Fundamental Equations

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

$$H= H_0 + H_1,$$ (910)

where $H_0$ is the Hamiltonian of a free particle of mass $m$ ,

$$H_0 = \frac{p^2}{2\,m},$$ (911)

and $H_1$ represents the non-time-varying source of the scattering. Let $\vert\phi\rangle$ be an energy eigenket of $H_0$ ,

$$H_0\, \vert\phi\rangle = E\, \vert\phi\rangle,$$ (912)

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

$$(H_0 + H_1)\, \vert\psi\rangle = E\,\vert\psi\rangle,$$ (913)

where $\vert\psi\rangle$ is an energy eigenstate of the total Hamiltonian whose wavefunction $\langle {\bf x}'\vert\psi\rangle$ is $\psi({\bf x}')$ . In general, both $H_0$ and $H_0+H_1$ have continuous energy spectra: i.e., their energy eigenstates are unbound. We require a solution of Equation (913) 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, (912), corresponding to the same energy eigenvalue.

Adopting the Schrödinger representation, we can write the scattering problem (913) in the form

$$(\nabla^2 + k^2)\,\psi({\bf x}) = \frac{2\,m}{\hbar^2}\, \langle {\bf x} \vert\,H_1\,\vert \psi\rangle,$$ (914)

where

$$E = \frac{\hbar^2 k^2}{2\,m}.$$ (915)

Equation (914) is called the Helmholtz equation, and can be inverted using standard Green's function techniques. Thus,

$$\psi({\bf x}) = \phi({\bf x}) + \frac{2\,m}{\hbar^2} \int d^3 x'\,G({\bf x}, {\bf x}') \,\langle {\bf x}' \vert\,H_1\,\vert\psi\rangle,$$ (916)

where

$$(\nabla^2 + k^2)\,G({\bf x}, {\bf x}') = \delta({\bf x} -{\bf x}').$$ (917)

Note that the solution (916) satisfies the boundary condition $\vert\psi\rangle \rightarrow \vert\phi\rangle$ as $H_1\rightarrow 0$ . As is well-known, the Green's function for the Helmholtz problem is given by

$$G({\bf x}, {\bf x}') = -\frac{\exp(\pm {\rm i}\,k\, \vert{\bf x} - {\bf x}'\vert\,)}{4\pi\,\vert{\bf x} - {\bf x}'\vert}.$$ (918)

Thus, Equation (916) becomes

$$\psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^2} \int d^3... ...rt{\bf x} - {\bf x}'\vert}\, \langle {\bf x}' \vert\,H_1\,\vert\psi^\pm\rangle.$$ (919)

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

$$\langle {\bf x}'\vert\,H_1\,\vert{\bf x}\rangle = V({\bf x})\, \delta({\bf x} -{\bf x}').$$ (920)

We can write

$$\langle {\bf x}'\vert\,H_1\,\vert \psi^\pm\rangle = \int d^3 x''\... ...e \langle {\bf x}'' \vert\psi^\pm\rangle\, = V({\bf x}') \,\psi^\pm ({\bf x}').$$ (921)

Thus, the integral equation (919) simplifies to

$$\psi^\pm({\bf x}) = \phi({\bf x}) - \frac{2\,m}{\hbar^2} \int d^3... ...\vert)}{4\pi\,\vert{\bf x} - {\bf x}'\vert}\, V({\bf x}')\, \psi^\pm({\bf x}').$$ (922)

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

$$\langle {\bf x} \vert {\bf k}\rangle = \frac{ \exp(\,{\rm i}\,{\bf k}\cdot{\bf x}) }{(2\pi)^{3/2}}.$$ (923)

The wavefunction is normalized such that

$$\langle {\bf k}\vert{\bf k}'\rangle =\int d^3 x\, \langle {\bf k}... ...\, {\bf x}\cdot({\bf k} -{\bf k}')]} {(2\pi )^3} = \delta ({\bf k} - {\bf k'}).$$ (924)

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

$$\vert{\bf x} - {\bf x}'\vert \simeq r - {\bf e}_r\cdot{\bf x}'$$ (925)

to first order in $r'/r$ , where

$${\bf e}_r = \frac{\bf x}{r}$$ (926)

is a unit vector that points from the scattering region to the observation point. Here, $r=\vert{\bf x}\vert$ and $r'=\vert{\bf x}'\vert$ . Let us define

$${\bf k}' = k\,{\bf e}_r.$$ (927)

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

$$\exp(\pm {\rm i}\, k\,\vert{\bf x} - {\bf x}' \vert\,) \simeq \exp(\pm {\rm i}\, k \,r) \exp(\mp {\rm i}\, {\bf k}' \cdot {\bf x}').$$ (928)

In the large-$r$ limit, Equation (922) reduces to

$$\psi^\pm({\bf x}) \simeq \frac{\exp(\,{\rm i}\,{\bf k}\cdot{\bf x... ...exp(\mp {\rm i} \,{\bf k}' \cdot {\bf x}')\, V({\bf x}')\, \psi^\pm ({\bf x}').$$ (929)

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 $\psi^\pm$ ) corresponds to a wave propagating away from the scattering region, whereas the minus sign corresponds to a wave propagating towards the scattering region. It is obvious that the former represents the physical solution. Thus, the wavefunction a long way from the scattering region can be written

$$\psi({\bf x}) = \frac{1}{(2\pi)^{3/2}} \left[\exp(\,{\rm i}\,{\bf k}\cdot{\bf x}) + \frac{\exp(\,{\rm i}\,k\,r)}{r} f({\bf k}', {\bf k}) \right],$$ (930)

where

$$f({\bf k}', {\bf k}) = - \frac{(2\pi)^2 \,m}{\hbar^2} \int d^3{\b... ... - \frac{(2\pi)^2 \,m}{\hbar^2}\, \langle {\bf k}'\vert\,H_1\,\vert\psi\rangle.$$ (931)

Let us define the differential 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 flux of particles. Recall, from Chapter 3, that the probability current (i.e., the particle flux) associated with a wavefunction $\psi$ is

$${\bf j} = \frac{\hbar}{m}\, {\rm Im}(\psi^\ast\, \nabla \psi).$$ (932)

Thus, the probability flux associated with the incident wavefunction,

$$\frac{ \exp(\,{\rm i} \,{\bf k}\cdot {\bf x})}{(2\pi)^{3/2}},$$ (933)

is

$${\bf j}_{\rm inc} = \frac{\hbar}{(2\pi)^{3}\,m} \,{\bf k}.$$ (934)

Likewise, the probability flux associated with the scattered wavefunction,

$$\frac{ \exp(\,{\rm i} \,k\,r)}{(2\pi)^{3/2}}\frac{ f({\bf k}', {\bf k})}{r},$$ (935)

is

$${\bf j}_{\rm sca}=\frac{\hbar}{(2\pi)^{3}\,m} \frac{\vert f( {\bf k}', {\bf k})\vert^{\,2}}{r^2} \, k\, {\bf e}_r.$$ (936)

Now,

$$\frac{d\sigma}{d {\mit\Omega}} \,d{\mit\Omega} = \frac{ r^2\,d{\mit\Omega} \, \vert{\bf j}_{\rm sca}\vert}{\vert{\bf j}_{\rm inc}\vert},$$ (937)

giving

$$\frac{d\sigma}{d {\mit\Omega}} = \vert f({\bf k}', {\bf k})\vert^{\,2}.$$ (938)

Thus, $\vert f({\bf k}', {\bf k})\vert^{\,2}$ gives 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 (920).