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Particle Physics HANDOUT V

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Particle Physics HANDOUT V 1 Particle Physics Dr M.A. Thomson ´­ ѵ ¼ µ- µ- e- e+ e- e+ µ+ µ+ µ- µ- e- e+ e- e+ µ+ µ+ Part II, Lent Term 2004 HANDOUT V Dr M.A. Thomson Lent 2004 2 Spin, Helicity and the Dirac Equation ★ Upto this point we have taken a hands-off approach to “spin”. ★ Scattering cross sections calculated for spin-less particles ★ To understand the WEAK interaction need to understand SPIN ★ Need a relativistic theory of quantum mechanics that includes spin ★ µ The DIRAC EQUATION Dr M.A. Thomson Lent 2004 3 ★ SPIN complicates things.... e- µ- The process γ + e+ µ represents the sum over all possible spin states µ- µ- µ- µ- e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ- e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ- e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ- e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ Å Å ¾ ¾ ¾ Å Å Å since ORTHOGONAL SPIN states. ½ ¾ ¾ Å ´ µ Cross-section : sum over all spin assignments, averaged over initial spin states. Dr M.A. Thomson Lent 2004 4 The Klein-Gordon Equation Revisited Schrodinger¨ Equation for a free particle can be written as ½ ¾ Ö Ø ¾Ñ Derivatives : 1st order in time and 2nd order in space coordinates µ not Lorentz invariant From Special Relativity: ¾ ¾ ¾ Ô · Ñ from Quantum Mechanics: Ô Ö Ø Combine to give the Klein-Gordon Equation: ¾ ¾ ¾ ´Ö Ñ µ ¾ Ø Second order in both space and time derivatives - by construction Lorentz invariant. ★ Negative energy solutions µ anti-particles ★ BUT negative energy solutions also give negative particle densities !? £ ¼ Try another approach..... Dr M.A. Thomson Lent 2004 5 Weyl Equations (The massless version of the Dirac Equation) Klein-Gordon Eqn. for massless particles: ¾ ¾ ´ Ö µ ¼ ¾ Ø ¾ ¾ ´ Ô µ ¼ Try to factorize 2nd Order KG equation ➔ equation linear in Ö AND : Ø Öµ´ · Öµ ¼ ´ Ø Ø with as yet undetermined constants Gives the two decoupled WEYL Equations ´ · · µ · Ü Ý Þ Ü Ý Þ Ø · · µ ´ Ý Þ Ü Ü Ý Þ Ø both linear in space and time derivatives. BUT must satisfy the Klein-Gordon Equation i.e. in operator form ´ Ôµ´ · Ôµ ¼ must satisfy ¾ ¾ ´ Ô µ ¼ Dr M.A. Thomson Lent 2004 6 Weyl equations give: ¾ ´ Ô Ô Ô Ô Ô Ô Ü Ü Ý Ý Þ Þ Ü Ü Ý Ý Þ Þ Ô Ô Ô Ô Ü Ý Ý Ü Ü Ý Ý Ü Ô Ô Ô Ô Ý Þ Þ Ý Ý Þ Þ Ý Ô Ô Ô Ô µ ¼ Þ Ü Ü Þ Þ Ü Ü Þ Therefore in order to recover the KG equation: ¾ ´ Ô Ô Ô Ô Ô Ô µ ¼ Ü Ü Ý Ý Þ Þ require: ¾ ¾ ¾ ½ Ü Ý Þ ´ · µ ¼ Ø Ü Ý Ý Ü µ Ü Þ , Ý , ANTI-COMMUTE The simplest choice for are the Pauli spin matrices ½ ¼ ¼ ½ ¼ - Þ Ý Ü ¼ ½ ¼ ½ ¼ - Hence solutions to the Klein-Gordon equation ´ Ôµ´ · Ôµ ¼ are given by the Weyl Equations: Ôµ ¼ ´ ´ · Ôµ ¼ ¾ ¢ ¾ Since are matrices, need 2 component wave-functions - WEYL SPINORS. ½ ´Ø Ô Öµ Æ ¾ The wave-function is forced to have a new degree of freedom - the spin of the fermion. Dr M.A. Thomson Lent 2004 7 Consider the FIRST Weyl Equation ´ Ôµ ¼ ´ · · · µ ¼ Ü Ý Þ Ø Ü Ý Þ For a plane wave solution: ½ ´Ø Ô Öµ Æ ¾ the first Weyl Equation gives ´ Ô Ô Ô µ ¼ Ü Ü Ý Ý Þ Þ ´ Ô µ ¼ where Ô Ô · Ô · Ô Ü Ü Ý Ý Þ Þ Ô Ô Ô Þ Ü Ý Ô · Ô Ô Ü Ý Þ Hence for the first WEYL equation, the SPINOR solutions of Ô µ ¼ ´ are given the coupled equations: µ Ô · ´Ô Ô µ Þ ½ Ü Ý ¾ ½ µ ´Ô · Ô µ Ô Ü Ý ½ Þ ¾ ¾ Dr M.A. Thomson Lent 2004 8 ★ Ô Þ Ô Ô Choose along axis, i.e. Þ : ´ Ôµ ¼ ½ ´ · Ôµ ¼ ¾ ½ ¼ There are two solutions · and . ¼ ½ ½ ★ ·Ô The first solution, · , requires , i.e. ¼ a positive energy (particle) solution. Similarly, the second ¼ Ô , requires , i.e. a negative energy ½ (anti-particle) solution. ★ Back to the FIRST WEYL equation ´ Ôµ ¼ Ô ´ ·½ ¼ Ô Ô Ô ½ ¼ ★ The solutions of the WEYL equations are Eigenstates of the HELICITY operator. Ô À Ô ½ with Eigenvalues ·½ and respectively. σ p HELICITY is the projection of a particle’s SPIN onto its flight direction. (Recall WEYL equations applicable for massless particles) Dr M.A. Thomson Lent 2004 9 ★ Interpret the two solutions of the FIRST WEYL equation ·½ as a RIGHT-HANDED À particle and a ½ LEFT-HANDED anti-particle À . E > 0, H = +1 E < 0, H = -1 RH particle LH anti-particle ★ The SECOND WEYL equation: ´ · Ôµ ¼ has LEFT-HANDED particle and RIGHT-HANDED anti-particle solutions. E < 0, H = +1 E > 0, H = -1 RH anti-particle LH particle SUMMARY: ★ By factorizing the Klein-Gordon equation into a form linear in the derivatives µ force particles to have a non-commuting degree of freedom, SPIN ! ★ Still obtain anti-particle solutions ★ Probability densities always positive ★ ‘Natural’ states are the Helicity Eigenstates ★ Weyl Equations are the ultra-relativistic (massless) limit of the Dirac Equation Dr M.A. Thomson Lent 2004 10 Spin in the Fundamental Interactions ★ The ELECTROMAGNETIC, STRONG, and WEAK interactions are all mediated by VECTOR (spin-1) fields. In the massless limit, the fundamental fermion states are eigenstates of the helicity operator. HANDEDNESS ★ (CHIRALITY) plays a central roleˆ in the interactions between the field bosons and the fermions; the only allowed couplings are: LH particle to LH particle RH particle to RH particle LH anti-particle to LH anti-particle RH anti-particle to RH anti-particle LH particle to RH anti-particle RH particle to LH anti-particle · · EXAMPLE Of the 16 possibilities ONLY the following SPIN combinations contribute to the cross-section µ- µ- e- e+ e- e+ µ+ µ+ µ- µ- e- e+ e- e+ µ+ µ+ ¾ Å All other SPIN combinations give zero Dr M.A. Thomson Lent 2004 11 Solutions of the Weyl Equations Consider the general case of a particle with travelling at an Þ angle with respect to the -axis Ô Ô Ó× Þ Ô Ô ×Ò Ü Ôµ ¼ WEYL 1 ´ ´ Ôµ µ Ô · Ô Þ ½ Ü ¾ ½ µ Ô Ô Ü ½ Þ ¾ ¾ ·Ô For the positive energy solution : µ Ó× · ×Ò ½ ¾ ½ µ ×Ò Ó× ½ ¾ ¾ ×Ò µ ½ ¾ ´½ Ó× µ Therefore ¾×Ò Ó× ½ ¾ ¾ ¾ ¾ ´½ Ó× · ×Ò µ ¾ ¾ ¾ Ó× ½ ¾ ×Ò ¾ ¾ ¾ ¾ · ½ ½ Normalizing such that ¾ gives: Ó× ½ ¾ ×Ò ¾ ¾ Dr M.A. Thomson Lent 2004 12 ★ So the positive energy solution to the first Weyl equation (RH particle) gives + Ó× ´Ø Ô Öµ ¾ Æ ÊÀ ÖÑÓÒ ÊÀ + ×Ò ¾ This is still a Eigenvalue of the helicity operator with À ·½µ ´ i.e. a RH particle but now referred to an external axis. The positive energy solution to the SECOND Weyl equation (LH particle) gives - ×Ò ´Ø Ô Öµ ¾ Æ ÄÀ ÖÑÓÒ ÄÀ + Ó× ¾ Not much more than the Quantum Mechanical rotation properties of spin- ½ . ¾ ★ The spin part of a RH particle/anti-particle wave-function can be written + Ó× ¾ ´ µ · ×Ò Ó× Ê ¾ ¾ + ×Ò ¾ ★ Similarly for a LH particle/anti-particle - ×Ò ¾ · Ó× ´ µ ×Ò Ä ¾ ¾ + Ó× ¾ Ô´ µ For particles/anti-particles with momentum , i.e. an · angle to the z-axis: · Ó× ´ · µ ×Ò Ê ¾ ¾ ×Ò ´ · µ Ó× Ä ¾ ¾ Dr M.A. Thomson Lent 2004 13 · · Application to µ- µ- e- e+ e- e+ ★ Four helicity combinations µ+ µ+ - - contribute to the cross-section. µ µ e- e+ e- e+ µ+ µ+ ★ Consider the first diagram µ- · · - + Ä Ê Ä Ê e e With the direction defin- µ+ ing the Þ axis: ★ The spin parts of the wave-functions : ´¼µ ´ µ · Ä = Ê = Ê Ä ´ µ ´ · µ · Ä = Ê Ä Ê ¾ ¾ Ó× Ó× µ ×Ò ×Ò ´ =- - + - ¾ ¾ ¾ ¾ ★ Giving the contribution the matrix element: ¾ ¾ ¾ Å · · ½ ¾ Ê Ä Ê Ä Õ ¾ ¾ Ó× ¾ Õ ¾ ½ ¾ ´½ · Ó× µ Õ ¾ Dr M.A. Thomson Lent 2004 14 ★ Perform same calculation for the four allowed helicity combinations µ- - + α 2 θ e e M1 cos ( /2) µ+ → 1 θ 2 (1+cos ) µ- - + α 2 θ e e M2 sin ( /2) µ+ → 1 θ 2 (1-cos ) µ- - + α 2 θ e e M3 sin ( /2) µ+ → 1 θ 2 (1-cos ) µ- - + α 2 θ e e M4 cos ( /2) µ+ → 1 θ 2 (1+cos ) ★ For unpolarized electron/positron beams : each of the above process contributes equally. Therefore SUM over all matrix elements and AVERAGE over initial spin states. Giving the total Matrix Element (remember spin-states are orthogonal so sum squared matrix elements) : ¨ © ½ ¾ ¾ ¾ ¾ ¾ Å Å · Å · Å · Å ½ ¾ ¿ ½ ½ ½ ¾ ¾ ¾ ´½ · Ó× µ · ´½ Ó× µ · Å Õ ½ ½ ¾ ¾ ´½ Ó× µ · ´½ · Ó× µ ¾ ¾ ´½ · Ó× µ Å Õ ★ Nothing more than the QM properties of a SPIN-1 particle decaying to two SPIN- ½ particles ¾ Dr M.A. Thomson Lent 2004 15 + f e µ f p2 γ θ √α √α - + Qf e e µ 1 p - 1 q2 e f f ¾ ¾ ´Õ ×µ Electron/Positron beams along Þ -axis Using the spin-averaged matrix element ¾ ¾ ´½ · Ó× µ Å Õ ¾ ´«µ ¾ ¾ Å ´½ · Ó× µ ¾ × ¾ ¾ ¾ Å ¿ ª ´¾ µ ¾ ´«µ × ½ ¾ ¾ ´½ · Ó× µ ¾ ¿ × ´¾ µ ¾ ¾ « É ¾ ´½ · Ó× µ × · ÕÕ for . Ó× The angle is determined from the measured directions of the Ó× jets. is plotted since it is not possible to uniquely identify which jet corresponds to the quark and which corresponds to the anti- quark. The curve shows the ex- ¾ · Ó× µ pected ´½ distribution. QUARKS are SPIN- ½ ¾ Dr M.A. Thomson Lent 2004 16 (yet again) · Total cross section for ª ª ¾ ¾ ¾ « É ¾ ´½ · Ó× µ×Ò × ¼ ¼ ¾ ¾ ·½ « É ¾ ´½ · Ý µÝ ´Ý Ó× µ ¾× ½ ¾ ¾ « É ¿× ¾ « · · ´ µ ¿× · · ´ µ · for collider data at centre-of-mass ener- gies 8-36 GeV ★ This is the complete lowest order calculation of the · · cross-section (in the limit of massless fermions).
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