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Next: Born Approximation Up: Scattering Theory Previous: Introduction

Fundamental Equations

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

$\displaystyle H = H_0 +H_1,$ (10.1)


$\displaystyle H_0 = \frac{p^{\,2}}{2\,m}$ (10.2)

is the Hamiltonian of a free particle of mass $ m$ , and $ H_1$ represents the non-time-varying source of the scattering. Let $ \vert\phi\rangle$ be an energy eigenket of $ H_0$ ,

$\displaystyle H_0\, \vert\phi\rangle = E\, \vert\phi\rangle,$ (10.3)

whose wavefunction is $ \phi({\bf x}')\equiv \langle {\bf x}'\vert\phi\rangle$ . This wavefunction is assumed to be a plane wave. Schrödinger's equation for the scattering problem is

$\displaystyle (H_0 + H_1)\, \vert\psi\rangle = E\,\vert\psi\rangle,$ (10.4)

where $ \vert\psi\rangle$ is an energy eigenstate of the total Hamiltonian whose wavefunction is $ \psi({\bf x}')\equiv \langle{\bf x}'\vert\psi\rangle$ . In general, both $ H_0$ and $ H_0+H_1$ have continuous energy spectra: that is, their energy eigenstates are unbound. We require a solution of Equation (10.4) that satisfies the boundary condition $ \vert\psi\rangle \rightarrow \vert\phi\rangle$ as $ H_1\rightarrow 0$ . Here, $ \vert\phi\rangle$ is a solution of the free-particle Schrödinger equation, (10.3), that corresponds to the same energy eigenvalue as $ \vert\psi\rangle$ .

Adopting the Schrödinger representation (see Section 2.4), we can write the scattering equation, (10.4), in the form

$\displaystyle (\nabla^{\,2} + k^{\,2})\,\psi({\bf x}) = \frac{2\,m}{\hbar^{\,2}}\, \langle {\bf x} \vert\,H_1\,\vert \psi\rangle,$ (10.5)


$\displaystyle E = \frac{\hbar^{\,2} \,k^{\,2}}{2\,m}.$ (10.6)

(See Exercise 1.) Equation (10.5) is known as the Helmholtz equation, and can be inverted using standard Green's function techniques [49]. Thus,

$\displaystyle \psi({\bf x}) = \phi({\bf x}) + \frac{2\,m}{\hbar^{\,2}} \int d^{...
...{\bf x}'\,G({\bf x}, {\bf x}') \,\langle {\bf x}' \vert\,H_1\,\vert\psi\rangle,$ (10.7)


$\displaystyle (\nabla^{\,2} + k^{\,2})\,G({\bf x}, {\bf x}') = \delta^{\,3}({\bf x} -{\bf x}').$ (10.8)

(See Exercise 2.) Here, $ \delta^{\,3}({\bf x})$ is a three-dimensional Dirac delta function. Note that the solution (10.7) satisfies the previously mentioned constraint $ \vert\psi\rangle \rightarrow \vert\phi\rangle$ as $ H_1\rightarrow 0$ . As is well known, the Green's function for the Helmholtz equation is given by

$\displaystyle G({\bf x}, {\bf x}') = -\frac{\exp(\pm {\rm i}\,k\, \vert{\bf x} - {\bf x}'\vert\,)}{4\pi\,\vert{\bf x} - {\bf x}'\vert}.$ (10.9)

(See Exercise 3.) Thus, Equation (10.7) becomes

$\displaystyle \psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^{\,2}} \int...
...rt{\bf x} - {\bf x}'\vert}\, \langle {\bf x}' \vert\,H_1\,\vert\psi^\pm\rangle.$ (10.10)

Let us suppose that the scattering Hamiltonian, $ H_1$ , is a function only of the position operators. This implies that

$\displaystyle \langle {\bf x}'\vert\,H_1\,\vert{\bf x}\rangle = V({\bf x})\, \delta^{\,3}({\bf x} -{\bf x}').$ (10.11)

We can write

$\displaystyle \langle {\bf x}'\vert\,H_1\,\vert \psi^\pm\rangle$ $\displaystyle = \int d^{\,3} {\bf x}''\,\langle {\bf x}'\vert\,H_1\,\vert{\bf x}''\rangle \langle {\bf x}'' \vert\psi^\pm\rangle$    
  $\displaystyle =\int d^{\,3}{\bf x}'' \,V({\bf x}')\,\delta^{\,3}({\bf x}'-{\bf x}'')\,\psi({\bf x}'') = V({\bf x}') \,\psi^\pm ({\bf x}'),$ (10.12)

where use has been made of the standard completeness relation $ \int d^{\,3} {\bf x}''\vert{\bf x}''\rangle \langle {\bf x}'' \vert=1$ . (See Section 1.15.) Thus, the integral equation (10.10) simplifies to give

$\displaystyle \psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^{\,2}} \int...
...\vert)}{4\pi\,\vert{\bf x} - {\bf x}'\vert}\, V({\bf x}')\, \psi^\pm({\bf x}').$ (10.13)

Suppose that the initial state, $ \vert\phi\rangle$ , possesses a plane-wave wavefunction with wavevector $ {\bf k}$ (i.e., it corresponds to a stream of particles of definite momentum $ {\bf p} = \hbar \,{\bf k}$ ). The ket corresponding to this state is denoted $ \vert{\bf k}\rangle$ . Thus,

$\displaystyle \phi({\bf x})\equiv \langle {\bf x} \vert {\bf k}\rangle = \frac{ \exp(\,{\rm i}\,{\bf k}\cdot{\bf x}) }{(2\pi)^{3/2}}.$ (10.14)

The preceding wavefunction is conveniently normalized such that

$\displaystyle \langle {\bf k}\vert{\bf k}'\rangle =\int d^{\,3} {\bf x}\, \lang...
...f x}\cdot({\bf k} -{\bf k}')]} {(2\pi )^3} = \delta^{\,3} ({\bf k} - {\bf k'}).$ (10.15)

(See Section 2.6 and Exercise 4.)

Suppose that the scattering potential, $ V({\bf x})$ , is non-zero only in some relatively localized region centered on the origin ( $ {\bf x} = {\bf0}$ ). Let us calculate the total wavefunction, $ \psi({\bf x})$ , far from the scattering region. In other words, let us adopt the ordering $ r\gg r'$ , where $ r=\vert{\bf x}\vert$ and $ r'=\vert{\bf x}'\vert$ . It is easily demonstrated that

$\displaystyle \vert{\bf x} - {\bf x}'\vert \simeq r - {\bf e}_r\cdot{\bf x}'$ (10.16)

to first order in $ r'/r$ , where $ {\bf e}_r={\bf x}/r$ is a unit vector that is directed from the scattering region to the observation point. Let us define

$\displaystyle {\bf k}' = k\,{\bf e}_r.$ (10.17)

Clearly, $ {\bf k}'$ is the wavevector for particles that possess the same energy as the incoming particles (i.e., $ k'=k$ ), but propagate from the scattering region to the observation point. Note that

$\displaystyle \exp(\pm {\rm i}\, k\,\vert{\bf x} - {\bf x}' \vert\,) \simeq \exp(\pm {\rm i}\, k \,r) \exp(\mp {\rm i}\, {\bf k}' \cdot {\bf x}').$ (10.18)

In the large-$ r$ limit, Equations (10.13) and (10.14) reduce to

$\displaystyle \psi^\pm({\bf x})$ $\displaystyle \simeq \frac{\exp(\,{\rm i}\,{\bf k}\cdot{\bf x})}{ (2\pi)^{3/2}}$    
  $\displaystyle \phantom{=} -\frac{m}{2\pi\,\hbar^{\,2}} \frac{\exp(\pm{\rm i}\,k...
...exp(\mp {\rm i} \,{\bf k}' \cdot {\bf x}')\, V({\bf x}')\, \psi^\pm ({\bf x}').$ (10.19)

The first term on the right-hand side of the previous equation is the incident wave. The second term represents a spherical wave centered on the scattering region. The plus sign (on $ \psi^\pm$ ) corresponds to a wave propagating away from the scattering region, whereas the minus sign corresponds to a wave propagating toward the scattering region. (See Exercise 5.) It is obvious that the former represents the physical solution. Thus, the wavefunction far from the scattering region can be written

$\displaystyle \psi({\bf x}) = \frac{1}{(2\pi)^{3/2}} \left[\exp(\,{\rm i}\,{\bf k}\cdot{\bf x}) + \frac{\exp(\,{\rm i}\,k\,r)}{r} f({\bf k}', {\bf k}) \right],$ (10.20)


$\displaystyle f({\bf k}', {\bf k})$ $\displaystyle = - \frac{(2\pi)^2 \,m}{\hbar^{\,2}} \int d^{\,3}{\bf x}' \, \fra...
...{\rm i}\,{\bf k}'\cdot {\bf x}' ) }{(2\pi)^{3/2}}\, V({\bf x}')\,\psi({\bf x}')$    
  $\displaystyle = - \frac{(2\pi)^2 \,m}{\hbar^{\,2}}\, \langle {\bf k}'\vert\,H_1\,\vert\psi\rangle.$ (10.21)

[See Equations (10.11) and (10.14).]

Let us define the differential scattering cross-section, $ d\sigma/d{\mit\Omega}$ , as the number of particles per unit time scattered into an element of solid angle $ d{\mit\Omega}$ , divided by the incident particle flux. Recall, from Chapter 3, that the probability current (which is proportional to the particle flux) associated with a wavefunction $ \psi$ is

$\displaystyle {\bf j} = \frac{\hbar}{m}\, {\rm Im}(\psi^\ast\, \nabla \psi).$ (10.22)

Thus, the particle flux associated with the incident wavefunction,

$\displaystyle \frac{ \exp(\,{\rm i} \,{\bf k}\cdot {\bf x})}{(2\pi)^{3/2}},$ (10.23)

is proportional to

$\displaystyle {\bf j}_{\rm incident} = \frac{\hbar\,{\bf k}}{(2\pi)^{3}\,m}.$ (10.24)

Likewise, the particle flux associated with the scattered wavefunction,

$\displaystyle \frac{ \exp(\,{\rm i} \,k\,r)}{(2\pi)^{3/2}}\frac{ f({\bf k}', {\bf k})}{r},$ (10.25)

is proportional to

$\displaystyle {\bf j}_{\rm scattered}=\frac{\hbar\,{\bf k}'}{(2\pi)^{3}\,m} \frac{\vert f( {\bf k}', {\bf k})\vert^{\,2}}{r^{\,2}}.$ (10.26)

Now, by definition,

$\displaystyle \frac{d\sigma}{d {\mit\Omega}} \,d{\mit\Omega} = \frac{ r^{\,2}\,...
...Omega} \, \vert{\bf j}_{\rm scattered}\vert}{\vert{\bf j}_{\rm incident}\vert},$ (10.27)


$\displaystyle \frac{d\sigma}{d {\mit\Omega}} = \vert f({\bf k}', {\bf k})\vert^{\,2}.$ (10.28)

Thus, $ \vert f({\bf k}', {\bf k})\vert^{\,2}$ is the differential cross-section for particles with incident momentum $ \hbar\,{ \bf k}$ to be scattered into states whose momentum vectors are directed in a range of solid angles $ d{\mit\Omega}$ about $ \hbar\,{ \bf 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 (10.11).

next up previous
Next: Born Approximation Up: Scattering Theory Previous: Introduction
Richard Fitzpatrick 2016-01-22