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Dynamical symmetry of the Kaluza-Klein monopole∗ L. Fehér† and P. A. Horva´thy‡ February 26, 2009 Abstract 9 0 The Kepler-type dynamical symmetries of the Kaluza-Klein monopole are reviewed. At 0 the classical level, the conservation of the angular momentum and of a Runge-Lenz vector 2 imply that the trajectories are conic sections. The o(4) algebra allows us to calculate the b bound-state spectrum, and the o(3,1) algebra yields the scattering matrix. The symmetry e algebra extends to o(4,2). F 6 2 1 SUMMARY OF KALUZA-KLEIN THEORY [1, 2] ] h One of the oldest and most enduring ideas regarding the unification of gravitation and gauge t - theory is Kaluza’s five dimensional unified theory. Kaluza’s hypothesis was that the world has p four spatial dimensions, but one of the dimensions has curled up to form a circle so small as to e h be unobservable. He showed that ordinary general relativity in five dimensions, assuming such [ a cylindrical ground state, contained a local U(1) gauge symmetry arising from the isometry 1 of the hidden fifth dimension. The extra components of the metric tensor constitute the gauge v fields of this symmetry and could be identified with the electromagnetic vector potential. 0 0 To be more specific, consider general relativity on a five dimensional space-time with the 6 Einstein-Hilbert action 4 1 (cid:90) √ S = − d5x −g R (1.1) . 5 5 2 16πG K 0 whereR isthefive-dimensionalcurvaturescalarofthemetricg ,g = det(g ),andG isthe 9 5 AB 5 AB K 0 five-dimensionalcouplingconstant. Ourconventionsare: uppercaseLatinlettersA,B,C denote : v five-dimensional indices 0,1,2,3,5; lower case Greek indices µ,ν... run over four dimensions, i 0,1,2,3, whereas lower case Latin indices run over four-dimensional spatial values 1,2,3,5. The X signature of g is (−,+,+,+,+) and the Riemann tensor is r AB a RK = ∂ ΓK −∂ ΓK +ΓK ΓJ −ΓK ΓJ LMN M LN N LM JM LN JN LM R = RK , R = RL , LM LKM 5 L and, except where indicated, h/2π = c = 1. In the absence of other fields the equations of motion are of course R = 0. AB ThebasicassumptionofKaluzaandKleinwasthatthecorrectvacuumisthespaceM4×S1, R the product of four dimensional Minkowski space with a circle of radius R. The radius of the ∗Talk given at the ’88 Schloss Hofen Meeting on Symmetries in Science III. Gruber B and Iachello F (eds). Plenum : New York. p. 399-417 (1989). †Bolyai Institute, University of Szeged. Present address: Research Institute for Particle and Nuclear Physics, Budapest, Hungary. e-mail: lfeher-at-rmki.kfki.hu ‡Dipartimento di Fisica, Universita` di Napoli, Italy. Present address: Laboratoire de Mathématiques et de Physique Théorique, Université de Tours, France. e-mail: horvathy-at-lmpt.univ-tours.fr 1 circleinthefifthdimensionisundeterminedbytheclassicalequationsofmotion, sinceanycircle is flat. If R is sufficiently small then all low-energy experiments will simply average over the fifth dimension. In fact the components of the metric, g (xµ,x5), can be expanded in Fourier AB series, g (xµ,x5) = (cid:88)g(n)(xµ)exp(cid:20)inx5(cid:21), (1.2) AB AB R n and all modes with n (cid:54)= 0 will have energies greater than hc/R. Thus the effective low-energy theory can be deduced by considering the metric g to be independent of x5. Under these AB assumptions the theory is invariant under general coordinate transformations that are independent of x5. In addition to ordinary four dimensional coordinate transformations xµ → xµ(xν), we have a U(1) local gauge transformation x5 → x5 +Λ(xµ), under which g transforms as a µ5 vector gauge field, g (x) → g (x)+∂ Λ. (1.3) µ5 µ5 µ Therefore the low-energy theory should be a theory of four dimensional gravity plus a U(1) gauge theory, i.e. electromagnetism, with the massless modes of g , g corresponding to the µν µ5 graviton (photon). The low-energy theory is also invariant under scale transformations in the fifth dimension, x5 → λx5, g → λ−2g , g → λ−1g . (1.4) 55 55 µ5 µ5 This global scale invariance is spontaneously broken by the Kaluza-Klein vacuum (since R is fixed) thus giving rise to a Goldstone boson, the dilaton. To exhibit the low-energy theory we write the metric as follows   g +A A A V µν µ ν µ gAB =  , A V V ν (1.5) g = detg = det(g )V = g V, 5 AB µν 4 ds2 = V(dx5+A dxµ)2+g dxµdxν. µ µν The five-dimensional curvature scalar can be expressed in terms of the four dimensional curvature, R , the field strength, F = ∂ A −∂ A , and the scalar field, V, 4 µν µ ν ν µ 2 √ R = R + 1VF Fµν − √ V . (1.6) 5 4 4 µν V Thus the effective low energy theory is described by the four dimensional action S = − 1 (cid:90) d4x√−g V1/2(cid:0)R + 1VF Fµν(cid:1), (1.7) 16πG 4 4 4 µν √ where we have dropped the terms in R involving V, since, when multiplied by g , these yield 5 5 a total derivative, and G K G = (1.8) 2πR √ √ is Newton’s constant, determined by hG/ 2πc3 ≈ 1.6×10−33 cm. This theory is recognizable as a variant of the Brans-Dicke theory [3] of gravity, with V1/2 identifiedasaBrans–Dickemasslessscalarfield, coupledtoelectromagnetism. V indeedsetsthe local scale of the gravitational coupling. In the vacuum V = 1. Also in the Brans–Dicke theory the coupling of V to matter is somewhat arbitrary, here it is totally fixed by five-dimensional 2 covariance. The radius R is determined by the electric charge. To see this consider a complex field Φ with action (cid:90) √ S = d5x −g (∂ Φ)(∂AΦ†). (1.9) Φ 5 A The Fourier component of Φ with non-trivial x5 dependence, Φ(n)(xµ)exp[inx5/R], will behave √ asaparticleofchargee = 16πGn/Randmassn/R,sinceforΦ(xµ,x5) = Φ(n)(xµ)exp[inx5/R] ∂AΦ∂AΦ† = (cid:12)(cid:12)(cid:12)(cid:16)∂µ+iRnAµ(cid:17)Φ(n)(cid:12)(cid:12)(cid:12)2+ VnR22(cid:12)(cid:12)Φ(n)(cid:12)(cid:12)2. (1.10) (Note that the properly normalized gauge field is (16πG)−1/2A ). Thus µ (cid:114) 2hG α = e2/2hc = 2hG/πc3R2, R = ≈ 3.7×10−32cm. (1.11) παc3 Consider a classical point test particle of unit mass. It has action (cid:114) (cid:90) dxAdxB S = dτ g (1.12) AB dτ dτ and the motion is given by a five-dimensional geodesic, d2xA dxB dxC +ΓA = 0. (1.13) dτ2 BC dτ dτ Now, since the space-time possesses a Killing vector, namely ∂ ∂ KA = , (1.14) ∂xA ∂x5 itisguaranteedthatK dxA/dτ isaconstantofthemotion. Indeedafirstintegralofeqn. (1.13) A is dxA dx5 dxµ K = V +A = q. (1.15) A µ dτ dτ dτ The remaining equations of motion then take the form d2xµ dxαdxβ dxν ∂µV +Γµ = qFµ +q2 . (1.16) dτ2 αβ dτ dτ ν dτ 2V2 Here Γµ is the four dimensional connection constructed from g . We recognize on the right αβ µν the Lorentz force if q is identified with the charge of the particle, plus an interaction with the scalar field V. 2 THE MONOPOLE OF GROSS, PERRY AND SORKIN [2] Now we are searching for solitons in the five-dimensional Kaluza - Klein theory sketched above. By a soliton we mean a non-singular solution to the classsical field equations which represent spatially localized lumps that are topologically stable. Such solutions are expected to have a large mass (of the order M/g, where M is the mass scale of the theory and g a dimensionless coupling constant). They are the starting points for the semi-classical construction of quantum mechanical particle states. Our goal is to construct solutions of the five dimensional field equations that approach the vacuum solution: V = 1, A = 0, g = η at spatial infinity. It is natural to consider static µ µν µν 3 metrics with ∂/∂t as Killing vector. It is also natural to look for solutions with g = δ . In 0A 0A this case the space-time is totally flat in the ‘time’ direction and the field equations are simply R = R = R = 0, (2.1) ij 5i 55 namely the four-dimensional, wholly space-like, manifold at each fixed t has a vanishing Ricci tensor. Thesearesimplytheequationsoffourdimensionaleuclideangravity,wherewecanthink of x5 as representing euclidean, periodic time. Our task is greatly simplified by the fact that the equations of four-dimensional euclidean gravity have been extensively studied. For example, the Kaluza-Klein monopole of Gross and Perry, and of Sorkin [2], which is the object of our considerations here, is obtained by imbedding the Taub-NUT gravitational instanton into five dimensional Kaluza-Klein theory. Its line element is expressed as ds2 = −dt2+(cid:18)1+ 4m(cid:19)(cid:0)dr2+r2(dθ2+sin2θdφ)(cid:1)2+ (dψ+4mcosθdφ)2 , (2.2) r 1+4m/r where r > 0, and the angles θ,φ,ψ, (0 ≤ θ ≤ π,0 ≤ φ ≤ 2π) parametrize S3 ≈ SU(2). The apparent singularity at the origin is unphysical [4] if ψ is periodic with period 16πm. Since we want our solution to approach the vacuum for large r, we must identify 16πm with 2πR. Thus √ R πG m = = . (2.3) 8 2e The gauge field, A , is clearly that of a monopole, A = 4mcosθ,B(cid:126) = ∇(cid:126) ×A(cid:126) = 4m(cid:126)r/r3, µ φ and has a string singularity along the whole z axis. As usual, this singularity is an artifact if and only in the period of x5 is equal to 16πm. The magnetic charge of our monopole is thus fixed by the radius of the Kaluza-Klein circle. If we scale the magnetic field so as to have the proper normalization, B(cid:126) → (16πG)−1/2B(cid:126), we find that the magnetic charge is 4m R 1 g = √ = √ = . (2.4) 16πG 2 16πG 2e Thus, as expected, our monopole has one unit of Dirac charge. The mass of a static, asymptotically flat spacetime can be defined. In our case it is (cid:90) M = − 2πR d3x∇(cid:126)2(cid:0)1(cid:1) = m . (2.5) 16πG V G K Since m is fixed by the radius of the vacuum circle, which in turn is fixed by e and G, the soliton mass is determined to be m2 M2 = P , (2.6) 16α (cid:112) where m = hc/2πG ≈ 2.17·10−5 g is the Planck mass. As it is costumary for solitons, the P monopole mass is 1/e times heavier than the mass scale of the theory. Remarkably, the Kaluza-Klein monopole has re-emerged recently [5] as the asymptotic limit of the curved manifold, whose geodesics describe the scattering of self-dual monopoles. 3 CLASSICAL DYNAMICS Let us consider the geodesic motion of a particle in the Kaluza – Klein monopole field, with Lagrangian (cid:32) (cid:33) 1 1 4m (ψ˙ +4mcosθφ˙)2 L = g x˙µx˙ν = (1+ )(cid:126)r˙2+ . (3.1) 2 µν 2 r 1+ 4m r 4 To the two cyclic variables ψ and t are associated the conserved quantities ψ˙ +4mcosθφ˙ q = , (3.2) 4m 1+ r 1 4m (cid:16) (cid:17) E = (1+ ) (cid:126)r˙2+q2 , (3.3) 2 r interpreted as the electric charge and the energy, respectively. It is convenient to introduce the mechanical 3-momentum (cid:0) 4m(cid:1)˙ p(cid:126)= 1+ (cid:126)r. (3.4) r The 5-dimensional geodesics are the solutions of the Euler-Lagrange equations associated to (3.1). The projection into 3-space of the motion is hence governed by the equation, ˙ dp(cid:126) (cid:126)r×(cid:126)r (cid:126)r (cid:126)r = −4mq +2mq2 −2m(cid:126)r˙2 . (3.5) dt r3 r3 r3 This complicated equation contains, in addition to the Dirac-monopole plus a Coulomb terms, also a velocity-square dependent term, typical for motion in curved space. Due to the manifest spherical symmetry, the monopole angular momentum, (cid:126)r J(cid:126)=(cid:126)r×p(cid:126)+4mq , (3.6) r is conserved. The presence of the velocity-square dependent force changes, when compared to the pure Dirac + Coulomb case, the situation dramatically. Energy conservation implies that (2E −q2)r2−(E −q2)r−J2 (r˙)2 = . (3.7) (r+4m)2 For positive m the particle cannot reach the center. Indeed, eqn. (3.3) shows that, for m > 0 the energy is at least q2/2, and hence 4mq2 r ≥ r = > 0. (3.8) 0 2E −q2 For m > 0 there are no bound motions. For m < 0 instead, the energy can be smaller than q2/2. In such a case the coefficient of r2 is negative and the system does admit bound motions (see below). Despite the complicated form of the equations of motion, the classical motions are surpris- ingly simple. The clue is the observation [6] that, in addition to the angular momentum, J(cid:126), there is also conserved ‘Runge-Lenz’ vector, namely (cid:126)r K(cid:126) = p(cid:126)×J(cid:126)−4m(E −q2) . (3.9) r This is verified by an explicit calculation, using the equations of motion. These conserved quantities allow for a complete description of the motion [6]. Indeed, eqns. (3.5) and (3.6) allow to prove that (cid:126)r J(cid:126)· = 4mq, (3.10) r (cid:32) (cid:33) (E −q2)J(cid:126) K(cid:126) + ·(cid:126)r = J2−(4mq)2. (3.11) q 5 Figure 1: The particle moves on a cone, whose axis is the conserved angular momentum, J(cid:126). The position belongs to a plane perpendicular to N(cid:126) = qK(cid:126) +(E −q2)J(cid:126). The trajectory is, therefore, a conic section. The first of these equations implies that, as it is usual in monopole interactions, the particle moves on a cone with axis J(cid:126) and opening angle cosα = |4mq|/J. The second implies in turn that the motions lie in the plane perpendicular to the vector N(cid:126) = qK(cid:126) +(E −q2)J(cid:126). They are therefore conic sections, see Fig. 1. Theformofthetrajectorydependsonβ, theplane’sinclination, beingsmallerorlargerthenthe complement of the cone’s opening angle, π/2−α. Now cosβ = N(cid:126) ·J(cid:126)/NJ. Using the relations K(cid:126) ·J(cid:126) = −(4m)2q(E −q2), (3.12) K(cid:126)2 = (2E −q2)(J(cid:126)2−(4mq)2)+4m2(E −q2)2, (3.13) a simple calculation yields that for   E < q2/2 ellipses       E = q2/2 the trajectories are parabolae     E > q2/2   hyperbolae Let us now study the unbound motions. Using the two conserved quantities J(cid:126) and K(cid:126) the classical scattering can also be described. Let ω be the scattering angle, γ the twist, and let (cid:126)v denotethevelocityatlargedistances. ThenE = ((cid:126)v2+q2)/2. Themodulusoftheorbitalangular momentum, L = |L(cid:126)|, is L = y|(cid:126)v|, where y is the impact parameter. From the conservation of the component of J(cid:126) which is parallel to the particle’s plane, we get, π−θ sinγ tan = −y|(cid:126)v| . (3.14) 2 4qm From the conservation of the parallel component of K(cid:126), we get in turn π−θ cosγ tan = −y|(cid:126)v| . (3.15) 2 4m(E −q2) 6 From these equations we deduce (cid:32) (cid:33) γ q θ 2m (cid:18) q (cid:19)2 tan = and tan = − 1+ . (3.16) 2 |(cid:126)v| 2 y |(cid:126)v| The classical cross-section is, therefore, (cid:32) (cid:33) dσ |dy/dθ| m2 (cid:18) q (cid:19)2 = y = 1+ . (3.17) dω sinθ sin4θ/2 |(cid:126)v| To clarify the structure of the conserved quantities, it is convenient to switch to the Hamil- tonian formalism. The momenta which are canonically conjugate to the coordinates are π = (1+4m/r)x˙ −qA , i = 1,2,3, and π = q, where A = A = 0, A = −4mcosθ. The i i i 4 r θ φ Poisson brackets are (cid:8)xi,π (cid:9) = δi , (cid:8)ψ,π (cid:9) = 1. j j 4 The mechanical momentum p(cid:126) = (cid:126)π +qA(cid:126) = (1+4m/r)(cid:126)r˙ satisfies therefore the Poisson bracket relations (cid:8)p ,p (cid:9) = −(4qm)(cid:15) xk, (cid:8)xi,p (cid:9) = δi , (cid:8)ψ,p (cid:9) = A . (3.18) i j ijkr3 j j j j The Hamiltonian is (cid:32) 2 (cid:33) 1 p(cid:126) 4m H = +(1+ )q2 . (3.19) 2 1+4m/r r The Poisson brackets of the angular momentum and the Runge-Lenz vector are (cid:8)J ,J (cid:9) = (cid:15) J , (cid:8)J ,K (cid:9) = (cid:15) K , (cid:8)K ,K (cid:9) = (q2−2H)(cid:15) J . (3.20) i k ikn n i k ikn n i k ikn n For each fixed value E set (q2−2E)−1/2K(cid:126)   E < q2/2       M(cid:126) = K(cid:126) for E = q2/2 . (3.21)     (2E −q2)−1/2K(cid:126)   E > q2/2 For energies lower then q2/2 the angular momentum, J(cid:126), and the rescaled Runge-Lenz vector, M(cid:126) , form an o(4) algebra [6]. For E > q2/2 the dynamical symmetry generated by J(cid:126) and M(cid:126) is rather o(3,1). In the parabolic case, E = q2/2, the algebra is o(3)⊕ R3. s 4 QUANTIZATION First, we find the quantum operators by DeWitt’s rules [7]. xˆ is multiplication by x , and the i i canonical momentum operator is πˆ = −i(cid:0)∂ + 1(∂ lng)(cid:1). i i 4 i √ xî and πˆ are hermitian with respect to L2( gd4x), and satisfy the basic commutation relations i [xî,xˆj] = [πˆ ,πˆ ] = 0, [xî,πˆ ] = iδi . i j j j It is, however, more useful to introduce the modified (with respect to the Taub-NUT volume element non-self-adjoint!) operator pˆ = −i∂ +qA , (4.1) k k k 7 with commutation relations xk [pˆ,pˆ ] = −i(4mq)(cid:15) , [xî,pˆ ] = iδi , [ψˆ,pˆ ] = iA . (4.2) i j ijkr3 j j j j The charge operator q = −i∂ has eigenvalues q = s/4m, s = 0,±1/2,±1,... Our 4-metric is ψ Ricci flat, so the quantum Hamiltonian is [7] H(cid:98) = 1g−1/4πˆµ√ggµνπˆνg−1/4 = − √1 ∂µ(cid:0)√g∂µ(cid:1) = −(cid:52), (4.3) 2 2 g 2 the covariant Laplacian on curved 4-space. Then a lengthy calculation yields the three dimensional expression (cid:18) (cid:19) H(cid:98) = 1 (1+4m/r)−1pˆkpˆk +(cid:0)1+ 4m(cid:1)q2 , (4.4) 2 r which is formally the same as the classical expression. Notice, however, that the order of the operators is important. The charge operator q commutes with the Hamiltonian, and is therefore conserved. The angular momentum operator is J(cid:126)ˆ=(cid:126)rˆ×p(cid:126)ˆ+4qm(cid:126)r. (4.5) r Inserting the square of the angular momentum, J2, into the Hamiltonian, we get Hˆ = 1 (cid:18)− 1 ∂ (r2∂ )+ J2−(4qm)2 +(cid:0)1+ 4m(cid:1)2q2(cid:19). (4.6) 4m r2 r r r2 r 2(1+ ) r Substituting here j(j +1) for J2, s/4m for q, introducing the new wave function ψ = rΨ and multiplying by 2r(1+4m/r), the eigenvalue equation HˆΨ = EΨ becomes (cid:20) d2 j(j +1) 8mE −s2/2m (cid:16)(cid:0) s (cid:1)2 (cid:17)(cid:21) − + − −2E ψ = 0. (4.7) dr2 r2 r 4m Choosingψ ∼ rj+1 asboundaryconditionnearr = 0,weobtainawell-definedproblemforwhich (4.6) is self-adjoint on the Hilbert space L2(R ,dr) over the half-line. Continuum solutions of + eqn. (4.7) which are bounded at r = 0 have the form ψ(r) = rj+1eikrF(iλ+j +1,2j +2,−2ikr), (4.8) k2 = 2E −s2, (4.9) q2−E λ = −4m , (4.10) (cid:112) 2E −q2 where F is the confluent hypergeometric function. Square-integrable bound states correspond to λ such that λ2 = −(k+j +1)2 = −n2, k = 0,1,2,..., i.e., 4m(q2−E) − = n, n = |s|+1,|s|+2,... (4.11) (cid:112) q2−2E For q2 > 2E i.e. for s2 > 32Em2 (which, by eqn. (4.4), is only possible for negative m), one gets the bound-state energy levels, 1 (cid:112) (cid:16) (cid:112) (cid:17) E = n2−s2 ±n− n2−s2 , n = |s|+1,|s|+2,... (4.12) n (4m)2 8 with degeneracy n2−s2. Observe that n2−s2 = (n+s)(n−s) is always an integer, since n and s are simultaneously integers or half-integers. The two signs correspond to the lightly bound states E > 0 and to the tightly bound states with E < 0. For E > q2/2 one gets scattering states. The quantum cross section can be calculated [6] by solving the problem in parabolic coordinates. It is identical to the classical expression (3.19) with q/|(cid:126)v| replaced by s/4mk, i.e., m2(cid:16)1+(cid:0)s/4mk(cid:1)2(cid:17)2 dσ = . (4.13) dω sin4 θ 2 QuantizingtheRunge-LenzvectorK(cid:126) isahardtask. Theclueis[8]thatK(cid:126) isassociatedwith three Killing tensors Ki , i = 1,2,3. Remember that a Killing tensor is a symmetric tensor µν K on(curved)space,suchthatK = 0(thesemicolondenotesmetric-covariantderivative, µν (µν;α) (·) = ∇ (·)). To a Killing tensor is associated a conserved quantity which is quadratic in the ; µ velocity, namely 1 K = K x˙µx˙ν. µν 2 For example, the metric tensor g itself satisfies these conditions; the corresponding conserved µν quantity is the energy. Following Carter [8], the quantum operator of K is 1 1 K(cid:98) = − ∇µKµν∇ν = − ∇µKµν∇ν. (4.14) 2 2 The classical expression (3.9) of the Runge-Lenz vector allows us to identify the components of the Killing tensor and use Carter’s prescription to get the quantum operators. A long and complicated calculation yields K(cid:126)ˆ = 1(cid:0)p(cid:126)ˆ×J(cid:126)ˆ−J(cid:126)ˆ×p(cid:126)ˆ(cid:1)−4m(cid:126)r(Hˆ −q2). (4.15) 2 r ˆ Again, the order of the operators is relevant. For notational convenience we drop the ‘hat’ {·} from our operators in what follows. 5 THE SPECTRUM FROM THE DYNAMICAL SYMMETRY Using the fundamental commutation relations (4.2) one verifies that the operators J(cid:126) and K(cid:126) satisfy the quantized version of (3.20), [J ,J ] = i(cid:15) Jn, [J ,K ] = i(cid:15) Kn, [K ,K ] = i(q2−2H)(cid:15) Jn. (5.1) i k ikn i k ikn i k ikn Analogously to (3.21), on the fixed-energy eigenspace HΨ = EΨ define the rescaled Runge Lenz operator M(cid:126) by (q2−2E)−1/2K(cid:126)   E < q2/2       M(cid:126) = K(cid:126) for E = q2/2 . (5.2)     (2E −q2)−1/2K(cid:126)   E > q2/2 M(cid:126) and J(cid:126) close, just like classically, to an o(4) algebra for E < q2/2, [J ,J ] = i(cid:15) Jn, [J ,M ] = i(cid:15) Mn, [M ,M ] = i(cid:15) Jn. (5.3) i k ikn i k ikn i k ikn 9 to o(3,1) for E > q2/2, [J ,J ] = i(cid:15) Jn, [J ,M ] = i(cid:15) Mn, [M ,M ] = −i(cid:15) Jn. (5.4) i k ikn i k ikn i k ikn and to o(3)⊕ R3 in the parabolic case, E = q2/2, s [J ,J ] = i(cid:15) Jn, [J ,M ] = i(cid:15) Mn, [M ,M ] = 0. (5.5) i k ikn i k ikn i k Following Pauli [9], this allows us to calculate the energy spectrum. We have indeed the con- straint equations, K(cid:126) ·J(cid:126) = −(4m)2q(E −q2), (5.6) K(cid:126)2 = (2E −q2)(cid:0)J(cid:126)2−(4mq)2+1(cid:1)+4m2(E −q2)2. (5.7) Observe that the quantum expressions differ from their classical counterparts, (3.12)-(3.13), in an important term, (h/2π)2 = 1 in our units. Those states Ψ with constant charge, q = s/4m, and energy E < q2/2 4mqΨ = sΨ, HΨ = EΨ (5.8) form a representation space for o(4). It is more convenient to consider the commuting operators A(cid:126) = 1(cid:0)J(cid:126)+M(cid:126) (cid:1), B(cid:126) = 1(cid:0)J(cid:126)−M(cid:126) (cid:1), (5.9) 2 2 which generate two independent o(3)’s, o(4) = o(3) ⊕ o(3). A common eigenvector Ψ of the commuting operators q,H,A(cid:126)2,A ,B(cid:126)2,B satisfies 3 3 A(cid:126)2Ψ = a(a+1)Ψ, A Ψ = a Ψ, B(cid:126) 2Ψ = b(b+1)Ψ, B Ψ = b Ψ, (5.10) 3 3 3 3 where a and b are half-integers, and a = −a,−a+1,...,a, b = −b,−b+1,...,b. 3 3 Consider the (so far non-negative real) number 4m(q2−E) n = − . (5.11) (cid:112) q2−2E The operator identities (5.6), (5.7) imply that a(a+1)+b(b+1) = (s2−1−n), a(a+1)−b(b+1) = sn2. Then some algebra yields 2a+1 = ±(n+s), 2b+1 = ±(n−s) ⇒ a−b = ±s, a+b+1 = ±n = n, since a and b are non-negative. Due to the first of these relations, n is integer or half-integer, dependingonsbeingintegerorhalf-integer. Eqn. (5.11)isthusidenticaltoeqn. (4.11),yielding the bound-state spectrum once more. 6 ALGEBRAIC CALCULATION OF THE S MATRIX [10, 11] The scattering states form rather a representation space for o(3,1). This can be used to derive algebraically the S-matrix. Let us indeed re-write the Runge-Lenz operator K(cid:126) as (cid:126)r K(cid:126) = (1+4m/r)(i(cid:126)v−J(cid:126)×(cid:126)v)−4m (H −q2), (6.1) r 10