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for s= 1. for EM- waves total energy density:. # of photons. the Klein-Gordon equation:. wave number vector:. (or any Dirac, i.e . spin ½ particle: muons , taus , quarks ). for FREE ELECTRONS. electrons. ±1 or 1,2. positrons. u. v. a spinor satisfying :. v. u.
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for s=1 for EM-waves total energy density: # of photons the Klein-Gordon equation: wave number vector:
(or any Dirac, i.e. spin ½ particle: muons, taus, quarks) for FREE ELECTRONS electrons ±1or 1,2 positrons u v a spinor satisfying: v u Note for each:with -s -E,-p for u3, u4 i.e. we write Notice: now we express all terms of the “physical” (positive) energy of positrons!
s s dk3 The most GENERAL solutions will be LINEAR COMBINATIONS k g h s linear expansion coefficients where gg(k,s), hh(k,s), Insisting{(r,t), (r´,t)}= {†(r,t), †(r´,t)}=0 (in recognition of the Pauli exclusion principal), or, equivalently: {(r,t), †(r´,t)} =d3(r – r) respecting the condition on Fourier conjugate fields: forces the g, h to obey the same basic commutation relation (in the “conjugate” momentum space) the gg(k,s), hh(k,s) “coefficients” cannot simple be numbers!
k s s s s dk3 dk3 k g h s † † g h s
We were able to solve Dirac’s (free particle) Equation by looking for solutions of the form: This form automatically satisfied the Klein-Gordon equation. But the appearance of the Dirac spinors means the factoring effort isolated what very specialclass of particles? (x) = ue-ixp/h u(p) cpz E-mc2 c(px+ipy) E-mc2 c(px-ipy) E-mc2 -cpz E-mc2 1 0 0 1 c(px-ipy) E+mc2 -cpz E+mc2 cpz E+mc2 c(px+ipy) E+mc2 1 0 1 0 u(p) a “spinor” describing either spin up or down components
The fundamental mediators of forces: the VECTOR BOSONS What about vector (spin 1) particles? Again try to look for solutions of the form (x) = m(p)e-ixp/h Polarization vector (againcharacterizing SPIN somehow) but by just returning to the Dirac-factored form of the Klein-Gordon equation, will we learn anything new? What about MASSLESS vector particles?(the photon!) the Klein-Gordon equation: 2 = 0 becomes: or Where the d’Alembertian operator: 2
is a differential equation you have already solved in Mechanics and E & M 2 = 0 Classical Electrodynamics J.D.Jackson (Wiley) derives the relativistic (4-vector) expressions for Maxwells’ equations can both be guaranteed by introducing the scalar V and vector A “potentials” which form a 4-vector: (V;A) along with the charge and current densities: (c;J) Then the single relation: completely summarizes:
( ) for example, the arbitrary assignment of zero gravitational potential energy Potentials can be changed by a constant or even leaving everything invariant. In solving problems this gives us the flexibility to “adjust” potentials for our convenience 0 0 The Lorentz Gauge The Coulomb Gauge In the Lorentz Gauge: 0 a “vector particle” with 4 components (V;A) and a FREE PHOTON satisfies: The VECTOR POTENTIAL from E&M is the wave function in quantum mechanics for the free photon!
so continuing (with our assumed form of a solution) (p) like the Diracu, a polarization vector characterizing spin Substituting into our specialized Klein-Gordon equation: (for massless vector particles) E2=p2c2 just as it should for a massless particle!
(p) Like we saw with the Dirac u before, has components! 4 but not all of them are independent! How many? A0 The Lorentz gauge constrains p m e m = 0 p0e 0-e .p = 0 which you should recognize as the familiar condition on em waves A = 0 p = 0 while the Coulomb Gauge only for free photons Obviously only 2 of these 3-dim vectors can be linearly independent such that p = 0 Why can’t we have a basis of 3 distinct polarization directions? We’re trying to describe spin 1 particles! (mspin = -1, 0, 1)
spin 1 particles: mspin = -1 , 0 , +1 anti- aligned aligned The m=0 imposes a harsher constraint (adding yet another zero to all the constraints on the previous page!) v =c The masslessness of our vector particle implies ??? In the photon’s own frame longitudinal distances collapse. How can you distinguishmspin = 1? Furthermore:with no frame traveling faster than c, can never change a ’s spin by changing frames. What 2 independent polarizations are then possible?
The most general solution: eswheres = 1, 2 ors = 1 moving forward moving backward Notice here: no separate ANTI-PARTICLE (just one kind of particle with 2 spin states) Massless force carriers have no anti-particles.
Finding a Klein-Gordon Lagrangian or The Klein-Gordon Equation L Provided we can identify the appropriate this should be derivable by The Euler-Lagrange Equation L L
I claim the expression L serves this purpose
L L L L L L L L
You can show (and will for homework!) show the Dirac Equation can be derived from: LDIRAC(r,t) We might expect a realistic Lagrangian that involves systems of particles L(r,t) = LK-G + LDIRAC but each term describes free non-interacting particles describes photons describes e+e-objects L + LINT But what does terms look like? How do we introduce the interactions the experience?
We’ll follow (Jackson) E&M’s lead: A charge interacts with a field through: current-field interactions the fermion (electron) the boson (photon) field from the Dirac expression for J particle state antiparticle (hermitian conjugate) state Recall the “state functions” Have coefficients that must satisfy anticommutation relations. They must involve operators! What does such a PRODUCT of states mean?
