for s=1 for EM-waves total energy density:  # of photons wave number vector: the Klein-Gordon equation:

for FREE ELECTRONS electronspositrons ±1 or 1,2 uv a spinor satisfying : u v Note for each: with i.e. we write ss E,pE,p for u 3, u 4 Notice: now we express all terms of the “physical” (positive) energy of positrons! (or any Dirac, i.e. spin ½ particle: muons, taus, quarks )

The most GENERAL solutions will be LINEAR COMBINATIONS  s s s dk 3 k  g h linear expansion coefficients Insisting {  (r,t),  (r ´,t)}= {  † (r,t),  † (r ´,t)}=0 (in recognition of the Pauli exclusion principal), or, equivalently: {  (r,t),  † (r ´,t)} =  3 (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) where g  g(k,s), h  h(k,s), the g  g(k,s), h  h(k,s) “coefficients” cannot simple be numbers!

 s ss dk 3 k  g h ††  s s s k  g h

 (x) = u e -ix  p  /h u(p)u(p) cp z E+mc 2 c(p x +ip y ) E+mc c(p x  ip y ) E+mc 2  cp z E  mc cp z E  mc 2 c(p x +ip y ) E  mc c(p x  ip y ) E  mc 2  cp z E  mc 2 This form automatically satisfied the Klein-Gordon equation. But the appearance of the Dirac spinors means the factoring effort isolated what very special class of particles? We were able to solve Dirac’s (free particle) Equation by looking for solutions of the form: u(p  ) a “spinor” describing either spin up or down components

What about vector (spin 1) particles? Again try to look for solutions of the form  (x) =   (p) e -i x  p  /h Polarization vector (again characterizing 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!) becomes : or 2  = 0 Where the d’Alembertian operator: 2 The fundamental mediators of forces: the VECTOR BOSONS the Klein-Gordon equation:

2  = 0 is a differential equation you have already solved in Mechanics and E & M 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:

Potentials can be changed by a constant for example, the arbitrary assignment of zero gravitational potential energy ( ) leaving everything invariant.or even In solving problems this gives us the flexibility to “adjust” potentials for our convenience The Lorentz GaugeThe Coulomb Gauge 00 In the Lorentz Gauge: 0 and a FREE PHOTON satisfies: The VECTOR POTENTIAL from E&M is the wave function in quantum mechanics for the free photon! a “vector particle” with 4 components (V;A)

so continuing (with our assumed form of a solution)  (p) (p) like the Dirac u, a polarization vector characterizing spin Substituting into our specialized Klein-Gordon equation: (for massless vector particles) E2=p2c2E2=p2c2 just as it should for a massless particle!

only for free photons  (p) (p) Like we saw with the Dirac u before,  has components! How many? 4   A = 0 The Lorentz gauge constrains A0A0 p    = 0 p 0  0 . p = 0 while the Coulomb Gauge   p = 0 which you should recognize as the familiar condition on em waves but not all of them are independent! 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! (m spin =  1, 0, 1)

spin 1 particles: m spin =  1, 0, +1 aligned anti- aligned The m=0 imposes a harsher constraint (adding yet another zero to all the constraints on the previous page!) The masslessness of our vector particle implies ??? v = c In the photon’s own frame longitudinal distances collapse. How can you distinguish m spin =  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:   s where s = 1, 2 or s =  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 The Klein-Gordon Equation or Provided we can identify the appropriate this should be derivable by The Euler-Lagrange Equation L L L

L I claim the expression serves this purpose

LL L LL L L L

You can show (and will for homework!) show the Dirac Equation can be derived from: L DIRAC (r,t) We might expect a realistic Lagrangian that involves systems of particles = L K-G + L DIRAC L (r,t) describes e + e  objects describes photons but each term describes free non-interacting particles L + L INT 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  antiparticle (hermitian conjugate) state particle state What does such a PRODUCT of states mean? Recall the “state functions” Have coefficients that must satisfy anticommutation relations. They must involve operators!