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Physics LibreTexts

9.2: Fundamental

( \newcommand{\kernel}{\mathrm{null}\,}\)

Consider time-independent scattering theory, for which the Hamiltonian of the system is written

H0 is the Hamiltonian of a free particle of mass H0=p22m, ???

and |ϕ be an energy eigenket of H0|ϕ=E|ϕ, ???

whose wavefunction ϕ(x) . This state is assumed to be a plane wave state or, possibly, a spherical wave state. Schrödinger's equation for the scattering problem is

|ψ is an energy eigenstate of the total Hamiltonian whose wavefunction ψ(x) . In general, both H0+H1 have continuous energy spectra: i.e., their energy eigenstates are unbound. We require a solution of Equation ??? that satisfies the boundary condition H10 . Here, (2+k2)ψ(x)=2m2x|H1|ψ, ???

where

ψ(x)=ϕ(x)+2m2d3xG(x,x)x|H1|ψ, ???

where

|ψ|ϕ
as G(x,x)=exp(±ik|xx|)4π|xx|. ???

Thus, Equation ??? becomes

$ \psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^2} \int d^3...
...rt{\bf x} - {\bf x}'\vert}\, \langle {\bf x}' \vert\,H_1\,\vert\psi^\pm\rangle.$ ???

Let us suppose that the scattering Hamiltonian, x|H1|x=V(x)δ(xx). ???

We can write

$ \langle {\bf x}'\vert\,H_1\,\vert \psi^\pm\rangle = \int d^3 x''\...
...e \langle {\bf x}'' \vert\psi^\pm\rangle\, = V({\bf x}') \,\psi^\pm ({\bf x}').$ ???

Thus, the integral equation ??? simplifies to

$ \psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^2} \int d^3...
...\vert)}{4\pi\,\vert{\bf x} - {\bf x}'\vert}\, V({\bf x}')\, \psi^\pm({\bf x}').$ ???

Suppose that the initial state k (i.e., a stream of particles of definite momentum |k . The associated wavefunction takes the form

x|k=exp(ikx)(2π)3/2. ???

The wavefunction is normalized such that

$ \langle {\bf k}\vert{\bf k}'\rangle =\int d^3 x\, \langle {\bf k}...
...\, {\bf x}\cdot({\bf k} -{\bf k}')]} {(2\pi )^3} = \delta ({\bf k} - {\bf k'}).$ ???

Suppose that the scattering potential x=0 ). Let us calculate the wavefunction rr . It is easily demonstrated that

r/r
, where

r=|x|
and k=ker. ???

Clearly, k is the wavevector for particles that possess the same energy as the incoming particles (i.e., exp(±ik|xx|)exp(±ikr)exp(ikx). ???

In the large-r limit, Equation ??? reduces to

$ \psi^\pm({\bf x}) \simeq \frac{\exp(\,{\rm i}\,{\bf k}\cdot{\bf x...
...exp(\mp {\rm i} \,{\bf k}' \cdot {\bf x}')\, V({\bf x}')\, \psi^\pm ({\bf x}').$ ???

The first term on the right-hand side is the incident wave. The second term represents a spherical wave centred on the scattering region. The plus sign (on ψ(x)=1(2π)3/2[exp(ikx)+exp(ikr)rf(k,k)], ???

where

$ f({\bf k}', {\bf k}) = - \frac{(2\pi)^2 \,m}{\hbar^2} \int d^3{\b...
... - \frac{(2\pi)^2 \,m}{\hbar^2}\, \langle {\bf k}'\vert\,H_1\,\vert\psi\rangle.$ ???

Let us define the differential cross-section, dΩ , divided by the incident flux of particles. Recall, from Chapter 3, that the probability current (i.e., the particle flux) associated with a wavefunction j=mIm(ψψ). ???

Thus, the probability flux associated with the incident wavefunction,

jinc=(2π)3mk. ???

Likewise, the probability flux associated with the scattered wavefunction,

jsca=(2π)3m|f(k,k)|2r2ker. ???

Now,

dσdΩ=|f(k,k)|2. ???

Thus, k to be scattered into states whose momentum vectors are directed in a range of solid angles k . Note that the scattered particles possess the same energy as the incoming particles (i.e., k=k ). This is always the case for scattering Hamiltonians of the form specified in Equation ???.

Contributors

  • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 9.2: Fundamental is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

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