aa r X i v : . [ h e p - ph ] F e b Non-forward scattering of twisted particles
Igor P. Ivanov
IFPA, Universit´e de Li`ege, All´ee du 6 Aoˆut 17, bˆatiment B5a, 4000 Li`ege, BelgiumSobolev Institute of Mathematics, Koptyug avenue 4, 630090, Novosibirsk, Russia
November 27, 2018
Abstract
Twisted photons (i.e. photons carrying non-zero orbital angular momentum) arewell-known in optics. Recently, it was suggested to use Compton backscattering off anultra-relativistic electron beam to boost optical twisted photons into the high energyrange. However, only the case of strictly forward/backward scattering has been studiedso far. Here, we consider generic kinematic features of processes in which a twistedparticle scatters with non-zero transverse momentum transfer.
In perturbative quantum field theory we assume that interaction among the fields can betreated as a perturbation of the free field theory. This perturbation leads to scattering be-tween asymptotically free multiparticle states, which are usually constructed from the planewave one-particle states. This choice greatly simplifies the calculations and represents a veryaccurate approximation to the real experimental situation in virtually all circumstances. How-ever, one can, in principle, choose any complete basis for the one-particle states other than theplane wave basis, provided that it is still made up of solutions of the free field equations. Suchstates can carry new quantum numbers absent in the plane wave choice and, if experimentallyrealized, they can offer new opportunities in high-energy physics.Thanks to the recent experimental progress in optics, it is now possible to create laserbeams carrying non-zero orbital angular momentum (OAM) [1], for a recent review see [2].The lightfield in such beams is made of states which are non-plane wave solutions of theMaxwell equations. Each photon in this lightfield, which we call a twisted photon , carries anon-zero OAM quantized in units of ~ . Several sets of solutions have been investigated, such asBessel beams or Gauss-Laguerre beams, but in all cases the spacial distribution of the lightfieldis necessarily non-homogeneous in the sense that the equal phase fronts are not planes buthelices. Such states form a complete basis which can be used to describe the initial and finalasymptotically free states. Moreover, it is the basis of choice for experimental situations whenthe initial states are prepared in a state of (more or less) definite OAM.Twisted photons have been produced in various wavelength domains, from radiowave [3]to optical, with prospects to create a brilliant X-ray beam of twisted light in the keV range[4]. Very recently it was suggested to use the Compton backscattering of twisted optical1hotons off an ultra-relativistic electron beam to create a beam of high-energy photons withnon-zero OAM [5, 6]. The technology of Compton backscattering is well established, andthe high-energy electron beams and the OAM optical laser beams are already available. Thesuggestion of [5, 6] paves the way for the twisted photons, and the twisted states in general,into the high-energy physics. The wealth of new physics opportunities related to this newdegree of freedom is yet to be understood (see [7] for initial steps in this direction).In this paper we begin this exploration by focusing on a technical question of how tocalculate the scattering of a one-particle twisted state off plane wave states:twisted + X ( p ) → twisted ′ + X ′ ( p ′ ) . (1)Here systems X and X ′ are one- or multi-particle states described by plane waves, p and p ′ being their respective total momenta. We consider the case of non-forward (and non-backward)scattering, that is, when the transverse momenta of systems X and X ′ with respect to theaxis used to define the twisted states are different, p = p ′ .In this work we focus on the general kinematical properties of the scattering matrix whichhold for all processes of type (1). The purpose of this exercise is twofold. The main goal is todevelop the formalism of treating the scattering processes with twisted particles in the initialand final states. The secondary goal is to extend the original calculation of [5, 6] of the strictlybackward Compton scattering of twisted optical photons to non-zero transverse momentumtransfer. The basic questions are: how to write the cross section of this process, and how theparameters of the final twisted photon depend on the momentum transfer.Since the kinematical details turn out to be rather unconventional, we describe them ina pedagogical manner with the simplest possible process: the decay of a massive twistedscalar into two massless scalar. In this way we separate the effect of non-homogeneous spatialdistribution from the possible effect of non-trivial polarization states, which the twisted photoncan have and which we postpone for a future study. We stress however that the kinematicalfeatures we explore with this example are pertinent to all the scattering processes of type (1).Additional process-specific properties arise on top of these kinematical features.The paper is organized as follows. In Section 2 we give an introduction into twisted particlesand describe some of its properties. In Section 3 we describe the general problem we tackle:scattering of a twisted particle accompanied by a non-zero momentum transfer. We pinpointthe key quantity to analyze, and then, in Section 4, we investigate this quantity in detail witha simple pedagogical example. Section 5 contain discussion of the results obtained and in thefinal Section we draw our conclusions. In Appendix we derive some properties of the Besselfunctions used in the main text. As mentioned in the introduction, we focus in this paper on twisted scalar particles only. Inthis Section we follow essentially [5, 6].We represent a state with non-zero OAM with a Bessel beam-type twisted state. Thisis a solution of the wave equation in the cylindric coordinates with definite energy ω andlongitudinal momentum k z along a fixed axis z , definite modulus of the transverse momentum2 k | (all transverse momenta will be written in bold) and a definite z -projection of OAM. Ifthe plane wave state | P W ( k ) i is | P W ( k ) i = e − iωt + ik z z · e i kr , (2)then a twisted scalar state | κ, m i is defined as the following superposition of plane waves: | κ, m i = e − iωt + ik z z Z d k (2 π ) a κm ( k ) e i kr , (3)where a κm ( k ) = ( − i ) m e imφ k √ π δ ( | k | − κ ) √ κ . (4)In the coordinate space, | κ, m i = e − iωt + ik z z · ψ κm ( r ) , ψ κm ( r ) = e imφ r √ π √ κJ m ( κr ) . (5)Here, following [5] we call κ the conical momentum spread, m is the z -projection of OAM,and the dispersion relation is k µ k µ = ω − k z − κ = M . We note in passing that the averagevalues of the four-momentum carried by a twisted state is h k µ i = ( ω, , k z ) , (6)so that h k µ ih k µ i = M + κ , which is larger than the true mass of the particle squared.The transverse spatial distribution is normalized according to Z d r ψ ∗ κ ′ m ′ ( r ) ψ κm ( r ) = δ m,m ′ √ κκ ′ Z rdrJ m ( κr ) J m ( κ ′ r ) = δ m,m ′ δ ( κ − κ ′ ) . (7)The plane wave can be recovered from the twisted states as follows: | P W ( k = 0) i = lim κ → r πκ | κ, i , (8) | P W ( k ) i = r πκ + ∞ X m = −∞ i m e − imφ k | κ, m i , κ = | k ⊥ | . (9)If needed, these two cases can be written as a single expression: | P W ( k ) i = lim κ →| k | r πκ + ∞ X m = −∞ i m e − imφ k | κ, m i . (10)From these expressions one sees that the twisted states with different m and κ representnothing but another basis for the transverse wave functions. When calculating cross sections and decay rates, we need to integrate the transition proba-bility over the phase space of the final particles. When calculating the density of states, we3onsider large but finite volume and count how many mutually orthogonal states with pre-scribed boundary conditions can be squeezed inside. In the present case due to the cylindricalsymmetry of the problem, we choose a cylinder of large radius R and length L z . In the caseof plane waves we have dn P W = πR L z dk z d k (2 π ) . (11)The full number of states with transverse momenta up to κ and longitudinal momenta | k z | ≤ k z is k z L z · R κ / π .To count the number of twisted states | κ, m i in the same volume, we specify the boundarycondition, e.g. ψ κm ( r = R ) = 0, which makes κ discrete such that κ i R is the i -th root of theBessel function J m . Then we note that the position of the first root of the Bessel function J m ( x ) is always at x > m , and as m grows x → m . For a given κ , the maximal m for whichthe wave can still be contained inside the cylindrical volume is m max = κR , which has a verynatural quasiclassical interpretation.If m is small and not growing with R , then one can use the well-known asymptotic formof the Bessel functions to count the number of states: dn tw = Rdκ L z dk z ∆ m π . (12)Here, ∆ m is written instead of just 1 to signal the presence of a discrete running parameter m . If m is not restricted to small values, this asymptotic form of J m ( x ) cannot be used sinceit requires m ∼ < x . Instead, the so-called approximation by tangents can be used, which givesthe following density of states: dn tw = p m max − m dκκ ∆ mπ L z dk z π . (13)In the limit m ≪ m max ≡ κR this expression reproduced (12). Alternatively, one can calculatethe radial part of the density of states via the adiabatic invariant as suggested in [6]. Thenumber of radial excitations n r for a fixed m is n r = Z Rm/κ k r ( r ) drπ , k r ( r ) = r κ − m r . (14)The density of states is then given by dn r = dn r dκ dκ = p m max − m dκκπ . (15)One important remark is in order. Effectively, switching from the plane wave to twistedstate basis for the final particles implies replacement d k → s − m m max κdκ ∆ mm max . (16)Note that the contribution of each “partial wave” with a fixed m vanishes in the infinitevolume limit as 1 /R . However, the number of partial waves grows ∝ R , and in order to get4 non-vanishing result for a physical observable, one must integrate over the full available m interval up to m max . This holds even if the transverse momenta stay small, and is related tothe fact that the plane wave contains contributions from all impact parameters with respectto any axis non-collinear to its momentum.Another expression one needs for the probability calculations is the normalization constantsfor the one-particle states. A usual plane wave one-particle state is normalized to 2 E · V ; torenormalize it to one particle per the entire volume, the plane wave should be multiplied by N P W , with N P W = 12
EV , V = πR L z . (17)For a twisted states the corresponding normalization factor N tw is N tw = 12 E πκ p m max − m L z , (18)which in the small- m case simplifies to N tw ≈ E πRL z , (19)which was also derived in [6]. Note however that even in the general case the product ofthe normalization constant squared and the density of states for each final twisted particle issimplified as N tw dn tw = dκdk z ∆ m E · π . (20) X’(p’), m ’, m’ κκ X(p)
Figure 1: A generic process of non-forward scattering of a twisted state off plane waves.Consider again the generic process (1) which is schematically depicted in Fig. 1. Here wehave one twisted particle in the initial and one in the final states. Since we deal here with thekinematical features of the process, it is inessential whether these particles are of the same typeor not. These two particles are described by the states | κ, m i and | κ ′ , m ′ i defined with respectto a common z -axis. All the other particles are assumed to be plane waves, which allows usto define the total transverse momenta p and p ′ of the systems X and X ′ , respectively. Thetransverse momentum transfer, q ≡ p − p ′ is also well-defined, and we can consider two cases:(i) strictly forward scattering, q = 0, (2) non-forward scattering, q = 0. Note that we use the5erm “forward” for any process with zero transverse momentum transfer, i.e. both for strictlyforward and strictly backward scattering.Suppose that we know the S -matrix for the same process with all particles represented bythe plane waves. If the would-be twisted particles in the initial and final states have momenta k and k ′ , the plane wave S -matrix has the familiar form: S P W = i (2 π ) δ (4) ( k + p − k ′ − p ′ ) · M ( k , k ′ , φ kk ′ ) . (21)The invariant amplitude M depends among other on the transverse momenta squared k and k ′ as well as on the azimuthal angle φ kk ′ between them, which is explicitly indicated in (21).Any other azimuthal angle, say, between k and p , can be written as φ kp = φ kq + φ qp , andtherefore it is expressible via φ kk ′ and internal angles in the systems X and X ′ .To get the S -matrix for the twisted particle scattering process (1), one just needs to apply(4) to the initial and final states [5]: S tw = Z d k (2 π ) d k ′ (2 π ) a ∗ κ ′ m ′ ( k ′ ) a κm ( k ) S P W . (22)The delta-functions present in (22) fix k = κ , k ′ = κ ′ , and, as we will see later, they specifythe angle φ kk ′ up to its sign. If the plane wave matrix element is not sensitive to the sign ofthe angle φ kk ′ , M ( κ , κ ′ , φ kk ′ ) = M ( κ , κ ′ , − φ kk ′ ), then it can be taken out of the integral,and we are left with the following transverse process-independent master integral: I m,m ′ ( κ, κ ′ , q ) = Z d k (2 π ) d k ′ (2 π ) a ∗ κ ′ m ′ ( k ′ ) a κm ( k ) δ (2) ( k − k ′ + q ) . (23)If the matrix element depends on the sign of φ kk ′ , then it can be decomposed into a symmetricand antisymmetric terms M s = M ( φ kk ′ ) + M ( − φ kk ′ )2 , M a = M ( φ kk ′ ) − M ( − φ kk ′ )2 , (24)then in the twisted scattering amplitude M s will be accompanied by the master integral(23), while M a will enter together with a slightly modified version of the master integral,¯ I m,m ′ ( κ, κ ′ , q ), in which the integrand of (23) is multiplied by the sign function ǫ ( φ kk ′ ). Infact, two interfering plane wave amplitudes appearing here might lead to novel observableeffects in twisted particle cross sections, see details in [7].In the next Section we compute the master integral (23) and explore the singularities of |I| which appear in the cross section/decay rate calculations. However, let us briefly summarizethe results right away: • In the strictly forward case, q = 0, I m,m ′ ∝ δ ( κ − κ ′ ) δ m,m ′ , that is, the quantum numbersof the twisted state are transferred to the final particles without any change. • In the non-forward case, q = 0, a distribution over κ ′ arises, and κ ′ can be large if themomentum transfer is large. • At any non-zero momentum transfer instead of δ m,m ′ we observe a distribution of m ′ over the entire possible range of values, − m ′ max ≤ m ′ ≤ m ′ max , where m ′ max = κ ′ R → ∞ . • |I| contains singularities at the border of the kinematically allowed region, which mustbe carefully dealt with. 6 Decay of a twisted scalar
We would like to explore the integral (23) in the context of the simplest possible problem:decay of a twisted scalar particle with mass M into a pair of massless distinguishable particlesdue to the cubic interaction g · Φ φ φ . To make the presentation more pedagogical, we willfirst calculate the decay rate when both particles in the final state are plane waves, then forthe plane wave plus twisted final state, and finally for the case when both final particles aretwisted. We will calculate the decay width in the center of mass frame defined by k z = 0. Thisis not the true rest frame because due to the transverse motion a twisted particle is never atrest. For future reference, let us first recall the calculation for the standard case when all theparticles including the initial one are plane waves. The S -matrix is given as usual by S = i (2 π ) δ (4) ( p − k − k ) · g . When squaring the delta-function, we use the standard prescription[ δ (4) ( p − k − k )] −→ δ (4) ( p − k − k ) · δ (4) (0) = δ (4) ( p − k − k ) V T (2 π ) . (25)We also use the plane wave normalization for all the particles, The decay probability per unittime for a particle at rest is: d Γ = (2 π ) g δ (4) ( p − k − k ) V TT · ( N P W ) · dn P W ( k ) dn P W ( k )= g (2 π ) δ ( M − ω − ω )8 M ω ω d k , (26)so that the total width is Γ = g πM . (27)Now we repeat this calculation for the initial twisted state | κ, m i , while keeping the planewave basis the final particles. The S -matrix is S = i (2 π ) g δ ( E − ω − ω ) δ ( k z + k z ) Z d k (2 π ) a κm ( k ) δ (2) ( k − k − k ) (28)= i (2 π ) g δ ( E − ω − ω ) δ ( k z + k z ) ( − i ) m (2 π ) / e imφ δ ( κ − k ) √ κ , (29)where k ≡ p k + k + 2 | k || k | cos( φ − φ ) and φ is the angle of the 2D vector k w.r.t.some axis x . The phase factor is inessential and m disappears in the decay rate.When squaring the above expression, we encounter the square of δ ( κ − k ), which is treatedas in [5, 6]: [ δ ( κ − k )] = δ ( κ − k ) δ (0) → δ ( κ − k ) Rπ . (30)This prescription comes from the observation that at large but finite R and at κ = k thedivergent integral in (28) is regularized as δ (0) = Z ∞ rdr [ J m ( κr )] → Z R rdr [ J m ( κr )] ≈ Rπ . (31)7hus, with the normalization factors plugged in, the decay rate has the form d Γ = (2 π ) g Eω ω · T δ ( E − ω − ω ) δ ( k z + k z ) T L z δ ( κ − k ) κ Rπ · V πRL z · V d k (2 π ) V d k (2 π ) = g (2 π ) δ ( κ − k ) κ δ ( E − ω − ω )8 Eω ω dk z d k d k . (32)For the transverse integral we write Z dφ δ ( κ − k ) κ = 2 Z dφ δ (cid:2) κ − k − k − | k || k | cos( φ − φ ) (cid:3) = 1∆ , (33)where ∆ = 14 q k k + κ k + κ k ) − κ − k − k (34)is the area of the triangle with sides κ , | k | and | k | . The angular integral can be non-zero k κκ Figure 2: The allowed kinematical region of the values of | k | and | k | for a fixed κ definedby the “triangle rules” (35).only if such a triangle can be formed, that is if | k | and | k | satisfy the “triangle rules”: κ ≤ | k | + | k | , | k | ≤ κ + | k | , | k | ≤ κ + | k | . (35)The allowed values of | k | and | k | for a fixed κ are shown in Fig. 2. Note that the angularintegral (33) receives equal contributions from two points: φ − φ = ± δ , where δ = arccos (cid:18) κ − k − k | k || k | (cid:19) . (36)As usual, the energy delta function can be killed by the k z integration Z dk z δ ( E − ω − ω ) ω ω = 2 Ek ∗ z , (37)where k ∗ z = 12 E q E + k + k − E k + E k + k k ) . (38)8he decay rate becomes d Γ = g π E k ∗ z | k | d | k | | k | d | k | ∆ . (39)The integration region over | k | and | k | is defined by the requirement that, in addition to(35), the longitudinal momentum k ∗ z is well-defined, which cuts the rectangular shape shownin Fig. 2 from above. It is conveniently described with variables x = | k | − | k | κ , z = | k | + | k | κ , x ∈ [ − , , z ∈ [1 , z max ] , z max ≡ Eκ > . (40)In these variables∆ = κ p ( z − − x ) , k ∗ z = E s(cid:18) − x z max (cid:19) (cid:18) − z z max (cid:19) , (41)and the decay rate takes form d Γ = g π E · ( z − x ) dzdx [( z max − z )( z max − x )( z − − x )] / . (42)This integral can be taken exactly, and it gives the decay width of the twisted scalar particleΓ = g πE , (43)which is a very natural result. In the limit κ →
0, we recover the plane wave decay width(27).
Let us now describe the final state as twisted state plus a plane wave: | κ, m i → | κ , m i + | P W ( k ) i . (44)Since the full decay width cannot depend on the basis we choose for the final particles, wemust recover the same result (43) in this basis. In addition to that, we also want to know howthe final twisted state parameters κ and m are related to the initial state parameters κ and m . The S -matrix is now expressed in terms of the master integral (23): S = i (2 π ) gδ ( E − ω − ω ) δ ( k z + k z ) · I m,m ( κ, κ , k ) . (45)Let us first calculate this integral in the strictly forward case k = 0: I m,m ( κ, κ ,
0) = i m − m (2 π ) Z d k d k e imφ − im φ δ ( | k | − κ ) √ κ δ ( | k | − κ ) √ κ δ (2) ( k − k )= i m − m (2 π ) √ κκ Z dφ dφ e imφ − im φ · δ ( k − k ) δ ( φ − φ )= 1(2 π ) δ ( κ − κ ) δ m,m , (46)9hich was first obtained in [5]. This result implies that the twisted state quantum number aretransferred from the initial to the final twisted particle without any change. The differentialdecay rate is d Γ = g (2 π ) δ ( E − ω − ω ) δ ( k z + k z )8 Eω ω δ m,m δ ( κ − κ ) · dκ dk z d k = g (2 π ) δ ( E − ω − ω )8 Eω ω d k = g (2 π ) d k E k ∗ z , (47)and we stress that this result is applicable only at k = 0. Now we consider the non-forward case. The master integral can be rewritten as I m,m ( κ, κ , k ) = i m − m (2 π ) √ κκ Z dφdφ e imφ − im φ · δ (2) ( k − k − k ) . (48)There are two ways to look at this integral. First, we can rewrite δ (2) ( k − k − k ) = 1(2 π ) Z d r e i rk − i rk − i rk (49)and represent the integral as I m,m ( κ, κ , k ) = i m − m e i ( m − m ) φ (2 π ) √ κκ · Z ∞ rdrJ m ( κr ) J m ( κ r ) J m − m ( | k | r ) , (50)where φ is the azimuthal angle of k . Note that although we are considering the case withtwo twisted particles (one in the initial and one in the final state), an integral over three Besselfunction arises automatically.The master integral can be also calculated by a direct integration over angles in (48). Werewrite the delta-function as δ ( k − k − k ) = 2 δ [ κ − ( k + k ) ] δ ( φ − φ k + k )= 2 δ [ κ − κ − k − κ | k | cos( φ − φ )] δ ( φ − φ k + k ) . (51)One sees that the integral can be non-zero only if it is possible to form a triangle with sides κ , κ , and | k | ; thus, κ and | k | satisfy the same triangle rules as in (35). Fixing κ and | k | means that the triangle can be formed only at two choices of relative azimuthal angles of k , k and k . Let us introduce δ = arccos (cid:18) κ + κ − k κκ (cid:19) , δ = arccos (cid:18) κ + k − κ κ | k | (cid:19) , (52)Then, the two configurations of transverse momenta correspond to φ − φ = ± δ , φ − φ = ∓ δ , (53)10o that the signs of φ − φ and φ − φ are always opposite. The master integral then becomes I m,m ( κ, κ , k ) = i m − m (2 π ) √ κκ e i ( m − m ) φ cos[ m δ − ( m − m ) δ ]∆ , (54)where ∆ is the same as in (34) with k replaced by κ . Comparison of (50) with (54) gives theresult for the integral of the triple Bessel function product. See also [8] for some mathematicsinvolved in evaluation of this and similar integrals.Let us also check the k → | k | ≪ κ , the distribution over κ spans from κ − | k | to κ + | k | , see Fig. 2. The angle δ →
0, while δ can still be arbitrary.However, in the k → m − m = 0 term survives due to J m − m ( | k | r ), so that thecosine in (54) approaches unity. The analysis of ∆ shows thatlim | k |→ √ κκ ∆ = 2 πδ ( κ − κ ) , (55)so that one indeed recovers the strictly forward result for the master integral given in (46).Note that although we wrote simply | k | →
0, the exact definition of this limit is involved andis discussed in Section 5.2.Note also that the modified master integral ¯ I m,m ′ ( κ, κ ′ , q ) mentioned in the previous Sec-tion has the same form as (54) but contains sine instead of cosine of m δ + ( m − m ) δ . Thismodified integral does not enter our particular calculation since the amplitude is symmetricunder the φ kk ′ sign flip. After integrating over k z , the decay rate can be written as d Γ = 4 π g E k ∗ z · |I m,m ( κ, κ , k ) | · dκ R d k . (56)Note that this decay rate is differential not only in κ and k but also in the discrete variable m ; the full decay width includes integrals over momenta and a summation over all possible m ’s.A close inspection shows that the immediate integration over κ or k cannot be donedue to singularities of |I| along the boundaries of the kinematically allowed region shown inFig. 2. In contrast to the plane wave case, the denominator now contains ∆ instead of just∆. Therefore, in terms of variables x and z one encounters singularities of the form Z − dx − x and Z z max dzz − . Clearly, this is an artefact of the infinite radial integration range. If instead we take R to belarge but finite, we expect that a trick similar to (30) should be at work, namely that afterregularization |I| would yield R times a less singular function.This trick does not seem to work for each m separately. However, as we prove in Appendix,it works for |I| summed over all possible m . In the limit R → ∞ we obtain + ∞ X m = −∞ |I m,m ( κ, κ , k ) | = 1(2 π ) Rκ π , (57)11ith the same ∆ as before. The regularization parameter R then disappears from the result,and the decay rate reads d Γ = g π E k ∗ z · κ dκ | k | d | k | ∆ . (58)Comparing (58) with the previous results (47) and (39) leads us to two conclusions. • The transition from the strictly forward to the non-forward cross section/decay rateconsists in replacement δ ( κ − κ ) κ → π ∆ . (59) • The result (58) for a twisted particle in the final state coincides with the result (39) forthe case when both final particles are plane waves.Although these conclusions were drawn for the specific process we consider, the way it isderived suggests that this might be a universal feature for many (or all) processes of type (1).
For completeness, let us also recalculate the decay rate in the basis when both final particlesare described by twisted states: | κ , m i and | κ , m i . We remind that all twisted states aredefined with respect to the same common z axis. It turns out that this calculation closelyfollows the case of twisted state plus plane wave just considered. This is not surprising becausethe appearance of the triple Bessel integral highlights the fact that when two particles (onein the initial and one in the final state) are twisted, the third one is automatically projectedfrom the plane wave onto an appropriately defined twisted state as well.The S -matrix takes the form S = ig (2 π ) / δ ( E − ω − ω ) δ ( k z + k z ) · δ m,m + m √ κκ κ Z rdr J m ( κr ) J m ( κ r ) J m ( κ r ) , (60)and we again encounter the triple Bessel-function integral. The decay rate is written as d Γ = g π E k ∗ z · κ dκ κ dκ ∆ . (61)This expression is identical to (58) up to the obvious replacement | k | → κ ; its integrationover all conical momenta spreads κ , κ gives again (43). κ Let us first discuss what typical values of κ and m essentially contribute to the decay rate.The differential decay rate (58) shows that at large | k | ≫ κ the conical momentum spread κ is limited to the interval from | k | − κ to | k | + κ with an inverse square root singularity atthe endpoints. This singularity is integrable, so that the entire interval more or less equallycontributes to the integral. Since | k | can be as high as E , the total decay width is therefore12ominated by large κ ≫ κ . In a more complicated process, the decay rate or the cross sectionwill include the amplitude squared which can serve as a cut-off function. For example, in theCompton scattering one expects that κ up to ∼ m e will contribute to the cross section. m Although we found a result for the decay rate summed over all m , we can trace the main m -region from the intermediate formulas. The result is that essentially all m from minusto plus infinity are important for the decay rate, which is in a strong contrast to the strictlyforward result m = m .Indeed, the distribution over final m can be seen in (56), where the master integral I m,m ′ is given by (54). For generic transverse momenta, growth of m leads to oscillations of thecosine function with constant amplitude. Thus, when averaging over the entire κ interval,one can approximate cosine squared by 1 /
2, and the dependence on m drops off. This resultholds for any non-zero transverse momentum transfer | k | → m -distributions are so dramatically different, a naturalquestion arises whether there is a continuous transition from the non-forward to the forwardscattering. The answer to this question involves an accurate treatment of two limits: | k | → R → ∞ . Let us keep R large but finite, and set | k | →
0; then the transition issmooth. Looking at the triple-Bessel representation of the master integral (50) with the upperlimit replaced by R , one sees that the result will begin to significantly decrease only when theposition of the first node of the last Bessel function J m − m ( | k | r ) falls outside of the integrationrange, that is for | k | ∼ < | m − m | /R , where m = m . If m = m , then at | k | ≪ /R the lastBessel function can be approximated by the unity. Therefore, the limit | k | → | k | → R → ∞ provided that | k | R ≪ . If instead R → ∞ at fixed | k | , then the transition of non-forward to forward results isdiscontinuous at | k | = 0.In [5] it is claimed with the specific example of the Compton cross-section that if thetransverse momentum transfer is small compared to κ (in our notation, finite | k | ≪ κ atinfinite R ), then the m -dependence has a narrow distribution peaked at m = m . Ouranalysis shows that this cannot be true, and this conclusion of [5] is likely to be the result ofan incorrect small- | k | approximation for I m,m ( κ, κ , k ). Looking back at the formalism used, we can conclude that our result that all m essentiallycontribute to the decay rate/cross section just reflects the unfortunate choice of the samecommon axis z for all the twisted states appearing in the process. It does not give a clue ofhow twisted the final particles are with respect to their own propagation axes defined by theiraverage values of the 3-momentum operator. Indeed, even a simple non-forward plane wavewhen expanded in the basis of twisted states contains all partial waves, see (9). Neverthelessit carries a zero orbital angular momentum with respect to its own direction of propagation.Therefore, it appears that a more physically reasonable quantity is the “orbital helicity” ,projection of orbital angular momentum on the axis of motion. The relevant question is then13ow this “orbital helicity”, not the OAM with respect to a fixed axis, is transferred from theinitial to the final twisted state. We postpone this question for future studies. Photons carrying non-zero orbital angular momentum (twisted photons) are well known inoptics. Thanks to the recent suggestion [5, 6] to use the Compton backscattering of opticaltwisted photons off a high-energy electron beam, twisted photons are now entering the high-energy physics. If the new degree of freedom they offer is effectively realized in experiment,twisted photons can bring novel opportunities to particle physics.In this paper we took a first look at the non-forward scattering of a twisted particle off asystem of plane waves: twisted + X ( p ) → twisted ′ + X ′ ( p ′ ) , p ′ = p , and investigated universal kinematical features pertinent to such processes.With the simple example of the decay of a scalar twisted particle, we evaluated the masterintegral appearing in such processes, discussed its singularities and explained how to writethe cross section/decay rate. These results can be now used, for example, to investigate theCompton backscattering of the twisted photons in the non-forward region, which was missingin the original suggestion [5, 6].Discussing the results obtained, we came to the conclusion that a more physically motivatedquantity to describe a twisted state would be the orbital angular momentum projection noton the reaction axis but on the direction of the outgoing twisted particle. Incorporation ofthis “orbital helicity” into the present formalism remains to be done. Acknowledgements
The author is grateful to V. Serbo for numerous useful discussions on this subject and toI. Ginzburg for comments. This work was supported by the Belgian Fund F.R.S.-FNRS viathe contract of Charg´e de recherche, and in part by grants RFBR No.08-02-00334-a and NSh-3810.2010.2.
A Regularization of |I | R behavior of the m -sum of the squares of the triple-Besselintegral: + m max X m = − m max (cid:20)Z R rdrJ m ( κr ) J m ( κ r ) J m − m ( κ r ) (cid:21) , (62)which appears in the decay rate (56). Evaluation of the integral itself with R → ∞ performedin the main text shows that it can be non-zero only if κ , κ , κ satisfy the triangle rules (35),i.e. a triangle with these sides can be constructed. Since m describes the initial state, we takeit small and not growing with R : m ≪ m max = κ R , while m can extend up to m max . Thefinal κ , κ can be much larger than κ . 14ince the expression (62) is regularized with large but finite R , the summation and inte-gration can be interchanged: Z rdr r ′ dr ′ J m ( κr ) J m ( κr ′ ) + m max X m = − m max J m ( κ r ) J m − m ( κ r ) J m ( κ r ′ ) J m − m ( κ r ′ ) . (63)Thanks to the properties of the Bessel functions, only m ’s up to min ( κ , r, κ , r ′ ) are ef-fectively contributing to this sum; for larger m the Bessel functions strongly decrease. But r, r ′ ≤ R , which means that the limits on the summation can in fact be safely extended to theinfinity. Then, the sum of the product of four Bessel functions is treated in the following way: + ∞ X m = −∞ J m ( κ r ) J m − m ( κ r ) J m ( κ r ′ ) J m − m ( κ r ′ ) (64)= + ∞ X m ,m ′ = −∞ J m ( κ r ) J m − m ( κ r ) J m ′ ( κ r ′ ) J m − m ′ ( κ r ′ ) · δ m ,m ′ = + ∞ X m ,m ′ = −∞ π Z π dαe i ( m − m ′ ) α J m ( κ r ) J m − m ( κ r ) J m ′ ( κ r ′ ) J m − m ′ ( κ r ′ )= 12 π Z π dα " + ∞ X m = −∞ e im α J m ( κ r ) J m − m ( κ r ) + ∞ X m ′ = −∞ e − im ′ α J m ′ ( κ r ′ ) J m − m ′ ( κ r ′ ) . The first sum in the square brackets is calculates as follows: + ∞ X m = −∞ e im α J m ( κ r ) J m − m ( κ r )= ( − i ) m (2 π ) Z dφ dφ e iκ r cos φ + iκ r cos φ + ∞ X m = −∞ e im α + im φ + i ( m − m ) φ = ( − i ) m π e imα Z dφ e imφ e iκ r cos φ + ik r cos( φ + α ) . (65)The combination of angles and momenta inside the exponential can be expressed as κ cos φ + κ cos( φ + α ) = | k + k | α cos( φ + δφ ) , (66)where | k + k | α ≡ q κ + κ + 2 κ κ cos α , tan δφ = κ sin ακ + κ cos α . (67)Geometrically, | k + k | α is the norm of the sum of two vectors of moduli κ and κ and therelative azimuthal angle α . Therefore, (65) is + ∞ X m = −∞ e im α J m ( κ r ) J m − m ( κ r ) = e im ( α − δφ ) J m ( | k + k | α r ) . (68)15his expression can be viewed as a 2D generalization of the well-known addition formula forthe Bessel functions + ∞ X m = −∞ J m ( x ) J m − m ( y ) = J m ( x + y ) . Now, the second sum differs only by α → − α (or, alternatively, complex conjugation) and r → r ′ . Therefore, the summation (64) is simplified to12 π Z π dα J m ( | k + k | α r ) J m ( | k + k | α r ′ ) . (69)Note that this integral involves only the Bessel functions of small order. We now plug thisexpression in (63) and get12 π Z π dα (cid:20)Z rdr J m ( κr ) J m ( | k + k | α r ) (cid:21) (cid:20)Z r ′ dr ′ J m ( κr ′ ) J m ( | k + k | α r ′ ) (cid:21) (70)As usual, we extend the integration range in one of the integrals to infinity, which gives adelta-function, and then we use it on the second integral calculated up to R . The originalexpression (62) then becomes12 π Z π dα δ ( κ − | k + k | α ) κ Rπκ = R π κ · , (71)where ∆ is, as always, the area of the triangle with sides κ , κ , κ . References [1] L. Allen et al., Phys. Rev.
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