Physics 125 Course Notes Scattering F. Porter. 3 The Differential Cross Section 5

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1 Physics 125 Course Notes Scattering F. Porter Contents 1 Introduction 1 2 The S Matrix 2 3 The Differential Cross Section 5 4 Partial Wave Expansion 11 5 Optical Theorem 13 6 Interim Remarks 15 7 Resonances 16 8 The Phase Shift Relation of Phase Shift to Logarithmic Derivative of the Radial Wave Function Low Energy Limit Example The Born Expansion, Born Approximation Born Approximation for Phase Shifts Born Approximation and Example of the Soft Sphere Potential Angular Distributions Exercises 44 1 Introduction Much of what we learn in physical investigations is performed by scattering objects off of one another. The quantum mechanical theory for scattering 1

2 processes is thus an important subject. We have already gotten a glimpse into how to deal with scattering in our discussion of time-dependent perturbation theory, and in our investigation of angular momentum. We carry this development further in this note and derive more of the important concepts and tools for dealing with scattering. Many of the results we obtain will in fact be applicable to relativistic processes. We start by developing the basic physical picture and mathematical formalism for a simple situation with the essential features. That is, we begin by considering the problem of the scattering of a spinless particle of mass m>0 from a fixed, static, spherically symmetric center of force, centered at the origin. Assume further that the force is of finite range far enough away, any wave packet acts as if there is no force. 1 In another view, considering the spreading of the wave function, as t ±, the wave packet is so spread out that little of it is affected by the force field. 2 The S Matrix Let φ(p,t) be the momentum space wave function of the scattering wave packet. Assume it is normalized: φ φ = 1. The time dependent Schrödinger equation is: i t φ(p,t)=hφ(p,t), (1) where H includes the influence of the force. Let E(p) = m 2 + p 2,orp 2 /2m in the non-relativistic case, where p = p. We restrict consideration to the subspace in which φ(p,t) is orthogonal to any bound state wave functions, if any exist. That is, we restrict ourselves to scattering, in which the wave function can be considered to be far away from the center of force at large times. Corresponding to φ(p,t) (an exact solution to the Schrödinger equation for a possible scattering event), define the notion of an initial wave function, φ i (p), by taking the following limit at asymptotically early times: φ i (p) = lim t φ(p,t)eite(p). (2) 1 The Coulomb force is thus excluded by this criterion. As seen in the exercises of the Approximate Methods note, the cross section for the Coulomb interaction is infinite. However, we have also seen how the Coulomb force may be treated as a limiting case. 2

3 That is, φ i (p)e ite(p) is a solution to the Schrödinger equation in the absence of the scattering force. Likewise, we define a final wave function according to: φ f (p) = lim φ(p,t)e ite(p). (3) t We should perhaps remark a bit further on the choice of phase factor in these asymptotic limits. Put most simply, these choices just correspond to the interaction picture. However, we may gain a bit of additional insight as follows: Recall that for a free particle the momentum space wave function behaves under a time translation by t 0 as: M(t 0 )ψ(p,t)=e it 0 p2 2m ψ(p,t). (4) Thus, our initial asymptotic limit, Eqn. 2, is like taking φ i (p) =M(t 0 )φ(p,t=0), (5) if the motion were strictly force-free for all times. A similar correspondence holds for the asymptotic final state wave function. Thus, our initial and final phases are matched to the t = 0 phase for the case of no force. Of course, even with a force, φ i (p) determines φ(p,t), and hence φ f (p). We note that it is convenient to deal with momentum space wave functions in this discussion of scattering, because the coordinate space wave functions in general become infinitely spread out as t. We are couching our discussion in terms of physically realizable wave packets here. Typically, textbooks resort to the use of ( unphysical ) planewave states, which simplifies the treatment somewhat, at the expense of requiring further justification. It is reassuring to find that we can obtain the same results by considering only physical states. Thus, our development follows closely the nice pedagogical treatment in Ref. [1]. We are interested in the problem of determining φ f, given φ i. In principle, this requires solving the Schrödinger equation with φ i specifying the initial boundary condition. Usually, however, we do not require the explicit detailed solution we are interested in only certain aspects of the relation between φ i and φ f, such as the force-induced phase difference. The transformation giving φ f, in terms of φ i is a linear transformation, which also preserves the normalization of the wave packet. Hence, it is a unitary transformation: φ f = S φ i, (6) 3

4 where SS = I. (7) The unitary operator S is called the S matrix. In general, we may represent the S matrix as an integral transform: φ f (p) = d 3 (q)s(p, q)φ i (q), (8) ( ) and the unitarity condition is represented as: d 3 (q)s(p, q)s (q, p ) = d 3 (q)s(p, q)s (p, q) ( ) ( ) = d 3 (q)s (q, p )S(q, p ) ( ) = δ (3) (p p ). (9) The scattering matrix S(p, q) contains precisely all the information which is available in an asymptotic experiment, as described by φ i, φ f, where the details of what happens near the origin are not observed. Consider briefly the classical analog, for the classical scattering of a particle of mass m on a static, spherically symmetric force centered at the origin. The motion is described by canonical variables x(t) and p(t), which may be obtained by solving the Hamilton equations of motion: ẋ i = pi H (10) ṗ i = xi H. (11) The quantities x(t) and p(t) are the classical analog of φ(p,t), since they describe the detailed behavior of the particle. As long as the force falls off sufficiently rapidly at large distances, the motion at large distances and times (presuming no bound states are involved) is uniform: { xi + p x(t) = i t/m, as t, x f + p f t/m, as t. (12) Thus, we can define the initial and final asymptotic values as: [ p i = lim p(t), x i = lim x(t) t ] t t m p i p f = lim t p(t), x f = lim t 4 [ x(t) t m p f (13) ]. (14)

5 In particular, x i, x f are (or rather, is) the positions that would occur at t =0if the motion were strictly uniform for all times, with the asymptotic momenta. Thus, (x i, p i ) is the analog of φ i (p), and (x f, p f ) is the analog of φ f (p). Once again, specification of (x i, p i ) uniquely determines (x f, p f ), and the transformation giving this relationship is the analog of the S matrix. Note that, in the absence of a force field, The analogous statement in quantum mechanics is x i = x f, p i = p f. (15) φ i (p) =φ f (p), (16) since φ(p,t)=φ i (p)e ite(p) is the solution of the Schrödinger equation for free particle motion: φ f (p) = lim φ(p,t)e ite(p) t (17) = lim φ i (p)e ite(p) e ite(p), (free particle) t (18) = φ i (p). (19) 3 The Differential Cross Section In the Approximate Methods note, we defined a differential cross section in the discussion of time-dependent perturbation theory. We now return to this, and develop the notion of a cross section with more care. We start with the S matrix formalism just introduced. Given φ i (p) and S, we obtain φ f (p) according to: φ f (p) = d 3 (q)s(p, q)φ i (q). (20) ( ) Now define the scattered wave by: or φ s (p) =φ f (p) φ i (p), (21) φ s = φ f φ i (22) = (S I) φ i. (23) 5

6 In the absence of a force, φ f (p) =φ i (p), hence S = I and φ s (p) = 0. Thus, the name scattered wave is appropriate for φ s (p). Since the scattering force is assumed to be stationary, energy is conserved in the scattering event (e.g., for an infinitely heavy source of force, any finite momentum transfer to the source yields zero kinetic energy transfer). Hence, the asymptotic initial and final magnitudes of momenta will be equal. We may thus write: S(p, q) δ (3) (p q) =δ(p q)t (p, q), (24) where p = p, q = q, and we have extracted the factor δ(p q) from the integral transform for S I. Thus, we have: φ s (p) = d 3 (q)δ(p q)t (p, q)φ i (q). (25) ( ) Note that T (p, q) is physical only for p = q ( on the energy shell ), though it is sometimes useful to extend its domain of definition analytically offshell. Let {φ 0 (p; α) α some index set} be a set of potential wave packets, such as might be available in a given experimental arrangement (e.g., a beam delivered by an accelerator). We assume that the particles in our beam all have approximately the same momentum p i, i.e., φ 0 (p; α) 0, unless p p i. (26) Assume further that φ 0 φ 0 = 1 for all α, and that p i is in the 3-direction: p i = p i ê 3. The index α is a shape parameter, since we use it to specify the shape of the incident wave packet of a particle in the beam. note that different particles may have different wave packets. We consider the packet φ i (p; α; x) =φ 0 (p; α)e ix p, (27) obtained by translating the packet φ 0 (p; α) by amount x in space (assumed to be transverse to the three direction, x 3 = 0). Fixing x and α, and taking the packet as our initial wave, we obtain the scattered wave as: φ s (p; α; x) = d 3 (q)δ(p q)t (p, q)φ 0 (q; α)e ix q. (28) ( ) We are interested in the probability, P (u; α; x)dω u, that the particle is in the scattered wave with momentum in dω u around unit vector u. At least if 6

7 u is not too close to the forward direction defined by p i, this probability is just the probability that the particle s momentum after scattering is in dω u : P (u; α; x) = = 0 0 p 2 dp φ s (pu; α; x) 2 (29) p 2 dp d 3 (q) d 3 (q )δ(p q)δ(p q ) (30) ( ) ( ) T (pu, q)t (pu, q )φ 0 (q; α)φ 0(q ; α)e ix (q q ). (31) We may integrate over p, setting p = q : P (u; α; x) = d 3 (q) d 3 (q )q 2 δ(q q ) (32) ( ) ( ) T (qu, q)t (q u, q )φ 0 (q; α)φ 0(q ; α)e ix (q q ), (33) where we have made use of the fact that q = q for non-vanishing contributions. The point of the parameters α, x is the following: We wish to make sure that our formalism is valid for real experimental situations, i.e., that we have not left out some essential feature in our theory for describing experimental situations. Thus, we assume that the real exdperimental beam is a statistical ensemble of wave packets φ 0 (p; α; x). Thus, the parameter x describes the displacement of a wave packet in this ensemble from some ideal (i.e., x = 0) position, and α describes the shape of the wave packet in the ensemble of shapes. The ensemble is described by specifying the probability distributions for x and α, which depends on the particular experimental arrangement. Note that we are interested here in scattering of a one particle beam on a one particle target, i.e., correlations among particles in the beam or in the target are assumed to be zero. Let P (u)dω u be the probability that a beam particle comes out in dω u about u. We compute P (u) by averaging P (u; α; x) over the ensemble probability distribution: P (u) = f(α)dα 1 d 2 (x)p (u; α; x), (34) {α} πr 2 x R where f(α) is the probability distribution for α, with f(α)dα =1. (35) {α} 7

8 The other integral is evaluated on a circular disk centered on the origin and perpendicular to the beam direction. We have assumed here that the probability distribution for x is uniform over the disk, and that the size of the disk (or, size of the uniform region of the beam) is large compared with relevant target dimensions (e.g., atomic or nuclear sizes). These assumptions may be modified as the situation warrants. beam R e^3 Figure 1: Illustration of a beam setup, with uniformity over a disk of radius R. We define an effective differential cross section by: dσ eff (u) = πr 2 P (u)dω u (36) = f(α)dα d 2 (x)p (u; α; x)dω u. (37) {α} x R This cross section is the differential cross section, per scattering center, which we would observe experimentally with our given beam (e.g., the observed event rate for a scattering is equal to dσ eff times the luminosity ). It depends on the properties of our beam (i.e., the probability distributions for α and x) and hence, is an effective cross section rather than a fundamental cross section depending only on the interaction. We wish to define such a fundamental cross section and relate it to the S-matrix. Consider the limit in which R, and the momentum in the beam is sharply defined. First, for R : dσ eff (u) = f(α)dα d 2 (x)p (u; α; x) (38) {α} ( ) = f(α)dα d 2 (x) d 3 (q) d 3 (q )q 2 δ(q q ) (39) {α} ( ) ( ) ( ) 8

9 T (qu, q)t (q u, q )φ 0 (q; α)φ 0(q ; α)e ix (q q ). (40) Noting that we have: dσ eff (u) = ( ) {α} d 2 (x)e ix (q q ) =(2π) 2 δ(q 1 q 1 )δ(q 2 q 2 ), (41) f(α)dα d 3 (q) d 3 (q )q 2 δ(q 1 q 1)δ(q 2 q 2)δ(q q ) ( ) ( ) (2π) 2 T (qu, q)t (q u, q )φ 0 (q; α)φ 0 (q ; α). (42) We next use the identity: δ(q 1 q 1)δ(q 2 q 2)δ(q q )= q [ δ (3) (q q )+δ(q 1 q q 3 1)δ(q 2 q 2)δ(q 3 + q 3) ]. (43) While substituting this in, let us also apply the assumption that the momentum in the ( incident ) beam is well-defined, i.e., Thus, the product φ 0 (p; α) 0, unless p p i = p i ê 3. (44) φ 0 (q; α)φ 0 (q ; α) 0, unless both q p i and q p i. (45) In particular, this product is small unless q q, and q 3 q 3. Thus, we may neglect the contribution in the integrand from the δ(q 3 + q 3 ) term, and obtain: dσ eff (u) = (2π) 2 f(α)dα d 3 (q) d 3 (q ) q3 {α} ( ) ( ) q 3 δ(3) (q q ) T (qu, q)t (q u, q )φ 0 (q; α)φ 0 (q ; α) (46) = (2π) 2 f(α)dα d 3 (q) q3 {α} ( ) q 3 T (qu, q) 2 φ 0 (q; α) 2. (47) Now, let us again impose our assumption of a well-defined beam momentum, in a more stringent way. Let us require that φ 0 (q; α) 2 is sufficiently sharply peaked about q = p i that the function q 3 q 3 T (qu, q) 2 (48) 9

10 does not vary significantly over the region where φ 0 (q; α) 2 is significant. With this assumption, we may take: q 3 q 3 T (qu, q) 2 p3 i p i3 T (p iu, p i ) 2 = p 2 i T (p i u, p i ) 2. (49) Then dσ eff (u) = (2πp i ) 2 T (qu, q) 2 f(α)dα d 3 (q) φ 0 (q; α) 2 (50) {α} ( ) = (2πp i ) 2 T (qu, q) 2. (51) Thus, in this limit, the effective cross section no longer depends on the precise shape of the beam distribution. We interpret this limiting form as the desired fundamental (differential) cross section, σ(p f ; p i ). We may rewrite the S matrix, as an integral transform, in the form: S(p f ; p i ) = δ (3) (p f p i )+δ(p f p i )T (p f ; p i ) (52) = δ (3) (p f p i )+ i 2πp i δ(p f p i )f(p f, p i ), (53) where we have defined the scattering amplitude: The cross section is given by: f(p f, p i ) 2πip i T (p f ; p i ). (54) σ(p f ; p i ) = (2πp i ) 2 T (p f ; p i ) 2 (55) = f(p f, p i ) 2. (56) That is, the scattering cross section, appropriately, is the square of the scattering amplitude. Note that the scattering amplitude has units of length. We have defined a differential cross section, and related the S matrix to the scattering amplitude. We have made some assumptions concerning the physical situation, and the results may need to be modified in certain cases. For example, there may be narrow resonances where the assumption that the beam is well-defined compared with the variation in T 2 breaks down, and hence the effective and fundamental cross sections differ (Fig. 2). 10

11 σ Figure 2: The fundamental cross section (narrow peak) may become smeared out to the effective cross section (wide peak). 4 Partial Wave Expansion For a spherically symmetric center-of-force, the symmetry may be used to obtain a useful expression for the scattering amplitude (and, hence, the S matrix), called the partial wave expansion. This form is useful in practice because it is often the case that only a few terms in the expansion contribute significantly to the description of the scattering process. In particular, if the wavelength of the scattered particle is large compared with the size of the force center, only a few phase shifts dominate this is low energy scattering. We may obtain the partial wave expansion as follows: By spherical symmetry, S(p, q) can only depend on rotationally invariant quantities. There are three such invariants in spinless scattering: p 2, q 2 and p q. Since p = q, we are left with only two distinct invariants. Thus, we may write the S matrix in the form: S(p, q) = 1 pq δ(p q)b(p; u p u q ), (57) where u p u q p q p. (58) 2 Notice that this form is such that B is dimensionless. 11

12 Now consider an eigenstate of J 2 and J 3 with eigenvalues j(j + 1) and m = 0 (the initial momentum is along ê 3, hence J 3 = L 3 = 0): φ j,0 (p) =φ(p)p j (u p u 3 ), (59) where P j is a Legendre polynomial. Let S operate on this initial wave function: φ j,0 (p) = d 3 (q)s(p, q)φ j,0 (q) (60) = d 3 (q) 1 pq δ(p q)b(p; u p u q )φ(q)p j (u q u 3 ) (61) = φ(p) dω u B(p; u p u )P j (u u 3 ). (62) 4π By conservation of angular momentum, φ j,0(p) is an eigenstate of J 2 and J 3 with eigenvalues j(j +1) and m = 0. Since S is unitary, φ j,0 (p) is normalized to one if φ j,0 (p) is. Thus, φ j,0(p)/φ j,0 (p) must be a function of p of modulus one, and φ j,0 (p) =φ (p)p j (u p u 3 ). (63) Hence, we may write φ j,0 (p) = e2iδ j(p) φ j,0 (p) (64) = e 2iδj(p) φ(p)p j (u p u 3 ), (65) where the phase shifts, δ j (p), are real functions of p. Substituting into Eqn. 62, we have e 2iδj(p) φ(p)p j (u p u 3 )= dω u B(p; u p u )P j (u u 3 ). (66) 4π We obtain an expression for δ j (p) by letting u p = u 3 and using the fact that P j (1) = 1: e 2iδj(p) φ(p) = dω u B(p; u 3 u )P j (u u 3 ) (67) = 2π 4π 1 1 d cos θb(p, cos θ)p j (cos θ). (68) The Legendre polynomials are complete, and obey the orthogonality relation: 1 dzp j (z)p j (z) = 2 1 2j +1 δ jj. (69) 12

13 Hence, we may expand: or B(p, cos θ) = S(p, q) = j=0 δ(p q) pq 2j +1 4π e2iδ j(p) P j (cos θ), (70) j=0 2j +1 4π e2iδ j (p) P j (cos θ), (71) where cos θ = u p u q. This result, Eqn. 71, is the partial wave expansion for the S matrix. When all phase shifts δ j =0,S = I, so we may write: S(p, q) δ (3) (p q) = δ(p q) 4πpq (2j +1) [ e 2iδj(p) 1 ] P j (cos θ). (72) j=0 But we also have, S(p, q) δ (3) (p q) = i δ(p q)f(p, q). (73) 2πp Comparing these, we obtain the partial wave expansion for the scattering amplitude: f(p, q) =f(p; cos θ) = 1 2ip (2j +1) [ e 2iδj(p) 1 ] P j (cos θ). (74) j=0 In order to have convergence, we suspect we ll have δ j (p) 0asj. The unitarity of the S matrix is contained in the reality of the functions δ j (p). 5 Optical Theorem Let us consider more generally than the partial wave expansion the unitarity of the S matrix: d 3 (q)s(p, q)s (p, q) =δ (3) (p p ). (75) Substitute in ( ) S(p, q) =δ (3) (p q)+ 13 i δ(p q)f(p, q), (76) 2πp

14 obtaining δ (3) (p p ) = δ (3) (p p ) (77) + i 2πp δ(p p )f(p, p ) i 2πp δ(p p )f (p, p ) + 1 q 2 dqdω 4π 2 p p u δ(p q)δ(p q)f(p, q)f (p, q). ( ) This simplifies to i δ(p p ) [f(p, p ) f (p, p )] = δ(p p ) dω 2π p 4π 2 u f(p, q)f (p, q), (4π) (78) with q = p. But for a symmetric central force we must have f(p, p )=f(p, p )=f(p ; u u ). (79) Thus, letting p = p = p, wefind If(p; u u )= p dω u f(p; u u)f (p; u u). (80) 4π 4π In particular, if we take u = u, then If(p;1) = p dω u f(p; u u) 2 (81) 4π 4π = p dω u σ(pu,pu ) (82) 4π 4π = p 4π σ T (p), (83) where σ T is the total cross section. We have just derived what is known as the optical theorem: σ T (p) = 4π p If(p;1). (84) The total cross section is equal to 4π/p times the imaginary part of the forward scattering amplitude. The total cross section may also be readily obtained in terms of the partial wave expansion: dσ(p, cos θ) σ T (p) = dω (85) (4π) dω 14

15 = f(p, cos θ) 2 dω (86) (4π) 1 = 2π dz 1 (2j + 1)(2j +1) [ e 2iδj(p) 1 ][ e 2iδ 1 4p 2 j (p) 1 ] P j (z)p j (z) j=0 j =0 = 4π (2j +1) [ e 2iδj(p) 1 ][ e 2iδj(p) 1 ] (87) p 2 j=0 = 4π (2j + 1) sin 2 δ p 2 j (p). (88) j=0 It is left as an exercise for the reader to show that this result agrees with the optical theorem. We conclude this section by remarking that the optical theorem is a rather general consequence of wave scattering in which there is a conservation property akin to the conservation of probability in quantum mechanics. For example, it also holds in the scattering of electromagnetic waves where energy and power flow are conserved. For a discussion, see, for example, J. D. Jackson s text Classical Electrodynamics. 6 Interim Remarks We pause here to make a few remarks concerning the nature of our results and possible extensions to them: 1. Our discussion has been pretty general, up to assumptions of spherical symmetry, and of the force falling off rapidly enough with distance. 2. In particular, the results are valid whether the particles are relativistic or non-relativistic. We nowhere made any assumptions concerning speed, only using general properties of waves and quantum mechanics. Any wave equation suffices, since we only use it in the discussion of asymptotic states. 3. The discussion also applies to the elastic collision of two particles in their center-of-mass system. Note that this system has spherical symmetry in this frame. Relativity is unimportant here as well (though we may need it to transform to a different frame of reference). We may also consider inelastic scattering in the CM frame, as long as we properly formulate our conservation of energy. 15

16 4. We may further extend the discussion to the elastic scattering of two particles with any spins in the CM frame. The essential change is that the asymptotic wave function for particles of spin s are (2s + 1)- component momentum space wave functions, and the scattering amplitude f(p f, p i ) now becomes also a matrix operator in spin space. Ref. [2] develops this theory for scattering of the form a + b c + d. We have actually already seen the essential aspects in our class discussions of the consequence of angular momentum conservation on the description of scattering scattering in the helicity basis. Note that we have not yet said much about how to determine the phase shifts δ j (p) in a problem of interest we have only shown that they are the key ingredients in the scattering problem. Our discussion will now turn towards this issue. 7 Resonances Consider the partial wave expansion: We note that f(p, q) =f(p; cos θ) = 1 2ip j=0 1 [ e 2iδ j (p) 1 ] = 2i 2j +1 4π This function has a maximum magnitude of one whenever [ e 2iδ j (p) 1 ] P j (cos θ). (89) 1 cot δ j i. (90) δ j = π 2 + nπ, n = integer. (91) When this occurs, we are said to have a resonance in the jth partial wave. Suppose now that the jth channel exhibits a resonance at energy E = E 0. Then cot δ j (E) (written as a function of energy) vanishes at E = E 0. To describe the scattering in the neighborhood of this resonance, we make an expansion of cot δ j (E) to linear order: cot δ j (E) = 2 Γ (E E 0)+O [ (E E 0 ) 2]. (92) Thus, in the neighborhood of a resonance, the contribution of this channel 16

17 5 3 j ) 1 _ a l t e d ( t o c delta_j(e)/pi Figure 3: Graph of cot δ j (E), showing linear behavior near π/2, with negative slope. to the scattering amplitude is: f j (p; cos θ) = 2j +1 1 p 2 (E E Γ 0) i P j(cos θ) (93) = Γ/2p (2j +1) E E 0 + iγ/2 P j(cos θ). (94) The contribution to the total cross section from this channel alone is (using the optical theorem): σ Tj (p) = 4π p If j(p; 1) (95) = (2j +1) 4π p I Γ/2p (96) E E 0 + iγ/2 4π(2j +1) Γ = p 2 2 I E E 0 iγ/2 (97) (E E 0 ) 2 +Γ 2 /4 = 4π p (2j +1) Γ 2 /4 2 (E E 0 ) 2 +Γ 2 /4. (98) We arrive at a Breit-Wigner form for the cross section in the neighborhood of a resonance. The choice of parameterization is such that Γ is the full 17

18 width at half maximum of the Breit-Wigner distribution. The cross section decreases from its peak value by 1/2 at E = E 0 ± Γ/2. Note, finally, that the maximum contribution to the cross section in partial wave j is just: σ Tj (E)max = σ Tj (E 0 )= 4π (2j +1). (99) p2 This maximum follows from the unitarity bound: 1 [ e 2iδ j 1 ] = e iδ j sin δ j 1. (100) 2i 8 The Phase Shift Intuitively, we may think of the phase shift as follows: Suppose, for example, we have a dominantly attractive potential. The wave will oscillate more rapidly in the region of the potential than it would if the potential were not present. A wave starting at phase zero at the origin will accumulate phase as it propagates out to large distances. The phase is accumulated faster in the presence of an attractive potential than if no potential is present. Thus, there will be a phase shift of the wave in the potential relative to the wave in no potential at large distances (and once outside the region of the potential, this shift is independent of distance). Fig. 4 provides an illustration of this effect. Asymptotically, the scattered wave will be positively phase shifted with respect to an unscattered wave. Similarly, a dominantly repulsive potential will yield negative phase shifts. To see how we may get at the phase shifts, let us consider plane wave solutions, and their partial wave expansion. The force-free Schrödinger equation is 2 ψ + k 2 ψ =0, (E = k 2 /2m). (101) The plane wave solution for a wave traveling in the +z direction (if k>0) is: ψ(x) =e ikz = e ikr cos θ. (102) The separation of variables expansion of such a function in spherical polar coordinates is: ψ(x) = l l=0 m= l [A lm j l (kr)+b lm n l (kr)] P m l (θ)e imφ, (103) 18

19 r V(r) Figure 4: Illustration of the production of a phase shift due to a potential. The dashed line illustrates the wave in the absence of a potential. At large r the accumulated phase (starting from r = 0) of the wave in the potential is larger than the accumulated phase of the wave without the potential. where π j l (x) = 2x J l+ 1 (x); 2 n l (x) =( ) l+1 π 2x J l 1 (x) 2 (104) are spherical Bessel functions. Since e ikr cos θ does not depend on φ (beam along the z-axis, hence no z-component of L), only the m = 0 terms contribute. Also, n l (kr) is not 19

20 regular at r =0, 2 so B lm = 0. Hence, e ikr cos θ = A l j l (kr)p l (cos θ). (107) l=0 It remains to determine the coefficients A l. We use the orthogonality relation 1 dxp l (x)p l (x) = 2 1 2l +1 δ ll, (108) to obtain 2 1 A l j l (kr) 2l +1 = dxe ikrx P l (x). (109) 1 We can match the two sides by considering large kr. For the left hand side, ( 1 j l (x) x x sin x lπ ). (110) 2 On the right side, we note that by partial integrations we may obtain an expansion in powers of 1/kr, e.g., after two such integrations: 1 dxe ikrx P l (x) = 1 1 ikr eikrx P l (x) 1 1 (111) 1 [ 1 ikr ikr eikrx P l(x) ] dxe ikrx P l (x). ikr 1 It appears (left to the reader to prove) that the first term is dominant for large kr. Using also P l (±1) = (±) l, we thus have: 1 dxe ikrx 1 [ P l (x) kr e ikr ( ) l e ikr]. (112) 1 ikr So, we must compare: ( 1 [ e ikr ( ) l e ikr] 1 2 = A l ikr kr 2l +1 sin kr lπ ) (113) = A [ l ikr 2l +1 e iπl/2 e ikr ( ) l e ikr]. (114) 2 For small x: n l (x) x 0 { 1 x, l =0 (2l 1)!! x l+1, l > 0; j l (x) x 0 x l (105) (2l + 1)!!. (106) 20

21 Solving for A l =(i) l (2l + 1), and therefore e ikr cos θ = (i) l (2l +1)j l (kr)p l (cos θ). (115) l=0 Now suppose we have a potential V (r) which vanishes everywhere outside a sphere of radius R. The scattering amplitude, in terms of the phase shifts, as we have seen, is: f(k; θ) = 1 2ik (2l +1) ( e 2iδ l 1 ) P l (cos θ). (116) l=0 Let us relate this to the scattered wave function. For r<r, we may write: ψ = (i) l (2l +1)R l (k; r)p l (cos θ), (117) l=0 where χ l = rr l, and [ ] χ l + k 2 l(l +1) 2mV (r) χ r 2 l =0. (118) Outside the sphere r = R we may write: [ ψ = (i) l (2l +1) l=0 j l (kr)+ 1 2 α lh (1) l ] (kr) P l (cos θ), (119) where h (1) l is the spherical Hankel function of the first kind: h (1) l (kr) = j l (kr)+in l (kr) (120) kr 1 e ikr (i) l+1 kr. (121) The asymptotic form shows that this corresponds to an asymptotically outgoing spherical wave. The spherical Hankel function of the second kind, h (2) l, would correspond to an asymptotically incoming spherical wave. We don t include such a term, because we have already included the incoming plane wave in the j l (kr) term. 21

22 The asymptotic behavior of the wave function is therefore: ψ (2l +1)i l sin ( ) kr lπ 2 + α l 1 e ikr P kr 2 (i) l+1 l (cos θ) (122) kr l=0 1 2ikr (2l +1) [ e ikr (1 + α l ) ( ) l e ikr] P l (cos θ). (123) l=0 This is now expressed in terms of an incoming spherical wave and an outgoing spherical wave. For elastic scattering, we must have conservation of the number of particles (unitarity), so 1+α l 2 = ( ) l 2, (124) where the left hand side corresponds to the outgoing wave, and the right hand side is the incoming. This is satisfied if α l is of the form: α l = e 2iδ l 1, (125) where δ l is real. Now consider the scattered wave function in the asymptotic limit: ψ S = ψ e ikr cos θ (126) 1 = (2l +1) [ (1 + α l )e ikr ( ) l e ikr] P l (cos θ) 2ikr l=0 (2l + 1)(i) l 1 ( l=0 kr sin kr lπ ) P l (cos θ) 2 = 1 (2l +1) [ (1 + α l )e ikr ( ) l e ikr e ikr +( ) l e ikr] P l (cos θ) 2ikr l=0 (127) = 1 (2l +1)α l e ikr P l (cos θ) 2ikr (128) l=0 = f(k; θ) eikr r. (129) We see that the scattering amplitude has the interpretation as the coefficient of the outgoing (scattered) spherical wave. 22

23 Notice that the outgoing part of the wave in Eqn. 127 is [ e ikr cos θ ] out 1 2ikr (2l +1)e ikr P l (cos θ), (130) l=0 while the outgoing part of the actual wave (which includes in addition the outgoing part of the unscattered wave) is ψ out kr 1 2ikr (2l +1)e 2iδ l e ikr P l (cos θ), (131) l=0 where e 2iδ l =1+αl. Comparing Eqns. 130 and 131, we see that 2δ l is the difference in phase between the outgoing parts of the actual wave function and the e ikr cos θ plane wave. It is perhaps useful to make a comment here concering interference among partial waves. Note that in the differential cross section, dσ dω = f(k; θ) 2, (132) the different partial waves can interfere, i.e., in general we cannot distinguish which partial angular momentum states contribute to the scattering in a given direction. On the other hand, in the total cross section: σ T = (4π) dω f(k; θ) 2 = 4π k 2 (2l + 1) sin 2 δ l ; (133) l=0 by virtue of the orthogonality of the P l (cos θ), there is no interference among the partial waves. The cross section decomposes into a sum of partial cross sections of definite angular momenta. 8.1 Relation of Phase Shift to Logarithmic Derivative of the Radial Wave Function Recall the radial wave equation for r<r: [ ] χ l + k 2 l(l +1) 2mV (r) χ r 2 l =0, χ l = rr l. (134) 23

24 We must have continuity of the wave function and its derivative on the sphere r = R. Hence, R l (k, R) = j l (kr)+ 1 2 α lh (1) l (kr) (135) dr l (k, r) [ = k j dr l(kr)+ 1 ] r=r 2 α lh (1) l (kr). (136) The factor of k on the right side of Eqn: 136 is because the prime notation denotes derivatives with respect to the argument of the function, i.e., with respect to kr. We divide Eqn. 136 by Eqn. 135, and let x = kr: dr l (k,r) [ dr j l (x)+ 1 = k α 2 lh (1) l (x) ] R l (k, r) r=r j l (x)+ 1α 2 lh (1) l (x). (137) With dr = rd log r, we may write this in the form: [ j l (x)+ 1α 2 lh (1) l (x) ] L l d log R l(k, r) d log r Solving for α l = e 2iδ l 1: = x r=r j l (x)+ 1 2 α lh (1) l (x). (138) α l = 2 L lj l (x) xj l(x), (139) L l h (1) l (x) xh (1) l where x = kr. Thus, if the logarithmic derivative L l is determined, then the phase shift is known. 8.2 Low Energy Limit Let us use the result just obtained to demonstrate that the partial wave expansion normally converges faster in the low energy limit, x = kr 1. We ll start with the power series expansions for the spherical Bessel/Hankel functions: h (1) l (x) = j l (x)+in l (x), (140) j l (x) = 2 l x l ( ) n (n + l)! n!(2n +2l + 1)! x2n, (141) n=0 n l (x) = ( ) l l x l+1 24 n=0 ( ) n l (n l)! x 2n. (142) n!(2n 2l)!

25 As a technical aside, we see that we really need to make sense of the last formula above in the case l>n. Guided by the identity: we define (n l)! (2n 2l)! z!( z)! = πz sin πz, (143) (2l 2n)! =( )n l, if n<l. (144) (l n)! For x 1 (the low energy limit), we keep only the n = 0 term in these expansions: Thus, j l (x) = l! (2l + 1)! 2l x l + O(x l+1 ) (145) n l (x) = ( )l+1 ( ) l (2l)! + O(x l+1 ). (146) 2 l x l+1 l! Therefore, L l j l (x) xj l(x) = 2 l l! (2l + 1)! (L l l)x l + O(x l+2 ) (147) L l h (1) l (x) xh (1) l = i [L l n l (x) xn l(x)] + O(x l ) (148) = i (2l)! 2 l l! (L l + l +1) 1 x + l+1 O(xl,x l+1 ). (149) α l 2i 2l +1 [ 2 l ] 2 l! L l l (2l)! L l + l +1 x2l+1. (150) Let us use this result to check the convergence of the partial wave expansion for small x. The coefficient of the P l+1 (cos θ) term divided by the coefficient of the P l (cos θ) term is: (2l +3)α l+1 (2l +1)α l = 1 L l+1 l 1 (2l +1) 2 L l+1 + l +2 L l + l +1 x 2. (151) L l l In general then, we should have very good convergence for x 1. In the low energy limit the l th scattering partial wave is proportional to: α l k 2l+1. (152) 25

26 That is, or Using we may also write: α l = e 2iδ l 1=2i sin δ l e iδ l, (153) sin δ l k 2l+1, as kr 0. (154) j l (x) = 1 2 [ h (1) l (x)+h (2) l (x) ], (155) 1+α l = L lh (2) l (x) xh (2) l (x) L l h (1) l (x) xh (1) l (x). (156) We may obtain exact expressions for the two lowest phase shifts with, Hence, h (2) l (x) = h (1) l (x), for real x, (157) h (1) 0 (x) = i x eix, (158) h (1) 1 (x) = 1 ( x eix 1+ i ). (159) x 1+α 0 = e 2ix 1+ix/ˆL 0 1 ix/ˆl 0 (160) 1+α 1 = e 2ix 1+ix x2 /(ˆL 1 +1) 1 ix x 2 /(ˆL 1 +1), (161) where x = kr and ˆL l d log χ l, (162) d log r r=r = d log R l +1. (163) d log r r=r 8.3 Example Suppose we scatter from the potential (see Fig. 5): { V0 > 0 r<r, V (r) = 0, r R. (164) 26

27 V(r) V0 R r Figure 5: Graph of the example scattering potential. We wish to determine, for example, the phase shift and partial cross section for l = 0 (S-wave) scattering. Since the potential is repulsive, we should find δ 0 < 0. Let The radial wave equation for r<ris In particular, for l =0: 3 [ χ l + k 2 k 2 = 2mE (165) k0 2 = 2mV 0 (166) = k0 2 k 2. (167) ] l(l +1) k 2 r 2 0 χ l =0. (168) χ 0 +(k2 k 2 0 )χ 0 =0. (169) 3 Recall that we may express the wave function for the spherically-symmetric central force problem in general in the form ψ(x) = A nlm R nl (r)y lm (θ, φ), n,l,m 27

28 If E<V 0, then > 0, and χ 0 = 2 χ 0,or χ 0 = A sinh r, r < R, E < V 0. (170) If instead E>V 0, then = i is purely imaginary, hence χ 0 = A sin r, r < R, E > V 0. (171) Outside the sphere bounded by r = R, we have χ 0 = k 2 χ 0, hence χ 0 (r) = sin(kr + δ), r > R (172) where we have chosen an arbitrary normalization. We compare this with our expansion: ψ(x) = l=0 [ i l (2l +1) j l (kr)+ 1 2 α lh (1) l ] (kr) P l (cos θ), r > R. (173) Consider the l = 0 term in particular: [ χ 0 (r) =r j 0 (kr)+ 1 ] 2 α 0h (1) 0 (kr). (174) Using, j 0 (x) = sin x/x, h (1) 0 (x) = ie ix /x, and α 0 = e 2iδ 0 1, we obtain (not worrying about the overall normalization), χ 0 (r) = sin kr i 2 α 0e ikr (175) = 1 ( e ikr e ikr + e ikr+2iδ 0 e ikr) 2i (176) = e iδ 0 sin(kr + δ 0 ), (177) or, simply χ 0 (r) =sin(kr + δ 0 ), absorbing the phase factor into the normaliztion. Comparing with Eqn. 172, we find δ = δ 0. where 1 2 [ ] l(l +1) 2mr r 2 [rr nl(r)] + 2mr 2 V (r) R nl (r) =E n R nl (r) (n may be a continuous index in general), and we have defined χ nl (r) rr nl (r). At least as long as V (r) is finite as r 0, R nl (0) must be finite, hence the boundary condition χ nl (0) = 0. 28

29 We determine δ 0 by matching χ 0 and χ 0 at r = R. The logarithmic derivative L 0 = Rχ 0 (R)/χ 0(R) contains the information about the parameter of interest, δ 0, and we need never determine A (or A ). For E<V 0 : L 0 = cosh R R = Rcoth R sinh R (178) = kr cos(kr + δ 0) sin(kr + δ 0 ) = kr cot(kr + δ 0). (179) That is, L 0 = kr cot(kr + δ 0 )= Rcoth( R), E < V 0. (180) Similarly, for E>V 0 : L 0 = kr cot(kr + δ 0 )= R cot( R), E > V 0. (181) Given V 0 and R, we can thus solve these equations for δ 0 (E). Noting our earlier result: 1+α 0 = e 2iδ 0 = e 2ikR 1+ikR/L 0, 1 ikr/l 0 (182) we have, for E<V 0 : e 2iδ 0 2ikR 1+i(k/ ) tanh R = e 1 i(k/ ) tanh R, E < V 0. (183) This is of the form e ( ) 2ikR 1+ia 1 ia = e 2ikR e iθ, where θ = tan 1 [2a/(1 a 2 )]. Therefore, we may write: δ 0 = kr k tanh R 2 tan 1 1 ( ) 2, E < V 0, (184) k tanh 2 R δ 0 = kr k tan R 2 tan 1 1 ( ) 2, k tan 2 R E > V 0. (185) Consider the low energy limit, k 0( k 0 ), E<V 0 : δ 0 kr k 2 tan 1 k 0 tanh R 1 ( ) 2 (186) k k 0 tanh 2 R = kr + k ( ) 2 tanh k 0 R + O k. (187) k 0 29 k 0

30 Alternatively, we may consider the hard sphere limit, k 0 (i.e., V 0 ). This is nearly the same limit, for any fixed energy. To lowest order in k/k 0, δ 0 = kr + O(k/k 0 ). (188) Thus, for the hard sphere the lowest partial wave phase shift depends linearly on k, as shown in Fig. 6. δ 0 0 kr Figure 6: The S wave phase shift for the hard sphere potential. Consider now the total cross section in the l = 0 channel: σ 0 = 4π k 2 sin2 δ 0. (189) In the low energy limit, this is σ 0 4π ( k 2 sin2 kr + k ) tanh k 0 R (190) k 0 4π k 2 ( kr + k k 0 tanh k 0 R) 2 (191) 30

31 ( = 4πR 2 1 tanh k ) 2 0R. (192) k 0 R In the hard sphere limit (k 0 ), this becomes σ 0 =4πR 2. (193) In the low energy limit, this partial wave dominates, so this is the total cross section as well. This result is larger than the result we might expect purely geometrically, i.e., for the scattering of a well-localized wave packet. The difference must be attributed to the wave nature of our probe, and the phenomenon of diffraction. We note in passing that when writing ( σ 0 =4πa 2 0, a 0 = R 1 tanh k ) 0R, (194) k 0 R a 0 is referred to as the scattering length of the potential. Now let us consider the case of finite V 0 (a soft sphere?). For example, in the high energy limit, k k 0,and k, hence: δ 0 kr tan kr 2 tan 1 (195) 1 tan 2 kr = 0. (196) This should be the expected result in the high energy limit, since the effect of a finite fixed potential becomes negligible as the energy increases. Thus, at very low energies and at very high energies, the phase shift approaches zero. In between, it must reach some maximum (negative) value. Fig. 7 shows the S wave phase shift for scattering on the soft sphere potential, and Fig. 8 shows the S wave cross section for scattering on the soft sphere potential. 9 The Born Expansion, Born Approximation We have mentioned the Born expansion in our note on resolvents and Green s functions, and the Born approximation in our note on approximate methods. We now revisit it explicitly in the context of scattering. We start by developing the approach in general, and then apply it to the example of the preceding section. 31

32 0-0.5 kr δ 0 Figure 7: The S wave phase shift for the soft sphere potential, with R =1, k 0 R =4. In the resolvent note, we saw the Born expansion as the Liouville-Neumann expansion of a perturbed Green s function. We will see this again, except that it is convenient (eliminates a minus sign) here to redefine the Green s function with the opposite sign from the resolvent note. Thus, we here take G(z) = 1 H z. (197) The Schrödinger equation we wish to solve is of the form: ( ) 2 2m + V ψ = Eψ. (198) It is convenient to let E = k 2 /2m and V = U/2m, so that this equation is of the form: ( 2 + k 2) ψ(x) =U(x)ψ(x). (199) We already found the Green s function for the free particle wave equation (Helmholtz equation), ( 2 + k 2 )ψ = 0 in the resolvent note: G(x, y; k) = e±ik x y 4π x y. (200) 32

33 8 7 6 σ kr Figure 8: The S wave cross section for the soft sphere potential, with R =1, k 0 R =4. The solution to the desired Schrödinger equation is then ψ(x) =ψ 0 (x)+ d 3 (x )G(x, x ; k)u(x ψ(x ), (201) ( ) where ψ 0 (x) is a solution to the Helmholtz equation. The reader is encouraged to verify that ψ is indeed a solution, by substituting into the Schrödinger equation. 4 Now we may apply this to the scattering of an incident plane wave, ψ 0 = e ik x, using the outgoing Green s function, to obtain the scattered wave: G + (x, y; k) = e+ik x y 4π x y, (202) ψ S (x) ψ(x) ψ 0 (x) (203) = d 3 (x )G + (x, x ; k)u(x ψ(x ). (204) ( ) 4 Recall that ( 2 x + k 2 )G(x, x ; k) =δ (3) (x x ). 33

34 Since ψ S (x) =f(k; θ) eikr r, (205) where θ is the polar angle to the observation point, hence equal to the polar angle of the scattered k-vector, we can extract the scattering amplitude f(k; θ) by considering the asymptotic situation, where the source size (nonnegligible region of U(x)) is small: G + (x, x ; k) x 1 e ik x e ik x. (206) 4π x In obtaining this expression, we have held x finite, and used x x x x x x. (207) Thus, f(k; θ) = 1 d 3 (x )e ik x U(x )ψ(x ). 4π ( ) (208) Let us return to the general integral equation for the wave function: ψ(x) =ψ 0 (x)+ d 3 (x )G(x, x ; k)u(x )ψ(x ). (209) ( ) We plug this expression into the integrand, obtaining: ψ(x) = ψ 0 (x)+ d 3 (x)g(x, x ; k)u(x )ψ 0 (x ) (210) ( ) + d 3 (x )d 3 (x )G(x, x ; k)u(x )G(x, x ; k)u(x )ψ(x ). ( ) Iteration gives the Neumann series: ψ(x) = ψ 0 (x)+ d 3 (x )d 3 (x n )G(x, x ; k)u(x ) (211) n=1 ( ) G(x, x ; k)u(x )...G(x n 1, x n ; k)u(x n )ψ 0 (x n ). (212) Typically, we use this by substituting a plane wave for ψ 0 (x). If the potential is weak, we expect this series to converge quickly. This expansion is sometimes called the Born expansion. To first order in U(x) we have: ψ(x) =ψ 0 (x)+ d 3 (x )G(x, x ; k)u(x )ψ 0 (x )+O(U 2 ). (213) ( ) 34

35 With ψ 0 (x) =e ik x, and G(x, x ; k) 1 e ikr e ik x, (214) 4π r we find: ψ(x) e ik x 1 e ikr d 3 (x )e ik x U(x )e ik x. (215) 4π r ( ) The interpretation of k is as the momentum vector of the incident plane wave, and of k is as the momentum vector of the scattered wave. Hence, ψ S (x) 1 e ikr d 3 (x )e i(k k ) x U(x ), (216) 4π r ( ) or f(k;ω ) 1 d 3 (x )e i(k k ) x U(x ), (217) 4π ( ) where Ω is the direction of the scattered wave, i.e., the direction of k. This result is known as the Born approximation. With this result, we have the differential cross section: dσ = f(k;ω ) 2 (218) dω 1 2 = 16π 2 d 3 (x )e i(k k ) x U(x ) (219) ( ) = m2 4π 2 ˆV (k k) 2, (220) where we have used U(x) =2mV (x), and ˆV is the Fourier transform of V. Recall that we already derived this result in our discussion of time-dependent perturbation theory (Approximate Methods course note). If U(x) = U(r) only, i.e., we have a spherically symmetric scattering potential, we may incorporate this into our result. Let θ be the scattering angle, i.e., the angle between k and k. Then, noting that k = k = k, we have k k =2ksin θ. Since U(x) =U(r), 2 f(k;ω )=f(k, θ) only, and f(k; θ) = 1 r 2 dr U(r ) d cos θ dφ e i2k(sin θ 2 )r cos θ, (221) 4π 0 4π 35

36 where θ is the angle between x and k k. Now let K = k k =2ksin θ. 2 We can do the integrals over angles, to obtain: f(k; θ) = r 2 dr U(r sin Kr ). (222) 0 Kr This is the Born approximation for a spherically symmetric potential. 9.1 Born Approximation for Phase Shifts Now let us consider what the Born approximation yields for the phase shifts in a partial wave expansion. Start with our earlier (exact asymptotic) result: ψ(x) = ψ 0 (x)+ψ S (x) (223) = ψ 0 (x)+ eikr f(k; θ), r (224) where f(k; θ) = 1 d 3 (x )e ik x U(x )ψ(x ). 4π ( ) (225) The expansion of ψ(x) in partial waves is ψ(x) = i l (2l +1)R l (r)p l (cos θ). (226) l=0 We insert this into the expression for f: f(k; θ) = 1 1 2π i l (2l+1) r 2 dr R l (r ) d cos θ dφ e ik x U(x )P l (cos θ ). 4π l= (227) If, again, we have a spherical potential, U(x) =U(r) only, then we can perform the angular integrals. Note that θ is the scattering angle, i.e., the angle between k and k.ifkis along the z axis, then θ is the polar angle of k. Thus, k is not necessarily along the z-axis, and we have: k x = kr [cos θ cos θ + sin θ sin θ cos(φ k φ x )]. (228) Now consider our partial wave expansion of a plane wave: e ik x = e ikr cos θ kx = i l (2l +1)j l (kr)p l (cos θ kx ) (229) l=0 = i l 4π(2l +1)j l (kr)y l0 (θ kx,φ kx ), (230) l=0 36

37 where θ kx is the angle between k and x, and we have used the identity 4π P l (cos θ) = 2l +1 Y l0(θ, φ). (231) We may re-express this in terms of the polar angles (θ x,φ x )ofx and (θ k,φ k ) of k using the addition theorem for spherical harmonics: 4π(2l +1)Y l0 (θ kx,φ kx )= l m= l Y lm (θ k,φ k )Y lm (θ x,φ x ), (232) where Thus, and hence, cos θ kx = cos θ k cos θ x + sin θ k sin θ x cos(φ k φ x ). (233) e ik x =4π i l j l (kr) l l=0 m= l Y lm(θ k,φ k )Y lm (θ x,φ x ), (234) f(k; θ) = 1 i l (2l +1) r 2 dr R l (r ) 4π l=0 0 (235) 4π i l jl (k r )U(r ) (236) 1 1 l =0 2π d cos θ dφ We may simplify this in several ways: j l (x) is real if x is real. 0 l m= l Y l m(θ, φ k )Y l m(θ,φ )P l (cos θ ).(237) We may relabel integration variable r as r. Also, k is equal to k. The only dependence on φ is in Y l m(θ,φ ), hence 2π 0 dφ Y l m (θ,φ ) = 2πδ m0 Y l 0 (θ, 0) (238) 2l = 2πδ m0 4π P l (cos θ ). (239) 37

38 The integral over cos θ may then be accomplished with: Finally, we may use 1 1 dxp l (x)p l (x) = 2 2l +1 δ ll. (240) Y l0 (θ, 0) = Including all of these points leads to: f(k; θ) = (2l +1)P l (cos θ) l=0 2l +1 4π P l(cos θ). (241) 0 r 2 drr l (r)j l (kr)u(r). (242) So far, we have made no approximation this result is exact as long as the potential falls off rapidly enough as r. However, it depends on knowing R l (r), and this is where we shall now make an approximation. Note first, that by comparison with the expansion in phase shifts: f(k; θ) = 1 (2l +1) ( e 2iδ l 1 ) P l (cos θ), (243) 2ik l=0 we obtain: 1 ( e 2iδ l 1 ) = e iδ l sin δ l = k r 2 drr l (r)j l (kr)u(r). (244) 2i 0 The Born approximation consists in approximating the wave function by its value for no potential, which should be a good approximation as long as the potential has a small effect on it. Here, this means we replace R l (r) with j l (kr), valid as long as R l (r) j l (kr): e iδ l sin δ l k 0 r 2 dr [j l (kr)] 2 U(r). (245) Note that the integral is real in this approximation. In this approximation, δ l is small, hence we have the Born approximation for the phase shifts: δ l k r 2 dr [j l (kr)] 2 U(r). (246) 0 38

39 9.2 Born Approximation and Example of the Soft Sphere Potential As an example, let us return to our earlier example of the soft sphere : V (r) = { V0, r < R, 0, r > R. (247) Now U(r) =2mV (r), and let k 2 0 =2mV 0. Then, R δ l kk0 2 r 2 dr [j l (kr)] 2 (248) 0 ( ) 2 k0 kr = [xj l (x)] 2 dx. (249) k 0 For instance, consider δ 0, with j 0 (x) = sin x x : δ 0 = ( ) 2 k0 kr k ( k0 k 0 sin 2 xdx (250) ) 2 1 (2kR sin 2kR). (251) 4 Note that the Born approximation is a high energy approximation there must be little effect on the scattering wave. Consider what happens at low energy, kr 1: ( ) 2 k0 1 δ 0 = k 4 (2kR sin 2kR) kr(k 0R) 3. (252) 3 This may be compared with our earlier low energy result: ( δ 0 kr 1 tanh k ) 0R. (253) k 0 R Other than the linear dependence, they are not very similar. Fig. 9 illustrates this. At low kr, the agreement with the exact calculation is poor, as the energy increases, the agreement improves. 39

40 0 kr δ Figure 9: The S wave phase shift for the soft sphere potential, with R =1, k 0 R = 4. The curve which goes further negative is the Born approximation; the other curve is the exact result. 10 Angular Distributions If the laws of physics are rotationally invariant, we have angular momentum conservation. As a consequence, the angular distribution of particles in a scattering process is constrained. Here, we consider how this may be applied in the case of a scattering reaction of the form: a + b c + d. (254) These particles may carry spins let j a denote the spin of particle a, etc. It is convenient to work in the center-of-mass frame, and to pick particle a to be incoming along the z axis. If a different frame is desired, then it may be reached from the center-of-mass frame with a Gallilean or Lorentz transformation, whichever is appropriate. It is furthermore convenient to work in a helicity basis for the angular momentum states. Along the helicity axis, the component of orbital angular momentum of a particle is zero. We ll typically label the helicity of a particle with the symbol λ. For example, the helicity of particle a is denoted λ a. 40

41 Let us also work with a plane wave basis for our particles. Physical states may be constructed as a superposition of such states. We ll omit the magnitude of momentum in our labelling, and concentrate on the angles (polar angles denoted by θ, and azimuth angles denoted by φ). We ll also omit the spins of the scattering particles in the labels, to keep things compact. Thus, we label the initial state according to: and the final state by, i = θ a =0,φ a,λ a,λ b, (255) f = θ c,φ c,λ c,λ d, (256) Note that θ d = π θ c and φ d = π + φ c in the center-of-mass. This basis may be called the plane wave helicity basis. We may write the transition amplitude from the initial to final state in the form: f T i θ c,φ c,λ c,λ d T 0,φ a,λ a,λ b, (257) and the scattering cross section is: dσ λaλ b λ cλ d (θ θ c,φ φ c ) θ, φ, λ c,λ d T 0,φ a,λ a,λ b 2. (258) This gives the scattering cross section for the specified helicity states. If we do not measure the final helicities, then we must sum over all possible λ c and λ d. Similarly, if the initial beams (a and b) are unpolarized, we must average over all possible λ a and λ b. Thus, to get the scattering cross section in such a case, we must evaluate 1 1 2j a +1 λ a 2j b +1 λ b λ c λ d θ, φ, λ c,λ d T 0,φ a,λ a,λ b 2. (259) If a(or b) is massless and not spinless, then the 1/(2j a + 1) factor must be replaced by 1/2. The scattering process may go via intermediate states of definite angular momentum, for example, scattering through an intermediate resonance. For example. in e + e scattering just below µ + µ threshold, we may expect a resonance due to the presence of muonium bound states. Such states have definite angular momenta. In particular, the resonance for scattering through the lowest 3 S 1 bound state of muonium might be expected to be a large contribution in the scattering amplitude. Thus, we are led to describing 41