Photons and Quantum Electrodynamics. H. A. Tanaka
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1 Photons and Quantum Electrodynamics H. A. Tanaka
2 The Photon Apart from we need some other particle/object with definite Lorentz transformation properties to make Lorentz invariants What would we do with the vector term µ to get a Lorentz scalar? Recall the photon: Classically, we have Maxwell s equations: E =4 B = E + c Ḃ = B c Ė = 4 c J Recall that we can re-express the Maxwell equations using potentials: E = B = A these can in turn be combined to make a 4 vector:a µ =(, A) Likewise for the source terms ρ and J: J µ =(c, J)
3 Maxwell s Equation in Lorentz Covariant Form All four equations can be expressed as: µ µ A The issue is that A is (far) from unique: Consider: ( µ A µ )= 4 c J A µ A µ + µ µ µ (A + ) ( µ (A µ + µ )= F µ = µ A A µ = E x E y E z E x B z B y E y B z B x E z B y B x µ µ A ( µ (A µ )+ µ µ µ µ the last terms cancel, so the new Aµ is also a solution to Maxwell s solution they are physically the same, so we can make some conventions: Lorentz gauge condition : Coulomb gauge µa µ = A = µ µ A = 4 c J A =
4 Solutions to the Maxwell Equation in Free Space: Free means no sources (charges, currents): J µ = Find solution as usual by ansatz: A µ (x) =ae ip x µ (p) Now check: µ A (x) = ip µ ae ip x (p) µ µ A = µ A µ = p µ µ (p) = µ µ A (x) =( i) 2 p µ p µ ae ip x (p) = p 2 = m 2 c 2 = A = Conclusions: Photon is massless Polarization ε is transverse to photon direction: it has two degrees of freedom/polarizations = p =
5 Making a scalar object: In the end, these spaces must collapse: In Lorentz space, this happens by contracting indices:g µ a µ b = a µ b µ In spinor space, products of adjoint spinors with spinors (with gamma matrices possibly in between): ū v 2 Г=(product of g matrices) but some expressions have structure in both: sum over μ collapses the Lorentz structure M = g 2 e (p + p 2 ) 2 [ū(3) µ v(4)] [ v(2) µ u()] Contracted in spinor space, but not in Lorentz Same here Product of 4x4 matrices in spinor space = S µ S x x µ sum over μ collapses the Lorentz structure
6 Reminder of Dirac Spinors We can now construct the column vector u: u = N p z c/(e + mc 2 ) (p x + ip y )c/(e + mc 2 ) Use positive energy solutions electrons u 2 = N (p x ip y )c/(e + mc 2 ) p z c/(e + mc 2 ) v 2 u 3 = N p z c/(e + mc 2 ) (p x + ip y )c/(e + mc 2 ) C A v u 4 = N (p x ip y )c/(e + mc 2 ) p z c/(e + mc 2 ) C A Use negative energy solutions positrons
7 A second look at Dirac spinors electrons (x) =ae (i/ )p x u s (p) positrons (x) =ae (i/ )p x v s (p) s labels the spin states (two for electrons/positrons) The exponential term sets the space/time = energy/momentum Let s look at the spinor part u,v which determines the Dirac structure : If we insert ψ into the Dirac equation, we get: (i µ µ mc) = ( µ p µ mc) = ( µ p µ mc)u s (p) = (i µ µ mc) = ( µ p µ mc) = ( µ p µ + mc)v s (p) = momentum space Dirac equations If we take the adjoint of these equations, we get: ū s ( µ p µ mc) = v s ( µ p µ + mc) =
8 Orthogonality and Completeness of Spinors: From the explicit form of our u/v spinors, we can also show: ū i u j =2mc ij v i v j = 2mc ij ū i v j = v i u j = We can also show: s=,2 u s ū s =( µ p µ + mc) s=,2 v s v s =( µ p µ mc) a a 2 a 3 a 4 (b,b 2,b 3,b 4 )= a b a b 2 a b 3 a b 4 a 2 b a 2 b 2 a 2 b 3 a 2 b 4 a 3 b a 3 b 2 a 3 b 3 a 3 b 4 a 4 b a 4 b 2 a 4 b 3 a 4 b 4
9 Photon Polarizations and Orthogonality: We showed that the polarization 4-vector ε μ with the Lorentz and Coloumb gauge conditions must satisfy: p = We noted that this allows two degrees of freedom corresponding to transversely polarized electromagnetic fields. We need to two orthogonal ε basis vectors to span the space For example, if the photon is moving in the z direction, we can choose: µ = 2 µ = The polarization vectors satisfy orthogonality/completeness relations: i µ µj = ij s=,2 s i s j = ij ˆp iˆp j
10 The Feynman Rules: External Lines First right down the Feynman diagram(s) for the process and label the momentum flow use p s for external lines, q s for internal (Griffiths convention). Note that there are two flows: particle/antiparticle momentum These are separate Now the components of the expression External Lines: Electrons: incoming Positrons: incoming Photons: incoming u s (p) v s (p) µ(p) outgoingū s (p) outgoingv s (p) outgoing p p2 µ(p) q p3 p4
11 The Feynman Rules: Vertices and Propagators: For each QED vertex: where as before, momentum is + incoming, - outgoing from vertex Internal lines: ig e µ electron/positron propagator (2) 4 4 (k + k 2 + k 3 ) i( µ q µ + mc) q 2 m 2 c 2 p2 q p3 Photon propagator indices match vertices/polarization Integral over momentum: ig µ q 2 d 4 q (2 ) 4 p p4 Finally: cancel the overall delta function, what remains is -im
12 Example: p2 q p3 p p4 Order matters due to Dirac matrix structure (photon part doesn t care) Griffiths: go backward through the fermion lines: In the final state : ū(3) ig e In the initial state : v(2) ig e µ v(4) u() Throw in the internal photon propagator: (2) 4 4 (q p 3 p 4 ) (2) 4 4 (p + p 2 q) (2 ) 4 d 4 q i(2) 4 4 ge 2 (p [ū(3) µ + p 2 p 3 p 4 ) v(4)] g µ [ v(2) (p + p 2 ) 2 ge 2 M = (p + p 2 ) 2 [ū(3) µ v(4)] [ v(2) µ u()] ig µ q 2 u()]
13 Example: e + + e - e + + e - e e p q p3 p2 q p4 p p2 p4 e e ū(3) ig e µ u() v(2) ig e v(4) ū(3) ig e v(4) v(2) ig e u() p3 ig µ (p p 3 ) 2 ig (p + p 2 ) 2 (2 ) 4 4 (p + p 2 p 3 p 4 )
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