Scattering Theory

Two ways of stating the scattering problem. Rate vs. cross-section

The first way to formulate the quantum-mechanical scattering problem issemi-classical: It deals with (i) wave packets describing quantum particlesbefore the collision and (ii) outgoing semiclassical fluxes of the products ofreaction. In this picture, it is only the collision that is essentially quantum,while the evolution before and after the collision is semiclassical. A naturalquantity to describe the outcome of the reaction is the (differential) cross-section.

The second—and, generically, the most “industrial” from quite a numberof perspectives—way to formulate the problem is in terms of the rate ofcorresponding scattering reaction in a large but finite box.

The disadvantage of the first way is that the cross-section is naturallydefined for one particle, but if three1 or more particles participate in thereaction, the notion of cross-section becomes less convenient. Furthermore,for describing kinetic processes, we usually care about the rates, not cross-sections per se. Also, the rate of the process is a more fundamental notion,applying to the decay of a singe particle in two or more particles, in whichcase the notions of scattering and cross-section are irrelevant. Finally, usingthe relation (recall Problem 23)

σ =WVvgr

(1)

[between the (total) scattering cross-section σ, the (total) scattering rate Win the system of the volume V , and the group velocity vgr of the particle beingscattered], one can always relate the rate and cross-section to each other.

A crucially important fact about the scattering is that it is always semi-perturbative in the sense that, up to replacing the microscopic Hamiltonianwith an effective one—acting in a truncated Hilbert space dealing with the

1In both classical and quantum mechanics, the elastic scattering of two particles reducesto a scattering of one particle from the central potential and, for that reason, is perfectlydescribed in terms of the cross-section.

1

energies infinitesimally close to the energy of the initial state—the result forthe rate of the reaction corresponds to the Golden Rule.2

The algorithm

Solving this or that scattering (or decay) problem involves the following steps:

1. In an arbitrarily large but finite system with periodic boundary condition,formulate the effective second-quantized3 Hamiltonian dealing with yet un-known matrix elements (i.e., coupling constants), and respecting kinematicsof the process: (i) all the single-particle modes participating in the process inboth initial and final states, (ii) statistics of particles (bosons or fermions),(iii) conservation of momentum, (iv) conservation of spin projection, otherconservation laws/constraints, if any. Set the volume equal to unity, becauseit can be always restored in the final answer. Same with Planck’s constant.

2. Write the Golden Rule for the rate (total or differential) in the form ofsummation over the momenta of the products of reaction. Restore ~.

3. Replace the summation over the momenta with integrals [see solution tothe Problem 4 for details]. Upon restoring ~, it is slightly more convenientto integrate over the wave vectors rather than momenta.4

4. If you need the cross-section, find it from the rate by the formula (1).

5. Restore V . For the cross-section, this step is irrelevant since the cross-

2A slight reservation should be made for resonant scattering of two particles, wherethe semi-perturbative treatment should be upgraded by applying the general theory ofresonant scattering developed in corresponding section of the lecture notes.

3The formalism of second quantization is extremely convenient/visual, even if we aretalking of one particle only. Furthermore, even for solving a purely classical problem ofkinetics of weakly-nonlinear classical Hamiltonian field, it makes a perfect sense to treatthe field as a bosonic field with large occupation numbers.

4Since at ~ = 1, the momentum and the wave vector is the same, yet another naturaloption is to delay restoring ~ till the very end of calculation.

2

section does not depend on the volume.

6. Relate the matrix elements of your effective Hamiltonian to the actualmicroscopic Hamiltonian.

Depending on the microscopics, the step 6 can be either trivial—no differencebetween the effective and microscopic Hamiltonian (Born approximation), orquite serious (elastic scattering of two particles in a strong enough potential),or even notoriously difficult.

Born approximation works when the microscopic Hamiltonian is appro-priately weak.

Scattering a particle by a potential: Born approximation

Let us see how the above-described algorithm works for the problem of scat-tering a particle by a weak potential. Let U be a typical value of the potentialand R0 be a typical range of the potential. Let the wave vector of the particlesatisfy the condition

k . R−10 . (2)

Under this condition, the criterion of applicability of the Born approximationcan be immediately established by dimensionless analysis. Indeed, for thescattering interaction to be appropriately small, U has to be much smallerthan a certain characteristic quantity having the dimensions of energy. Sincewe have only three parameters to construct such a quantity: particle mass m,the range of the potential R0, and the Planck’s constant ~. The combinationof the three is unique, and we get:

U ~2

mR20

. (3)

Obviously, at k R−10 , the criterion gets only milder. Hence, condition (3)

is sufficient for the applicability of Born approximation at any k.In the second-quantized form, the interaction part of our Hamiltonian

3

reads5

V =

∫d3r Ψ†(r)U(r)Ψ(r). (4)

We are interested in the momentum representation where our non-interactingHamiltonian is diagonal. To get the momentum representation for V , wesubstitute

Ψ(r) =∑k

ak eik·r, Ψ†(r) =

∑k′

a†k′ e−ik′·r

into (4) and perform integration over r. This results in

V =∑k,k′

Uk′−k a†k′ ak, (5)

where

Uq =

∫e−iq·r U(r) d3r (6)

is the Fourier transform of the potential. This completes Step 1 of the algo-rithm.Step 2 gives the following result for the (total) rate:

W =2π

~∑k′

|Uk′−k|2 δ[~2(k′)2

2m− ~2k2

2m

].

Here k and k′ are the wave vectors before and after the scattering, respec-tively.Step 3 brings us to

W =2π

~

∫|Uk′−k|2 δ

[~2(k′)2

2m− ~2k2

2m

]d3k′

(2π)3

=m

2π2~3

∫|Uk′−k|2 δ[(k′)2 − k2] d3k′.

Recalling thatd3k′ = (k′)2dk′dΩk′ ,

5For definiteness, we work in 3D.

4

where Ωk′ is the solid angle of the vector k′, we see that we can integrateover k′ and get

W =mk

4π2~3

∫|Uk′−k|2 dΩk′ , (k′ = k).

Incidentally, in many cases we are interested in the differential rather thantotal rate. For the differential rate (of scattering into an infinitesimal solidangle dΩk′ along the vector k′) we have

dWk′ =mk

4π2~3|Uk′−k|2 dΩk′ (k′ = k).

Step 4 yields

dσk′ =m2

4π2~4|Uk′−k|2dΩk′ (k′ = k) (7)

(for the differential cross-section) and

σ =m2

4π2~4

∫|Uk′−k|2 dΩk′ (k′ = k) (8)

(for the total cross-section).Step 5 proves trivial for the cross-section, since the cross-section is not sup-posed to depend on the volume. The volume appears only if we go back fromthe cross-section to the rate, using (1). The algorithm is completed.

A dramatic simplification of the scattering picture takes place at

k R−10 .

In this limit, Uk′−k ≈ U0, where

U0 ≡ Uq=0.

The momentum-independence of Uk′−k means that the scattering is isotropic(the so-called s-scattering):

dσ =m2|U0|2

4π2~4dΩk′ (k R−1

0 ). (9)

The integral over the solid angle becomes trivial, and for the total cross-section we get

σ =m2|U0|2

π~4(k R−1

0 ). (10)

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Problem 24. A particle of the mass m and the wave vector k interactswith a two-level system (TLS). In its eigenstate basis, the non-interactingHamiltonian of TLS is

HTLS =

[∆ 00 0

](∆ > 0).

In the same basis, the interaction with the particle is described by the inter-action Hamiltonian

V = U(r)

[0 11 0

],

where U(r) is a short-ranged weak (i.e. Born) potential (r is the particle’scoordinate). The range R0 of the potential is such that

kR0 1.

Use the Golden Rule—justified by the weakness of the potential—to find thescattering cross-section for the particle in two characteristic cases:(i) When the initial state of TLS is the state with the energy ∆,(ii) When the initial state of TLS is the state with the zero energy.Observe that—and explain why (!)—in the case (ii) [but not necessarily inthe case (i)] the actual potential U(r) can be replaced with U0 δ(r). In bothcases (i) and (ii), discuss two characteristic limits: (a) k2/m ∆ and (b)∆ k2/m.

Scattering a particle by a potential: summing the perturbative se-ries

Consider the whole perturbative series using the following technical trick.Introduce time dependence of the perturbation by V → V eλt, with λ > 0,and work in the interaction picture starting from t0 = −∞. The idea behindthe trick is the quasi-instant rate, Wλ(t), taking place at small λ:

Wλ(t) = e2λtW [1 +O(λt∗)],

where W is the (constant) rate at λ = 0 and t∗ is the characteristic timegiven by

t−1∗ ∼ max

[1

mR20

,k2

m

].

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The leading effect of the exponential pre-factor is simply rescaling the am-plitude of the perturbation. In the end of calculations, we are supposed totake the limit λ→ +0.

We have

Vt = eλt∑k,k′

Uk′−k a†k′ ak (Schrodinger’s picture) (11)

and (recall Problem 13)

V (t) = eλt∑k,k′

ei(εk′−εk)t Uk′−k a†k′ ak (interaction picture), (12)

εk =k2

2m. (13)

For the evolution operator, U(t, t0), in the interaction picture we have (seeProblem 14; set t0 = −∞):

U(t,−∞) = 1+(−i)1

∫ t

−∞dt1V (t1)+(−i)2

∫ t

−∞dt1

∫ t1

−∞dt2 V (t1)V (t2)+ . . .

+ (−i)n∫ t

−∞dt1

∫ t1

−∞dt2 · · ·

∫ tn−1

−∞dtn V (t1)V (t2) · · ·V (tn) + . . . .

Let |q〉 be the state in the Fock space with only one particle in the systemoccupying the mode with the momentum q. Using (12), for the matrixelement of the operator U(t,−∞) between two states, |k〉 and |k′〉, we find

〈k′| U(t,−∞)|k〉 = δk′,k − iUk′−k

∫ t

−∞dt1e

(iεk′−iεk+λ)t1+

+ (−i)2∑p1

∫ t

−∞dt1

∫ t1

−∞dt2 Uk′−p1e

(iεk′−iεp1+λ)t1Up1−ke(iεp1−iεk+λ)t2 + . . .

The integration over time can be readily performed. In the n-th term, weintegrate sequentially, starting from tn, and observing that each next inte-gration deals with (and results in) a similar exponential function:

(−i)n∫ t

−∞dt1

∫ t1

−∞dt2 · · ·

∫ tn−1

−∞dtn e

(iεk′−iεp1+λ)t1e(iεp1−iεp2+λ)t2 · · · e(iεpn−1−iεk +λ)tn

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=e(iεk′−iεk+λ)t

εk − εk′ + iλ

1

εk − εp1 + iλ

1

εk − εp2 + iλ· · · 1

εk − εpn−1 + iλ.

We see that each of the terms of the series—apart from the very first delta-functional one—share one and the same factor

e(iεk′−iεk+λ)t

εk − εk′ + iλ.

All by itself, this proves the semi-perturbative character of the process. In-deed, pulling out this factor, we get

〈k′| U(t,−∞)|k〉 = δk′,k +e(iεk′−iεk+λ)t

εk − εk′ + iλF (k′,k), (14)

where F (k′,k) is a certain time-independent function given by the sum ofthe series.6 If we confined ourselves to the first perturbative correction (Bornapproximation), we would have

〈k′| U(t,−∞)|k〉 = δk′,k +e(iεk′−iεk+λ)t

εk − εk′ + iλUk′−k. (15)

Hence, the exact answer has the same structure as the Golden Rule, providedthe Born amplitude Uk′−k is replaced with the function F (k′,k). For thedifferential and total cross-sections, we thus have direct analogs of Eqs. (7)and (8), up to the replacement Uk′−k → F (k′,k):

dσk′ =m2

4π2~4|F (k′,k)|2dΩk′ (k′ = k), (16)

σ =m2

4π2~4

∫|F (k′,k)|2 dΩk′ (k′ = k). (17)

One can even introduce a (k-dependent) pseudo-potential for which F (k′,k)will formally look like the Born amplitude.

Problem 25. Derive (16) directly from (14), without resorting to the GoldenRule analogy.7 Hint. Start with the relation (1) between the cross-section

6Note the macroscopic smallness—controlled by an arbitrarily large system volume—ofthe second term in the r.h.s. of (14).

7Other way around, the analogy between (14) and (15) means that here we automati-cally re-derive the Golden Rule (Born) result (7).

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and the rate. Then calculate the rate—via probability—from (14) and takethe limit λ→ +0.

In accordance with our results, the series for F (k′,k) reads8:

F (k′,k) = Uk′−k +Uk′−p1G(p1)Up1−k +Uk′−p1G(p1)Up1−p2G(p2)Up2−k + . . .

. . . + Uk′−p1G(p1)Up1−p2G(p2)Up2−p3 · · ·G(pn)Upn−k + . . . , (18)

with

G(p) =1

εk − εp + iλ. (19)

[For clarity, we do not show the dependence of G on k, since k, as opposedto p, is a fixed parameter.] In some cases, this series is convergent and canbe used for practical calculations of F (k′,k).

Algebraically, the structure of the expansion is similar to the geometricseries. In particular, we have:

Uk′−p1G(p1)Up1−k + Uk′−p1G(p1)Up1−p2G(p2)Up2−k + . . . =

= Uk′−p1G(p1)[Up1−k + Up1−p2G(p2)Up2−k + . . .] =

= Uk′−p1G(p1)F (p1,k).

We thus conclude that F (k′,k) satisfies the equation (here we restore thesummation sign)

F (k′,k) = Uk′−k +∑p

Uk′−pG(p)F (p,k).

Replacing summation with integration, using explicit form for G, and alsorecalling that we need to take the limit λ → +0, we arrive at the followingintegral equation for F (k′,k):

F (k′,k) = Uk′−k + limλ→+0

∫d3p

(2π)3

Uk′−p F (p,k)

εk − εp + iλ. (20)

8To compactify the expression, we omit the summation signs, adopting the conventionthat there is a summation over all the momenta different from k and k′.

9

A very instructive case is the k → 0 limit for a short-ranged potential.Here we have

F (k′, 0) = Uk′ −∫

d3p

(2π)3

Uk′−p F (p, 0)

εp. (21)

Note that the singularity of in the denominator becomes integrable allowingus to set λ = 0. If, for clarity, we introduce corresponding pseudo-potential

U (psd)q = F (q, 0),

then Eq. (21) will look like a relationship between the genuine and pseudopotentials:

U (psd)q = Uq −

∫d3p

(2π)3

Uq−p U(psd)p

εp. (22)

The q → 0 limit is well defined (i.e., it is direction-independent no matter

what is the direction dependence of Uq and thus U(psd)q ):

limq→0

U (psd)q ≡ U

(psd)0 = U0 −

∫d3p

(2π)3

U−p U(psd)p

εp.

We conclude that the k → 0 limit corresponds to the s-scattering where wehave direct analogs of Eqs. (9) and (10), up to the replacement U0 → U

(psd)0 :

dσ =m2|U (psd)

0 |2

4π2~4dΩk′ (k → 0), (23)

σ =m2|U (psd)

0 |2

π~4(k → 0). (24)

Scattering a particle: general analysis

We will be dealing with one particle, but it is important to remember thatthe theory applies to two particles as well, by going to the center-of-massframe and introducing the reduced mass.

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Rather than dealing with the scattering rate and, correspondingly, theevolution operator, we will start with the wave packet formulation and showthat the problem of finding/analyzing scattering cross-section reduces to un-derstanding the asymptotic behavior of a stationary solution of the Schrodingerequation with appropriate boundary conditions. In this formulation, we startwith a large plane-wave packet as an initial state and a superposition (withasymptotically vanishing spatial overlap ) of the plane-wave packet and aspherical-wave packet in the limit of t → ∞. Let dpr be the probability forthe particle to get scattered into an infinitesimal solid angle of the magnitudedΩ and the radial direction r. From the properties of the Schrodinger equa-tion, we know that dpr can be represented as an integral of the probabilityflux through an element of the spherical surface corresponding to the solidangle in question. The radius of the surface has to be appropriately large(and otherwise arbitrary). The flux is a product of the local flux densityjout(r, t) and the area r2 dΩ of the element of the sphere. We thus have

dpr = r2dΩ

∫ ∞−∞

dt jout(r, t).

The subscript “out” is to remind that here we are talking of the outgoingspherical wave created as a result of scattering of the plane-wave packet.By definition, the differential cross-section dσr is an infinitesimal area of thesurface (normal to the wave-vector of the incident plane wave), such thatthe probability dpr is equal to the integral over the time of the plane-waveprobability flux through this area. Hence

dpr = dσr

∫ ∞−∞

dt jin(t) (defenition of cross-section),

where jin(t) is the (spatially uniform) flux density in the incident plane wavein the vicinity of the scattering center.

The size of the wave packet has to be appropriately large to guaranteethe quasi-steady-state regime when jout and jin are related to each other as

jout(r, t) = jin(t− t0)Q(r)

r2.

Here t0 is a certain time delay that has no effect on the integral over time.The denominator r2 takes into account the decrease of the flux density in aspherical wave so that the numerator Q(r) depends only on the direction but

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not on r. Crucially important is the fact that Q(r) does not depend on theshape of the packet and can actually be found from the stationary solution.

Equating two expressions for dpr, we conclude that

dσr = Q(r) dΩ

and thus can be found from the solution of a stationary problem.The relevant stationary problem is looking for the solution of the station-

ary Schrodinger equation with the energy k2/2m and the following asymp-totic form of the solution away from the scattering center (the z axis is alongthe wave vector k of the incident wave):

ψk(r) → eikz +f(r)

reikr (r →∞). (25)

The initial wave packet at the time moment t = 0 has the form

ψpacket(r, t = 0) =

∫ ∞−∞

Λ(q)ψk+q(r) e−iqz0 dq, (26)

where z0 is an appropriately large negative9 coordinate—the initial positionthe packet and Λ(q) is an arbitrary envelope function obeying the requirementthat it decays with |q| at |q| k, so that the uncertainty of momentum andenergy is negligibly small. In view of the requirement on z0, the contributionof the second term in the r.h.s. of (25) is negligible at t = 0 and remainsnegligible till the propagating packet hits the scattering center. After thepacket passes the center, its structure becomes dramatically different. Bothterms in the r.h.s. of (25) become important. The first term is responsiblefor the plane-wave packet that keeps moving with the constant linear velocityv = k/m while the second term generates a spherical wave packet expandingwith the radial velocity equal to the linear velocity of the plane-wave partof the wave function. At large enough time, the overlap of the radial andplane-wave parts of the wave function gets progressively smaller and we can

9In the case of positive z0, both terms in the r.h.s. of (25) contribute. The first termgenerates a plane-wave packet at the coordinate z = z0 while the second term generatesa spherical wave packet with the radius r = z0. The role of the variable z0 in the wavepacket is prescribed by the exponentials eiq(z−z0) and eiq(r−z0) for the plane and sphericalwaves. The conditions z = z0 and r = z0 correspond to maximal possible constructiveinterference and thus define the z-position of the plane-wave packet and the radius of thespherical packet, respectively.

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totally neglect it in the t→∞ limit. The desired ratio of (probability) fluxdensities, Q(r), is thus given by the the flux density corresponding to thesecond term in the r.h.s. of (25)over the flux density corresponding to thefirst term in the r.h.s. of (25). Recalling that, for any wave function ψ, thevector of the probability flux density is given by (here we restore ~)

j =i~2m

(ψ∇ψ∗ − ψ∗∇ψ), (27)

which, for the plane-wave part of the stationary solution (25) yields k/m,and r|f(r)|2k/mr2 for the radial part, we conclude that

Q(r) = |f(r)|2.

We arrive at the formuladσr = |f(r)|2 dΩ (28)

relating the differential elastic cross-section to the asymptotic form (25) ofcorresponding stationary wave function.

In the case of scattering by a potential, the function f(r) can be directlyrelated to the earlier-introduced function F (k′,k) (note that here k′ = rk):

f(r) = − m

2π~2F (rk,k) (scattering by a potential). (29)

We will not be deriving this formula because, up to a phase factor, therelation is obvious by comparing (28) to (16).

In what follows, we will be assuming spherical symmetry. In this case,f(r) ≡ f(θ) and dΩ = 2π sin θ dθ, with θ the polar angle, and

dσθ = 2π sin θ |f(θ)|2 dθ, σ = 2π

∫ π

0

|f(θ)|2 sin θ dθ. (30)

The solution is then naturally expanded in terms of spherical harmonics.Since the incident plane wave has only them = 0 components (no dependenceon the azimuthal angle), so does the rest of the wave function and we have

ψk(r) →1

r

∞∑l=0

Pl(cos θ)[Al e

−ikr + Bl eikr]

(r →∞).

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In accordance with (25), all the incoming spherical waves ∝ e−ikr are sup-posed to belong to the plane-wave part of the solution. This fixes the valuesof all the coefficients Al. With the known expansion for the plane wave

eikz → 1

2ikr

∞∑l=0

(2l + 1)Pl(cos θ)[(−1)l+1 e−ikr + eikr

](r →∞), (31)

we get

ψk(r)→1

2ikr

∞∑l=0

(2l+1)Pl(cos θ)[(−1)l+1 e−ikr + Sle

ikr]

(r →∞). (32)

Subtracting (31) from (32) and multiplying the result by r, we get

f(θ) =1

2ik

∞∑l=0

(Sl − 1)(2l + 1)Pl(cos θ). (33)

For further manipulations, recall the orthonormalization relation for theLegendre polynomials:∫ π

0

Pl(cos θ)Pl′(cos θ) sin θdθ =2δll′

2l + 1. (34)

The flux of the probability into a sphere (of a sufficiently large radius)divided by the flux density in the incident plane wave yields, by definition,the total cross-section of inelastic processes, σi. Explicitly performing thecalculation for the solution (32), using (27) and (34), we find

σi =π

k2

∞∑l=0

(2l + 1)(1− |Sl|2). (35)

On the other hand, calculating the elastic cross-section, σe, for the elasticchannel—in accordance with (30), using (33) and (34)—we get

σe =π

k2

∞∑l=0

(2l + 1)|1− Sl|2. (36)

Combining (35) and (36), for the total cross-section we have

σtot = σi + σe =2π

k2

∞∑l=0

(2l + 1)(1− ReSl). (37)

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We see that inelastic scattering, if present, inevitably implies elastic one, bythe condition |Sl| 6= 1. Another generic observation is that the cross-sectionsσe and σi (and thus σtot) are sums of independents contributions—called par-tial cross-sections—for each l-channel.

Problem 26. Perform all the calculations leading to Eqs. (35) through (37).For each partial cross-section, find its maximal possible value.

For purely elastic scattering—from now on we will be dealing only withthis case—we have

|Sl| = 1, ∀l (purely elastic scattering),

suggesting convenient parameterization in terms of the phase shifts

Sl = e2iδl(k). (38)

σ =4π

k2

∞∑l=0

(2l + 1) sin2 δl(k). (39)

In the case of a potential scattering,10 the phase shifts can be found (in-dependently for each l) by solving the standard one-dimensional differentialequation for the radial part, Rkl(r), of the wave function:

1

r2

d

dr

(r2 dRkl

dr

)+

[k2 − l(l + 1)

r2− 2m

~2U(r)

]Rkl = 0, r ∈ [0,∞).

(40)By linearity of Eq. (40), one of the two free constants of its solution is justa global factor. The phase shift does not depend on this factor. Indeed, inaccordance with (32) and (38), the asymptotic behavior of the Rkl(r) is

Rkl(r) →const

r

[(−1)l+1 e−ikr + e2iδl(k)eikr

](r →∞), (41)

and it is precisely from this asymptotic behavior we ultimately find δl(k),provided the other integration constant is fixed. And the latter is done by

10Note that (39) does not imply that the scattering is by a potential. Just the absenceof inelastic channels is sufficient.

15

realizing that at r → 0, a generic solution of (40) diverges, while the physicalsolution has to be finite. This leads to the boundary condition at r = 0requiring that Rkl(r) be finite at r → 0.

Problem 27. Find the l = 0 partial cross-section (aka s-scattering cross-section) for the following potential

U(r) =

U∗, r ≤ R0,0, r > R0.

Here U∗ is a certain positive or negative constant. Specially address the hard-sphere limit U∗ → +∞. Hint. It is convenient to use the parametrizationR(r) = χ(r)/r.

Coulomb scattering: Rutherford formula

To interpret his famous experiment revealing the existence of nuclei,Rutherford derived the formula for the scattering cross-section of two chargedparticles interacting via Coulomb potential U(r) = q1q2/r ( q1 and q2 are thetwo electric charges). For the target particle (the particle at rest in thelaboratory frame; a nucleus in the context of the experiment), Rutherfordobtained

dσ =(q1q2

4E

)2 dΩ

sin4(θ/2), (42)

where E is the kinetic energy of the incident particle (say, α-particle like inRutherford’s experiment). Rutherford analysis was based on classical me-chanics. Amazingly enough, solving the purely quantum problem (40)–(41)for the Coulomb potential leads to precisely the same formula (42).11 This,in particular, implies that the Born formula has to yield12 (and does yieldas can be immediately checked) precisely the same exact result. Note, how-ever, that for ~ to disappear from the answer for the Coulomb scattering

11To be precise, for the two particles one has to go from the laboratory frame to thecenter-of-mass frame, solve there the problem (40)–(41) for the fictitious particle with thereduced mass m∗ = m1m2/(m1 +m2), and then return back to the laboratory frame.

12Born (perturbative) limit can be viewed as an ultra-quantum limit corresponding to~→∞. Indeed, the potential term in (40) becomes infinitesimally small at ~→∞.

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cross-section, the result should be expressed in terms of either the energy, ormomentum, or velocity, but not the wave vector.

Breit-Wigner resonances

For the theory of Breit-Wigner resonances, leading to the formula

σ =4π(2l + 1)

k2

(Γ/2)2

(εk − E0)2 + (Γ/2)2, (43)

see the end of the section on perturbation theory.

Small momenta in the case of short-ranged interaction

Let the momentum of the particle be small comapred to the inverse range ofinteraction:

k R−10 . (44)

Then, in the range of distances such that R0 r k−1, equation (40)simplifies. The potential and the k2 terms can be neglected:

1

r2

d

dr

(r2 dRkl

dr

)− l(l + 1)

r2Rkl = 0 (R0 r k−1), (45)

leading to the solution

Rkl = c(l)1 r

l +c

(l)2

rl+1(R0 r k−1), (46)

where c(l)1 and c

(l)2 are certain constants. Then, solving the equation13

1

r2

d

dr

(r2 dRkl

dr

)+

[k2 − l(l + 1)

r2

]Rkl = 0 (r R0). (47)

13Because of the absence of the external potential, the solutions are simply the sphericalwaves, the eigenstates of the radial part of the Laplace operator.

17

with the boundary condition (46), one finds14

tan δl =c

(l)2

c(l)1

k2l+1

(2l − 1)!!(2l + 1)!!. (48)

Dimensionally, we have (the sign here is a matter of convention)

c(l)2

c(l)1

1

(2l − 1)!!(2l + 1)!!= −a2l+1

l ,

where al is a certain length. An important—and very natural under thecondition (44)—case corresponds to the inequality15

k−1 max(|al|, R0). (49)

Under condition (49), we have

|δl| ≈ |alk|2l+1 1 [k−1 max(|al|, R0)], (50)

and, applying (39) for the partial cross-section in the l-th channel we have

σl ≈ 4πa2l (2l + 1)|alk|4l [k−1 max(|al|, R0)]. (51)

In particular, for the s-scattering (l = 0) cross-section, we have (σs ≡ σl=0,as ≡ al=0):

σs ≈ 4πa2s [k−1 max(|as|, R0)]. (52)

Generically, al’s are k-independent under the condition (49), and, once thiscondition is satisfied, the s-scattering cross-section saturates to a constantvalue defined by the s-scattering length as, while the partial cross-sectionsin the l > 0 channels rapidly vanish with decreasing k, in accordance with(51). The only exception from this rule are previously discussed Breit-Wignerresonances. The inequality (44) is not directly relevant to Breit-Wigner res-onances. However, if (44) is satisfied, the prediction of the Breit-Wignerformula should be consistent with Eqs. (46) through (52). This implies—in

14For an even integer, (2n)!! = 2 · 4 · 6 · · · (2n) = 2nn!; for an odd integer, (2n + 1)!! =1 · 3 · 5 · · · (2n+ 1).

15In the absence of resonances, this inequality is essentially equivalent to (44), because,at the order-of-magnitude level, |al| . R0 (with the condition |al| R0 taking place foreither perturbatively weak (Born), or specially fine-tuned potential).

18

a remarkable contrast to the generic case—a very sharp dependence of |al|on k, with |al| R0 in the vicinity of the resonance, and, in particular,|al| = 1/k at the resonance.

Broad resonances, resonant s-scattering

Crucially different from Breit-Wigner resonances are the so-called broad res-onances, when the regime

|al| R0 (53)

occurs in a certain channel l, with k-independent al. Such an anomalousincrease of |al| takes place in the vicinity of the threshold of forming a boundstate in corresponding channel, with |al| =∞ right at the threshold. For theresonant condition (53) to take place, the side with respect to the binding

threshold16 is not important. The threshold corresponds to c(l)1 = 0, the sign

of the ratio c(l)1 /c

(l)2 defining the side with respect to the threshold. For |al|

to be anomalously large, the coefficient c(l)1 has to be appropriately close to

zero, irrespectively of its sign.With the condition (53), we have two typical regions of k. The region of

very small k’s is defined by the condition (49). Here Eqs. (50) through (52)apply. A new type of behavior, specific for the broad resonances only, takesplace in the interval

|al|−1 k R−10 .

Here from (48) we find

| tan δl| = |alk|2l+1 1 ⇒ sin2 δl ≈ 1, (54)

so that the partial cross-section in the l-th channel is equal to its maximalpossible value:

σl =4π(2l + 1)

k2(|al|−1 k R−1

0 ). (55)

To trace how the regime (55) crosses over to the asymptotic regime (51)–(52), and also to explicitly see the role of the binding threshold, let us explore

16That is, the presence or absence of a weakly bound state.

19

in more detail the most important case of resonant s-scattering. With thesubstitution R(r) = χ(r)/r, we cast Eq. (47) into

χ′′ + k2χ = 0. (56)

The boundary condition (46) is conveniently parameterized as

χ′

χ

∣∣∣∣r→0

= −κ, (57)

with

κ = −c1

c2

≡ 1

as. (58)

Solving the problem (56)–(58) one finds the expression for the phase shift:

cot δ0 = −κk,

and then the expression for the scattering amplitude17

f =1

2ik

(e2iδ0 − 1

)=

1

k(cot δ0 − i)= − 1

κ + ik.

For the total cross-section we thus get

σ = 4π|f |2 =4π

κ2 + k2.(59)

Equation (59) is consistent with Eqs. (55) and (52) in the regions of applica-bility of the latter, and is accurate in the crossover region k|as| = k/|κ| ∼ 1.

At κ > 0, there is a bound state χ(r) = e−κr. Its energy is readily found

χ′′ +2mE

~2χ = 0 ⇒ E = −~2κ2

2m.

The resonant condition |as| R0 implies that the wave function of thebound state is localized mostly outside the range of interaction. Consistentwith that, we have |E| ~2/(mR2

0), meaning that the binding is weak com-pared to the characteristic energy scale ~2/(mR2

0).

17The total scattering amplitude here essentially coincides with the s-scattering ampli-tude, because the other channels are suppressed by the smallness of k.

20

Scattering of identical particles

Consider scattering of two identical spinless18 bosons or fermions. The inter-acting Hamiltonian reads

V =1

2

∑k1,k2,q

Uq a†k1−qa

†k2+qak2 ak1 . (60)

In the center-of-mass frame, in the initial state, |i〉, there are two particleswith the opposite momenta, k and −k. In the final state, |f〉, there are twoparticles with momenta k′ and −k′. In the Hamiltonian (60), there are fourterms leading from |i〉 to |f〉:

Term 1: k1 = k, k2 = −k, q = k− k′,

Term 2: k1 = −k, k2 = k, q = k′ − k,

Term 3: k1 = k, k2 = −k, q = k + k′,

Term 4: k1 = −k, k2 = k, q = −k− k′.

Term 1 equals Term 2 and Terms 3 equals Term 4, so that we are left withonly two different terms:

U|k−k′| a†k′ a†−k′ a−kak + U|k+k′| a

†−k′ a

†k′ a−kak.

Taking into account (anti)commutation relations, these two terms can becombined into one (

U|k−k′| ± U|k+k′|)a†k′ a

†−k′ a−kak , (61)

where the sign plus (minus) is for bosons (fermions). Going from microscopicinteraction Hamiltonian (61) to the effective pseudo-perturbative Hamilto-nian amounts to the replacement

U|k−k′| → F (k′,k), U|k+k′| → F (−k′,k).

18For particles with spins/non-zero internal momentum, see next section.

21

The resulting effective Hamiltonian reads

[F (k′,k) ± F (−k′,k)] a†k′ a†−k′ a−kak . (62)

Adapting relation (63) to our case of two particles by m → m/2, i.e., bytaking into account that the mass in (63) is supposed to be replaced withthe reduced mass, we have

f(r) = − m

4π~2F (rk,k) (k′ = rk). (63)

This allows us to express the effective scattering term (61) as

−4π~2

m[f(θ) ± f(π − θ)] a†k′ a

†−k′ a−kak , (64)

where θ is the angle between the vectors k′ and k. Equation (64) is the mostgeneral form of the effective elastic-scattering term for two spinless identicalparticles (in the center-of-mass frame). It applies equally well to the (mostlyacademic) case of potential scattering of two point particles and the caseof elastic scattering of two identical composite bosons or fermions, providedeach of the two particles has zero total spin.

A comment is in order here concerning the (somewhat annoying) nega-tive sign in (63) and, correspondingly, in (64). We could easily get rid ofthe negative sign by simply redefining f(r) in Eq. (25): f(r) → −f(r). Aconventional way of achieving essentially the same goal is, however, slightlydifferent. One writes Eq. (25) the way it was written, but ultimately intro-duces the notion of scattering length as nothing but negative f . Note thatthe sing in our definition of partial scattering lengths, al, is consistent withthis convention. In the case when f is dominated by the s-scattering, wesimply have f = −as and the effective scattering term for bosons becomes

2π~2asm

a†k′ a†−k′ a−kak (bosons, low energy). (65)

For fermions, corresponding effective scattering term is essentially zero, upto small corrections from l > 0 terms.

Comparing (65) to (61), we conclude that the low-energy effective inter-action Hamiltonian for a dilute low-temperature Bose gas reads

Veff =U∗2

∑k1,k2,q

a†k1−qa†k2+qak2 ak1 (bosons, low energy), (66)

22

with

U∗ =4π~2asm

(67)

playing the role of effective coupling constant.

Elastic scattering of particles with spin or internal momentum

If our particles, in addition to being identical, also feature a non-zero spin orinternal momentum,19 our interaction Hamiltonian becomes

V =1

2

∑σ

∑k1,k2,q

Uq(σ4, σ3;σ2, σ1) a†(k1−q)σ4a†(k2+q)σ3

ak2σ2 ak1σ1 . (68)

In the center-of-mass frame, in the initial state, |i〉, there are two particleswith the opposite momenta, k and −k. The particle with the momentumk is in the spin state σa, and the particle with the momentum −k is in thespin state σb. The spin states can be the same or different.20

In the final state, |f〉, there are two particles with momenta k′ and −k′,and the same spin states.21 Without loss of generality, we will assume thatthe momentum k′ is associated with the σa spin state, and the momentum−k′ is associated with the σb spin state. Similarly to the case of spinlessparticles, in the Hamiltonian (68), there are four terms leading from |i〉 to|f〉:

Term 1: k1 = k, k2 = −k, q = k−k′, σ1 = σ4 = σa, σ2 = σ3 = σb,

Term 2: k1 = −k, k2 = k, q = k′−k, σ1 = σ4 = σb, σ2 = σ3 = σa,

Term 3: k1 = k, k2 = −k, q = k+k′, σ1 = σ3 = σa, σ2 = σ4 = σb,

Term 4: k1 = −k, k2 = k, q = −k−k′, σ1 = σ3 = σb, σ2 = σ4 = σa.

19Which, for our purposes, is essentially the same, so that we will be loosely use theterm “spin.”

20At σb 6= σa, the indistinguishability is still relevant, as long as the interaction canswap the spin states of two particles.

21We confine ourselves to the elastic scattering only.

23

As previously, Term 1 equals Term 2 and Terms 3 equals Term 4,22 and weare left with only two different terms:

U|k−k′|(σa, σb;σb, σa) a†k′σa

a†−k′σb a−kσb akσa

+U|k+k′|(σb, σa;σb, σa) a†−k′σb a

†k′σa

a−kσb akσa ,

which, analogously to (61), combine into one scattering term[U|k−k′|(σa, σb;σb, σa) ± U|k+k′|(σb, σa;σb, σa)

]a†k′σa a

†−k′σb a−kσb akσa . (69)

So far, there were not substantial difference with spinless case. But thenext step—going from Fourier components of microscopic potential to scat-tering amplitudes—becomes less straightforward. Now reducing the two-body problem to a single-particle one is possible only if the motion in thesubspace of interparticle distance separates not only from the center-of-masscoordinate (which is always the case), but from the spin degrees of freedomas well. The latter is normally possible, but requires that we work in therepresentation of the eigenstates of the total spin of two particles. Hence, wehave to decompose Uq(σ4, σ3;σ2, σ1) as

Uq(σ4, σ3;σ2, σ1) =∑S

U (S)q PS(σ4, σ3;σ2, σ1),

where S stands for the total spin of the two particles (say, S = 0 and S = 1,

if our two particles are spin-1/2 particles), U(S)q is the Fourier component of

the potential in this channel,23 and

PS(σ4, σ3;σ2, σ1)

is the matrix of the projector onto the S-channel. After separating out spindegrees of freedom, going from microscopic interaction Hamiltonian to theeffective pseudo-perturbative Hamiltonian amounts to the replacement

U(S)|k−k′| → F (S)(k′,k), U

(S)|k+k′| → F (S)(−k′,k),

leading to the effective Hamiltonian∑S

[F (S)(k′,k)PS(σa, σb;σb, σa)±

22Here we also take into account that the indistinguishability of the particles impliesUq(σ4, σ3;σ2, σ1) = Uq(σ3, σ4;σ1, σ2) .

23Normally, the potential or, speaking generally, interaction, is S-dependent.

24

±F (S)(−k′,k)PS(σb, σa;σb, σa)]a†k′σa a

†−k′σb a−kσb akσa . (70)

In terms of the scattering amplitudes f (S)(θ), we have

− m

4π~2

∑S

[f (S)(θ)PS(σa, σb;σb, σa)±

± f (S)(π − θ)PS(σb, σa;σb, σa)]a†k′σa a

†−k′σb a−kσb akσa . (71)

25