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The World Particle content Interactions e, q e, q Schrodinger Wave Equation He started with the energy-momentum relation for a particle Expecting them to act on plane waves e iEt ipr e he made the quantum mechanical replacement: How about a relativistic particle? e ipx The Quantum mechanical replacement can be made in a covariant form. Just remember the plane wave can ipx be written in a covariant form: iEt ipr e e e As a wave equation, it does not work. It doesn’t have a conserved probability density. It has negative energy solutions. The proper way to interpret KG equation is it is actually a field equation just like Maxwell’s Equations. Consider we try to solve this eq as a field equation with a source. m2 j ( x) We can solve it by Green Function. G( x, x' ) m2 ( x, x' ) x x' G is the solution for a point-like source at x’. By superposition, we can get a solution for source j. ( x ) 0 ( x ) d 4 x ' G ( x, x ' ) j ( x ' ) Green Function for KG Equation: G( x, x' ) m2G( x, x' ) 4 x x' By translation invariance, G is only a function of coordinate difference: G ( x, x ' ) G ( x x ' ) The Equation becomes algebraic after a Fourier transformation. d 4 p ip( x x ') ~ G ( x x' ) e G ( p) 4 2 d 4 p ip( x x ') ( x x' ) e 2 4 4 ~ p m G ( p) 1 2 2 ~ G ( p) 1 p 2 m2 This is the propagator! x' KG Propagation x Green function is the effect at x of a source at x’. That is exactly what is represented in this diagram. The tricky part is actually the boundary condition. For those amplitude where time 2 is ahead of time 1, propagation is from 2 to 1. For those amplitude where time 1 is ahead of time 2, propagation is from 1 to 2. B B B 2 B 1 1 C C A A A 2 A is actually the sum of the above two diagrams! ~ To accomplish this, G ( p) 1 p 2 m2 ~ G ( p) 1 p 2 m 2 i Blaming the negative energy problem on the second time derivative of KG Eq., Dirac set out to find a first order differential equation. This Eq. still needs to give the proper energy momentum relation. So Dirac propose to factor the relation! For example, in the rest frame: Made the replacement i mc t First order diff. Eq. Now put in 3-momenta: Suppose the momentum relation can be factored into linear combinations of p’s: Expand the right hand side: We get and we need: It’s easier to see by writing out explicitly: Oops! What! 0 1 1 0 0 or No numbers can accomplish this! Dirac propose it could be true for matrices. 0 1 1 0 2 by 2 Pauli Matrices come very close 0 i 1 0 3 0 0 1 0 1 2 1 1 0 i , 0 1 i i j , i j ij i j Dirac find it’s possible for 4 by 4 matrices We need: that is He found a set of solutions: Dirac Matrices Dirac find it’s possible for 4 by 4 matrices Pick the first order factor: Make the replacement and put in the wave function: If γ’s are 4 by 4 matrices, Ψ must be a 4 component column: mc 0 i 0 1 2 3 t x y z It consists of 4 Equations. The above could be done for 2 by 2 matrices if there is no mass. Massless fermion contains only half the degrees of freedom. Now put in 3-momenta: Suppose the momentum relation can be factored into linear combinations of p’s: Expand the right hand side: We get and we need: k 1 1 2 3 交叉項抵銷 k 1 - 1 - 2 - 3 Plane wave solutions for KG Eq. ( x) a e ipx ae ip0t ip x 2 p02 a e ipx p a e ipx m 2 a e ipx 2 p p m2 p 2 m2c 2 0 2 0 E p0 There are two solutions for each 3 momentum p (one for +E and one for –E ) ( x) a e p ipx a e ipx a e p iEt ip x a e iEt ip x Expansion of a solution by plane wave solutions for KG Eq. ( x) a e ipx p b e ipx a e iEt ip x p iEt ip x iEt ip x b e If Φ is a real function, the coefficients are related: ( x) a e p ipx a e ipx a e p iEt ip x a e Plane wave solutions for Dirac Eq. ( x) a e ipx u a e Multiply on the left with p 2 m2c 2 u 0 ip0t ip x u A1 u u u A A 2 u B u B1 u B2 u p mc p 2 m2c 2 0 p0 E There are two sets of solutions for each 3 momentum p (one for +E and one for –E ) How about u? p0 p0 p0 p0 p0 p0 p0 We need: p0 p0 You may think these are two conditions, but no. Multiply the first by p p0 p 2 p 0 0 2 p0 1 p 2 p p0 0 2 So one of the above is not independent if p 2 m2c 2 0 We need: p 0 or How many solutions for every p? Go to the rest frame! p (m,0) or (m,0) p0 p ( m ,0 ) p0 p0 p0 p0 0 uA 0 0 0 2m u B uB 0 uA is arbitrary 1 0 u A , 0 1 Two solution (spin up spin down) p (m,0) p0 p0 p0 p0 2m 0 u A 0 0 uB 0 uA 0 uB is arbitrary 1 0 u B , 0 1 Two solution (spin down and spin up antiparticle) There are four solutions for each 3 momentum p (two for particle and two for antiparticle) p or 0 p0 It’s not hard to find four independent solutions. p0 p0 p0 p0 p0 p0 - We got two positive and two negative energy solutions! Negative energy is still here! In fact, they are antiparticles. Electron solutions: ae ipx u ae iEt ipr u Positron solutions: p mc u (E, p) 0 p mc v (E, p) 0 ae iEt ipr 3, 4 1, 2 u ( 3, 4 ) ae ipx (1, 2 ) v ae iEt ipr v Expansion of a solution by plane wave solutions for KG Eq. ( x) a e p ipx b e ipx a e iEt ip x p b e iEt ip x Expansion of a solution by plane wave solutions for Dirac Eq. ( x) a u e ipx b v eipx p Bilinear Covariants Ψ transforms under Lorentz Transformation: Interaction vertices must be Lorentz invariant. The weak vertices of leptons coupling with W W W -ig -ig e νe g e W e e e W ? e e μ νμ Bilinear Covariants Ψ transforms under Lorentz Transformation: Interaction vertices must be Lorentz invariant. How do we build invariants from two Ψ’s ? A first guess: Maybe you need to change some of the signs: It turns out to be right! We can define a new adjoint spinor: is invariant! In fact all bilinears can be classified according to their behavior under Lorentz Transformation: A ( x) b v e ipx a u eipx ( x) a u e ipx b v eipx p p u 0 e( p1 ) u p3 a p3 u p1 0 u p3uupa1 pe( pe3()p1 ) u 0 1 Feynman Rules for external lines How about internal lines? Find the Green Function of Dirac Eq. G( x, x' ) m2G( x, x' ) 4 x x' i G( x, x' ) mcG( x, x' ) I 4 x x' i G( x, x' ) mcG( x, x' ) I 4 x x' Now the Green Function G is a 4 ˣ 4 matrix i G( x, x' ) mcG( x, x' ) i I 4 x x' Using the Fourier Transformation d 4 p ip( x x ') ~ G ( x x' ) e G ( p) 4 2 d 4 p ip( x x ') ( x x' ) e 4 2 4 ~ p m G ( p ) i ~ G ( p) i p m i p m p m 2 Fermion Propagator 2 p Photons: It’s easier using potentials: forms a four vector. 4-vector again Charge conservation Now the deep part: E and B are observable, but A’s are not! A can be changed by a gauge transformation without changing E and B the observable: So we can use this freedom to choose a gauge, a condition for A: 4 A J c For free photons: A 0 Almost like 4 KG Eq. A ( x) a e ipx Energy-Momentum Relation Polarization needs to satisfy Lorentz Condition: Lorentz Condition does not kill all the freedom: We can further choose then Coulomb Guage The photon is transversely polarized. For p in the z direction: For every p that satisfy there are two solutions! Massless spin 1 particle has two degrees of freedom. A A ( x) a p ,i 1, 2 (i ) ipx (i ) ipx e a e p p p p (q, ) (i ) 0 Feynman Rules for external photon lines Gauge Invariance Classically, E and B are observable, but A’s are not! A can be changed by a gauge transformation without changing E and B the observable: A' A (t , x ) Transformation parameter λ is a function of spacetime. A ' A ' A A A A But in Qunatum Mechanics, it is A that appear in wave equation: In a EM field, charged particle couple directly with A. Classically it’s force that affects particles. EM force is written in E, B. But in Hamiltonian formalism, H is written in terms of A. 2 1 e H p A x , t e x ,t 2m c Quantum Mechanics or wave equation is written by quantizing the Hamiltonian formalism: 2 1 i ieA e t 2m Is there still gauge invariance? B does not exist outside. Gauge invariance in Quantum Mechanics: 2 1 i ieA e t 2m In QM, there is an additional Phase factor invariance: i ( x , t ) e ( x , t ) It is quite a surprise this phase invariance is linked to EM gauge invariance when the phase is time dependent. A A ( x, t ) ie ( x ,t ) ( x , t ) e ( x , t ) This space-time dependent phase transformation is not an invariance of QM unless it’s coupled with EM gauge transformation! A A ( x, t ) ie ( x ,t ) ( x , t ) e ( x , t ) e ie ( x ,t ) e ie ( x ,t ) e ie ( x ,t ) e ie ( x ,t ) e ie ( x ,t ) Derivatives of wave function doesn’t transform like wave function itself. 1 2 i t 2m Wave Equation is not invariant! But if we put in A and link the two transformations: ieA e ie ( x ,t ) ieA e ieA ie e ie ( x ,t ) ie ( x ,t ) ie e ieA This “derivative” transforms like wave function. ie ( x ,t ) In space and time components: ie ( x ,t ) ieA e ieA ie ( x ,t ) ie e ie t t The wave equation: 2 1 i ieA e t 2m can be written as 2 1 i ie ieA t 2m 2 ie 1 i e ie e ieA t 2m ie It is invariant! A A ( x, t ) ie ( x ,t ) ( x , t ) e ( x , t ) This combination will be called “Gauge Transformation” It’s a localized phase transformation. Write your theory with this “Covariant Derivative”. D ieA e ie ( x ,t ) Your theory would be easily invariant. ieA There is a duality between E and B. Without charge, Maxwell is invariant under: Maybe there exist magnetic charges: monopole Magnetic Monopole The curl of B is non-zero. The vector potential does not exist. If A exists, A 0 3 B da dx A 0 there can be no monopole. But quantum mechanics can not do without A. Maybe magnetic monopole is incompatible with QM. But Dirac did find a Monopole solution: g 1 cos A r sin 1 A sin A A r r sin 1 1 Ar 1 Ar rA rA r sin r r r Dirac Monopole g 1 cos A r sin It is singular at θ = π. Dirac String It can be thought of as an infinitely thin solenoid that confines magnetic field lines into the monopole. But a monopole is rotationally symmetric. A g 1 cos Dirac String doesn’t seem to observe the symmetry r sin It has to! In fact we can also choose the string to go upwards (or any direction): g 1 cos They are related by a gauge transformation! A' r sin Since the position of the string is arbitrary, it’s unphysical. A g 1 cos r sin Using any charge particle, we can perform a Aharonov like interference around the string. The effects of the string to the phase is just like a thin solenoid: e da A e da B eg 4 Since the string is unphysical. e ieg 4 e 1 n 2g Charge Quantization Finally…. Feynman Rules for QED e e e e 4-rows 4-columns 4 ˣ 4 matrices 4 ˣ 4 matrices Dirac index flow, from left to right! 1 ˣ 1 in Dirac index pk p k' Numerator simplification p m u( p, s) 0 using p m u 0 The first term vanish! In the Lab frame of e Photon polarization has no time component. The third term vanish! Denominator simplification Assuming low energy limit: in low energy k 0 2nd term: 1 k Finally Amplitude: Amplitude squared 0 0 0 0 0 0 k k k2 0 旋轉帶電粒子所產生之磁偶極 磁偶極矩與角動量成正比 iA i r 2 e L 2m e 2 r v r2 erv e r p e L 2 2m 2m 帶電粒子自旋形成的磁偶極 e e s S g S m 2m Anomalous magnetic moment 1 g theory 1 1159652187.9 8.810 12 2 1 g experi 1 1159652188.4 4.310 12 2