Scattering Theory

Consider scattering of two particles in the center of mass frame, or equivalently scattering of a single particle from a potential V (r), which becomes zero suciently fast as r → ∞. The initial state is |ki, and the nal state after scattering is |k0i. The scattering matrix (S-matrix) describes probabilities that scattering events between dierent pairs of channels happen:

(±) (±) (±) Hψk (r) = (H0 + V (r))ψk (r) = Ekψk (r) which take the asymtotic form far away from the scattering center:

f(Ω) ψ(±)(r) = φ (r) + e±ikr k k r where

ikr φk(r) = e

The function f(Ω) of the solid angle is called scattering amplitude. It can be related to the scattering matrix in the following way. Start from the Lippmann-Schwinger equation:

(±) 1 (±) ψk (r) = φk(r) + + V ψk (r) Ek − H0 ± i0

which can be easily derived from the full Schrodinger equation and Hφk = Ekφk. By recursive substitution of the left-hand side into the right-hand side, one obtains: (±) 1 P |ψk i = 1 + + V |ki = 1 + V ∓ iπδ(Ek − H)V |ki Ek − H ± i0 Ek − H Thus:

(+) (−) |ψk i − |ψk i = −2πiδ(Ek − H)V |ki Now, using this expression we nd that the S-matrix takes form:

S = 1 − 2πiδ(Ek0 − Ek)T Next, we explicitly rewrite the Lippmann-Schwinger equation in the coordinate representation. To this end, + −1 2 we need to express the operator (Ek − H0 ± i0 ) in the coordinate representation (Ek = k /2m):

1 ikr 2me Z 0 0 ψ(+)(r) = φ (r) − d3r0e−ik r V (r0)ψ(+)(r0) k k 4πr k

Here, the vector k0 has the same magnitude as k, but a dierent direction - the same as r, which is that of the incident wave. Therefore, we can identify the scattering amplitude as: Z m 3 0 ∗ 0 0 (+) 0 m 0 (+) m 0 f(Ω) = − d r φ 0 (r )V (r )ψ (r ) = − hk |V |ψ i = − hk |T |ki 2π k k 2π k 2π Finally, the relationship between the scattering matrix and the scattering amplitude is: 0 3 0 2π hk |S|ki = (2π) δ(k − k) − 2πiδ(E 0 − E ) × − f(Ω) k k m

Using 0 2 −1 0 0 0 and m 0 we can write: δ(k − k) = (k sin θ) δ(k − k)δ(θ − θ)δ(φ − φ) δ(Ek0 − Ek) = k δ(k − k) δ(k0 − k) δ(θ0 − θ)δ(φ0 − φ) ik hk0|S|ki = (2π)3 + f(Ω) k2 sin θ 2π which facilitates derivation of the fowrard-scattering matrix elements - these will be diagonal elements of the S-matrix. The diagonal S-matrix elements are readily uncovered by integrating out k0, θ0 and φ0 at Ω = 0 (this removes the Dirac-function normalization and isolates the magnitude of the matrix elements):

hk|S|ki = (1 + 2ikf(0)) δ(0) At the end, it is worth noting that unitarity of the S-matrix implies the following equation for the T-matrix:

† † T − T = 2πiδ(Ef − Ei)T T from which follows the optical theorem (see below)

Scattering Cross-Section

In scattering, a dierential cross section is dened by the probability to observe a scattered particle in a given quantum state per solid angle unit, such as within a given cone of observation, if the target is irradiated by a ux of one particle per surface unit. Again, for a single-particle scattering on a potential, the asymptotic wavefunction is: f(Ω) ψ(r) ∝ eikz + eikr r where the rst term is an incomming plane-wave, and the second term is an outgoing radial wave, the scattered s-wave. The scattering amplitude f(Ω) is related to the dierential scattering cross-section (by denition):

dσ = |f(Ω)|2 dΩ The scattering amplitude for a spherically symmetric potential can be expanded in powers of the angular momentum l: ∞ −1 X 2iδl f(θ) = (2ik) (2l + 1)(e − 1)Pl(cos θ) l=0

where δl are phase shifts (phase dierences between the scattered waves at angular momenta l and the incident wave at innity), and Pl(x) are Legendre polynomials.

2 The total cross-section is the solid-angle integral of the dierential cross-section:

∞ Z dσ 4π X σ = dΩ = (2l + 1)(e2iδl − 1) dΩ k2 l=0 and satises the optical theorem:

4π σ = Im{f(0)} k These expressions establish the following relationships:

2 dσ m 2 = |T 0 | dΩ 2π k k

(2π)2 S 0 = δ 0 − 2πiδ(E 0 − E )T 0 = δ 0 + i δ(E 0 − E )f(Ω) k k k k k k k k k k m k k

k2σ Re{S } = 1 − 2kIm{f(0)} = 1 − kk 2π Scattering on a hard-sphere; scattering length

If the scattering potential is innite for r < a and zero for r > a, then two limits can be identied. For ka 1 (ν ≡ −1/a), (ka)2l+1 tan δ ≈ − l (2l + 1)((2l − 1)!!)2

1 f(θ) ≈ k2 2 θ ν − ik − ν 2 cos ( 2 )

4πa2 σ ≈ 1 − (ka)2(1 + 2 cos θ)

This limit is obtained for a general short-ranged (decreasing with r fast enough) potential at suciently small momenta. It can be used as a denition of the scattering length a even when potentials are not hard- core. Perhaps these expressions are also used to dene the scattering length in general in atomic physics; depending on the scattering physics, f(Ω) derived from the S-matrix may give a negative, or arbitrarily large value for a, and this is controlled by the detuning ν from the Feshbach resonance. For ka 1, σ ≈ 2πa2 (twice the classical result due to diraction eects)

Scattering on a quantum well; resonant scattering and unitarity limit A quantum well potential represents attractive interactions between particles in the center of mass reference frame, and can be idealized by for and for (with ). Wavenumber of V (r) = −V0 r < a V (r) = 0 r > a V0 > 0 √ a free particle with momentum will be p 2 2 while inside the quantum well, where k αk = α0 + k α0 = 2mV0 (we set ). After solving for the phase shifts the following approximate results for the scattering ~ = 1 δl cross-section are obtained. In the limit ka 1: tan(α a) tan(α a) σ ≈ 4πa2 1 − k ≈ 4πa2 1 − 0 αka α0a

3 The scattering cross-section is mostly independent of k in this limit, but by changing V0 the quantity α0a can approach an odd-integer multiple of π/2 and give rise to a resonance. The expression for σ has to be rederived then, whenever the energy of an incomming particle is close to the energy of a bound (or metastable) state in the quantum well.

For α0a ≈ π/2 or so (the partial wave l = 0 in resonance with a bound state at zero energy): 4π 4π σ ≈ 2 2 ≈ 2 k + α0 cot(α0a) k This is the situation in which the scattering cross-section acquires a signicant momentum dependence, and loses dependence on the microscopic scales that characterize the scattering potential. The microscopic details of the interactions between scattering particles cease to matter at the resonance, and a degree of universality is obtained, called the unitarity limit. The forward S-matrix in the unitarity limit is:

kσ 1 −if(0) = Im{f(0)} = ≈ 4π k

S = (1 + 2ikf(0)) δ(0) = −δ(0)

Note that the scattering amplitude f(0) is purely imaginary at the resonance, reecting the tendency of the attractive potential to bind the scattering particle. This is formally shown by calculating the l = 0 phase shift δl, which turns out to be π/2 exactly at the resonance (if other values of l do not lie exactly at the resonance, their contribution to σ will be negligible for k → 0 compared to the contribution of the l = 0 wave). If one insisted to dene the scattering length even in the vicinity of the resonance, which is not necessarily a natural thing to do, then one could use the expressions for the hard-sphere scattering as denitions of the scattering length a and nd that it diverges (ν = −1/a → 0): 1 1 f(θ) ≈ → ν − ik −ik This is measured in ultra-cold atom experiments as a function of magnetic eld H by tting to the following empirical formula: 1 Hw a = − = abg 1 − ν H − H0

where abg is the background (far-o resonance) scattering length, and Hw the resonance width (both measured experimentally; H0 is a eld that characterizes the microscopic aspects of fermion interactions). −1 The width of the resonance is characterized by a parameter γ ∝ (kF r0) , where r0 is an eective range (of the pairing interactions?).

Scattering in one dimension In one spatial dimension the previous denitions of the scattering cross-section make no sense, but it is

still possible to dene the S-matrix and scattering amplitude. Consider a potential barrier V (x) = V0 for 0 < x < a, and V (x) = 0 otherwise (take V0 > 0). An incomming plane wave partially reects and partially passes through the barrier, which we can write asymptotically:

ikx ikx f+e , x → ∞ ψ(x) = e + −ikx f−e , x → −∞ The solution of the Schrodinger equation gives:

i(α −k)a 4kαke k f+ = 2 2 2iα a − 1 (k + αk) − (k − αk) e k

4 2 2 2iαka k − αk 1 − e f− = 2 2 2iα a (k + αk) − (k − αk) e k where p and 2 . When (low energies - barrier bottom), or αk = 2m(Ek − V0) Ek = k /2m k |αk| k |αk| (energies close to barrier top, but only in the limit V0 → ∞ so that k → ∞), and sin(αka) 6= 0, a universal regime is obtained: f+ → −1, f− → 1 corresponding to full backward scattering. At high energies αk ≈ k, i(α −k)a so that f+ → e k − 1 and f− → 0, meaning that only forward scattering occurs albeit with a phase shift . (αk − k)a → ka − amV0/k