Scattering Theory

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

The first way to formulate the quantum-mechanical scattering problem is semi-classical: It deals with (i) wave packets describing quantum particles before the collision and (ii) outgoing semiclassical fluxes of the products of reaction. In this picture, it is only the collision that is essentially quantum, while the evolution before and after the collision is semiclassical. A natural quantity to describe the outcome of the reaction is the (differential) cross- section. The second—and, generically, the most “industrial” from quite a number of perspectives—way to formulate the problem is in terms of the rate of corresponding scattering reaction in a large but finite box. The disadvantage of the first way is that the cross-section is naturally defined for one particle, but if three1 or more particles participate in the reaction, 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 which case the notions of scattering and cross-section are irrelevant. Finally, using the relation (recall Problem 23)

W V σ = (1) vgr

[between the (total) scattering cross-section σ, the (total) scattering rate W in the system of the volume V, and the group velocity vgr of the particle being scattered], 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 Hamiltonian with an eﬀective one—acting in a truncated Hilbert space dealing with the

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

1 energies inﬁnitesimally close to the energy of the initial state—the result for the 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 kinematics of the process: (i) all the single-particle modes participating in the process in both initial and final states, (ii) statistics of particles (bosons or fermions), (iii) conservation of momentum, (iv) conservation of spin projection, other conservation laws/constraints, if any. Set the volume equal to unity, because it 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 of summation over the momenta of the products of reaction. Restore ~.

3. Replace the summation over the momenta with integrals [see solution to the Problem 4 for details]. Upon restoring ~, it is slightly more convenient to 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, where the semi-perturbative treatment should be upgraded by applying the general theory of resonant scattering developed in corresponding section of the lecture notes. 3The formalism of second quantization is extremely convenient/visual, even if we are talking of one particle only. Furthermore, even for solving a purely classical problem of kinetics of weakly-nonlinear classical Hamiltonian field, it makes a perfect sense to treat the field as a bosonic field with large occupation numbers. 4Since at ~ = 1, the momentum and the wave vector is the same, yet another natural option 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 actual microscopic Hamiltonian.

Depending on the microscopics, the step 6 can be either trivial—no difference between the effective and microscopic Hamiltonian (Born approximation), or quite serious (elastic scattering of two particles in a strong enough potential), or even notoriously difficult. Born approximation works when the microscopic Hamiltonian is appropriately weak.

Scattering a particle by a potential: Born approximation

Let us see how the above-described algorithm works for the problem of scattering a particle by a weak potential. Let U be a typical value of the potential and R0 be a typical range of the potential. Let the wave vector of the particle satisfy the condition −1 k . R0 . (2) Under this condition, the criterion of applicability of the Born approximation can be immediately established by dimensionless analysis. Indeed, for the scattering interaction to be appropriately small, U has to be much smaller than a certain characteristic quantity having the dimensions of energy. Since we have only three parameters to construct such a quantity: particle mass m, the range of the potential R0, and the Planck’s constant ~. The combination of the three is unique, and we get:

~2 U 2 . (3) mR0

−1 Obviously, at k R0 , 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 5 reads Z V = d3r Ψˆ †(r)U(r)Ψ(ˆ r). (4)

We are interested in the momentum representation where our non-interacting Hamiltonian is diagonal. To get the momentum representation for V , we substitute

ˆ X ik·r ˆ † X † −ik0·r Ψ(r) = aˆk e , Ψ (r) = aˆk0 e k k0 into (4) and perform integration over r. This results in

X † V = Uk0−k aˆk0 aˆk, (5) k,k0 where Z −iq·r 3 Uq = e U(r) d r (6) is the Fourier transform of the potential. This completes Step 1 of the algorithm. Step 2 gives the following result for the (total) rate:

2 0 2 2 2 2π X 2 ~ (k ) ~ k W = |U 0 | δ − . k −k 2m 2m ~ k0 Here k and k0 are the wave vectors before and after the scattering, respectively. Step 3 brings us to

Z 2 0 2 2 2 3 0 2π 2 ~ (k ) ~ k d k W = |Uk0−k| δ − ~ 2m 2m (2π)3 Z m 2 0 2 2 3 0 = |Uk0−k| δ[(k ) − k ] d k . 2π2~3 Recalling that 3 0 0 2 0 d k = (k ) dk dΩk0 ,

5For deﬁniteness, we work in 3D.

4 0 where Ωk0 is the solid angle of the vector k , we see that we can integrate over k0 and get Z mk 2 0 W = |Uk0−k| dΩk0 , (k = k). 4π2~3 Incidentally, in many cases we are interested in the differential rather than total rate. For the differential rate (of scattering into an infinitesimal solid 0 angle dΩk0 along the vector k ) we have

mk 2 0 dWk0 = |Uk0−k| dΩk0 (k = k). 4π2~3 Step 4 yields

2 m 2 0 dσk0 = |Uk0−k| dΩk0 (k = k) (7) 4π2~4 (for the diﬀerential cross-section) and 2 Z m 2 0 σ = |Uk0−k| dΩk0 (k = k) (8) 4π2~4 (for the total cross-section). Step 5 proves trivial for the cross-section, since the cross-section is not supposed to depend on the volume. The volume appears only if we go back from the cross-section to the rate, using (1). The algorithm is completed. A dramatic simpliﬁcation of the scattering picture takes place at

−1 k R0 .

In this limit, Uk0−k ≈ U0, where

U0 ≡ Uq=0.

The momentum-independence of Uk0−k means that the scattering is isotropic (the so-called s-scattering):

2 2 m |U0| −1 dσ = dΩk0 (k R0 ). (9) 4π2~4 The integral over the solid angle becomes trivial, and for the total cross- section we get 2 2 m |U0| −1 σ = (k R0 ). (10) π~4 5 Problem 24. A particle of the mass m and the wave vector k interacts with a two-level system (TLS). In its eigenstate basis, the non-interacting Hamiltonian of TLS is ∆ 0 H = (∆ > 0). TLS 0 0 In the same basis, the interaction with the particle is described by the interaction Hamiltonian 0 1 V = U(r) , 1 0 where U(r) is a short-ranged weak (i.e. Born) potential (r is the particle’s coordinate). The range R0 of the potential is such that

kR0 1. Use the Golden Rule—justiﬁed by the weakness of the potential—to ﬁnd the scattering 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 in the case (i)] the actual potential U(r) can be replaced with U0 δ(r). In both cases (i) and (ii), discuss two characteristic limits: (a) k2/m ∆ and (b) ∆ k2/m.

Scattering a particle by a potential: summing the perturbative series

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 behind the trick is the quasi-instant rate, Wλ(t), taking place at small λ: 2λt Wλ(t) = e W [1 + O(λt∗)], where W is the (constant) rate at λ = 0 and t∗ is the characteristic time given by 2 −1 1 k t∗ ∼ max 2 , . mR0 m

6 The leading eﬀect of the exponential pre-factor is simply rescaling the amplitude of the perturbation. In the end of calculations, we are supposed to take the limit λ → +0. We have

λt X † Vt = e Uk0−k aˆk0 aˆk (Schr¨odinger’spicture) (11) k,k0 and (recall Problem 13)

λt X i(k0 −k)t † V (t) = e e Uk0−k aˆk0 aˆk (interaction picture), (12) k,k0

k2 = . (13) k 2m

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

Z t Z t Z t1 1 2 U(t, −∞) = 1+(−i) dt1V (t1)+(−i) dt1 dt2 V (t1)V (t2)+... −∞ −∞ −∞

Z t Z t1 Z tn−1 n + (−i) dt1 dt2 ··· dtn V (t1)V (t2) ··· V (tn) + .... −∞ −∞ −∞ Let |qi be the state in the Fock space with only one particle in the system occupying the mode with the momentum q. Using (12), for the matrix element of the operator U(t, −∞) between two states, |ki and |k0i, we ﬁnd

Z t 0 (i 0 −ik+λ)t1 hk | U(t, −∞)|ki = δk0,k − iUk0−k dt1e k + −∞

Z t Z t1 2 X (i 0 −ip +λ)t1 (ip −ik+λ)t2 0 k 1 1 + (−i) dt1 dt2 Uk −p1 e Up1−ke + ... −∞ −∞ p1 The integration over time can be readily performed. In the n-th term, we integrate sequentially, starting from tn, and observing that each next integration deals with (and results in) a similar exponential function:

Z t Z t1 Z tn−1 n (i 0 −ip +λ)t1 (ip −ip +λ)t2 (ip −ik +λ)tn (−i) dt1 dt2 ··· dtn e k 1 e 1 2 ··· e n−1 −∞ −∞ −∞

7 e(ik0 −ik+λ)t 1 1 1 = ··· . 0 k − k + iλ k − p1 + iλ k − p2 + iλ k − pn−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(ik0 −ik+λ)t . k − k0 + iλ All by itself, this proves the semi-perturbative character of the process. In- deed, pulling out this factor, we get

(ik0 −ik+λ)t 0 e 0 hk | U(t, −∞)|ki = δk0,k + F (k , k), (14) k − k0 + iλ where F (k0, k) is a certain time-independent function given by the sum of the series.6 If we conﬁned ourselves to the ﬁrst perturbative correction (Born approximation), we would have

(ik0 −ik+λ)t 0 e hk | U(t, −∞)|ki = δk0,k + Uk0−k. (15) k − k0 + iλ Hence, the exact answer has the same structure as the Golden Rule, provided 0 the Born amplitude Uk0−k is replaced with the function F (k , k). For the diﬀerential and total cross-sections, we thus have direct analogs of Eqs. (7) 0 and (8), up to the replacement Uk0−k → F (k , k):

2 m 0 2 0 dσk0 = |F (k , k)| dΩk0 (k = k), (16) 4π2~4 2 Z m 0 2 0 σ = |F (k , k)| dΩk0 (k = k). (17) 4π2~4 One can even introduce a (k-dependent) pseudo-potential for which F (k0, k) will formally look like the Born amplitude.

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

6Note the macroscopic smallness—controlled by an arbitrarily large system volume—of the 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).

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

In accordance with our results, the series for F (k0, k) reads8:

0 0 0 0 F (k , k) = Uk −k + Uk −p1 G(p1)Up1−k + Uk −p1 G(p1)Up1−p2 G(p2)Up2−k + ...

0 ... + Uk −p1 G(p1)Up1−p2 G(p2)Up2−p3 ··· G(pn)Upn−k + ..., (18) with 1 G(p) = . (19) k − p + iλ [For clarity, we do not show the dependence of G on k, since k, as opposed to p, is a ﬁxed parameter.] In some cases, this series is convergent and can be used for practical calculations of F (k0, k). Algebraically, the structure of the expansion is similar to the geometric series. In particular, we have:

0 0 Uk −p1 G(p1)Up1−k + Uk −p1 G(p1)Up1−p2 G(p2)Up2−k + ... =

0 = Uk −p1 G(p1)[Up1−k + Up1−p2 G(p2)Up2−k + ...] =

0 = Uk −p1 G(p1)F (p1, k). We thus conclude that F (k0, k) satisﬁes the equation (here we restore the summation sign)

0 X F (k , k) = Uk0−k + Uk0−pG(p)F (p, k). p

Replacing summation with integration, using explicit form for G, and also recalling that we need to take the limit λ → +0, we arrive at the following integral equation for F (k0, k):

Z 3 0 d p Uk0−p F (p, k) 0 F (k , k) = Uk −k + lim 3 . (20) λ→+0 (2π) k − p + iλ

8To compactify the expression, we omit the summation signs, adopting the convention that there is a summation over all the momenta diﬀerent from k and k0.

9 A very instructive case is the k → 0 limit for a short-ranged potential. Here we have Z 3 0 d p Uk0−p F (p, 0) 0 F (k , 0) = Uk − 3 . (21) (2π) p Note that the singularity of in the denominator becomes integrable allowing us to set λ = 0. If, for clarity, we introduce corresponding pseudo-potential

(psd) Uq = F (q, 0), then Eq. (21) will look like a relationship between the genuine and pseudo potentials: Z 3 (psd) (psd) d p Uq−p Up Uq = Uq − 3 . (22) (2π) p The q → 0 limit is well deﬁned (i.e., it is direction-independent no matter (psd) what is the direction dependence of Uq and thus Uq ):

Z 3 (psd) (psd) (psd) d p U−p Up lim Uq ≡ U0 = U0 − 3 . q→0 (2π) p We conclude that the k → 0 limit corresponds to the s-scattering where we (psd) have direct analogs of Eqs. (9) and (10), up to the replacement U0 → U0 :

2 (psd) 2 m |U0 | dσ = dΩk0 (k → 0), (23) 4π2~4 m2|U (psd)|2 σ = 0 (k → 0). (24) π~4

Scattering a particle: general analysis

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

10 Rather than dealing with the scattering rate and, correspondingly, the evolution operator, we will start with the wave packet formulation and show that the problem of finding/analyzing scattering cross-section reduces to un- derstanding the asymptotic behavior of a stationary solution of the Sch¨rodinger equation with appropriate boundary conditions. In this formulation, we start with a large plane-wave packet as an initial state and a superposition (with asymptotically vanishing spatial overlap ) of the plane-wave packet and a spherical-wave packet in the limit of t → ∞. Let dprˆ be the probability for the particle to get scattered into an infinitesimal solid angle of the magnitude dΩ and the radial directionr ˆ. From the properties of the Sch¨rodinger equation, we know that dprˆ can be represented as an integral of the probability flux through an element of the spherical surface corresponding to the solid angle in question. The radius of the surface has to be appropriately large (and otherwise arbitrary). The flux is a product of the local flux density 2 jout(r, t) and the area r dΩ of the element of the sphere. We thus have Z ∞ 2 dprˆ = r dΩ dt jout(r, t). −∞ The subscript “out” is to remind that here we are talking of the outgoing spherical wave created as a result of scattering of the plane-wave packet. By definition, the differential cross-section dσrˆ is an infinitesimal area of the surface (normal to the wave-vector of the incident plane wave), such that the probability dprˆ is equal to the integral over the time of the plane-wave probability flux through this area. Hence Z ∞ dprˆ = dσrˆ dt jin(t) (defenition of cross-section), −∞ where jin(t) is the (spatially uniform) flux density in the incident plane wave in the vicinity of the scattering center. The size of the wave packet has to be appropriately large to guarantee the quasi-steady-state regime when jout and jin are related to each other as Q(ˆr) j (r, t) = j (t − t ) . out in 0 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 a spherical wave so that the numerator Q(ˆr) depends only on the direction but

11 not on r. Crucially important is the fact that Q(ˆr) does not depend on the shape of the packet and can actually be found from the stationary solution. Equating two expressions for dprˆ, we conclude that

dσrˆ = Q(ˆr) dΩ and thus can be found from the solution of a stationary problem. The relevant stationary problem is looking for the solution of the stationary Schr¨odinger equation with the energy k2/2m and the following asymptotic form of the solution away from the scattering center (the z axis is along the wave vector k of the incident wave):

f(ˆr) ψ (r) → eikz + eikr (r → ∞). (25) k r The initial wave packet at the time moment t = 0 has the form Z ∞ −iqz0 ψpacket(r, t = 0) = Λ(q)ψk+q(r) e dq, (26) −∞

9 where z0 is an appropriately large negative coordinate—the initial position the packet and Λ(q) is an arbitrary envelope function obeying the requirement that it decays with |q| at |q| k, so that the uncertainty of momentum and energy is negligibly small. In view of the requirement on z0, the contribution of the second term in the r.h.s. of (25) is negligible at t = 0 and remains negligible till the propagating packet hits the scattering center. After the packet passes the center, its structure becomes dramatically diﬀerent. Both terms in the r.h.s. of (25) become important. The ﬁrst term is responsible for the plane-wave packet that keeps moving with the constant linear velocity v = k/m while the second term generates a spherical wave packet expanding with the radial velocity equal to the linear velocity of the plane-wave part of the wave function. At large enough time, the overlap of the radial and plane-wave parts of the wave function gets progressively smaller and we can

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

12 totally neglect it in the t → ∞ limit. The desired ratio of (probability) flux densities, Q(ˆr), is thus given by the the flux density corresponding to the second term in the r.h.s. of (25)over the flux density corresponding to the first term in the r.h.s. of (25). Recalling that, for any wave function ψ, the vector of the probability flux density is given by (here we restore ~) i j = ~ (ψ∇ψ∗ − ψ∗∇ψ), (27) 2m which, for the plane-wave part of the stationary solution (25) yields k/m, andr ˆ|f(ˆr)|2k/mr2 for the radial part, we conclude that

Q(ˆr) = |f(ˆr)|2.

We arrive at the formula 2 dσrˆ = |f(ˆr)| dΩ (28) relating the diﬀerential elastic cross-section to the asymptotic form (25) of corresponding stationary wave function. In the case of scattering by a potential, the function f(ˆr) can be directly related to the earlier-introduced function F (k0, k) (note that here k0 =rk ˆ ): m f(ˆr) = − F (ˆrk, k) (scattering by a potential). (29) 2π~2 We will not be deriving this formula because, up to a phase factor, the relation is obvious by comparing (28) to (16). In what follows, we will be assuming spherical symmetry. In this case, f(ˆr) ≡ f(θ) and dΩ = 2π sin θ dθ, with θ the polar angle, and

Z π 2 2 dσθ = 2π sin θ |f(θ)| dθ, σ = 2π |f(θ)| sin θ dθ. (30) 0 The solution is then naturally expanded in terms of spherical harmonics. Since the incident plane wave has only the m = 0 components (no dependence on the azimuthal angle), so does the rest of the wave function and we have

∞ 1 X ψ (r) → P (cos θ) A e−ikr + B eikr (r → ∞). k r l l l l=0

13 In accordance with (25), all the incoming spherical waves ∝ e−ikr are supposed to belong to the plane-wave part of the solution. This ﬁxes the values of all the coeﬃcients Al. With the known expansion for the plane wave ∞ 1 X eikz → (2l + 1)P (cos θ) (−1)l+1 e−ikr + eikr (r → ∞), (31) 2ikr l l=0 we get

∞ 1 X ψ (r) → (2l +1)P (cos θ) (−1)l+1 e−ikr + S eikr (r → ∞). (32) k 2ikr l l l=0 Subtracting (31) from (32) and multiplying the result by r, we get

∞ 1 X f(θ) = (S − 1)(2l + 1)P (cos θ). (33) 2ik l l l=0 For further manipulations, recall the orthonormalization relation for the Legendre polynomials: Z π 2δll0 Pl(cos θ)Pl0 (cos θ) sin θdθ = . (34) 0 2l + 1 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 the calculation for the solution (32), using (27) and (34), we find

∞ π X σ = (2l + 1)(1 − |S |2). (35) i k2 l l=0

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

∞ π X σ = (2l + 1)|1 − S |2. (36) e k2 l l=0 Combining (35) and (36), for the total cross-section we have

∞ 2π X σ = σ + σ = (2l + 1)(1 − Re S ). (37) tot i e k2 l l=0

14 We see that inelastic scattering, if present, inevitably implies elastic one, by the condition |Sl|= 6 1. Another generic observation is that the cross-sections σe and σi (and thus σtot) are sums of independents contributions—called partial 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 with this case—we have

|Sl| = 1, ∀l (purely elastic scattering), suggesting convenient parameterization in terms of the phase shifts

2iδl(k) Sl = e . (38)

∞ 4π X σ = (2l + 1) sin2 δ (k). (39) k2 l l=0 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 differential equation for the radial part, Rkl(r), of the wave function: 1 d 2 dRkl 2 l(l + 1) 2m r + k − − U(r) Rkl = 0, r ∈ [0, ∞). r2 dr dr r2 ~2 (40) By linearity of Eq. (40), one of the two free constants of its solution is just a global factor. The phase shift does not depend on this factor. Indeed, in accordance with (32) and (38), the asymptotic behavior of the Rkl(r) is const R (r) → (−1)l+1 e−ikr + e2iδl(k)eikr (r → ∞), (41) kl r 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 absence of inelastic channels is suﬃcient.

15 realizing that at r → 0, a generic solution of (40) diverges, while the physical solution has to be ﬁnite. This leads to the boundary condition at r = 0 requiring 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 ≤ R , U(r) = ∗ 0 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 parametrization R(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 charged particles interacting via Coulomb potential U(r) = q1q2/r ( q1 and q2 are the two electric charges). For the target particle (the particle at rest in the laboratory frame; a nucleus in the context of the experiment), Rutherford obtained q q 2 dΩ dσ = 1 2 , (42) 4E sin4(θ/2) where E is the kinetic energy of the incident particle (say, α-particle like in Rutherford’s experiment). Rutherford analysis was based on classical mechanics. 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 yield as can be immediately checked) precisely the same exact result. Note, however, 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 the center-of-mass frame, solve there the problem (40)–(41) for the ﬁctitious particle with the reduced 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 ~ → ∞.

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

Breit-Wigner resonances

For the theory of Breit-Wigner resonances, leading to the formula 4π(2l + 1) (Γ/2)2 σ = 2 2 2 , (43) k (k − E0) + (Γ/2) 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 of interaction: −1 k R0 . (44) −1 Then, in the range of distances such that R0 r k , equation (40) simpliﬁes. The potential and the k2 terms can be neglected: 1 d dR l(l + 1) r2 kl − R = 0 (R r k−1), (45) r2 dr dr r2 kl 0 leading to the solution

c(l) R = c(l)rl + 2 (R r k−1), (46) kl 1 rl+1 0

(l) (l) 13 where c1 and c2 are certain constants. Then, solving the equation 1 d dR l(l + 1) r2 kl + k2 − R = 0 (r R ). (47) r2 dr dr r2 kl 0

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

17 with the boundary condition (46), one ﬁnds14

c(l) k2l+1 tan δ = 2 . (48) l (l) (2l − 1)!!(2l + 1)!! c1 Dimensionally, we have (the sign here is a matter of convention)

c(l) 1 2 = −a2l+1, (l) (2l − 1)!!(2l + 1)!! l c1 where al is a certain length. An important—and very natural under the condition (44)—case corresponds to the inequality15

−1 k max(|al|,R0). (49)

Under condition (49), we have

2l+1 −1 |δl| ≈ |alk| 1 [k max(|al|,R0)], (50) and, applying (39) for the partial cross-section in the l-th channel we have

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

In particular, for the s-scattering (l = 0) cross-section, we have (σs ≡ σl=0, as ≡ al=0): 2 −1 σs ≈ 4πas [k max(|as|,R0)]. (52)

Generically, al’s are k-independent under the condition (49), and, once this condition is satisfied, the s-scattering cross-section saturates to a constant value defined by the s-scattering length as, while the partial cross-sections in the l > 0 channels rapidly vanish with decreasing k, in accordance with (51). The only exception from this rule are previously discussed Breit-Wigner resonances. The inequality (44) is not directly relevant to Breit-Wigner resonances. However, if (44) is satisfied, the prediction of the Breit-Wigner formula 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 for either 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 resonances, when the regime |al| R0 (53) occurs in a certain channel l, with k-independent al. Such an anomalous increase of |al| takes place in the vicinity of the threshold of forming a bound state in corresponding channel, with |al| = ∞ right at the threshold. For the resonant condition (53) to take place, the side with respect to the binding 16 (l) threshold is not important. The threshold corresponds to c1 = 0, the sign (l) (l) of the ratio c1 /c2 defining the side with respect to the threshold. For |al| (l) to be anomalously large, the coefficient c1 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, takes place in the interval −1 −1 |al| k R0 . Here from (48) we find

2l+1 2 | tan δl| = |alk| 1 ⇒ sin δl ≈ 1, (54) so that the partial cross-section in the l-th channel is equal to its maximal possible value:

4π(2l + 1) σ = (|a |−1 k R−1). (55) l k2 l 0 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 the substitution R(r) = χ(r)/r, we cast Eq. (47) into

χ00 + k2χ = 0. (56)

The boundary condition (46) is conveniently parameterized as

0 χ = −κ, (57) χ r→0 with c1 1 κ = − ≡ . (58) c2 as Solving the problem (56)–(58) one ﬁnds the expression for the phase shift:

cot δ = −κ , 0 k and then the expression for the scattering amplitude17 1 1 1 f = e2iδ0 − 1 = = − . 2ik k(cot δ0 − i) κ + ik For the total cross-section we thus get 4π σ = 4π|f|2 = (59) κ2 + k2. Equation (59) is consistent with Eqs. (55) and (52) in the regions of applicability 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 2mE 2 2 χ00 + χ = 0 ⇒ E = −~ κ . ~2 2m

The resonant condition |as| R0 implies that the wave function of the bound state is localized mostly outside the range of interaction. Consistent 2 2 with that, we have |E| ~ /(mR0), meaning that the binding is weak com- 2 2 pared to the characteristic energy scale ~ /(mR0).

17The total scattering amplitude here essentially coincides with the s-scattering amplitude, 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 interacting Hamiltonian reads

1 X † † V = Uq aˆ aˆ aˆk aˆk . (60) 2 k1−q k2+q 2 1 k1,k2,q

In the center-of-mass frame, in the initial state, |ii, there are two particles with the opposite momenta, k and −k. In the ﬁnal state, |fi, there are two particles with momenta k0 and −k0. In the Hamiltonian (60), there are four terms leading from |ii to |fi:

0 Term 1: k1 = k, k2 = −k, q = k − k , 0 Term 2: k1 = −k, k2 = k, q = k − k, 0 Term 3: k1 = k, k2 = −k, q = k + k , 0 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 with only two diﬀerent terms:

† † † † U|k−k0| aˆk0 aˆ−k0 aˆ−kaˆk + U|k+k0| aˆ−k0 aˆk0 aˆ−kaˆk. Taking into account (anti)commutation relations, these two terms can be combined into one

† † U|k−k0| ± U|k+k0| aˆk0 aˆ−k0 aˆ−kaˆk , (61) where the sign plus (minus) is for bosons (fermions). Going from microscopic interaction Hamiltonian (61) to the eﬀective pseudo-perturbative Hamilto- nian amounts to the replacement

0 0 U|k−k0| → F (k ,k),U|k+k0| → F (−k , k).

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

21 The resulting eﬀective Hamiltonian reads

0 0 † † [F (k ,k) ± F (−k , k)]a ˆk0 aˆ−k0 aˆ−kaˆk . (62) Adapting relation (63) to our case of two particles by m → m/2, i.e., by taking into account that the mass in (63) is supposed to be replaced with the reduced mass, we have m f(ˆr) = − F (ˆrk, k)(k0 =rk ˆ ). (63) 4π~2 This allows us to express the eﬀective scattering term (61) as

2 4π~ † † − [f(θ) ± f(π − θ)]a ˆ 0 aˆ 0 aˆ aˆ , (64) m k −k −k k where θ is the angle between the vectors k0 and k. Equation (64) is the most general form of the effective elastic-scattering term for two spinless identical particles (in the center-of-mass frame). It applies equally well to the (mostly academic) case of potential scattering of two point particles and the case of elastic scattering of two identical composite bosons or fermions, provided each of the two particles has zero total spin. A comment is in order here concerning the (somewhat annoying) negative sign in (63) and, correspondingly, in (64). We could easily get rid of the negative sign by simply redefining f(ˆr) in Eq. (25): f(ˆr) → −f(ˆr). A conventional way of achieving essentially the same goal is, however, slightly different. One writes Eq. (25) the way it was written, but ultimately intro- duces the notion of scattering length as nothing but negative f. Note that the sing in our definition of partial scattering lengths, al, is consistent with this convention. In the case when f is dominated by the s-scattering, we simply have f = −as and the effective scattering term for bosons becomes

2 2π~ as † † aˆ 0 aˆ 0 aˆ aˆ (bosons, low energy). (65) m k −k −k k For fermions, corresponding eﬀective scattering term is essentially zero, up to small corrections from l > 0 terms. Comparing (65) to (61), we conclude that the low-energy eﬀective interaction Hamiltonian for a dilute low-temperature Bose gas reads

U∗ X † † Veﬀ = aˆ aˆ aˆk aˆk (bosons, low energy), (66) 2 k1−q k2+q 2 1 k1,k2,q

22 with 4π 2a U = ~ s (67) ∗ m 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 or internal momentum,19 our interaction Hamiltonian becomes

1 X X † † V = Uq(σ4, σ3; σ2, σ1)a ˆ aˆ aˆk σ aˆk σ . (68) 2 (k1−q)σ4 (k2+q)σ3 2 2 1 1 {σ} k1,k2,q

In the center-of-mass frame, in the initial state, |ii, there are two particles with the opposite momenta, k and −k. The particle with the momentum k is in the spin state σa, and the particle with the momentum −k is in the 20 spin state σb. The spin states can be the same or diﬀerent. In the ﬁnal state, |fi, there are two particles with momenta k0 and −k0, and the same spin states.21 Without loss of generality, we will assume that 0 the momentum k is associated with the σa spin state, and the momentum 0 −k is associated with the σb spin state. Similarly to the case of spinless particles, in the Hamiltonian (68), there are four terms leading from |ii to |fi:

0 Term 1: k1 = k, k2 = −k, q = k−k , σ1 = σ4 = σa, σ2 = σ3 = σb, 0 Term 2: k1 = −k, k2 = k, q = k −k, σ1 = σ4 = σb, σ2 = σ3 = σa, 0 Term 3: k1 = k, k2 = −k, q = k+k , σ1 = σ3 = σa, σ2 = σ4 = σb, 0 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 the term “spin.” 20 At σb 6= σa, the indistinguishability is still relevant, as long as the interaction can swap 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 we are left with only two diﬀerent terms:

† † U 0 (σ , σ ; σ , σ )a ˆ 0 aˆ 0 aˆ aˆ |k−k | a b b a k σa −k σb −kσb kσa † † + U 0 (σ , σ ; σ , σ )a ˆ 0 aˆ 0 aˆ aˆ , |k+k | b a b a −k σb k σa −kσb kσa which, analogously to (61), combine into one scattering term † † U 0 (σ , σ ; σ , σ ) ± U 0 (σ , σ ; σ , σ ) aˆ 0 aˆ 0 aˆ aˆ . (69) |k−k | a b b a |k+k | b a b a k σa −k σb −kσb kσa So far, there were not substantial diﬀerence with spinless case. But the next step—going from Fourier components of microscopic potential to scattering amplitudes—becomes less straightforward. Now reducing the two- body problem to a single-particle one is possible only if the motion in the subspace of interparticle distance separates not only from the center-of-mass coordinate (which is always the case), but from the spin degrees of freedom as well. The latter is normally possible, but requires that we work in the representation of the eigenstates of the total spin of two particles. Hence, we have to decompose Uq(σ4, σ3; σ2, σ1) as

X (S) Uq(σ4, σ3; σ2, σ1) = Uq PS(σ4, σ3; σ2, σ1), S where S stands for the total spin of the two particles (say, S = 0 and S = 1, (S) if our two particles are spin-1/2 particles), Uq 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 spin degrees of freedom, going from microscopic interaction Hamiltonian to the eﬀective pseudo-perturbative Hamiltonian amounts to the replacement

(S) (S) 0 (S) (S) 0 U|k−k0| → F (k ,k),U|k+k0| → F (−k , k), leading to the eﬀective Hamiltonian

X (S) 0 F (k ,k)PS(σa, σb; σb, σa) ± S 22Here we also take into account that the indistinguishability of the particles implies Uq(σ4, σ3; σ2, σ1) = Uq(σ3, σ4; σ1, σ2). 23Normally, the potential or, speaking generally, interaction, is S-dependent.

24 (S) 0 † † ± F (−k , k)P (σ , σ ; σ , σ ) aˆ 0 aˆ 0 aˆ aˆ . (70) S b a b a k σa −k σb −kσb kσa In terms of the scattering amplitudes f (S)(θ), we have m X − f (S)(θ)P (σ , σ ; σ , σ ) ± 4π 2 S a b b a ~ S

(S) † † ± f (π − θ)P (σ , σ ; σ , σ ) aˆ 0 aˆ 0 aˆ aˆ . (71) S b a b a k σa −k σb −kσb kσa