A Restricted angular integrals

# Colliding particles carrying non-zero orbital angular momentum

## Abstract

Photons carrying non-zero orbital angular momentum (twisted photons) are well-known in optics. Recently, it was suggested to use Compton backscattering to boost optical twisted photons to high energies. Twisted electrons in the intermediate energy range have also been produced recently. Thus, collisions involving energetic twisted particles seem to be feasible and represent a new tool in high-energy physics. Here we discuss some generic features of scattering processes involving twisted particles in the initial and/or final state. In order to avoid additional complications arising from non-trivial polarization states, we focus here on scalar fields only. We show that processes involving twisted particles allow one to perform a Fourier analysis of the plane wave cross section with respect to the azimuthal angles of the initial particles. In addition, using twisted states one can probe the autocorrelation function of the amplitude, which is inaccessible in the plane wave collisions. Finally, we discuss prospects for experimental study of these effects.

## 1 Introduction

In perturbative quantum field theory we assume that interaction among the fields can be treated as a perturbation of the free field theory. This perturbation leads to scattering between asymptotically free multiparticle states, which are usually constructed from the plane wave one-particle states. This choice greatly simplifies the calculations and represents a very accurate approximation to the real experimental situation in virtually all circumstances. However, one can, in principle, choose any complete basis for the one-particle states other than the plane wave basis, provided that it is still made up of solutions of the free field equations. Such states can carry new quantum numbers absent in the plane wave choice and, if experimentally realized, they can offer new opportunities in high-energy physics.

Thanks to the progress in optics made in the last two decades, it is now possible to create laser beams carrying non-zero orbital angular momentum (OAM) [1], for a recent review see [2]. The lightfield in such beams is described via non-plane wave solutions of the Maxwell equations. Each photon in this lightfield, which we call a twisted photon, carries a non-zero OAM quantized in units of . Several sets of solutions have been investigated, such as Bessel beams or Gauss-Laguerre beams, but in all cases the spatial distribution of the lightfield is necessarily non-homogeneous in the sense that the equal phase fronts are not planes but helices. Such states form a complete basis which can be used to describe the initial and final asymptotically free states. Moreover, it is the basis of choice for experimental situations when the 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 optical photons off an ultra-relativistic electron beam to create a beam of high-energy photons with non-zero OAM [5, 6]. The technology of Compton backscattering is well established [7], and the high-energy electron beams and the OAM optical laser beams are already available. In addition, in the last months several groups have reported successful creation of twisted electrons, first using phase plates [8] and then with computer-generated holograms [9]. Twisted electrons carried the energy as high as 300 keV and the orbital quantum number up to . With all these achievements, creating high-energy particles in a controlled orbital angular momentum state and colliding them seems now feasible. It is therefore very timely to ask what new insights into the properties of particles and their interactions one can gain with this new degree of freedom.

In this paper we begin this exploration by studying several generic scattering processes involving twisted particles in the initial and/or final states. Namely, we consider three specific cases:

• single-twisted scattering: collision of a twisted state with a plane wave,

• double-twisted scattering: collision of two twisted states,

• two-particle decay of an unstable twisted particle.

In the first two cases we assume that the final system is described by plane waves, while in the last case we consider three choices for the final two-particle state: when both particles are plane waves, when one is twisted, and when both are twisted. The single- and double-twisted scattering will give some hints at new physical opportunities that can be expected in collisions of high-energy particles carrying OAM, while the calculation of a twisted particle decay clarifies various technical details involved in passage from plane waves to twisted states.

The OAM and spin are two forms of angular momentum, and the problem of gauge-invariant separation of these two objects has a long history. It is still being debated both in the optics community, see for example [10], and in the HEP community, especially in the context of the notorious proton spin puzzle, [11]. For the problems we consider here it is sufficient to note that all experimental situations which seem to be realizable are well described by the paraxial approximation. It is known that in the paraxial approximation spin can be well separated from the -component of OAM of light, [12]. The same applies to fermions as well. Therefore, incorporation of both spin and OAM degrees of freedom, leading to non-trivial polarization fields, does not seem to pose any problem in the paraxial approximation.

However in the present paper we would like to focus specifically on the OAM component and to understand what new physical opportunities are offered by the non-trivial spatial dependence rather than unusual polarization states. Therefore we limit ourselves to scattering of scalar particles only. In this simple case all the non-zero angular momentum is definitely due to the orbital part. Additional features arising from the polarization parameter fields will be considered separately.

The paper is organized as follows. In Section 2 we introduce scalar twisted states and describe some of their properties. In Sections 3 and 4 we derive expressions for the cross section in the single-twisted and double-twisted cases, respectively. Section 5 gives a thorough discussion of kinematical features arising in the case when the final state contains twisted particles. In Section 6 we discuss the results obtained and draw our conclusions. In two Appendices we derive some technical results used in the paper.

## 2 Describing twisted states

### 2.1 Spatial distribution

As mentioned in the introduction, we focus in this paper on twisted scalar particles with mass . In their description we follow essentially [5, 6].

We represent a state with a non-zero OAM with a Bessel beam-type twisted state. This is a solution of the wave equation in the cylindric coordinates with a definite energy and a longitudinal momentum along a fixed axis , a definite modulus of the transverse momentum (all transverse momenta will be written in bold) and a definite -projection of OAM. If the plane wave state is

 |PW(k)⟩=e−iωt+ikzz⋅eikr, (1)

then a twisted scalar state is defined as the following superposition of plane waves:

 |κ,m⟩=e−iωt+ikzz∫d2k(2π)2aκm(k)eikr,aκm(k)=(−i)meimϕk√2πδ(|k|−κ)√κ. (2)

In the coordinate space,

 |κ,m⟩=e−iωt+ikzz⋅ψκm(r),ψκm(r)=eimϕr√2π√κJm(κr). (3)

Here, following [5] we call the conical momentum spread, is the -projection of OAM, and the dispersion relation is . We note in passing that the average values of the four-momentum carried by a twisted state is

 ⟨kμ⟩=(ω,0,kz), (4)

so that , which is larger than the true mass of the particle squared.

The transverse spatial distribution is normalized according to

 ∫d2rψ∗κ′m′(r)ψκm(r)=δm,m′√κκ′∫rdrJm(κr)Jm(κ′r)=δm,m′δ(κ−κ′). (5)

The plane wave can be recovered from the twisted states as follows:

 |PW(k=0)⟩ = limκ→0√2πκ|κ,0⟩, (6) |PW(k)⟩ = √2πκ+∞∑m=−∞ime−imϕk|κ,m⟩,κ=|k⊥|. (7)

If needed, these two cases can be written as a single expression:

 |PW(k)⟩=limκ→|k|√2πκ+∞∑m=−∞ime−imϕk|κ,m⟩. (8)

From these expressions one sees that the twisted states with different and represent nothing but another basis for the transverse wave functions.

### 2.2 Density of states

When calculating cross sections and decay rates, we need to integrate the transition probability over the phase space of the final particles. When calculating the density of states, we consider a large but finite volume and count how many mutually orthogonal states with prescribed boundary conditions can be squeezed inside. In the present case due to the cylindrical symmetry of the problem, we choose a cylinder of a large radius and a length . In the case of plane waves we have

 dnPW=πR2Lzdkzd2k(2π)3. (9)

The full number of states with transverse momenta up to and longitudinal momenta is .

To count the number of twisted states in the same volume, we specify the boundary condition, e.g. , which makes discrete such that is the -th root of the Bessel function . Then we note that the position of the first root of the Bessel function is always at , and as grows . For a given , the maximal for which the wave can still be contained inside the cylindrical volume is , which has a very natural quasiclassical interpretation.

If is small and not growing with , then one can use the well-known asymptotic form of the Bessel functions to count the number of states:

 dntw=RdκLzdkzΔm2π2. (10)

Here, is written instead of just 1 to signal the presence of a discrete running parameter .

If is not restricted to small values, this asymptotic form of cannot be used since it requires . Instead, the so-called approximation by tangents can be used, which gives the following density of states:

 dntw=√m2max−m2dκκΔmπLzdkz2π. (11)

In the limit this expression reproduced (10). Alternatively, one can calculate the radial part of the density of states via the adiabatic invariant as suggested in [6]. The number of radial excitations for a fixed is

 nr=∫Rm/κkr(r)drπ,kr(r)=√κ2−m2r2. (12)

The density of states is then given by

 dnr=dnrdκdκ=√m2max−m2dκκπ. (13)

One important remark is in order. Effectively, switching from the plane wave to twisted state basis for the final particles implies replacement

 d2k→4√1−m2m2maxκdκΔmmmax. (14)

Note that the contribution of each “partial wave” with a fixed vanishes in the infinite volume limit as . However, the number of partial waves grows , and in order to get a non-vanishing result for a physical observable, one must integrate over the full available interval up to . This remains true even if the transverse momenta stay small, and it is related to the fact that the plane wave contains contributions from all impact parameters with respect to any axis non-collinear to its propagation direction.

Another expression one needs for the probability calculations is the normalization constants for the one-particle states. A usual plane wave one-particle state is normalized to ; to renormalize it to one particle per the entire volume, the plane wave should be multiplied by , with

 N2PW=12EV,V=πR2Lz. (15)

For a twisted state the corresponding normalization factor is

 N2tw=12Eπκ√m2max−m2Lz, (16)

which in the small- case simplifies to

 N2tw≈12EπRLz, (17)

also derived in [6]. Note however that even in the general case the product of the normalization constant squared and the density of states for each final twisted particle is simplified as

 N2twdntw=dκdkzΔm2E⋅2π. (18)

### 2.3 Flux and the cross section

The definitions of the flux factor and the cross section have to be reevaluated when a collision of non-plane wave states is considered, which involves subtle issues described in [6]. By definition, the cross section is the transition probability per unit time divided by flux. In the plane wave case, when the four-momenta of colliding particles are fixed, both the flux and probability are constant across any chosen plane, so the proportionality between them holds locally. For a head-on collision one gets , which together with the energies of the incoming particles and one volume factor combine to the familiar Lorentz invariant expression

 IPW=√(pk)2−p2k2. (19)

This formula can of course be used in any frame, including cases when the collision is not head-on.

In the single-twisted case, when a twisted state collides with a plane wave, both the flux and the transition probability are not constant but change across the transverse plane. As noted in [6], one therefore needs to redefine the notion of the cross section to adapt to this situation. One introduces the averaged cross section defined as the transition probability integrated over all divided by the flux again integrated over all . While the definition of the former quantity is clear (this is what we calculate in the following two Sections), the proper definition of the integrated flux is more intricate and apparently not unique. A definition of the flux should however be correlated with a definition of the (generalized) luminosity, so that the observable event rate remains uniquely defined.

In [6] the forward electron-photon collision was considered (the 3-momentum of the plane wave electron was directed exactly along the axis ), and the following procedure was suggested: the total flux is just sum of the -components of the two fluxes (note that we actually assume the absolute values of the fluxes):

 ⟨j⟩=jez+⟨jγz⟩=v+cosαkV, (20)

where . We think that a more appropriate definition of flux should take into account not only the -components of the individual fluxes but also the relative lateral motion of the two waves. Indeed, the sole purpose of calculating the flux factor is, classically speaking, to derive the volume swept by one particle in the “gas” of opposing particles and find how many “attempts” at collision are made per unit time. Therefore, we propose the following general definition of the integrated flux factor for the single-twisted case:

 Itw=∫dϕk2πIPW(k,p). (21)

Note that it is well defined for any transverse momentum of the opposing plane wave. In the specific case considered above this definition gives , which differs from (20). This definition can be also generalized to the double-twisted case:

 I2tw=∫dϕk2πdϕp2πIPW(k,p). (22)

## 3 Single-twisted cross section

We start with a usual collision in which both incoming particles are described by plane waves with definite four-momenta and . The final system is also treated as a collection of plane waves with the total momentum . The invariant amplitude of this process is denoted by , where the transverse momenta of the initial particles are indicated explicitly.

The cross section of this process is calculated according to the standard rules. When squaring the scattering matrix element

 SPW=i(2π)4δ(4)(k+p−pX)⋅M(k,p), (23)

we re-interpret the square of the delta-function as

 [δ(4)(k+p−pX)]2=δ(4)(k+p−pX)⋅πR2LzT(2π)4. (24)

Using the plane-wave normalization factors (15) for all the initial and final particles, we get the cross section

 dσPW(k,p)=(2π)4δ(Ei−Ef)δ(pzi−pzf)δ(2)(k+p−pX)4IPW|M(k,p)|2⋅dΓX. (25)

For future convenience the delta-function is explicitly broken into the longitudinal and transverse parts. As usual, we can extract from the final phase space integration measure the integral over the total transverse momentum , , and write the cross section as

 dσPW(k,p)=(2π)4δ(Ei−Ef)δ(pzi−pzf)4IPW|M(k,p)|2⋅dΓ′X. (26)

Now we recalculate the cross section for the case when the first particle is in the twisted state . We apply prescription (2) to the -matrix element:

 Stw=∫d2k(2π)2aκm(k)SPW(k,p), (27)

as originally suggested in [5]. The plane-wave -matrix (23) contains a delta-function of transverse momenta, which makes it possible to simplify the square of the twisted -matrix element as

 |Stw|2 = ∫d2k(2π)2d2k′(2π)2aκm(k)a∗κm(k′)SPW(k,p)S∗PW(k′,p) (28) ∝ ∫d2k(2π)2d2k′(2π)2aκm(k)a∗κm(k′)δ(2)(k+p−pX)δ(2)(k′+p−pX)M(k,p)M∗(k′,p) = ∫d2k(2π)4aκm(k)a∗κm(k)δ(2)(k+p−pX)|M(k,p)|2.

The square of contains a radial delta-function squared, which is reinterpreted as

 [δ(κ−|k|)]2=δ(κ−|k|)⋅δ(0)⟶δ(κ−|k|)Rπ. (29)

This prescription comes from the observation that at large but finite and at the radial delta-function is regularized as

 δ(0)=∫∞0rdr[Jm(κr)]2→∫R0rdr[Jm(κr)]2≈Rπ. (30)

see [5, 6]. Therefore, with all the normalization factors (15) and (17) the single-twisted cross section takes form

 dσtw=∫d2k2πIPW(k,p)Itwδ(κ−|k|)κ⋅dσPW(k,p)=∫dϕk2πIPW(k,p)ItwdσPW(k,p), (31)

see Section 2.3 for the definition of the fluxes. Note that in the paraxial approximation one can replace the ratio of the fluxes by the unity.

A couple of remarks concerning this result are in order. In the usual case of a plane wave collision, the total initial and, therefore, final momenta are fixed. This is highlighted by the presence of the transverse delta-function in (25) and by implicit correlations among transverse momenta of the final particles inside in (26). In the process involving a twisted particle, the total final momentum is not fixed, making the final particles less correlated. Although all the plane waves which constitute the initial twisted state are summed up coherently at the amplitude level, the final states with different momenta do not interfere, and this breaks the coherence. As a result, the twisted particle cross section is expressed via an incoherent superposition of the cross sections induced by each initial plane wave. This is precisely what (31) displays.

The previous paragraph contains one additional subtlety. When saying that the final states with different momenta do not interfere, we implicitly assume that the final state is detected by a usual detector, which measures the linear momentum but not the OAM. On the contrary, if the final particles were detected by a hypothetical “coherent detector” sensitive to a coherent superposition of final states with distinct momenta, or if the production of particles in the reaction we consider is followed by another process which is OAM-selective, then the coherence would be restored. Thus, the coherence is not actually destroyed in the scattering process, but remains hidden.

The second remark concerns the angular region contributing to the integral (31). In the fully differential case, that is when we fix the momenta of all the final particles, the transverse delta-function inside assures that there is only one value of that contributes to the integral. At this level of consideration, representing the cross section as an angular integral might look somewhat misleading. However if this delta-function is killed by the integration over all allowed , or equivalently by integrating over one of the final particles, then the integral receives contributions from all angles . Note that the angular integral representation is also justified even in the fully differential case if the detector resolution of the final particles’ momenta is worse than .

Let us also show a slightly different derivation of (31). We first explicitly perform the integration in (27), which is effectively killed by the transverse part of the delta-function, keeping the scattering amplitude essentially unchanged:

 ∫d2k(2π)2aκm(k)δ(2)(k+p−pX)⋅M(k,p)=(−i)m(2π)3/2eimϕqδ(κ−q)√κ⋅M(q,p), (32)

where . Note that after the integration the transverse momentum cannot appear in the amplitude any more. However, since the other particles are plane waves with well defined momenta, the vector with modulus and azimuthal angle is also well defined and can be used instead of everywhere. Squaring (32), we again encounter the square of the radial delta-function which is reinterpreted as in (29). The cross section then becomes

 dσtw=(2π)4δ(Ei−Ef)δ(pzi−pzf)|M(q,p)|24Itw⋅dΓX⋅δ(κ−q)2πκ. (33)

The integral over the overall transverse momentum present in kills the radial delta-function:

 ∫d2pXδ(κ−q)2πκ=∫d2qδ(κ−q)2πκ=∫dϕq2π, (34)

and one recovers the result (31).

One can make several observations concerning (31):

• The cross section is -independent.

• The initial twisted particles effectively perform the angular averaging of the plane wave cross-section.

• There is no smallness associated with non-zero .

• There is no small factor associated with small .

Let us now consider the case when the twisted particle is not in an eigenstate of the OAM operator, but in a superposition of states with equal but different . For example, consider the twisted state of the form , with . Repeating the same calculation we encounter an interference term in the cross section:

 dσΔmtw=∫dϕk2πIPW(k,p)Itwcos(Δmϕk+α)dσPW(k,p), (35)

where and is the relative phase between the two complex coefficients and . The cross section for such an initial state takes the following form:

 dσ=dσtw+2|aa′|dσΔmtw. (36)

Results (31) and (35) mean that if the initial particle can be prepared in a twisted state with an adjustable superposition of different , a Fourier analysis of the cross section with respect to the initial azimuthal angle can be performed. In principle, the same analysis can be done with plane waves, but it would require making several experiments with different angles of the initial particle and then extracting the Fourier components via the partial wave analysis. From the experimental view, it is likely that systematics of the two schemes can be different, which makes them complementary to each other.

## 4 Double-twisted cross section

Let us now consider collision of two initial twisted particles. In this work we do not aim at a systematic study of this case, but rather outline some new features can be expected in such circumstances. Therefore we consider the simplest set-up, in which the two colliding particles are described by twisted states and defined with respect to the same quantization axis :

 |κ,m⟩=∫d2k(2π)2aκm(k)|PW1(k)⟩,|η,n⟩=∫d2p(2π)2aηn(p)|PW2(p)⟩, (37)

with the same functional form of the projectors as before. Here, the subscripts and refer to the first and second colliding particles. The “double-twisted” version of (27) is

 S2tw=∫d2k(2π)2d2p(2π)2aκm(k)aηn(p)SPW(k,p), (38)

and its square is proportional to

 ∫d2kd2pd2k′d2p′(2π)8aκm(k)aηn(p)a∗κm(k′)a∗ηn(p′) × δ(2)(k+p−pX)δ(2)(k′+p′−pX)M(k,p)M∗(k′,p′). (39)

Let us consider the kinematical restrictions imposed by the delta-functions entering this expression and compare them with the result found in the previous Section. Here, too, the moduli of the momenta are fixed and pairwise equal: and . Besides, the two pairs of momenta sum up to a well-defined . For each pair there are two possibilities satisfying these conditions, shown in Fig. 1, which are mirror reflections of each other with respect to the direction of . Therefore, the integral (39) receives contributions from two kinematical configurations:

 direct: k′=k, p′=p, reflected: k′=k∗≡−k+2(knX)nX, p′=p∗≡−p+2(pnX)nX, (40)

with . As we will see below, in contrast to the “single-twisted” case, here the existence of two configuration for any given leads to a residual coherence between different plane wave components in the twisted state.

Our goal now is to see whether the double-twisted cross section can be expressed as an angle-averaged plane wave cross section similarly to (31). To this end let us first study a generic expression

 J=∫dϕkdϕpδ(2)(k+p−pX)⋅f(k,p), (41)

where and are fixed and is a continuous function of the incoming particles’ momenta. Due to the delta-function, the integral receives contributions only from two points, shown in Fig. 1. At these points the azimuthal angles and take specific values:

 ϕk=ϕX±δk,ϕp=ϕX∓δp, (42)

where is the azimuthal angle of and

 δk=arccos(p2X+κ2−η22|pX|κ),δp=arccos(p2X−κ2+η22|pX|η) (43)

are functions of the absolute values of the momenta. Let us denote the values of at these two points as and , respectively. In Appendix A we derive the following result

 J=f++f−2Δ=∫dϕkdϕpδ(2)(k+p−pX)⋅f++f−2, (44)

where is the area of the triangle of sides , , . It allows us to rewrite the square of the integral as

 |J|2 = ∫dϕkdϕpδ(2)(k+p−pX)f(k,p)f∗++f∗−2Δ (45) = 14Δ∫dϕkdϕpδ(2)(k+p−pX)[|f+|2+|f−|2+2Re(f+f∗−)] = 12Δ∫dϕkdϕpδ(2)(k+p−pX)[|f(k,p)|2+Ref(k,p)f∗(k∗,p∗)],

where and are given by (40).

These results can be directly applied to the integral (39) if we note that it has the form of with the function

 f(k,p)=1(2π)3√κηeimϕk+inϕpM(k,p).

The integral (39) then takes form

 1(2π)6sin(δk+δp)∫dϕkdϕpδ(2)(k+p−pX) ×{|M(k,p)|2+Re[e2im(ϕk−ϕX)+2in(ϕp−ϕX)M(k,p)M∗(k∗,p∗)]}. (46)

Bringing together all the normalization coefficients, we finally arrive at the following representation for the double-twisted cross section:

 dσ2tw=18πsin(δk+δp)∫dϕkdϕpIPW(k,p)I2tw[dσPW(k,p)+dσ′(k,p)], (47)

where the flux is given by (22) and

 Missing or unrecognized delimiter for \right (48)

We see that the “direct” contribution yields the standard cross section, while the “reflected” contribution gives rise to the novel quantity . This quantity describes the auto-correlation of the amplitude and is absent in the plane wave case.

Let us also see how (47) simplifies in the case when . The novel quantity reduces to the usual cross section, , and we obtain

 dσ2tw=cos2(mδk−nδp)4πsin(δk+δp)∫dϕkdϕpIPW(k,p)I2twdσPW(k,p). (49)

It is only in this case that the double-twisted cross section is expressed via the angular integral of the plane wave cross section.

As in the single-twisted case, we note that if all the final momenta were fixed and if the detector had an infinitely good momentum resolution, there would be just two points contributing to the angular integral (47). If at least one of these conditions is broken, the integrand extends over the full integration range.

Similarly to the single-twisted case, the double-twisted cross section stays finite even when the conical momentum spreads and are very small. However in contrast to the single-twisted case, the cross section now depends also on the orbital angular momenta and . This dependence comes solely from the correlation term in (47). If the cross section is averaged with a sufficiently smooth function of the initial or final momenta, then in the limit of large and the correlation term gets suppressed by rapid oscillations, and the double-twisted cross section stays approximately and independent.

Certainly, one can also consider double-twisted cross sections with twisted states in superposition of different and . We leave a detailed study of this situation for a future work.

Finally we reiterate the point that for this particular version of double-twisted scattering we chose the simplest possible set-up, in which both twisted states are defined with respect to the same axis. One can also study what would happen if two different axes, either parallel or not, were used. Answering this question requires a more elaborate formalism.

## 5 Two-particle decay of a twisted scalar

### 5.1 Preliminary remarks

Previous two Sections were devoted to the cases when the twisted particles appeared only in the initial state, while the final state was assumed to be describable with the plane waves. Let us now consider the case when at least one of the final particles is also twisted. This is exactly the situation that arises in the original suggestion of [5, 6] to use Compton backscattering of the twisted optical photons to produce final high-energy OAM photons. The key question here is to what extent the twisted parameters of the initial state determine the parameters of the final twisted state . Here again both twisted states are defined with respect to the same -axis. In the case of strictly backward scattering the conservation of the orbital momentum ensures that , , [5]. In this Section we study how this result changes for the non-forward scattering. Note that we use the term “forward” for any process with zero transverse momentum transfer, i.e. both for strictly forward and strictly backward kinematics.

In order to focus on the final state kinematics and to avoid possible complications coming from a non-trivial matrix element, we would like to answer the above question in the context of the simplest possible problem: the decay of a twisted scalar particle with mass into a pair of massless distinguishable particles due to the cubic interaction . The momenta of the initial and final particles are , and , respectively. We will calculate the decay width in the center of mass frame defined by . This is not the true rest frame because due to the transverse motion a twisted particle is never at rest. To make the presentation more pedagogical, we will first calculate the decay rate when both particles in the final state are plane waves, then for the plane wave plus twisted final state, and finally for the case when both final particles are twisted. Although the total decay width must be the same in all these cases, the differential decay rates will be rather different.

### 5.2 Two plane waves

Again, let us first recall the standard calculation for the case when all the particles including the initial one are plane waves. The -matrix is given as usual by . With the plane wave normalization coefficients for all the particles, the differential decay rate for a particle at rest is

 dΓ = (2π)4g2δ(4)(p−k1−k2)VTT⋅(N2PW)3⋅dnPW(k1)dnPW(k2) (50) = g2(2π)2δ(M−ω1−ω2)8Mω1ω2d3k1,

so that the total width is

 Γ=g216πM. (51)

Now we repeat this calculation for the initial twisted state , while keeping the plane wave basis the final particles. The -matrix is

 S = i(2π)4gδ(E−ω1−ω2)δ(k1z+k2z)∫d2k(2π)2aκm(k)δ(2)(k−k1−k2) (52) = i(2π)4gδ(E−ω1−ω2)δ(k1z+k2z)(−i)m(2π)3/2eimϕ12δ(κ−k12)√κ, (53)

where and is the angle of the 2D vector w.r.t. some axis . Similaly to the single scattering cross section, this phase factor is inessential and disappears in the decay rate.

The square of is treated as in (29), and with the appropriate normalization factors taken into account, the decay rate has the form

 dΓ = (2π)3g28Eω1ω2⋅Tδ(E−ω1−ω2)δ(k1z+k2z)TLzδ(κ−k12)κRπ⋅1V2πRLz⋅Vd3k1(2π)3Vd3k2(2π)3 (54) = g2(2π)3δ(κ−k12)κδ(E−ω1−ω2)8Eω1ω2dkzd2k1d2k2.

For the transverse integral we write

 ∫dϕ2δ(κ−k12)κ=2∫dϕ2δ[κ2−k21−k22−2|k1||k2|cos(ϕ1−ϕ2)]=1Δ, (55)

where is the area of the triangle with sides , and , see Appendix A. As usual, the energy delta function can be killed by the integration

 ∫dkzδ(E−ω1−ω2)ω1ω2=2Ek∗z, (56)

where

 k∗z=12E√E4+k41+k42−2(E2k21+E2k22+k21k22). (57)

The decay rate becomes

 dΓ=g24π214E2k∗z|k1|d|k1||k2|d|k2|Δ. (58)

The integration region over and is defined by the requirement that, in addition to (81), the longitudinal momentum is well-defined, which cuts the rectangular shape shown in Fig. 2 in Appendix A. It is conveniently described with variables

 x=|k1|−|k2|κ,z=|k1|+|k2|κ,x∈[−1,1],z∈[1,zmax],zmax≡Eκ>1. (59)

In these variables

 Δ=κ24√(z2−1)(1−x2),k∗z=E2√(1−x2z2max)(1−z2z2max), (60)

and the decay rate takes form

 dΓ=g216π2E⋅(z2−x2)dzdx[(z2max−z2)(z2max−x2)(z2−1)(1−x2)]1/2. (61)

This integral can be taken exactly, and it gives the decay width of the twisted scalar particle

 Γ=g216πE, (62)

which is a very natural result. In the limit , we recover the plane wave decay width (51).

Of course, one could also arrive at (62) just by using our previous results concerning the single-twisted cross section modified to the case of one initial particle. However the detailed derivation given here is needed to understand what changes when the final state includes twisted particles.

### 5.3 Twisted state plus plane wave

Let us now describe the final state as twisted state plus a plane wave:

 |κ,m⟩→|κ1,m1⟩+|PW(k2)⟩. (63)

Since the full decay width cannot depend on the basis we choose for the final particles, we must recover the same result (62) in this basis. In addition to that, we also want to know how the final twisted state parameters and are related to the initial state parameters and .

The -matrix is now

 S=∫d2k(2π)2d2k1(2π)2a∗κ1m1(k1)aκm(k)SPW. (64)

It can be written as

 S=i(2π)4gδ(E−ω1−ω2)δ(kz1+kz2)⋅Im,m1(κ,κ1,k2). (65)

where the master integral is defined as

 Im,m1(κ,κ1,k2)=∫d2k(2π)2d2k1(2π)2a∗κ1m1(k1)aκm(k)δ(2)(k−k1−k2). (66)

Let us first calculate this integral in the strictly forward case :

 Im,m1(κ,κ1,0) = im1−m(2π)3∫d2kd2k1eimϕ−im1ϕ1δ(|k|−κ)√κδ(|k1|−κ1)√κ1δ(2)(k−k1) (67) = im1−m(2π)3√κκ1∫dϕdϕ1eimϕ−im1ϕ1⋅2δ(k2−k21)δ(ϕ−ϕ1) = 1(2π)2δ(κ