Klein-Gordon Equation ● The existence of plane waves φ(r, t) exp(ik r iωt) ∝ · − satisfying de Broglie and Einstein relations p = !k , E = !ω implies the quantum operator interpretation ∂ p i! , E i! . → − ∇ → ∂t ● Then the relativistic energy-momentum equation E2 = p2c2 + m2c4 implies the Klein-Gordon equation 2 2 ∂ φ 2 2 2 2 4 ! = ! c φ + m c φ − ∂t2 − ∇ 1 ● In covariant notation (see handout) 2 µ mc ∂µ∂ + φ = 0 ! ! " # $ where 1 ∂2 ∂ ∂µ = 2 µ c2 ∂t2 − ∇ ● KG wave function φ is a Lorentz-invariant (scalar) function; Lorentz transformation r, t r!, t! implies φ φ! where → → φ!(r!, t!) = φ(r, t) . Hence it must represent a spin-zero particle (no orientation). ● Since φ 2 is also invariant, this cannot represent a probability density. A | | density transforms as time-like (0-th) component of a 4-vector, due to Lorentz contraction of volume element. 2 ● Correct definition of density follows from the continuity equation: ∂ρ = J ∂t −∇ · (J = corresponding current vector), i.e. µ ∂µJ = 0 where J µ = (cρ, J) is the 4-current. ● We can obtain an equation of this form from the KG equations for φ and φ∗, 2 2 ∂ φ ∂ φ∗ 2 2 2 i! φ∗ φ = i!c φ∗ φ φ φ∗ , ∂t2 − ∂t2 ∇ − ∇ % & ' ( ∂ ∂φ ∂φ∗ 2 i! φ∗ φ = i!c (φ∗ φ φ φ∗) . ∂t ∂t − ∂t ∇ · ∇ − ∇ % & Hence ∂φ ∂φ∗ ρ = i! φ∗ φ ∂t − ∂t % & 2 J = i!c (φ∗ φ φ φ∗) − ∇ − ∇ µ 2 µ µ i.e. J = i!c (φ∗∂ φ φ∂ φ∗). − 3 iEt/ ● Normalization is such that the energy eigenstate φ = Φ(r)e− ! has ρ = 2E Φ 2. Thus Φ = 1 corresponds to 2E particles per unit volume | | | | (relativistic normalization). ● Compare with Schro¨dinger current S i! 1 KG J = (φ∗ φ φ φ∗) = J −2m ∇ − ∇ 2mc2 which thus has in fact E/mc2 particles per unit volume. 4 Problems with Klein-Gordon Equation 1. Density ρ is not necessarily positive (unlike φ 2) equation was rejected | | ⇒ initially. 2. Equation is second-order in t need to know both φ and ∂φ at t = 0 in order ⇒ ∂t to solve for φ at t > 0. Thus there is an extra degree of freedom, not present in the Schr¨odinger equation. 3. The equation on which it is based (E2 = p2c2 + m2c4) has both positive and negative solutions for E. iEt/ ● Actually these problems are all related, since a solution φ = Φ(r)e∓ ! has ρ = 2E Φ 2, and so for the general solution ± | | iEt/! +iEt/! φ = Φ+(r)e− + Φ (r)e − i! ∂φ both φ(t = 0) = Φ+ + Φ and E ∂t = Φ+ Φ are needed in order to − t=0 − − specify Φ+ and Φ . ) − ) ) 5 Electromagnetic Waves ● In units where %0 = µ0 = c = 1 (‘Heaviside-Lorentz’) Maxwell’s equations are ∂B E = ρ , E = ∇ · em ∇ × − ∂t ∂E B = 0 , B = J + ∇ · ∇ × em ∂t µ where (ρem, J em) = Jem is the electromagnetic 4-current. ● In terms of the scalar and vector potentials V and A, ∂A E = V , B = A . − ∂t − ∇ ∇ × So we find ∂2A ∂V ( A) ( A) 2A = J ∇ × ∇ × ≡ ∇ ∇ · − ∇ em − ∂t2 − ∇ ∂t ● In terms of the 4-potential Aµ = (V, A) (∂ ∂ν )Aµ ∂µ(∂ Aν ) ∂ F νµ = J µ ν − ν ≡ ν em 6 where the electromagnetic field-strength tensor is F νµ = ∂ν Aµ ∂µAν = F µν . − − ● E and B, and hence Maxwell’s equations, are invariant under gauge transformations µ µ µ µ A A! = A + ∂ χ → where χ(r, t) is an arbitrary scalar function. µ µ ● Therefore we can always choose A such that ∂µA = 0 (Lorenz gauge). If µ µ µ µ µ ∂ A = f = 0, we can change to A! = A + ∂ χ where ∂ ∂ χ = f. µ ( µ − µ ν µ ● Then in free space (J = 0) we have ∂ν ∂ A = 0. ❖ Massless KG equation for each component of Aµ ❖ Aµ is ‘wave function’ of photon ❖ Aµ is a 4-vector photon has spin 1. ⇒ µ µ µ ik x µ ● Plane wave solutions A = ε exp(ik r iωt) ε e− · where ε = · − ≡ polarization 4-vector, k x kµx , kµ = (ω, k) = wave 4-vector. · ≡ µ 7 ● From wave equation k k = 0 hence ω2 = k2, i.e. E2 = p2c2 (massless · photons). ● From Lorenz gauge condition ε k = 0 ε0 = ε k/ω. · ⇒ · µ µ µ µ ● Polarization 4-vector ε! = ε + ak is equivalent to ε for any constant a. Hence we can always choose ε0 = 0. Then Lorenz condition becomes transversity condition: ε k = 0. · ● E.g. for k along z-axis we can express εµ in terms of plane polarization states µ µ εx = (0, 1, 0, 0) , εy = (0, 0, 1, 0) , or circular polarization states εµ = (0, 1, i, 0)/√2. R,L ± N.B. only 2 polarization states for real photons. 8 Electromagnetic Interactions ● As in classical (and non-relativistic quantum) physics, we introduce e.m. interactions via the minimal substitution in the equations of motion: E E eV , p p eA → − → − i.e. pµ pµ eAµ , ∂µ ∂µ + ieAµ → − → ● The Klein-Gordon equation becomes µ µ 2 (∂µ + ieAµ)(∂ + ieA )φ + m φ = 0 , (∂ ∂µ + m2)φ = ie[∂ (Aµφ) + A (∂µφ)] + e2A Aµφ µ − µ µ µ The conserved current is now (! = c = 1) µ µ µ µ J = i(φ∗∂ φ φ ∂ φ∗) 2eA φ∗φ − − 9 Klein Paradox ● Consider KG plane waves incident on electrostatic barrier, height V , width a V e ipx−iEt A eip’x−iEt T e ipx−iEt R e−ipx−iEt B e−ip’x−iEt x=0 x=a KG equation for x < 0, x > a gives E2 = p2 + m2 p = + E2 m2 ⇒ − (sign from B.C.). * µ 2 2 2 ● In 0 < x < a, A = (V, 0) and so (E eV ) = p! + m − p! = + (E eV m)(E eV + m) ⇒ − − − (sign choice is arbitrary since*we include p!). ± 10 ● Matching φ and ∂φ/∂x at x = 0 and a gives (as for Schro¨dinger equation) 2 2 i p p! − T = cos p!a + sin p!a | | − 2 p p ) % ! & ) ) ) ) ) ) ) ● Now consider behaviour as V is increased: ❖ eV < E m: p! is real, T < 1 ( T = 1 when p!a = nπ). − | | | | ❖ E m < eV < E + m: p! is imaginary, T < 1, transmission by tunnelling. − | | ❖ eV > E + m: p! is real again! T = 1 when p!a = nπ!? | | ● Note that when eV > E + m density inside barrier is negative: 2 2 ρ! = 2(E eV ) φ < 2m φ − | | − | | 2 ● Meanwhile, the current inside remains positive, J ! = 2p T (current x | | conservation). Hence when eV > E + m there is a negative density flowing from right to left, giving a positive current. We interpret this as a flow of µ µ antiparticles: Jem = eJ always. 11 ● When eV > E + m and T = 1, antiparticles created at the back of the barrier | | (x = a) travel to x = 0 and annihilate the incident particles. At the same time, particles created at x = a travel to x > a, replacing the incident beam. E E=0 E π+ π π+ ● Antiparticles are trapped inside the barrier, but field is zero there, so there can be perfect transmission for any thickness. 12 ● Antiparticles are like particles propagating backwards in time π+ t t π+ − + π+ π+ π π x x eV < E−m eV > E+m 13 Charge Conjugation ● If φ is a negative-energy plane-wave solution of the KG equation, with momentum p, φ = exp(ip r + iEt) (E > 0), then φ∗ = exp( ip r iEt) is a · − · − positive-energy wave with momentum p. Furthermore, in e.m. fields, φ∗ − behaves as a particle of charge e: − µ µ 2 (∂µ + ieAµ)(∂ + ieA )φ + m φ = 0 µ µ 2 (∂ ieA )(∂ ieA )φ∗ + m φ∗ = 0 ⇒ µ − µ − ● Thus if φ is a negative-energy solution, we take it to represent an antiparticle with wave function φ∗ (and hence positive energy, opposite charge and momentum). ● Correspondingly, KG equation is invariant w.r.t. φ φ∗, e e. This is → → − called charge conjugation, C. N.B. Under C, J µ J µ as expected. → − 14 Electromagnetic Scattering ● We assume (for the moment) the same formula as in NRQM for the scattering amplitude in terms of the first-order perturbation due to e.m. field: µ µ 4 = i φ∗ ie[∂ (A φ ) + A (∂ φ )] d x Afi − f { µ i µ i } + µ µ 4 by parts = e A [φ∗ (∂ φ ) (∂ φ∗ )φ ]d x µ f i − f i + = ie A J µ d4x − µ fi + µ µ µ µ where J = i[φ∗ (∂ φ ) (∂ φ∗ )φ ] is generalization of J to φ = φ fi f i − f i f ( i (transition current). Note that to get to order e1 we only need J µ to Afi fi order e0. Similarly, for Aµ we can use the free-field form µ µ ik x A = ε e− · ipf,i x ● For plane waves, φf,i = e− · , µ µ i(pf pi) x Jfi = (pi + pf ) e − · 15 Hence µ i(p p k) x 4 = ieε (p + p ) e f − i− · d x Afi − µ i f + = ie(2π)4ε (p + p ) δ4(p p k) − · i f f − i − ● This corresponds to the Feynman rules for the diagram p i p k f ❖ An overall factor of (2π)4 δ4(p p k) (momentum conservation) f − i − ❖ εµ for an external photon line ❖ ie(p + p )µ for a vertex involving a spin-0 boson of charge e. − i f N.B. 4-momentum cannot be conserved in this process for free particles! But we shall see that it can occur as part of a more complicated process, e.g.