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4.2 Quantum mechanical scattering

We consider scattering from a finite range potential, which can be expressed in the more precise mathematical relation rV (r) → 0 as r →∞. We use the “time-independent approach”, (see Mandl, Quantum Mechanics, Chapter 11).

The idea is that a beam of particles, that is not very dense so that they don’t significantly interact which each other, impinges on a target described by the the potential V (r). If we observe the particles far away, we are really interested in their energy which must be positive for scattering states. Thus we write

ψ(r,t) = ϕ(r){e}^{−iEt∕ℏ}\quad ,

where ϕ satisfies the time-independent Schrödinger equation

−{{ℏ}^{2}\over 2m}{\mathop{∇}}^{2}ϕ(r) + V (r)ϕ(r) = Eϕ(r).

For positive energy we introduce a wave number k, E = {ℏ}^{2}{k}^{2}∕(2m), and find

({\mathop{∇}}^{2} + {k}^{2})ϕ(r) = {2m\over {ℏ}^{2}} V (r)ϕ(r) = ρ(r).

Here we replaced the right-hand side temporarily by an independent function, for reasons that will become apparent below.

As usual we still need to determine the boundary conditions. We have two processes that describe those: there is an incoming beam that outside the range of interaction becomes a plane wave, ϕ(r) → {e}^{ikz}, and we do have the scattered particles, which are a result of the interaction with the potential. These are most naturally described in terms of spherical waves.

Spherical waves are solutions to the radial part of the Laplace operator,
−{1\over { r}^{2}} {d\over dr}{r}^{2} {d\over dr}f(r) = {k}^{2}f(r).

Solutions are {{e}^{±ikr}\over r} . Using the radial part of the momentum operator {p}_{r} ={ e}_{r}{ℏ\over i} {d\over dr}, we find that the plus sign in the exponent corresponds to outward travelling ways; the minus sign is thus for an incoming wave.

Outgoing spherical waves are of the form

f(θ,ϕ){{e}^{ikr}\over r} ,\quad \text{$r$ large}\quad ,

so we have the “asymptotic” boundary condition

ϕ(r) → {e}^{ikz} + f(θ,ϕ){{e}^{ikr}\over r} \qquad \text{for $r →∞$}\quad .

Also, the scattering amplitude f(θ,ϕ) goes to zero as V goes to zero.

If we ignore the fact that ρ depends on ϕ (for the time being!) and write ϕ(r) = {e}^{ikz} + χ(r), we find we have to solve the equation

({\mathop{∇}}^{2} + {k}^{2})χ(r) = ρ(r),

subject to the boundary condition

χ(r) → f(θ,ϕ){{e}^{ikr}\over r} \qquad \text{for $r →∞$}.

Solve this by the Green’s function method,

({\mathop{∇}}^{2} + {k}^{2})G(r,r') = δ(r −r')\quad ,

or using translational invariance,

({\mathop{∇}}^{2} + {k}^{2})G(r) = δ(r)\quad .

Actually, since we know the solution to

{\mathop{∇}}^{2}{G}_{ 0}(r) = δ(r)

is

{G}_{0} = − {1\over 4π} {1\over r}\quad ,

we know that the G must approach this solution for r ↓ 0. With the fact that

({\mathop{∇}}^{2} + {k}^{2}){{e}^{±ikr}\over r} = 0

for r\mathrel{≠}0, and that the boundary conditions require a plus sign in the exponent, we conclude that

G(r,r') = − {1\over 4π} {{e}^{ik|r−r'|}\over |r −r'|} .

Using Stoke’s theorem, it is not hard to show that

\begin{eqnarray*} {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{\text{sphere}}dV \kern 1.66702pt ({\mathop{∇}}^{2} + {k}^{2}){{e}^{ikr}\over r} & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{\text{sphere}}dS ⋅\mathop{∇}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{\text{sphere}} + {k}^{2}{\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{\text{sphere}}dV \kern 1.66702pt {{e}^{ikr}\over r} & %& \\ & = 4π{R}^{2}{\left [ {d\over dr} {{e}^{ikr}\over r} \right ]}_{r=R} + {k}^{2}4π{\mathop{\mathop{\mathop{∫}\nolimits }}\nolimits }_{ 0}^{R}{e}^{ikr}r\kern 1.66702pt dr & %& \\ & = 4π{e}^{ikR}(ikR − 1) + 4π\left (−1 − {e}^{ikR}(ikR − 1)\right ) = −4π.& %& \\ \end{eqnarray*}

This also shows we really have a δ function.

We thus find, substituting the solution for χ and its relation to ϕ,
ϕ(r) = {e}^{ikz} +\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'G(r,r')ρ(r').

If we also remember how ρ is related to ϕ, we get (the coordinate representation of) the integral equation for scattering,

ϕ(r) = {e}^{ikz} + {2m\over {ℏ}^{2}} \mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'G(r,r')V (r')ϕ(r')\quad .
(4.13)

This equation is called the Lipmann-Schwinger equation.

The Born approximation

One way to tackle the scattering problem for a weak potential is to solve the problem by iteration, i.e., each time a ϕ appears we replace it by the right-hand side of (4.13). This results in the equation

\begin{eqnarray*} ϕ(r)& =& {e}^{ikz} + {2m\over {ℏ}^{2}} \mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'G(r,r')V (r'){e}^{ikz'} + %& \\ & &{ \left ({2m\over {ℏ}^{2}} \right )}^{2}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'{d}^{3}r''G(r,r')V (r')G(r',r'')V (r''){e}^{ikz''} + \mathop{\mathop{…}}%& \\ \end{eqnarray*}

For weak potentials we can truncate at first order:

ϕ(r) = {e}^{ikz} − {m\over 2π{ℏ}^{2}}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'{{e}^{ik|r−r'|}\over |r −r'|} V (r'){e}^{ikz'}\quad .

To extract f(θ,ϕ) we only need to know the behaviour for large r. Write k = k\hat{z}, kz = k ⋅r. Also k' ≡ k(r∕r). For r ≫ r',

|r −r'|≈ r\left [1 −r ⋅r'∕{r}^{2} + \mathop{\mathop{…}}\right ]

so

ϕ(r) → {e}^{ikz} + f(θ,ϕ){{e}^{ikr}\over r} \quad ,

with

f(θ,ϕ) = − {m\over 2π{ℏ}^{2}}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {d}^{3}r'V (r'){e}^{i(k−k')⋅r}\quad .

This is called the Born approximation.

This is a good approximation in electron scattering from atoms, for example.

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