Physics 125Course Notes

Scattering040407 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 188.1 Relation of Phase Shift to Logarithmic Derivative of the Radial

Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . 238.2 Low Energy Limit . . . . . . . . . . . . . . . . . . . . . . . . . 248.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

9 The Born Expansion, Born Approximation 319.1 Born Approximation for Phase Shifts . . . . . . . . . . . . . . 369.2 Born Approximation and Example of the “Soft Sphere” Potential 39

10 Angular Distributions 40

11 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

processes is thus an important subject. We have already gotten a glimpseinto how to deal with scattering in our discussion of time-dependent pertur-bation theory, and in our investigation of angular momentum. We carry thisdevelopment further in this note and derive more of the important conceptsand tools for dealing with scattering. Many of the results we obtain will infact be applicable to relativistic processes.

We start by developing the basic physical picture and mathematical for-malism for a simple situation with the essential features. That is, we beginby considering the problem of the scattering of a spinless particle of massm > 0 from a fixed, static, spherically symmetric center of force, centered atthe origin. Assume further that the force is of “finite range” – far enoughaway, any wave packet acts as if there is no force.1 In another view, consid-ering the spreading of the wave function, as t → ±∞, the wave packet is sospread 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 wavepacket. Assume it is normalized: 〈φ|φ〉 = 1. The time dependent Schrodingerequation is:

i∂tφ(p, t) = Hφ(p, t), (1)

where H includes the influence of the force. Let E(p) =√m2 + p2, or p2/2m

in the non-relativistic case, where p = |p|. We restrict consideration to thesubspace 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 wavefunction can be considered to be far away from the center of force at largetimes.

Corresponding to φ(p, t) (an exact solution to the Schrodinger equationfor a possible scattering event), define the notion of an “initial” wave function,φi(p), by taking the following limit at asymptotically early times:

φi(p) = limt→−∞

φ(p, t)eitE(p). (2)

1The Coulomb force is thus excluded by this criterion. As seen in the exercises ofthe 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

That is, φi(p)e−itE(p) is a solution to the Schrodinger equation in the absenceof the scattering force. Likewise, we define a “final” wave function accordingto:

φf(p) = limt→∞

φ(p, t)eitE(p). (3)

We should perhaps remark a bit further on the choice of phase factor inthese asymptotic limits. Put most simply, these choices just correspond tothe interaction picture. However, we may gain a bit of additional insight asfollows: Recall that for a free particle the momentum space wave functionbehaves under a time translation by t0 as:

M(t0)ψ(p, t) = eit0p2

2mψ(p, t). (4)

Thus, our initial asymptotic limit, Eqn. 2, is like taking

φi(p) = M(t0 → −∞)φ(p, t = 0), (5)

if the motion were strictly force-free for all times. A similar correspondenceholds for the asymptotic final state wave function. Thus, our initial and finalphases are “matched” to the t = 0 phase for the case of no force. Of course,even with a force, φi(p) determines φ(p, t), and hence φf(p). We note that itis convenient to deal with momentum space wave functions in this discussionof scattering, because the coordinate space wave functions in general becomeinfinitely “spread out” as t→ ∞.

We are couching our discussion in terms of physically realizable wavepackets here. Typically, textbooks resort to the use of (“unphysical”) plane-wave states, which simplifies the treatment somewhat, at the expense ofrequiring further justification. It is reassuring to find that we can obtainthe same results by considering only physical states. Thus, our developmentfollows closely the nice pedagogical treatment in Ref. [1].

We are interested in the problem of determining φf , given φi. In principle,this requires solving the Schrodinger equation with φi specifying the initialboundary condition. Usually, however, we do not require the explicit detailedsolution – we are interested in only certain aspects of the relation betweenφi and φf , such as the force-induced phase difference.

The transformation giving φf , in terms of φi is a linear transformation,which also preserves the normalization of the wave packet. Hence, it is aunitary transformation:

|φf〉 = S|φi〉, (6)

3

whereSS† = I. (7)

The unitary operator S is called the S matrix.In general, we may represent the S matrix as an integral transform:

φf(p) =∫

(∞)d3(q)S(p,q)φi(q), (8)

and the unitarity condition is represented as:∫

(∞)d3(q)S(p′,q)S†(q,p′′) =

∫

(∞)d3(q)S(p′,q)S∗(p′′,q)

=∫

(∞)d3(q)S∗(q,p′)S(q,p′′)

= δ(3)(p′ − p′′). (9)

The scattering matrix S(p,q) contains precisely all the information which isavailable in an “asymptotic” experiment, as described by φi, φf , where thedetails of what happens near the origin are not observed.

Consider briefly the classical analog, for the classical scattering of a parti-cle 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 beobtained by solving the Hamilton equations of motion:

xi = ∂piH (10)

pi = −∂xiH. (11)

The quantities x(t) and p(t) are the classical analog of φ(p, t), since theydescribe the detailed behavior of the particle. As long as the force falls offsufficiently rapidly at large distances, the motion at large distances and times(presuming no bound states are involved) is uniform:

x(t) =

xi + pit/m, as t→ −∞,xf + pf t/m, as t→ ∞.

(12)

Thus, we can define the “initial” and “final” asymptotic values as:

pi = limt→−∞

p(t), xi = limt→−∞

[x(t) − t

mpi

](13)

pf = limt→∞

p(t), xf = limt→∞

[x(t) − t

mpf

]. (14)

4

In particular, xi, xf are (or rather, is) the positions that would occur att = 0 if the motion were strictly uniform for all times, with the asymptoticmomenta. Thus, (xi,pi) is the analog of φi(p), and (xf ,pf) is the analog ofφf(p).

Once again, specification of (xi,pi) uniquely determines (xf ,pf), and thetransformation giving this relationship is the analog of the S matrix. Notethat, in the absence of a force field,

xi = xf , pi = pf . (15)

The analogous statement in quantum mechanics is

φi(p) = φf(p), (16)

since φ(p, t) = φi(p)e−itE(p) is the solution of the Schrodinger equation forfree particle motion:

φf(p) = limt→∞

φ(p, t)eitE(p) (17)

= limt→∞

φi(p)e−itE(p)eitE(p), (free particle) (18)

= φi(p). (19)

3 The Differential Cross Section

In the Approximate Methods note, we defined a differential cross sectionin the discussion of time-dependent perturbation theory. We now return tothis, and develop the notion of a cross section with more care. We start withthe S matrix formalism just introduced. Given φi(p) and S, we obtain φf (p)according to:

φf(p) =∫

(∞)d3(q)S(p,q)φi(q). (20)

Now define the scattered wave by:

φs(p) = φf(p) − φi(p), (21)

or

|φs〉 = |φf〉 − |φi〉 (22)

= (S − I)|φi〉. (23)

5

In the absence of a force, φf (p) = φi(p), hence S = I and φs(p) = 0. Thus,the name “scattered wave” is appropriate for φs(p).

Since the scattering force is assumed to be stationary, energy is conservedin the scattering event (e.g., for an infinitely heavy source of force, any finitemomentum transfer to the source yields zero kinetic energy transfer). Hence,the asymptotic initial and final magnitudes of momenta will be equal. Wemay 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 theintegral transform for S − I. Thus, we have:

φs(p) =∫

(∞)d3(q)δ(p− q)T (p,q)φi(q). (25)

Note that T (p,q) is “physical” only for p = q (“on the energy shell”), thoughit is sometimes useful to extend its domain of definition analytically “off-shell”.

Let φ0(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 beamdelivered by an accelerator). We assume that the particles in our “beam” allhave approximately the same momentum pi, i.e.,

φ0(p;α) ≈ 0, unless p ≈ pi. (26)

Assume further that 〈φ0|φ0〉 = 1 for all α, and that pi is in the 3-direction:pi = pie3. The index α is a “shape parameter”, since we use it to specifythe shape of the incident wave packet of a particle in the beam. note thatdifferent particles may have different wave packets. We consider the packet

φi(p;α;x) = φ0(p;α)e−ix·p, (27)

obtained by translating the packet φ0(p;α) by amount x in space (assumedto be transverse to the three direction, x3 = 0). Fixing x and α, and takingthe packet as our initial wave, we obtain the scattered wave as:

φs(p;α;x) =∫

(∞)d3(q)δ(p− q)T (p,q)φ0(q;α)e−ix·q. (28)

We are interested in the probability, P (u;α;x)dΩu, that the particle is inthe scattered wave with momentum in dΩu around unit vector u. At least if

6

u is not too close to the forward direction defined by pi, this probability isjust the probability that the particle’s momentum after scattering is in dΩu:

P (u;α;x) =∫ ∞

0p2dp |φs(pu;α;x)|2 (29)

=∫ ∞

0p2dp

∫

(∞)d3(q)

∫

(∞)d3(q′)δ(p− q)δ(p− q′) (30)

T (pu,q)T ∗(pu,q′)φ0(q;α)φ∗0(q

′;α)e−ix·(q−q′). (31)

We may integrate over p, setting p = q′:

P (u;α;x) =∫

(∞)d3(q)

∫

(∞)d3(q′)q2δ(q − q′) (32)

T (qu,q)T ∗(q′u,q′)φ0(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 contribu-tions.

The point of the parameters α,x is the following: We wish to make surethat our formalism is valid for real experimental situations, i.e., that we havenot left out some essential feature in our theory for describing experimen-tal situations. Thus, we assume that the real exdperimental “beam” is astatistical ensemble of wave packets φ0(p;α;x). Thus, the parameter x de-scribes 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 theensemble of shapes. The ensemble is described by specifying the probabil-ity distributions for x and α, which depends on the particular experimentalarrangement. Note that we are interested here in scattering of a “one particlebeam” on a “one particle target”, i.e., correlations among particles in thebeam or in the target are assumed to be zero.

Let P (u)dΩu be the probability that a beam particle comes out in dΩu

about u. We compute P (u) by averaging P (u;α;x) over the ensemble prob-ability distribution:

P (u) =∫

αf(α)dα

1

πR2

∫

|x|≤Rd2(x)P (u;α;x), (34)

where f(α) is the probability distribution for α, with

∫

αf(α)dα = 1. (35)

7

The other integral is evaluated on a circular disk centered on the originand perpendicular to the beam direction. We have assumed here that theprobability distribution for x is uniform over the disk, and that the size ofthe disk (or, size of the uniform region of the beam) is large compared withrelevant target dimensions (e.g., atomic or nuclear sizes). These assumptionsmay be modified as the situation warrants.

Re3

beam

Figure 1: Illustration of a beam setup, with uniformity over a disk of radiusR.

We define an “effective” differential cross section by:

dσeff(u) = πR2P (u)dΩu (36)

=∫

αf(α)dα

∫

|x|≤Rd2(x)P (u;α;x)dΩu. (37)

This cross section is the differential cross section, per scattering center, whichwe would observe experimentally with our given beam (e.g., the observedevent rate for a scattering is equal to dσeff times the “luminosity”). It de-pends on the properties of our beam (i.e., the probability distributions for αand x) and hence, is an “effective” cross section rather than a “fundamental”cross section depending only on the interaction. We wish to define such afundamental cross section and relate it to the S-matrix.

Consider the limit in which R → ∞, and the momentum in the beam issharply defined. First, for R → ∞:

dσeff(u) =∫

αf(α)dα

∫

(∞)d2(x)P (u;α;x) (38)

=∫

αf(α)dα

∫

(∞)d2(x)

∫

(∞)d3(q)

∫

(∞)d3(q′)q2δ(q − q′) (39)

8

T (qu,q)T ∗(q′u,q′)φ0(q;α)φ∗0(q

′;α)e−ix·(q−q′). (40)

Noting that∫

(∞)d2(x)e−ix·(q−q′) = (2π)2δ(q1 − q′1)δ(q2 − q′2), (41)

we have:

dσeff(u) =∫

αf(α)dα

∫

(∞)d3(q)

∫

(∞)d3(q′)q2δ(q1 − q′1)δ(q2 − q′2)δ(q − q′)

(2π)2T (qu,q)T ∗(q′u,q′)φ0(q;α)φ∗0(q

′;α). (42)

We next use the identity:

δ(q1−q′1)δ(q2−q′2)δ(q−q′) =q

|q3|[δ(3)(q − q′) + δ(q1 − q′1)δ(q2 − q′2)δ(q3 + q′3)

].

(43)While substituting this in, let us also apply the assumption that the momen-tum in the (“incident”) beam is well-defined, i.e.,

φ0(p;α) ≈ 0, unless p ≈ pi = pie3. (44)

Thus, the product

φ0(q;α)φ∗0(q

′;α) ≈ 0, unless both q ≈ pi and q′ ≈ pi. (45)

In particular, this product is small unless q ≈ q′, and q3 ≈ q′3. Thus, wemay neglect the contribution in the integrand from the δ(q3 + q′3) term, andobtain:

dσeff(u) = (2π)2∫

αf(α)dα

∫

(∞)d3(q)

∫

(∞)d3(q′)

q3

|q3|δ(3)(q − q′)

T (qu,q)T ∗(q′u,q′)φ0(q;α)φ∗0(q

′;α) (46)

= (2π)2∫

αf(α)dα

∫

(∞)d3(q)

q3

|q3||T (qu,q)|2|φ0(q;α)|2. (47)

Now, let us again impose our assumption of a well-defined beam momen-tum, in a more stringent way. Let us require that |φ0(q;α)|2 is sufficientlysharply peaked about q = pi that the function

q3

|q3||T (qu,q)|2 (48)

9

does not vary significantly over the region where |φ0(q;α)|2 is significant.With this assumption, we may take:

q3

|q3||T (qu,q)|2 → p3

i

|pi3||T (piu,pi)|2 = p2

i |T (piu,pi)|2. (49)

Then

dσeff(u) = (2πpi)2|T (qu,q)|2

∫

αf(α)dα

∫

(∞)d3(q)|φ0(q;α)|2 (50)

= (2πpi)2|T (qu,q)|2. (51)

Thus, in this limit, the effective cross section no longer depends on the preciseshape of the beam distribution. We interpret this limiting form as the desired“fundamental” (differential) cross section, σ(pf ;pi).

We may rewrite the S matrix, as an integral transform, in the form:

S(pf ;pi) = δ(3)(pf − pi) + δ(pf − pi)T (pf ;pi) (52)

= δ(3)(pf − pi) +i

2πpi

δ(pf − pi)f(pf ,pi), (53)

where we have defined the scattering amplitude:

f(pf ,pi) ≡ −2πipiT (pf ;pi). (54)

The cross section is given by:

σ(pf ;pi) = (2πpi)2|T (pf ;pi)|2 (55)

= |f(pf ,pi)|2. (56)

That is, the scattering cross section, appropriately, is the square of the scat-tering amplitude. Note that the scattering amplitude has units of length.

We have defined a differential cross section, and related the S matrix tothe scattering amplitude. We have made some assumptions concerning thephysical situation, and the results may need to be modified in certain cases.For example, there may be narrow “resonances” where the assumption thatthe 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

σ

Figure 2: The fundamental cross section (narrow peak) may become smearedout to the effective cross section (wide peak).

4 Partial Wave Expansion

For a spherically symmetric center-of-force, the symmetry may be used toobtain a useful expression for the scattering amplitude (and, hence, the Smatrix), called the partial wave expansion. This form is useful in practicebecause it is often the case that only a few terms in the expansion contributesignificantly to the description of the scattering process. In particular, ifthe 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 sym-metry, S(p,q) can only depend on rotationally invariant quantities. Thereare three such invariants in spinless scattering: p2, q2 and p · q. Since|p| = |q|, we are left with only two distinct invariants. Thus, we may writethe S matrix in the form:

S(p,q) =1

pqδ(p− q)B(p;up · uq), (57)

whereup · uq ≡

p · qp2

. (58)

Notice that this form is such that B is dimensionless.

11

Now consider an eigenstate of J2 and J3 with eigenvalues j(j + 1) andm = 0 (the initial momentum is along e3, hence 〈J3〉 = 〈L3〉 = 0):

φj,0(p) = φ(p)Pj(up · u3), (59)

where Pj is a Legendre polynomial. Let S operate on this initial wave func-tion:

φ′j,0(p) =

∫d3(q)S(p,q)φj,0(q) (60)

=∫d3(q)

1

pqδ(p− q)B(p;up · uq)φ(q)Pj(uq · u3) (61)

= φ(p)∫

4πdΩu′B(p;up · u′)Pj(u

′ · u3). (62)

By conservation of angular momentum, φ′j,0(p) is an eigenstate of J2 and J3

with eigenvalues j(j+1) and m = 0. Since S is unitary, φ′j,0(p) is normalized

to one if φj,0(p) is. Thus, φ′j,0(p)/φj,0(p) must be a function of p of modulus

one, andφ′

j,0(p) = φ′(p)Pj(up · u3). (63)

Hence, we may write

φ′j,0(p) = e2iδj(p)φj,0(p) (64)

= e2iδj(p)φ(p)Pj(up · u3), (65)

where the phase shifts, δj(p), are real functions of p. Substituting intoEqn. 62, we have

e2iδj(p)φ(p)Pj(up · u3) =∫

4πdΩu′B(p;up · u′)Pj(u

′ · u3). (66)

We obtain an expression for δj(p) by letting up = u3 and using the factthat Pj(1) = 1:

e2iδj(p)φ(p) =∫

4πdΩu′B(p;u3 · u′)Pj(u

′ · u3) (67)

= 2π∫ 1

−1d cos θB(p, cos θ)Pj(cos θ). (68)

The Legendre polynomials are complete, and obey the orthogonality relation:

∫ 1

−1dzPj(z)Pj′(z) =

2

2j + 1δjj′. (69)

12

Hence, we may expand:

B(p, cos θ) =∞∑

j=0

2j + 1

4πe2iδj(p)Pj(cos θ), (70)

or

S(p,q) =δ(p− q)

pq

∞∑

j=0

2j + 1

4πe2iδj (p)Pj(cos θ), (71)

where cos θ = up · uq. This result, Eqn. 71, is the partial wave expansionfor the S matrix.

When all phase shifts δj = 0, S = I, so we may write:

S(p,q) − δ(3)(p − q) =δ(p− q)

4πpq

∞∑

j=0

(2j + 1)[e2iδj(p) − 1

]Pj(cos θ). (72)

But we also have,

S(p,q) − δ(3)(p − q) =i

2πpδ(p− q)f(p,q). (73)

Comparing these, we obtain the partial wave expansion for the scatter-ing amplitude:

f(p,q) = f(p; cos θ) =1

2ip

∞∑

j=0

(2j + 1)[e2iδj(p) − 1

]Pj(cos θ). (74)

In order to have convergence, we suspect we’ll have δj(p) → 0 as j → ∞.The unitarity of the S matrix is contained in the reality of the functionsδj(p).

5 Optical Theorem

Let us consider more generally than the partial wave expansion the unitarityof the S matrix:

∫

(∞)d3(q)S(p′,q)S∗(p′′,q) = δ(3)(p′ − p′′). (75)

Substitute in

S(p,q) = δ(3)(p − q) +i

2πpδ(p− q)f(p,q), (76)

13

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

4π2p′p′′

∫

(∞)q2dqdΩuδ(p

′ − q)δ(p′′ − q)f(p′,q)f ∗(p′′,q).

This simplifies to

− i

2π

δ(p′ − p′′)

p′[f(p′,p′′) − f ∗(p′′,p′)] =

δ(p′ − p′′)

4π2

∫

(4π)dΩuf(p′,q)f ∗(p′′,q),

(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, we find

=f(p;u′ · u′′) =p

4π

∫

4πdΩuf(p;u′ · u)f ∗(p;u′′ · u). (80)

In particular, if we take u′ = u′′, then

=f(p; 1) =p

4π

∫

4πdΩu|f(p;u′ · u)|2 (81)

=p

4π

∫

4πdΩuσ(pu, pu′) (82)

=p

4πσT (p), (83)

where σT is the total cross section.We have just derived what is known as the optical theorem:

σT (p) =4π

p=f(p; 1). (84)

The total cross section is equal to 4π/p times the imaginary part of theforward scattering amplitude.

The total cross section may also be readily obtained in terms of the partialwave expansion:

σT (p) =∫

(4π)

dσ(p, cos θ)

dΩdΩ (85)

14

=∫

(4π)|f(p, cos θ)|2dΩ (86)

= 2π∫ 1

−1dz

1

4p2

∞∑

j=0

∞∑

j′=0

(2j + 1)(2j ′ + 1)[e2iδj(p) − 1

] [e−2iδj′ (p) − 1

]Pj(z)Pj′(z)

=4π

p2

∞∑

j=0

(2j + 1)[e2iδj(p) − 1

] [e−2iδj(p) − 1

](87)

=4π

p2

∞∑

j=0

(2j + 1) sin2 δj(p). (88)

It is left as an exercise for the reader to show that this result agrees with theoptical theorem.

We conclude this section by remarking that the optical theorem is a rathergeneral consequence of wave scattering in which there is a conservation prop-erty akin to the conservation of probability in quantum mechanics. For exam-ple, it also holds in the scattering of electromagnetic waves where energy andpower flow are conserved. For a discussion, see, for example, J. D. Jackson’stext “Classical Electrodynamics”.

6 Interim Remarks

We pause here to make a few remarks concerning the nature of our resultsand possible extensions to them:

1. Our discussion has been pretty general, up to assumptions of sphericalsymmetry, and of the force falling off rapidly enough with distance.

2. In particular, the results are valid whether the particles are relativis-tic or non-relativistic. We nowhere made any assumptions concerningspeed, only using general properties of waves and quantum mechanics.Any “wave equation” suffices, since we only use it in the discussion ofasymptotic states.

3. The discussion also applies to the elastic collision of two particles intheir center-of-mass system. Note that this system has spherical sym-metry in this frame. Relativity is unimportant here as well (thoughwe may need it to transform to a different frame of reference). Wemay also consider inelastic scattering in the CM frame, as long as weproperly formulate our conservation of energy.

15

4. We may further extend the discussion to the elastic scattering of twoparticles with any spins in the CM frame. The essential change isthat the asymptotic wave function for particles of spin s are (2s + 1)-component momentum space wave functions, and the scattering am-plitude f(pf ,pi) 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 dis-cussions of the consequence of angular momentum conservation on thedescription of scattering scattering in the helicity basis.

Note that we have not yet said much about how to determine the phaseshifts δj(p) in a problem of interest – we have only shown that they arethe key ingredients in the scattering problem. Our discussion will now turntowards this issue.

7 Resonances

Consider the partial wave expansion:

f(p,q) = f(p; cos θ) =1

2ip

∞∑

j=0

2j + 1

4π

[e2iδj(p) − 1

]Pj(cos θ). (89)

We note that1

2i

[e2iδj (p) − 1

]=

1

cot δj − i. (90)

This function has a maximum magnitude of one whenever

δj =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 = E0.

Then cot δj(E) (written as a function of energy) vanishes at E = E0. Todescribe the scattering in the neighborhood of this resonance, we make anexpansion of cot δj(E) to linear order:

cot δj(E) = − 2

Γ(E − E0) +O

[(E − E0)

2]. (92)

Thus, in the neighborhood of a resonance, the contribution of this channel

16

-5

-3

-1

1

3

5

0 0.2 0.4 0.6 0.8 1

delta_j(E)/pi

c

o

t

(

d

e

l

t

a

_

j

)

Figure 3: Graph of cot δj(E), showing linear behavior near π/2, with negativeslope.

to the scattering amplitude is:

fj(p; cos θ) =2j + 1

p

1

− 2Γ(E − E0) − i

Pj(cos θ) (93)

= (2j + 1)−Γ/2p

E − E0 + iΓ/2Pj(cos θ). (94)

The contribution to the total cross section from this channel alone is (usingthe optical theorem):

σTj(p) =4π

p=fj(p; 1) (95)

= (2j + 1)4π

p= −Γ/2p

E − E0 + iΓ/2(96)

= −4π(2j + 1)

p2

Γ

2= E − E0 − iΓ/2

(E − E0)2 + Γ2/4(97)

=4π

p2(2j + 1)

Γ2/4

(E − E0)2 + Γ2/4. (98)

We arrive at a Breit-Wigner form for the cross section in the neighborhoodof a resonance. The choice of parameterization is such that Γ is the full

17

width at half maximum of the Breit-Wigner distribution. The cross sectiondecreases from its peak value by 1/2 at E = E0 ± Γ/2.

Note, finally, that the maximum contribution to the cross section in par-tial wave j is just:

σTj(E)max = σTj(E0) =4π

p2(2j + 1). (99)

This maximum follows from the unitarity bound:∣∣∣∣1

2i

[e2iδj − 1

]∣∣∣∣ =∣∣∣eiδj sin δj

∣∣∣ ≤ 1. (100)

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 morerapidly in the region of the potential than it would if the potential were notpresent. A wave starting at phase zero at the origin will accumulate phaseas it propagates out to large distances. The phase is accumulated fasterin 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 thewave in no potential at large distances (and once outside the region of thepotential, this shift is independent of distance). Fig. 4 provides an illustrationof this effect. Asymptotically, the scattered wave will be positively phaseshifted with respect to an unscattered wave. Similarly, a dominantly repulsivepotential will yield negative phase shifts.

To see how we may get at the phase shifts, let us consider plane wave solu-tions, and their partial wave expansion. The force-free Schrodinger equationis

∇2ψ + k2ψ = 0, (E = k2/2m). (101)

The plane wave solution for a wave traveling in the +z direction (if k > 0)is:

ψ(x) = eikz = eikr cos θ. (102)

The separation of variables expansion of such a function in spherical polarcoordinates is:

ψ(x) =∞∑

`=0

∑

m=−`

[A`mj`(kr) +B`mn`(kr)]Pml (θ)eimφ, (103)

18

V(r)

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 larger the accumulated phase (starting from r = 0) of the wave in the potentialis larger than the accumulated phase of the wave without the potential.

where

j`(x) =

√π

2xJ`+ 1

2(x); n`(x) = (−)`+1

√π

2xJ−`− 1

2(x) (104)

are spherical Bessel functions.Since eikr 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`(kr) is not

19

regular at r = 0,2 so B`m = 0. Hence,

eikr cos θ =∞∑

`=0

A`j`(kr)Pl(cos θ). (107)

It remains to determine the coefficients A`. We use the orthogonalityrelation ∫ 1

−1dxP`(x)P`′(x) =

2

2`+ 1δ``′ , (108)

to obtain

A`j`(kr)2

2`+ 1=∫ 1

−1dxeikrxP`(x). (109)

We can match the two sides by considering large kr. For the left hand side,

j`(x) −→x→∞1

xsin

(x− `π

2

). (110)

On the right side, we note that by partial integrations we may obtain anexpansion in powers of 1/kr, e.g., after two such integrations:∫ 1

−1dxeikrxP`(x) =

1

ikreikrxP`(x)|1−1 (111)

− 1

ikr

[1

ikreikrxP ′

`(x)|1−1 −1

ikr

∫ 1

−1dxeikrxP ′′

` (x)].

It appears (left to the reader to prove) that the first term is dominant forlarge kr. Using also P`(±1) = (±)`, we thus have:

∫ 1

−1dxeikrxP`(x) →kr→∞

1

ikr

[eikr − (−)`e−ikr

]. (112)

So, we must compare:

1

ikr

[eikr − (−)`e−ikr

]= A`

1

kr

2

2`+ 1sin

(kr − `π

2

)(113)

= A`1

ikr

1

2`+ 1e−iπ`/2

[eikr − (−)`e−ikr

]. (114)

2For small x:

n`(x) →x→0

− 1

x , ` = 0− (2`−1)!!

x`+1 , ` > 0;(105)

j`(x) →x→0x`

(2`+ 1)!!. (106)

20

Solving for A` = (i)`(2`+ 1), and therefore

eikr cos θ =∞∑

`=0

(i)`(2`+ 1)j`(kr)Pl(cos θ). (115)

Now suppose we have a potential V (r) which vanishes everywhere outsidea sphere of radius R. The scattering amplitude, in terms of the phase shifts,as we have seen, is:

f(k; θ) =1

2ik

∞∑

`=0

(2`+ 1)(e2iδ` − 1

)P`(cos θ). (116)

Let us relate this to the scattered wave function. For r < R, we may write:

ψ =∞∑

`=0

(i)`(2`+ 1)R`(k; r)P`(cos θ), (117)

where χ` = rR`, and

χ′′` +

[k2 − `(`+ 1)

r2− 2mV (r)

]χ` = 0. (118)

Outside the sphere r = R we may write:

ψ =∞∑

`=0

(i)`(2`+ 1)[j`(kr) +

1

2α`h

(1)` (kr)

]P`(cos θ), (119)

where h(1)` is the spherical Hankel function of the first kind:

h(1)` (kr) = j`(kr) + in`(kr) (120)

→kr→∞1

(i)`+1

eikr

kr. (121)

The asymptotic form shows that this corresponds to an asymptotically out-going spherical wave. The spherical Hankel function of the second kind, h

(2)` ,

would correspond to an asymptotically incoming spherical wave. We don’tinclude such a term, because we have already included the incoming planewave in the j`(kr) term.

21

The asymptotic behavior of the wave function is therefore:

ψ ∼∞∑

`=0

(2`+ 1)i`

sin

(kr − `π

2

)

kr+α`

2

1

(i)`+1

eikr

kr

P`(cos θ) (122)

∼ 1

2ikr

∞∑

`=0

(2`+ 1)[eikr(1 + α`) − (−)`e−ikr

]P`(cos θ). (123)

This is now expressed in terms of an incoming spherical wave and an outgoingspherical wave.

For elastic scattering, we must have conservation of the number of parti-cles (unitarity), so

|1 + α`|2 = |(−)`|2, (124)

where the left hand side corresponds to the outgoing wave, and the righthand side is the incoming. This is satisfied if α` is of the form:

α` = e2iδ` − 1, (125)

where δ` is real.Now consider the scattered wave function in the asymptotic limit:

ψS = ψ − eikr cos θ (126)

=1

2ikr

∞∑

`=0

(2`+ 1)[(1 + α`)e

ikr − (−)`e−ikr]P`(cos θ)

−∞∑

`=0

(2`+ 1)(i)` 1

krsin

(kr − `π

2

)P`(cos θ)

=1

2ikr

∞∑

`=0

(2`+ 1)[(1 + α`)e

ikr − (−)`e−ikr − eikr + (−)`e−ikr]P`(cos θ)

(127)

=1

2ikr

∞∑

`=0

(2`+ 1)α`eikrP`(cos θ) (128)

= f(k; θ)eikr

r. (129)

We see that the scattering amplitude has the interpretation as the coefficientof the outgoing (scattered) spherical wave.

22

Notice that the outgoing part of the wave in Eqn. 127 is

[eikr cos θ

]out

∼ 1

2ikr

∞∑

`=0

(2`+ 1)eikrP`(cos θ), (130)

while the outgoing part of the actual wave (which includes in addition theoutgoing part of the unscattered wave) is

ψout ∼kr→∞1

2ikr

∞∑

`=0

(2`+ 1)e2iδ`eikrP`(cos θ), (131)

where e2iδ` = 1 + α`. Comparing Eqns. 130 and 131, we see that 2δ` is thedifference in phase between the outgoing parts of the actual wave functionand the eikr cos θ plane wave.

It is perhaps useful to make a comment here concering interference amongpartial 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 distinguishwhich partial angular momentum states contribute to the scattering in agiven direction. On the other hand, in the total cross section:

σT =∫

(4π)dΩ|f(k; θ)|2 =

4π

k2

∞∑

`=0

(2`+ 1) sin2 δ`; (133)

by virtue of the orthogonality of the P`(cos θ), there is no interference amongthe partial waves. The cross section decomposes into a sum of partial crosssections 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′′` +

[k2 − `(`+ 1)

r2− 2mV (r)

]χ` = 0, χ` = rR`. (134)

23

We must have continuity of the wave function and its derivative on the spherer = R. Hence,

R`(k, R) = j`(kR) +1

2α`h

(1)` (kR) (135)

dR`(k, r)

dr

∣∣∣∣∣r=R

= k[j ′`(kR) +

1

2α`h

(1)′` (kR)

]. (136)

The factor of k on the right side of Eqn: 136 is because the prime notationdenotes derivatives with respect to the argument of the function, i.e., withrespect to kr.

We divide Eqn. 136 by Eqn. 135, and let x = kR:

dR`(k,r)dr

R`(k, r)

∣∣∣∣∣r=R

= k

[j ′`(x) + 1

2α`h

(1)′` (x)

]

j`(x) + 12α`h

(1)` (x)

. (137)

With dr = rd log r, we may write this in the form:

L` ≡d logR`(k, r)

d log r

∣∣∣∣∣r=R

= x

[j ′`(x) + 1

2α`h

(1)′` (x)

]

j`(x) + 12α`h

(1)` (x)

. (138)

Solving for α` = e2iδ` − 1:

α` = −2L`j`(x) − xj ′`(x)

L`h(1)` (x) − xh

(1)′`

, (139)

where x = kR. Thus, if the logarithmic derivative L` is determined, then thephase shift is known.

8.2 Low Energy Limit

Let us use the result just obtained to demonstrate that the partial waveexpansion normally converges faster in the low energy limit, x = kR 1.We’ll start with the power series expansions for the spherical Bessel/Hankelfunctions:

h(1)` (x) = j`(x) + in`(x), (140)

j`(x) = 2`x`∞∑

n=0

(−)n(n+ `)!

n!(2n+ 2`+ 1)!x2n, (141)

n`(x) = (−)`+1 1

2`x`+1

∞∑

n=0

(−)n−`(n− `)!

n!(2n− 2`)!x2n. (142)

24

As a technical aside, we see that we really need to make sense of the lastformula above in the case ` > n. Guided by the identity:

z!(−z)! =πz

sin πz, (143)

we define(n− `)!

(2n− 2`)!= (−)n−` (2`− 2n)!

(`− n)!, if n < `. (144)

For x 1 (the low energy limit), we keep only the n = 0 term in theseexpansions:

j`(x) =`!

(2`+ 1)!2`x` +O(x`+1) (145)

n`(x) =(−)`+1

2`x`+1

(−)−`(2`)!

`!+O(x−`+1). (146)

Thus,

L`j`(x) − xj ′`(x) = 2` `!

(2`+ 1)!(L` − `)x` +O(x`+2) (147)

L`h(1)` (x) − xh

(1)′` = i [L`n`(x) − xn′

`(x)] +O(x`) (148)

= −i(2`)!2``!

(L` + `+ 1)1

x`+1+O(x`, x−`+1). (149)

Therefore,

α` ≈ − 2i

2` + 1

[2``!

(2`)!

]2L` − `

L` + `+ 1x2`+1. (150)

Let us use this result to check the convergence of the partial wave ex-pansion for small x. The coefficient of the P`+1(cos θ) term divided by thecoefficient of the P`(cos θ) term is:

(2`+ 3)α`+1

(2`+ 1)α`

=1

(2`+ 1)2

L`+1 − `− 1

L`+1 + `+ 2

L` + `+ 1

L` − `x2. (151)

In general then, we should have very good convergence for x 1.

In the low energy limit the `th scattering partial wave is proportional to:

α` ∝ k2`+1. (152)

25

That is,α` = e2iδ` − 1 = 2i sin δ`e

iδ` , (153)

or| sin δ`| ∼ k2`+1, as kR → 0. (154)

Using

j`(x) =1

2

[h

(1)` (x) + h

(2)` (x)

], (155)

we may also write:

1 + α` = −L`h(2)` (x) − xh

(2)′` (x)

L`h(1)` (x) − xh

(1)′` (x)

. (156)

We may obtain exact expressions for the two lowest phase shifts with,

h(2)` (x) = h

(1)∗` (x), for real x, (157)

h(1)0 (x) = − i

xeix, (158)

h(1)1 (x) = −1

xeix(1 +

i

x

). (159)

Hence,

1 + α0 = e−2ix 1 + ix/L0

1 − ix/L0

(160)

1 + α1 = e−2ix 1 + ix− x2/(L1 + 1)

1 − ix− x2/(L1 + 1), (161)

where x = kR and

L` ≡ d logχ`

d log r

∣∣∣∣∣r=R

, (162)

=d logR`

d log r

∣∣∣∣∣r=R

+ 1. (163)

8.3 Example

Suppose we scatter from the potential (see Fig. 5):

V (r) =V0 > 0 r < R,0, r ≥ R.

(164)

26

V(r)

V

Rr

0

Figure 5: Graph of the example scattering potential.

We wish to determine, for example, the phase shift and partial crosssection for ` = 0 (S-wave) scattering. Since the potential is repulsive, weshould find δ0 < 0. Let

k2 = 2mE (165)

k20 = 2mV0 (166)

∆ =√k2

0 − k2. (167)

The radial wave equation for r < R is

χ′′` +

[k2 − `(`+ 1)

r2− k2

0

]χ` = 0. (168)

In particular, for ` = 0:3

χ′′0 + (k2 − k2

0)χ0 = 0. (169)

3Recall that we may express the wave function for the spherically-symmetric centralforce problem in general in the form

ψ(x) =∑

n,`,m

An`mRn`(r)Y`m(θ, φ),

27

If E < V0, then ∆ > 0, and χ′′0 = ∆2χ0, or

χ0 = A sinh ∆r, r < R, E < V0. (170)

If instead E > V0, then ∆ = i∆′ is purely imaginary, hence

χ0 = A′ sin ∆′r, r < R, E > V0. (171)

Outside the sphere bounded by r = R, we have χ′′0 = −k2χ0, hence

χ0(r) = sin(kr + δ), r > R (172)

where we have chosen an arbitrary normalization. We compare this with ourexpansion:

ψ(x) =∞∑

`=0

i`(2`+ 1)[j`(kr) +

1

2α`H

(1)` (kr)

]P`(cos θ), r > R. (173)

Consider the ` = 0 term in particular:

χ0(r) = r[j0(kr) +

1

2α0h

(1)0 (kr)

]. (174)

Using, j0(x) = sin x/x, h(1)0 (x) = −ieix/x, and α0 = e2iδ0 − 1, we obtain (not

worrying about the overall normalization),

χ0(r) = sin kr − i

2α0e

ikr (175)

=1

2i

(eikr − e−ikr + eikr+2iδ0 − eikr

)(176)

= eiδ0 sin(kr + δ0), (177)

or, simply χ0(r) = sin(kr + δ0), absorbing the phase factor into the normal-iztion. Comparing with Eqn. 172, we find δ = δ0.

where1

2mr∂2

∂r2[rRn`(r)] +

[`(`+ 1)2mr2

− V (r)]Rn`(r) = EnRn`(r)

(n may be a continuous index in general), and we have defined χn`(r) ≡ rRn`(r). At leastas long as V (r) is finite as r → 0, Rn`(0) must be finite, hence the boundary conditionχn`(0) = 0.

28

We determine δ0 by matching χ0 and χ′0 at r = R. The logarithmic

derivative L0 = Rχ′0(R)/χ0(R) contains the information about the parameter

of interest, δ0, and we need never determine A (or A′). For E < V0:

L0 = ∆Rcosh ∆R

sinh ∆R= ∆R coth ∆R (178)

= kRcos(kR + δ0)

sin(kR + δ0)= kR cot(kR + δ0). (179)

That is,L0 = kR cot(kR + δ0) = ∆R coth(∆R), E < V0. (180)

Similarly, for E > V0:

L0 = kR cot(kR + δ0) = ∆′R cot(∆′R), E > V0. (181)

Given V0 and R, we can thus solve these equations for δ0(E).Noting our earlier result:

1 + α0 = e2iδ0 = e−2ikR 1 + ikR/L0

1 − ikR/L0

, (182)

we have, for E < V0:

e2iδ0 = e−2ikR 1 + i(k/∆) tanh ∆R

1 − i(k/∆) tanh∆R, E < V0. (183)

This is of the form e−2ikR(

1+ia1−ia

)= e−2ikReiθ, where θ = tan−1[2a/(1 − a2)].

Therefore, we may write:

δ0 = −kR +1

2tan−1

2 k∆

tanh ∆R

1 −(

k∆

)2tanh2 ∆R

, E < V0, (184)

δ0 = −kR +1

2tan−1

2 k∆′ tan∆′R

1 −(

k∆′

)2tan2 ∆′R

, E > V0. (185)

Consider the low energy limit, k → 0 (∆ → k0), E < V0:

δ0 → −kR +1

2tan−1

2 kk0

tanh ∆R

1 −(

kk0

)2tanh2 ∆R

(186)

= −kR +k

k0tanh k0R +O

(k

k0

)2 . (187)

29

Alternatively, we may consider the “hard sphere” limit, k0 → ∞ (i.e., V0 →∞). This is nearly the same limit, for any fixed energy. To lowest order ink/k0,

δ0 = −kR +O(k/k0). (188)

Thus, for the hard sphere the lowest partial wave phase shift depends linearlyon k, as shown in Fig. 6.

0 kR

0δ

Figure 6: The S wave phase shift for the hard sphere potential.

Consider now the total cross section in the ` = 0 channel:

σ0 =4π

k2sin2 δ0. (189)

In the low energy limit, this is

σ0 → 4π

k2sin2

(−kR +

k

k0

tanh k0R

)(190)

→ 4π

k2

(−kR +

k

k0tanh k0R

)2

(191)

30

= 4πR2

(1 − tanh k0R

k0R

)2

. (192)

In the hard sphere limit (k0 → ∞), this becomes

σ0 = 4πR2. (193)

In the low energy limit, this partial wave dominates, so this is the totalcross section as well. This result is larger than the result we might expectpurely 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 thephenomenon of diffraction.

We note in passing that when writing

σ0 = 4πa20, a0 = R

(1 − tanh k0R

k0R

), (194)

a0 is referred to as the “scattering length” of the potential.Now let us consider the case of finite V0 (a “soft sphere”?). For example,

in the high energy limit, k k0, and ∆′ → k, hence:

δ0 → −kR +1

2tan−1 2 tan kR

1 − tan2 kR(195)

= 0. (196)

This should be the expected result in the high energy limit, since the effectof a finite fixed potential becomes negligible as the energy increases.

Thus, at very low energies and at very high energies, the phase shiftapproaches 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 spherepotential.

9 The Born Expansion, Born Approximation

We have mentioned the Born expansion in our note on resolvents and Green’sfunctions, and the Born approximation in our note on approximate methods.We now revisit it explicitly in the context of scattering. We start by de-veloping the approach in general, and then apply it to the example of thepreceding section.

31

05 10 15 20

kR

-0.5

-1.0

-1.5

-2.0

-2.5

-3.0δ0

Figure 7: The S wave phase shift for the soft sphere potential, with R = 1,k0R = 4.

In the resolvent note, we saw the Born expansion as the Liouville-Neumannexpansion of a perturbed Green’s function. We will see this again, exceptthat it is convenient (eliminates a minus sign) here to redefine the Green’sfunction with the opposite sign from the resolvent note. Thus, we here take

G(z) = − 1

H − z. (197)

The Schrodinger equation we wish to solve is of the form:(−∇2

2m+ V

)ψ = Eψ. (198)

It is convenient to let E = k2/2m and V = U/2m, so that this equation isof the form: (

∇2 + k2)ψ(x) = U(x)ψ(x). (199)

We already found the Green’s function for the free particle wave equation(Helmholtz equation), (∇2 + k2)ψ = 0 in the resolvent note:

G(x,y; k) = − e±ik|x−y|

4π|x − y|. (200)

32

0

1

2

3

4

5

6

7

8

0 202 4 6 8 10 12 14 16 18

kR

0σ

Figure 8: The S wave cross section for the soft sphere potential, with R = 1,k0R = 4.

The solution to the desired Schrodinger equation is then

ψ(x) = ψ0(x) +∫

(∞)d3(x′)G(x,x′; k)U(x′ψ(x′), (201)

where ψ0(x) is a solution to the Helmholtz equation. The reader is encour-aged to verify that ψ is indeed a solution, by substituting into the Schrodingerequation.4

Now we may apply this to the scattering of an incident plane wave, ψ0 =eik·x, using the “outgoing” Green’s function,

G+(x,y; k) = − e+ik|x−y|

4π|x − y| , (202)

to obtain the scattered wave:

ψS(x) ≡ ψ(x) − ψ0(x) (203)

=∫

(∞)d3(x′)G+(x,x′; k)U(x′ψ(x′). (204)

4Recall that (∇2x + k2)G(x,x′; k) = δ(3)(x − x′).

33

Since

ψS(x) = f(k; θ)eikr

r, (205)

where θ is the polar angle to the observation point, hence equal to the po-lar angle of the scattered k-vector, we can extract the scattering amplitudef(k; θ) by considering the asymptotic situation, where the source “size” (non-negligible region of U(x)) is small:

G+(x,x′; k) →|x|→∞ − 1

4π

eik|x|

|x|e−ik′·x′

. (206)

In obtaining this expression, we have held x′ finite, and used

|x− x′| ≈ |x| − x · x′

x. (207)

Thus,

f(k; θ) = − 1

4π

∫

(∞)d3(x′)e−ik′·x′

U(x′)ψ(x′). (208)

Let us return to the general integral equation for the wave function:

ψ(x) = ψ0(x) +∫

(∞)d3(x′)G(x,x′; k)U(x′)ψ(x′). (209)

We plug this expression into the integrand, obtaining:

ψ(x) = ψ0(x) +∫

(∞)d3(x)G(x,x′; k)U(x′)ψ0(x

′) (210)

+∫

(∞)d3(x′)d3(x′′)G(x,x′; k)U(x′)G(x′,x′′; k)U(x′′)ψ(x′′).

Iteration gives the Neumann series:

ψ(x) = ψ0(x) +∞∑

n=1

∫

(∞)d3(x′)d3(xn′)G(x,x′; k)U(x′) (211)

G(x′,x′′; k)U(x′′) . . . G(xn−1′,xn′; k)U(xn′)ψ0(xn′). (212)

Typically, we use this by substituting a plane wave for ψ0(x). If thepotential is “weak”, we expect this series to converge quickly. This expansionis sometimes called the Born expansion. To first order in U(x) we have:

ψ(x) = ψ0(x) +∫

(∞)d3(x′)G(x,x′; k)U(x′)ψ0(x

′) +O(U2). (213)

34

With ψ0(x) = eik·x, and

G(x,x′; k) ∼ − 1

4π

eikr

re−ik′·x′

, (214)

we find:

ψ(x) ∼ eik·x − 1

4π

eikr

r

∫

(∞)d3(x′)e−ik′·x′

U(x′)eik·x′. (215)

The interpretation of k is as the momentum vector of the incident planewave, and of k′ is as the momentum vector of the scattered wave. Hence,

ψS(x) ≈ − 1

4π

eikr

r

∫

(∞)d3(x′)ei(k−k′)·x′

U(x′), (216)

or

f(k; Ω′) ≈ − 1

4π

∫

(∞)d3(x′)ei(k−k′)·x′

U(x′), (217)

where Ω′ is the direction of the scattered wave, i.e., the direction of k′. Thisresult is known as the Born approximation.

With this result, we have the differential cross section:

dσ

dΩ′ = |f(k; Ω′)|2 (218)

=1

16π2

∣∣∣∣∣

∫

(∞)d3(x′)ei(k−k′)·x′

U(x′)

∣∣∣∣∣

2

(219)

=m2

4π2

∣∣∣V (k′ − k)∣∣∣2, (220)

where we have used U(x) = 2mV (x), and V is the Fourier transform of V .Recall that we already derived this result in our discussion of time-dependentperturbation theory (Approximate Methods course note).

If U(x) = U(r) only, i.e., we have a spherically symmetric scatteringpotential, we may incorporate this into our result. Let θ be the “scatteringangle”, i.e., the angle between k and k′. Then, noting that |k| = |k′| = k,we have |k − k′| = 2k sin θ

2. Since U(x) = U(r), f(k; Ω′) = f(k, θ) only, and

f(k; θ) = − 1

4π

∫ ∞

0r′2dr′U(r′)

∫

4πd cos θ′dφ′ei2k(sin θ

2)r′ cos θ′, (221)

35

where θ′ is the angle between x′ and k−k′. Now let K = |k−k′| = 2k sin θ2.

We can do the integrals over angles, to obtain:

f(k; θ) = −∫ ∞

0r′2dr′U(r′)

sinKr′

Kr′. (222)

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 shiftsin a partial wave expansion. Start with our earlier (exact asymptotic) result:

ψ(x) = ψ0(x) + ψS(x) (223)

= ψ0(x) +eikr

rf(k; θ), (224)

where

f(k; θ) = − 1

4π

∫

(∞)d3(x′)e−ik′·x′

U(x′)ψ(x′). (225)

The expansion of ψ(x) in partial waves is

ψ(x) =∞∑

`=0

i`(2`+ 1)R`(r)P`(cos θ). (226)

We insert this into the expression for f :

f(k; θ) = − 1

4π

∞∑

`=0

i`(2`+1)∫ ∞

0r′2dr′R`(r

′)∫ 1

−1d cos θ′

∫ 2π

0dφ′e−ik′·x′

U(x′)P`(cos θ′).

(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., theangle between k and k′. If k is along the z axis, then θ is the polar angle ofk′. Thus, k′ is not necessarily along the z-axis, and we have:

k′ · x′ = kr′ [cos θ cos θ′ + sin θ sin θ′ cos(φk′ − φx′)] . (228)

Now consider our partial wave expansion of a plane wave:

eik·x = eikr cos θkx =∞∑

`=0

i`(2`+ 1)j`(kr)P`(cos θkx) (229)

=∞∑

`=0

i`√

4π(2`+ 1)j`(kr)Y`0(θkx, φkx), (230)

36

where θkx is the angle between k and x, and we have used the identity

P`(cos θ) =

√4π

2`+ 1Y`0(θ, φ). (231)

We may re-express this in terms of the polar angles (θx, φx) of x and (θk, φk)of k using the addition theorem for spherical harmonics:

√4π(2`+ 1)Y`0(θkx, φkx) =

∑

m=−`

Y ∗`m(θk, φk)Y`m(θx, φx), (232)

wherecos θkx = cos θk cos θx + sin θk sin θx cos(φk − φx). (233)

Thus,

eik·x = 4π∞∑

`=0

i`j`(kr)∑

m=−`

Y ∗`m(θk, φk)Y`m(θx, φx), (234)

and hence,

f(k; θ) = − 1

4π

∞∑

`=0

i`(2`+ 1)∫ ∞

0r′2dr′R`(r

′) (235)

4π∞∑

`′=0

i`′j∗`′(k

′r′)U(r′) (236)

∫ 1

−1d cos θ′

∫ 2π

0dφ′ ∑

m=−`

Y`′m(θ, φk′)Y ∗`′m(θ′, φ′)P`(cos θ′).(237)

We may simplify this in several ways:

• j`(x) is real if x is real.

• We may relabel integration variable r′ as r. Also, k′ is equal to k.

• The only dependence on φ′ is in Y ∗`′m(θ′, φ′), hence

∫ 2π

0dφ′Y ∗

`′m(θ′, φ′) = 2πδm0Y∗`′0(θ

′, 0) (238)

= 2πδm0

√2`′

4πP`′(cos θ′). (239)

37

• The integral over cos θ′ may then be accomplished with:

∫ 1

−1dxP`(x)P`′(x) =

2

2`+ 1δ``′ . (240)

• Finally, we may use

Y`0(θ, 0) =

√2`+ 1

4πP`(cos θ). (241)

Including all of these points leads to:

f(k; θ) = −∞∑

`=0

(2`+ 1)P`(cos θ)∫ ∞

0r2drR`(r)j`(kr)U(r). (242)

So far, we have made no approximation – this result is “exact” as longas the potential falls off rapidly enough as r → ∞. However, it depends onknowing R`(r), and this is where we shall now make an approximation. Notefirst, that by comparison with the expansion in phase shifts:

f(k; θ) =1

2ik

∞∑

`=0

(2`+ 1)(e2iδ` − 1

)P`(cos θ), (243)

we obtain:

1

2i

(e2iδ` − 1

)= eiδ` sin δ` = −k

∫ ∞

0r2drR`(r)j`(kr)U(r). (244)

The Born approximation consists in approximating the wave function by itsvalue for no potential, which should be a good approximation as long as thepotential has a small effect on it. Here, this means we replace R`(r) withj`(kr), valid as long as R`(r) ≈ j`(kr):

eiδ` sin δ` ≈ −k∫ ∞

0r2dr [j`(kr)]

2 U(r). (245)

Note that the integral is real in this approximation. In this approximation,δ` is small, hence we have the Born approximation for the phase shifts:

δ` ≈ −k∫ ∞

0r2dr [j`(kr)]

2 U(r). (246)

38

9.2 Born Approximation and Example of the “SoftSphere” 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 k20 = 2mV0. Then,

δ` ≈ −kk20

∫ R

0r2dr [j`(kr)]

2 (248)

= −(k0

k

)2 ∫ kR

0[xj`(x)]

2 dx. (249)

For instance, consider δ0, with j0(x) = sin xx

:

δ0 ≈ −(k0

k

)2 ∫ kR

0sin2 xdx (250)

= −(k0

k

)21

4(2kR− sin 2kR). (251)

Note that the Born approximation is a “high energy” approximation –there must be little effect on the scattering wave. Consider what happens atlow energy, kR 1:

δ0 = −(k0

k

)21

4(2kR− sin 2kR) ≈ −kR(k0R)3

3. (252)

This may be compared with our earlier low energy result:

δ0 ≈ −kR(

1 − tanh k0R

k0R

). (253)

Other than the linear dependence, they are not very similar. Fig. 9 illustratesthis. At low kR, the agreement with the exact calculation is poor, as theenergy increases, the agreement improves.

39

2 4 6 8 10 12 14 16 18 200

-1

-2

-3

-4

-5

-6

kR

0δ

Figure 9: The S wave phase shift for the soft sphere potential, with R = 1,k0R = 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 momentumconservation. As a consequence, the angular distribution of particles in ascattering process is constrained. Here, we consider how this may be appliedin the case of a scattering reaction of the form:

a+ b→ c+ d. (254)

These particles may carry spins – let ja denote the spin of particle a, etc.It is convenient to work in the center-of-mass frame, and to pick particlea to be incoming along the z axis. If a different frame is desired, then itmay be reached from the center-of-mass frame with a Gallilean or Lorentztransformation, whichever is appropriate.

It is furthermore convenient to work in a helicity basis for the angularmomentum states. Along the helicity axis, the component of orbital angularmomentum of a particle is zero. We’ll typically label the helicity of a particlewith the symbol λ. For example, the helicity of particle a is denoted λa.

40

Let us also work with a plane wave basis for our particles. Physicalstates may be constructed as a superposition of such states. We’ll omit themagnitude of momentum in our labelling, and concentrate on the angles(polar angles denoted by θ, and azimuth angles denoted by φ). We’ll alsoomit the spins of the scattering particles in the labels, to keep things compact.Thus, we label the initial state according to:

|i〉 = |θa = 0, φa, λa, λb〉, (255)

and the final state by,|f〉 = |θc, φc, λc, λd〉, (256)

Note that θd = π− θc and φd = π+φc in the center-of-mass. This basis maybe called the “plane wave helicity basis”.

We may write the transition amplitude from the initial to final state inthe form:

〈f |T |i〉 ∝ 〈θc, φc, λc, λd|T |0, φa, λa, λb〉, (257)

and the scattering cross section is:

dσλaλbλcλd(θ ≡ θc, φ ≡ φc) ∝ |〈θ, φ, λc, λd|T |0, φa, λa, λb〉|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

λc and λd. Similarly, if the initial beams (a and b) are unpolarized, we mustaverage over all possible λa and λb. Thus, to get the scattering cross sectionin such a case, we must evaluate

1

2ja + 1

∑

λa

1

2jb + 1

∑

λb

∑

λc

∑

λd

|〈θ, φ, λc, λd|T |0, φa, λa, λb〉|2. (259)

If a(or b) is massless and not spinless, then the 1/(2ja + 1) factor must bereplaced by 1/2.

The scattering process may go via intermediate states of definite angularmomentum, for example, scattering through an intermediate resonance. Forexample. in e+e− scattering just below µ+µ− threshold, we may expecta resonance due to the presence of “muonium” bound states. Such stateshave definite angular momenta. In particular, the resonance for scatteringthrough the lowest 3S1 bound state of muonium might be expected to be alarge contribution in the scattering amplitude. Thus, we are led to describing

41

the transition amplitude most naturally in a basis which specifies the totalangular momentum state. We may label a basis state as:

|j,m, λa, λb〉, or |j,m, λc, λd〉, (260)

where j specifies the total angular momentum, and m is its component alonga chosen quantization axis. This basis is sometimes called the “sphericalhelicity basis”.

For a given pair of initial and final helicity states, we are thus interestedin computing matrix elements of the form:

〈f |T |i〉 ∝∑

j,m

〈θc, φc, λc, λd|j,m, λc, λd〉〈j,m, λc, λd|T∑

j′,m′|j ′, m′, λa, λb〉〈j ′, m′, λa, λb|0, φa, λa, λb〉.

(261)The terms of the form 〈θ, φ, λ1, λ2|j,m, λ′1, λ′2〉 are just scalar products ofvectors expressed in different bases. We thus need to learn how to do thebasis transformation.

We may write the basis transformation in the form:

|θ, φ, λ1, λ2〉 =∑

j,m,λ′1,λ′

2

|j,m, λ′1, λ′2〉〈j,m, λ′1, λ′2|θ, φ, λ1, λ2〉 (262)

=∑

j,m

|j,m, λ1, λ2〉〈j,m, λ1, λ2|θ, φ, λ1, λ2〉, (263)

where we have used the fact that states with different helicities are orthogo-nal:

〈j,m, λ′1, λ′2|θ, φ, λ1, λ2〉 ∝ δλ′1λ1δλ′

2λ2. (264)

Consider now the state

|θ = 0, φ, λ1, λ2〉 =∑

j,m

|j,m, λ1, λ2〉〈j,m, λ1, λ2|θ = 0, φ, λ1, λ2〉. (265)

For this state, the helicity axis is parallel(or antiparallel) to the quantizationaxis for the third component of angular momentum, and hence λ1 = m1,λ2 = −m2. Thus, m = λ1 − λ2 ≡ α. The coefficient

〈j,m, λ′1, λ′2|θ, φ, λ1, λ2〉 = cjδmα, (266)

where cj is a constant depending only on j, to be determined (see exercises).The expansion of this state in the spherical helicity basis is thus:

|θ = 0, φ, λ1, λ2〉 =∑

j,m

cj|j, α, λ1, λ2〉. (267)

42

Let us now perform a rotation on our special state to an arbitrary state:

|θ, φ, λ1, λ2〉 = R3(φ)R2(θ)R3(−φ)|θ = 0, φ, λ1, λ2〉. (268)

This product of rotations on a state of angular momentum j is given by theEuler angle parameterization of the Dj rotation matirces:

R3(φ)R2(θ)R3(−φ) = Dj(φ, θ,−φ). (269)

The matrix elements for rotation u are:

Djm1m2

(u) = 〈j,m1|Dj(u)|j,m2〉. (270)

Hence,

Dj(u)|j, α, λ1, λ2〉 =∑

m

|j,m, λ1, λ2〉〈j,m, λ1, λ2|Dj(u)|j, α, λ1, λ2〉, (271)

andketθ, φ, λ1, λ2 =

∑

j,m

|j,m, λ1, λ2〉Djmα(φ, θ,−φ). (272)

If we anticipate the result of the exercises for cj, namely cj =√

2j + 1/4π,we have the desired result:

〈θ, φ, λ1, λ2|j,m, λ′1, λ′2〉 =

√2j + 1

4πDj∗

mα(φ, θ,−φ)δλ′1λ1δλ′

2λ2, (273)

where α ≡ λ1 − λ2. Putting this into our expression for the transitionamplitude, we obtain (defining αf ≡ λc − λd and αi ≡ λa − λb):

〈f |T |i〉 ∝∑

j,m

∑

j′,m′

√2j + 1

4π

√2j ′ + 1

4πDj∗

mαf(φ, θ,−φ)Dj′

m′αi(φa, 0,−φa)〈j,m, λc, λd|T |j ′, m′, λa, λb〉.

(274)Suppose the interaction is one which conserves angular momentum. In

this case,

〈j,m, λc, λd|T |j ′, m′, λa, λb〉 = δjj′δmm′〈j,m, λc, λd|T |j,m, λa, λb〉, (275)

and further m = λa−λb ≡ αi. Thus, the non-zero transition matrix elementsare numbers of the form T j

λaλbλcλd, which may be called “helicity amplitudes”.

We have:

〈f |T |i〉 ∝∑

j

2j + 1

4πT j

λaλbλcλdDj∗

αiαf(φ, θ,−φ)Dj

αiαi(φa, 0,−φa). (276)

43

The “big-D” functions give us the angular distribution for each intermediatej value and helicity state. Using

Djm1m2

(φ, θ,−φ) = e−i(m1−m2)φdjm1m2

(θ), (277)

and djmm(0) = 1, we have

〈f |T |i〉 ∝∑

j

2j + 1

4πT j

λaλbλcλdei(αi−αf )φdj

αiαf(θ). (278)

Note that the azimuth dependence is a phase, depending only on the helici-ties.

Squaring, we obtain the scattering angular distribution:

dσλaλbλcλd

dΩ(θ, φ) ∝

∣∣∣∣∣∣∑

j

(2j + 1)T jλaλbλcλd

djαiαf

(θ)

∣∣∣∣∣∣

2

, (279)

where αf ≡ λc − λd and αi ≡ λa − λb.

11 Exercises

1. Show that the total cross section we computed in the partial waveexpansion,

σT (p) =4π

p2

∞∑

j=0

(2j + 1) sin2 δj(p), (280)

is in agreement with the optical theorem.

2. We have discussed the “central force problem”. Consider a particle ofmass m under the influence of the following potential:

V (r) =V0, 0 ≤ r ≤ a0, a < r,

(281)

where V0 is a constant.

(a) Write down the Schrodinger equation for the wave function ψ(xxx).Consider solutions which are simultaneous eigenvectors of H, LLL2,and Lz. Solve the angular dependence, and reduce the remainingproblem to a problem in one variable. [You’ve done this alreadyfirst quarter, so you may simply retrieve that result here.]

44

(b) Let E be the eigenvalue of the Hamiltonian, H. Consider the casewhere E > V0. Solve the Schrodinger equation for eigenstatesψ(xxx). It will probably be convenient to use the quantity k =√

2m(E − V0). Consider the limit as r → ∞ for your solutions,and give an interpretation in terms of spherical waves.

(c) Repeat the solution for the case where E < V0. It will probably be

convenient to use the quantity K =√

2m(V0 − E). Again, con-sider the limit as r → ∞ and give an interpretation, contrastingwith the previous case.

Hint: You will probably benefit by thinking about solutions in theform of spherical Bessel/Neumann functions, and/or spherical Hankelfunctions.

3. When we calculated the density of states for a free particle, we used a“box” of length L (in one dimension), and imposed periodic boundaryconditions to ensure no net flux of particles into or out of the box.We have in mind, of course, that we can eventually let L → ∞, andare really interested in quantities per unit length (or volume). Let usjustify more carefully the use of periodic boundary conditions, i.e., wewish to convince ourselves that the intuitive rationale given above iscorrect. To do this, consider a free particle in a one-dimensional “box”from −L/2 to L/2. Remembering that the Hilbert space of allowedstates is a linear space, show that the periodic boundary condition:

ψ(−L/2) = ψ(L/2),

is required for acceptable wave functions. “Acceptable” here meansthat the probability to find a particle in the box must be constant.

4. In our discussion of scattering theory, we supposed we had a beam ofparticles from some ensemble of wave packets, and obtained an “effec-tive” (observed) differential cross-section:

σeff(u) =∫

α

f(α)dα∫

|x|≤R

d2(x)P (µ;∝;x)

This formula assumed that the beam particles were distributed uni-formly in a disk of radius R centered at the origin in the e1 − e2 plane,

45

and that the distribution of the shape parameter was uncorrelated withposition in this disk.

(a) Try to obtain an expression for σeff(u) without making these as-sumptions.

(b) Using part (a), write down an expression for σeff(u) appropriateto the case where the beam particles are distributed according toa Gaussian of standard deviation ρ in radial distance from theorigin (in the e2 − e3 plane), and where the wave packets are alsodrawn from a Gaussian distribution in the expectation value ofthe magnitude of the momentum. Let the standard deviation ofthis momentum distribution be α = α(x), for beam position x.

(c) For your generalized result of part (a), try to repeat our limitingcase argument to obtain the “fundamental” cross section. Discuss.

5. Let us briefly consider the consequences of reflection invariance (parityconservation) for the scattering of a particle with spin s on a spinlesstarget. [We consider elastic scattering only here]. Thus, assume theinteraction is reflection invariant:

(a) How does the S matrix transform under parity, i.e., what is P−1SP ,where P is the parity operator?

(b) What is the condition on the helicity amplitudes Ajλµ(pi) (corre-

sponding to scattering with total angular momentum j) imposedby parity conservation?

(c) What condition is imposed on the orbital angular momentum am-plitudes Bj

``′(pi)? You may use “physical intuition” if you like,but it should be convincing. In any event, be sure your answermakes intuitive sense.

6. We consider the resonant scattering of light by an atom. In particular,let us consider sodium, with 2P1/2 ↔2 S1/2 resonance at λ = λ0 =5986A. Let σ0T be the total cross section at resonance, for a mono-chromatic light source (i.e., σ0T is the “fundamental” cross section).

(a) Ignoring spin, estimate σ0T , first in terms of λ0/2π, and thennumerically in cm2. Compare your answer with a typical atomicsize.

46

(b) Suppose that we have a sodium lamp source with a line widthgoverned by the mean life of the excited 2P1/2 state (maybe noteasy to get this piece of equipment!). The mean life of this stateis about 10−8 second. Suppose that this light is incident on an ab-sorption cell, containing sodium vapor and an inert (non-resonant)buffer gas. Let the temperature of the gas in the absorption cellbe 200C. Obtain an expression for the effective total cross sec-tion, σeffT which an atom in the cell presents to the incident light.Again, make a numerical calculation in cm2.

(c) Using your result above, find the number density of Na atoms (#of atoms/cm3) which is required in the cell in order that intensityof the incident light is reduced by a factor of two in a distanceof 1 cm. It should be noted (and your answer should be plausiblehere) that such a gas will be essentially completely transparent tolight of other (non-resonant) wavelengths.

7. Consider scattering from the simple potential:

V (x) =V0 r = |x| < R0 r > R.

In the low energy limit, we might only look at S-wave ` = 0 scatter-ing. However, in the high energy limit, we expect scattering in otherpartial waves to become significant. For simplicity, let us here considerscattering on a hard sphere, V0 → ∞.

(a) For a hard sphere potential, calculate the total cross section inpartial wave `. Give the exact result, i.e., don’t take the highenergy limit yet. You may quote your answer in terms of thespherical Bessel functions.

(b) Find a simple expression for the phase shift δ` in the high energylimit (kR `). Keep terms up to O(1) in your result.

(c) Determine the total cross section (including all partial waves) inthe high energy limit, kR → ∞. [This is the only somewhat trickypart of this problem to calculate. One approach is as follows:Write down the total cross section in terms of your results for part(a). Then, for fixed k, consider which values of `may be importantin the sum. Neglect the other values of `, and make the high energy

47

approximation to your part (a) result. Finally, evaluate the sum,either directly, or by turning it into an appropriate integral.]

8. Consider the graph in Fig. 10.

0 50 100 2001500

40

80

120

160

0

1δ

δ

δ

πT (MeV)

PhaseShift(degrees)

Figure 10: Made-up graph of phase shifts δ0 and δ1 for elastic π+p scattering(neglecting spin).

Assume that the other phase shifts are negligible (e.g., “low energy”is reasonably accurate). The pion mass and energy here are suffi-ciently small that we can at least entertain the approximation of aninfinitely heavy proton at rest – we’ll assume this to be the case,in any event. Note that Tπ is the relativistic kinetic energy of the

π+ : Tπ =√P 2

π +m2π −mπ.

(a) Is the π+p force principally attractive or repulsive (as shown inthis figure)?

(b) Plot the total cross section in mb (millibarns) as a function ofenergy, from Tπ=40 to 200 MeV.

48

(c) Plot the angular distribution of the scattered π+ at energies of120, 140 and 160 MeV.

(d) What is the mean free path of 140 MeV pions in a liquid hydrogentarget, with these “protons”?

9. We now start to consider the possibility of “inelastic scatting”. Forexample, let us suppose there is a “multiplet” of N non-identical parti-cles, all of mass m. We consider scattering on a spherically symmetriccenter-of-force, with the properly that the interaction can change aparticle from one number of the multiplet to another member. Wemay in this case express the scattering amplitude by fαβ(k; cos θ), withα, β = 1, . . . , N , corresponding to a unitary S-matrix on Hilbert spaceL2(R

3) ⊗ VN :

Sαβ(kf ;ki) = δαβδ(3)(kf − ki) +

i

2πkiδ(kf − ki)fαβ(kf ;ki)

β is here to be interpreted as identifying the initial particle, and α thefinal particle. The generalization of our partial wave expansion to thissituation is clearly:

fαβ(k; cos θ) =1

2ik

∞∑

`=0

(2`+ 1)(A(`)αβ − δαβ)P`(cos θ)

Where A(`) is an N ×N unitary matrix:

A(`) = exp[2i4`(k)]

with 4` an N ×N Hermitian “phase shift matrix” note that fαα is theelastic scattering amplitude for particle α.

(a) Find expressions, in terms of A(`)(k), for the following total crosssections, for an incident particle α: (integrated over angles)

i. σelαTOT, the total elastic cross section

ii. σinelαTOT, the total inelastic cross section (sometimes called the

“reaction” cross section).

iii. σαTOT, the total cross section.

(b) Try to give the generalization of the optical theorem for this scat-tering of particles in a multiplet.

49

10. In the previous problem you considered the scattering of particles in amultiplet. You determined the total elastic (sometimes called “scatter-ing”) cross section and the total inelastic (“reaction”) cross sections in

terms of the A(`)αβ matrix in the partial wave expansion. Consider now

the graph in Fig. 11.

4

00 1

Allowed

NotAllowed

2

k

k

2

2

el(l)

inel(l)

(2l+1)

(2l+1)

TOT

TOTπ

π

σ

σ

α

α

Figure 11: The allowed and forbidden regions for possible elastic and inelasticcross sections for the scattering of particles in a multiplet.

This graph purports to show the allowed and forbidden regions for thetotal elastic and inelastic cross sections in a given partial wave `. Derivethe formula for the allowed region of this graph. Make sure to checkthe extreme points.

50

11. In the angular distribution section, we discussed the transformationbetween two different types of “helicity bases”. In particular, we con-sidered a system of two particles, with spins j1 and j2, in their CMframe.

One basis is the “spherical helicity basis”, with vectors of the form:

|j,m, λ1, λ2〉, (282)

where j is the total angular momentum, m is the total angular mo-mentum projection along the 3-axis, and λ1, λ2 are the helicities of thetwo particles. We assumed a normalization of these basis vectors suchthat:

〈j ′, m′, λ′1, λ′2|j,m, λ1, λ2〉 = δjj′δmm′δλ1λ′

1δλ2λ′

2. (283)

The other basis is the “plane-wave helicity basis”, with vectors of theform:

|θ, φ, λ1, λ2〉, (284)

where θ and φ are the spherical polar angles of the direction of particleone. We did not specify a normalization for these basis vectors, but anobvious (and conventional) choice is:

〈θ′, φ′, λ′1, λ′2|θ, φ, λ1, λ2〉 = δ(2)(Ω′ − Ω)δλ1λ′

1δλ2λ′

2, (285)

where d(2)Ω refers to the element of solid angle for particle one.

In the section on angular distributions, we obtained the result for thetransformation between these bases in the form:

|θ, φ, λ1, λ2〉 =∑

j,m

cj|j,m, λ1, λ2〉Djmα(φ, θ,−φ), (286)

where α ≡ λ1 − λ2. Determine the numbers cj.

References

[1] Eyvind H. Wichmann, “Scattering of Wave Packets”, American Journalof Physics, 33 (1965) 20-31.

[2] M. Jacob and G. C. Wick, “On the General Theory of Collisions forParticles with Spin”, Annals of Physics, 7 (1959) 404.

51