Particle Physics Dr M.A. Thomson Part II, Lent Term 2004 HANDOUT V

Transcription

1 Particle Physics Dr M.A. Thomson (ifl μ m)ψ = Part II, Lent Term 24 HANDOUT V Dr M.A. Thomson Lent 24

2 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 ) The DIRAC EQUATION Dr M.A. Thomson Lent 24

3 SPIN complicates things... The process e - e + γ represents the sum over all possible spin states 3 M jmj 2! X i! j X i M i M i j 2 = since ORTHOGONAL " SPIN states. X 2ß ff = 4 i X i jm i j 2 ρ(e f ) jm i j 2 Cross-section : sum over all spin assignments, averaged over initial spin states. # Dr M.A. Thomson Lent 24

4 The Klein-Gordon Equation Revisited 4 Schrödinger Equation for a free particle can be written = 2m r2 ψ Derivatives : st order in time and 2nd order in space coordinates ) not Lorentz invariant From Special Relativity: E 2 = p 2 + m 2 from Quantum Mechanics: ^E = ; ^p = Combine to give the Klein-Gordon 2 2 = (r 2 m 2 )ψ Second order in both space and time derivatives - by construction Lorentz invariant. Negative energy solutions ) anti-particles BUT negative energy solutions also give negative particle densities!? ψ Λ ψ < Try another approach... Dr M.A. Thomson Lent 24

5 Weyl Equations 5 (The massless version of the Dirac Equation) Klein-Gordon Eqn. for massless particles: 2 r 2 )ψ = i:e: (^E 2 ^p 2 )ψ = Try to factorize 2nd Order KG equation equation linear in @t + ~ff:r)ψ = with as yet undetermined constants ~ff Gives the two decoupled WEYL Equations (ff x )ψ both linear in space and time derivatives. BUT must satisfy the Klein-Gordon Equation i.e. in operator form must satisfy (^E ~ff:^p)(^e + ~ff:^p)ψ = (^E 2 ^p 2 )ψ = Dr M.A. Thomson Lent 24

6 Weyl equations give: 6 (^E 2 ff x^p x ff x^p x ff y^p y ff y^p y ff z^p z ff z^p z ff x^p x ff y^p y ff y^p y ff x^p x ff y^p y ff z^p z ff z^p z ff y^p y ff z^p z ff x^p x ff x^p x ff z^p z )ψ = Therefore in order to recover the KG equation: (^E 2 ^p x^p x ^p y^p y ^p z^p z ) = ; require: ff x 2 = ff y 2 = ff z 2 = (ff x ff y + ff y ff x ) = etc: ) ff x, ff y, ff z ANTI-COMMUTE The simplest choice for ff are the Pauli spin matrices ff x = ; ff y = -i i ; ff z = Hence solutions to the Klein-Gordon equation (^E ~ff:^p)(^e + ~ff:^p)ψ = are given by the Weyl Equations: (^E ~ff:^p)ffi = (^E + ~ff:^p)χ = Since ff i are 2 2 matrices, need 2 component wave-functions - WEYL SPINORS. ffi = N ffi e ffi 2 i(et ~p:~r) - The wave-function is forced to have a new degree of freedom - the spin of the fermion. Dr M.A. Thomson Lent 24

7 Consider the FIRST Weyl Equation 7 (^E ~ff:^p)ffi )ffi = For a plane wave solution: ffi = N ψ the first Weyl Equation gives ffi ffi 2! e i(et ~p:~r) (E ff x p x ff y p y ff z p z )ffi = (E ~ff:~p)ffi = where ~ff:~p = ff x p x + ff y p y + ff z p z p = z p x ip y p x + ip y p z Hence for the first WEYL equation, the SPINOR solutions of (E ~ff:~p)ffi = are given the coupled equations: ) ) p zffi + (p x ip y )ffi 2 = Effi (p x + ip y )ffi p z ffi 2 = Effi 2 Dr M.A. Thomson Lent 24

8 Choose ~p along z axis, i.e. p z = jpj: 8 (E jpj)ffi = (E + jpj)ffi 2 = There are two solutions ffi + = The first solution, ffi + = and ffi =., requires E = +jpj, i.e. a positive energy (particle) solution. Similarly, the second ffi =, requires E = jpj, i.e. a negative energy (anti-particle) solution. Back to the FIRST WEYL equation (E ~ff:^p)ffi = ~ff:^pffi = Effi ~ff:^p jpj ffi = E jpj ffi = ( + E > E < The solutions of the WEYL equations are Eigenstates of the HELICITY operator. ^H = ~ff:^p jpj with Eigenvalues + and respectively. 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 24 σ p

9 Interpret the two solutions of the FIRST WEYL equation as a RIGHT-HANDED H = + particle and a LEFT-HANDED anti-particle H =. 9 E >, H = + E <, H = - RH particle LH anti-particle The SECOND WEYL equation: (^E + ~ff:^p)χ = has LEFT-HANDED particle and RIGHT-HANDED anti-particle solutions. E <, H = + E >, H = - RH anti-particle LH particle SUMMARY: By factorizing the Klein-Gordon equation into a form linear in the derivatives ) 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 24

10 Spin in the Fundamental Interactions The ELECTROMAGNETIC, STRONG, and WEAK interactions are all mediated by VECTOR (spin-) fields. In the massless limit, the fundamental fermion states are eigenstates of the helicity operator. HANDEDNESS (CHIRALITY) plays a central rôle 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 e + e! μ + μ Of the 6 possibilities ONLY the following SPIN combinations contribute to the cross-section All other SPIN combinations give zero jm i j 2 Dr M.A. Thomson Lent 24

11 Solutions of the Weyl Equations Consider the general case of a particle with travelling at an angle with respect to the z-axis p z = jpj cos p x = jpj sin WEYL (^E ^ff:^p)ffi = (^ff:^p)ffi = ^Effi ) ) p zffi + p x ffi 2 = Effi p x ffi p z ffi 2 = Effi 2 For the positive energy solution E = +jpj: ) Therefore ) ffi cos + ffi 2 sin = ffi ffi sin ffi 2 cos = ffi 2 ) ffi = ffi 2 sin ( cos ) ffi ffi 2 = ffi = cos ffi 2 sin 2 2sin 2 cos 2 ( cos sin2 2 ) Normalizing such that ffi 2 + ffi 2 2 = gives: ffi = cos 2 ffi 2 = sin 2 Dr M.A. Thomson Lent 24

12 So the positive energy solution to the first Weyl equation (RH particle) gives ψ + cos 2! ffi RH = N + sin e i(et ~p:~r) RH fermion 2 This is still a Eigenvalue of the helicity operator with (H = +) i.e. a RH particle but now referred to an external axis. 2 The positive energy solution to the SECOND Weyl equation (LH particle) gives ψ - sin! χ LH = N 2 + cos e i(et ~p:~r) LH fermion 2 Not much more than the Quantum Mechanical rotation properties of spin- 2. The spin part of a RH particle/anti-particle wave-function can be written ψ R ( ) = ψ + cos 2 + sin 2! Similarly for a LH particle/anti-particle ψ L ( ) = ψ - sin! 2 + cos 2 = cos 2 " + sin 2 # = sin 2 " + cos 2 # For particles/anti-particles with momentum ~p( ), i.e. an angle + ß to the z-axis: ψ R ( + ß) = sin 2 " + cos 2 # ψ L ( + ß) = cos 2 " sin 2 # Dr M.A. Thomson Lent 24

13 Application to e + e! μ + μ 3 Four helicity combinations contribute to the cross-section. Consider the first diagram e R e+ L! μ R μ+ L With the e direction defining the z axis: The spin parts of the wave-functions : ψ e e+ R L ψ μ μ+ R L = ψ R ()ψ L (ß)="" = ψ R ( )ψ L ( + ß) =-cos 2 2 ""- cos 2 sin 2 ("#+#")-sin2 2 ## Giving the contribution the matrix element: jm j 2 = jhψ μ μ+ j e2 R L q jψ 2 e e+ ij 2 R L = e4 q 4 jcos2 2 j2 = e4 2 ( + cos ) 2 q 4 2 Dr M.A. Thomson Lent 24

14 Perform same calculation for the four allowed helicity combinations 4 M α cos 2 (θ/2) 2 (+cosθ) M 2 α sin 2 (θ/2) 2 (-cosθ) M 3 α sin 2 (θ/2) 2 (-cosθ) M 4 α cos 2 (θ/2) 2 (+cosθ) 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) : jmj 2 = 4 Φ jm j 2 + jm 2 j 2 + jm 3 j 2 + jm 4 j 2Ψ jmj 2 = 4 e 4 q 4 f 4 ( + cos )2 + 4 ( cos )2 + 4 ( cos )2 + 4 ( + cos )2 g jmj 2 = e4 4q 4 ( + cos2 ) Nothing more than the QM properties of a SPIN- particle decaying to two SPIN- 2 particles Dr M.A. Thomson Lent 24

15 e + e - α p µ 2 p µ γ q 2 f Q f α f e - θ e + Electron/Positron beams along z-axis (q 2 = 4E 2 = s) f f 5 Using the spin-averaged matrix element jmj 2 = e4 4q 4 ( + cos2 ) jmj 2 = (4ßff)2 4s 2 ( + cos 2 ) dff dω = 2ßjMj2 E2 (2ß) 3 = 2ß (4ßff)2 4s 2 ( + cos 2 ) s 4 = ff2 Q 2 f 4s ( + cos2 ) (2ß) 3 dff dj cos j for e + e! qq. The angle is determined from the measured directions of the jets. j cos j is plotted since it is not possible to uniquely identify which jet corresponds to the quark and which corresponds to the antiquark. The curve shows the expected ( + cos 2 ) distribution. QUARKS are SPIN- 2 Dr M.A. Thomson Lent 24

16 (yet again) Total cross section for e + e! ff ff = = Z Z 2ß dff dω dω Z ß = ßff2 Q 2 f 2s ff 2 Q 2 f 4s ( + cos2 )sin d dffi Z + ( + y 2 )dy (y = cos ) 6 = 4ßff2 Q 2 f 3s ff(e + e! μ + μ ) = 4ßff2 3s ff(e + e! μ + μ ) for e + e collider data at centre-of-mass energies 8-36 GeV This is the complete lowest order calculation of the e + e! μ + μ cross-section (in the limit of massless fermions). Dr M.A. Thomson Lent 24

17 The Dirac Equation 7 NON-EXAMINABLE WEYL Equations describe massless SPIN- particles. But 2 all known fermions are MASSIVE. Again start from the KG 2 2 = (r 2 m 2 )ψ ^H 2 ψ = (^p 2 + m 2 )ψ Write down equation LINEAR in space and time derivatives ^Hψ = (~ff:^p + fim)ψ and require it to be a solution of the KG equation: ^Hψ = (ff x :^p x + ff y :^p y + ff z :^p z + fi:m)ψ ^H 2 ψ = ff 2 i :^p 2 x + ::: +(ff x ff y + ff y ff x )^p x^p y + ::: +(ff x fi + fiff x )^p x m + ::: +fi 2 m 2 For this to satisfy Klein-Gordon equation: ^H 2 ψ = (^p x 2 + ^p y 2 + ^p z 2 + m 2 )ψ require ff i 2 = fi 2 = (i = x; y; z) ff i ff j + ff j ff i = (i 6= j) ff i fi + fiff i = Now require 4 anti-commuting matrices. Dr M.A. Thomson Lent 24

18 The Dirac Equation: 8 ^Hψ = (~ff:^p + fim)ψ Can be written in a slightly different = ( i~ff:r + = ( ifi~ff:r + + ifi~ff:r- fi2 m)ψ fi 2 m)ψ = with fl μ = (fi; fi~ff) Giving (ifl μ m)ψ = with (fl ) 2 = (fl ) 2 = (fl 2 ) 2 = (fl 3 ) 2 = (fl i fl j fl j fl i ) = (i 6= j) Identify the fl μ as matrices which must satisfy the anti-commutation relations above. The Pauli spin matrices provide only 3 anti-commuting matrices and the lowest dimension matrices satisfying these requirements are 4 4. The fl-matrices are closely related to the 2 2 Pauli spin matrices. Dr M.A. Thomson Lent 24

19 fl = fl = fl 2 = fl 3 = also i i - - fl 5 = ifl fl fl 2 fl 3 = A = - A = A = A ffx -ff x ffy -ff y ffz -ff z A = Solutions to the Dirac Equation are written as four-component Dirac SPINORs ψ = ψ 2 A ψ 3 ψ 4 NOTE: this is not the only possible representation of the fl matrices - just the most commonly used Dr M.A. Thomson Lent 24 9

20 Rest Frame Solutions of the Dirac Equation 2 Dirac Equation: (ifl μ m)ψ m)ψ = Consider a particle at REST: p x = ψ =, Dirac Equation becomes: i@ - m)ψ =@t 2 3 =@t A = 4 =@t = mψ ; 2 = Giving two orthogonal E = +m solutions: u (t) Ae imt ; u 2 (t) ψ ψ 2 ψ 3 ψ 4 A Ae imt i.e. positive energy spin-up and spin-down PARTICLES The two other equations = mψ 3 ; = mψ 4 give two orthogonal E = m solutions: u 3 (t) Ae +imt ; u 4 (t) Ae +imt i.e. ve energy spin-up and spin-down ANTI-PARTICLES Dr M.A. Thomson Lent 24

21 The DIRAC equation 2 gives PARTICLE/ANTI-PARTICLE solutions requires the particles/anti-particles to have an additional degree of freedom (SPIN)! in the massless limit, the DIRAC equation reduces to the two uncoupled WEYL equations In general the Dirac Equation gives FOUR simultaneous equations for the components of the SPINOR. e.g. more general solutions u = N p z (E+m) (p x +ip y ) (E+m) C A ; u 2 = N (p x ip y ) (E+m) p z (E+m) C A For (u ; u 2 ) E = p p 2 + m 2 u 3 = N p z (E m) (p x +ip y ) (E m) C A ; u 4 = N (p x ip y ) (E m) p z (E m) C A For (u 3 ; u 4 ) E = p p 2 + m 2 The DIRAC equation lives in the realm of PART III Particle Physics. Dr M.A. Thomson Lent 24

22 Lorentz Structure of Interactions 22 NON-EXAMINABLE e - e - e γ q 2 Electron Current Propagator p e p Proton Current Matrix element M factorises into 3 terms : im = hu e jiefl μ ju e i Electron Current igμν q 2 Photon Propagator hu p jiefl ν ju p i Proton Current Fermions are 4-component SPINORS. ) interaction enters as 4 4 matrices. Lorentz invariance allows only five possible forms for the interaction: SCALAR uu, PSEUDO-SCALAR ufl 5 u, VECTOR ufl μ u, AXIAL-VECTOR ufl μ fl 5 u, TENSOR uff μν u Electro-magnetic and Strong forces are VECTOR interactions - which determines the HELICITY structure. Treats helicity states symmetrically ) PARITY CONSERVATION The WEAK interaction has a different form: (V-A) i.e. fl μ ( fl 5 ). Projects out a single helicity combination : ) PARITY VIOLATION The WEAK interaction is the subject of next lecture Dr M.A. Thomson Lent 24