Quantization of the Photon Field QED

Transcription

1 Quantization of the Photon Field QED Reminder: Classical Electrodynamics Before we start quantizing the hoton field, let us reflect on classical electrodynamics. The Hamiltonian is given by H = d 3 x ( E 2 + B 2) For the electric field, we find E = 0 E = B t (for now) For the magnetic field, we find B = A B = 0 B = E t Define Maxwell s Field Tensor: F µν = µ A ν ν A µ F 0i = 0 A i i A 0 F i j = i A j j A i so that Maxwell s equations are given by µ F µν = 0 µ ɛ µνρσ F ρσ = 0 (= j ν for sources) 1 Quantization of the Photon Field The A field describes a sin one boson. The simlest equation that it must fulfill is the Klein Gordon equation so let us start with this. Call L = 1 4 F µν(x)f µν (x) the Lagrangian density for a connection field (U(1)) with µ F µν = A ν µ ν A µ = 0 If A µ is a solution, then A µ = A µ + µ χ(x) is also a solution because of gauge freedom. Call Π µ the canonical conjugate momentum to A µ. Π 0, the canonical conjugate momentum to A 0, vanishes. 1

2 1.1 Gauge constraints In order to simlify future calculations, constaints can be imosed ( fixing the gauge ). Lorenz gauge The Lorenz gauge, after Ludvig Lorenz (and not Hendrik Antoon Lorentz after whom Lorentz transformations are named), states that µa µ vanishes. Fixing the Lorenz gauge is always ossible. Imagine, µ A µ were not zero, but µ A µ = Θ(x). Then we could use gauge freedom in A µ to shift it to A µ + µ χ with χ = Θ(x). We will always be able to construct A µ in a way that enables the Lorenz gauge. What are the advantages of the Lorenz gauge? The most obvous advantage is that Maxwell s equations boil down to the Klein Gordon equation in each comonent, µ F µν = A ν µ ν A µ = Aν = 0 One could think of quantizing every comonent indeendently like a scalar field. Then each comonent will fulfill the Klein Gordon equation, but we know that the hoton has two olarizations, therefore only two degrees of freedom. But if we quantize the way we just roosed, we would have two extra degrees of freedom. This is a roblem. Let us try and quantize a bit differently. Focus on the wavelike structure: A µ = Nɛ µ e ikx,k 2 = 0 Here, ɛ µ denotes the olarization vector. e ikx emhasizes the wavelike structure. Even with the Lorenz condition, k ɛ = kµɛ µ = 0 there is still a remaining gauge freedom. χ(x) with χ(x) = 0 is free to choose. Shifting ɛ µ to ɛ µ + βk µ, we can make k 0 disaear. For that reason, the Lorenz condition is equivalent to ɛ k = 0. Thus, the olarization can be chosen to be in a sacelike lane, orthogonal to the direction of roagation of the A field. For examle: ɛ 1 = (1,0,0) ɛ 2 = (0,1,0) for k = (0, 0, k ) Obviously, there are exactly two indeendent olarizations allowed, just as we wanted. Because of the sacelike character of the olarization vector, the scalar roduct is negativedefinite: ɛ (λ)ɛ(λ ) = δλλ λ,λ {1,2} With this ansatz, let us quantize the hoton field. A µ = d 3 k (2π) 3 2ω k The Lagrangian density gives 2 { ɛ µ ( k,λ)a( k,λ)e ikx + ɛ µ ( k,λ)a ( k,λ)e +ikx} λ=1 L = 1 4 F µνf µν 1 2 ( µ A µ) 2 2

3 The canonical conjugate momentum is given by Π 0 = µa µ Π i = i A 0 0 A i = E i Finally, we have arrived at the equal time commutator relations: [ Aµ(x),Πν (y) ] x = igµνδ (4) (x y) 0 =y 0 The equal time commutator relation (ECTR) builds a commutator which transforms as a twotensor under the Lorentz grou. g µν is the metric tensor. Even though it might look like a well-derived ECTR, it rovokes a serious roblem: Because of the Lorenz condition, µ [ Aµ(x),Πν (y) ] should vanish, but it does not. Instead, it roduces [ Aµ(x),Πν (y) ] = i ν δ (4) (x y) which is some distribution, but certainly not generally zero. Quantizing the hoton field the way we did it is roblematic! It does not work the usual way. Allowing a olarization vector with only two ossible olarizations obviously leads to no good. 2 Quantization of the Photon Field 2.0 This time, let us try four instead of two olarizations. For k in the 3 direction, k = (ω k,0,0,±ω k ), we choose One timelike olarization: (1,0,0,0) T Two transversal olarizations: (0,1,0,0) T and (0,0,1,0) T One longitudinal olarization: (0,0,0,1) T The scalar roduct for olarizations is Our new ansatz for the hoton field is ɛ (k,λ) ɛ(k,λ ) = g λλ g λλ ɛ µ (k,λ)ɛ ν (k,λ) = g µν λ A µ (x) = d 3 k (2π) 3 2ω k 3 { ɛ µ (k,λ)a(k,λ)e ikx + ɛ µ (k,λ)a (k,λ)e +ikx} λ=0 The new ETCR are [ a(k,λ),a (k,λ ) ] = g λλ (2π) 3 2ω k δ (3) ( k k ) The right hand side is negative for timelike olarizations. This would mean a negative norm. This is a roblem! Maybe we can fix this by keeing two out of the four degrees of freedom which we have, and arrange the other two in such a way that they erase each other so they will not lay a role. 3

4 Let us now imose Lorenz s constraint, µ A µ, in a strong sense. Start by keeing in mind that for all hysical states φ, ψ, µ A µ (+) ψ = 0 = φ µ A ( ) µ where µ A (+) µ denotes the ositive frequency art, and µ A ( ) µ the negative frequency art. Therefore, the exectation value of µ A µ is also vanishing, φ µ A µ ψ = 0 Also, for longitudinal wave modes, which are in the 3-direction if k is in the 3-direction, ( a( k,0) a( k,3) ) ψ = 0 Therefore, the term 1 2 d 3 k (2π) 3 { a (k,0)a(k,0) a (k,3)a(k,3) } = 0 vanishes and does not contribute to the Hamiltonian density, consequently. Only the 1- and 2-comonents aear in the Hamiltonian! In other words, only transverse modes generate dynamics. 2.1 The covariant ξ gauge Instead of adding ( µ A µ ) 2 to the Lagrangian as the kinetic term for the hoton field, one could add the more generally scaled exression 1 2ξ ( µa µ ) 2. L = 1 4 F µνf µν 1 2ξ ( µa µ ) 2 The Klein-Gordon equation becomes { ( g µν 1 1 } ) µ ν A ν 0 ξ This is the equation of motion for a free connection field in a covariant ξ gauge. For the interactions, relace µ with D µ = µ + iea µ and A µ with A µ i µα(x) to get ψ/ ψ ψ /Dψ = ψ/ ψ + ie ψ /Aψ The last term, ie ψ /Aψ, gives the form of the only allowed vertex in quantum electrodynamics. 3 Feynman Rules of QED Last, let us summarize all Feynman rules of QED. The Lagrangian is given by L = 1 4 F µνf µν + ψ(i /D m)ψ 4

5 3.1 Internal lines There are two different kinds of roagators; on configuration sace, the fermionic roagator S αβ F (x y) deends on the sin orientations α and β of the fermion fields created at y and annihilated at x. The hoton roagator G µν is gauge deendent. If we choose ξ = 0, we get a transversal roagator. This is called the Landau gauge. If we choose ξ = 1, the Feynman gauge, the calculations become very easy. _ q ψ β ψ α = S αβ F (x y) = 0 Tψα (x) ψ β (y) [ ] i 0 = /q m + iɛ q A µ A ν = G µν = 0 T A µ (x)a µ (y) g µν (1 ξ) qµqν q 0 = i 2 q 2 + iɛ The vertex, or interaction diagram, with sinor indices α and β, corresonds to β αβ µ = (ieγ µ ) βα 3.2 External lines Here come the Feynman rules for incoming articles. 3.3 Comuting the Amlitude α incoming hoton: k = ɛ µ ( k,λ) outgoing hoton: k = ɛ µ ( k,λ) incoming electron: = u α (, s) outgoing electron: = ū α (, s) incoming ositron: = v α (, s) outgoing ositron: = u α (, s) In order to comute the amlitude, one must do more than just write down the exressions for each external article, roagator, and vertex. Moreover, one must Multily with ( 1) for each closed fermion loo. Take into accound a relative sign between diagrams: There is a (-1) factor for each ermutation or exchange of external lines. 5

6 Examle Consider the grah in a ξ gauge. Ignoring the external line contribution and any trivial overall factors, the loo integral gives q k µ k+q ν q = 1 γ µ /k + /q m + iɛ γ g µν (1 ξ) kµ k ν k 2 ν k 2 + iɛ /k + /q + m 1 (k + q) 2 m 2 + iɛ γµ k 2 + iɛ d4 k γ µ } {{ } Feynman gauge d 4 k (1 ξ) /k(/k + /q mi)/k ( (k + q) 2 m + iɛ ) ( 2 +iɛ) 2 d4 k In general, the hoton roagators is gauge deendent. If we are not working in the Feynman gauge, every hoton roagator contributes two terms. Thus, terms with several hotons become nightmares. 6