 𝑢 2 𝑠 2 = 1+ cos 2 𝜃 2 You find: and:

Measured e+e + cross section 𝝈 𝒕𝒐𝒕 = 𝟒𝝅 𝜶 𝟐 𝟑𝒔 ≈ 𝟖𝟕 𝒔 𝐧𝐛 𝜶≈𝟏/𝟏𝟑𝟕 𝐡𝐜 𝟐 ≈𝟑𝟖𝟗𝟑𝟕𝟗 𝐆𝐞𝐕 𝟐 nb Z  ee

Measured angular distributions ee ee

And … e+e + gives you e+e qq with three colours u, c, t d, s, b s udsc uds Rudsc = 3[4/9+1/9+1/9+4/9]=3.3 udsc no color Rudsc = 4/9+1/9+1/9+4/9]=1.1

And … e+e + gives you e+e  + Price to pay: can not neglect the -mass near threshold ….  m=1.7768 GeV -mass determined via very accurate threshold scan @ BES/Bejing!

* Other processes  Compton scattering: e  e k k’ p p’ q=p+k q=p-k’ Compton scattering: e  e p k k’ p’ e+ e  Pair creation: e+e  

Lecture 8 Study Thomson 6