Fundamental Equations

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

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

where

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

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

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

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

where

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

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

where

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

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

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

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

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

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

We can write

$$\langle {\bf x}'\vert\,H_1\,\vert \psi^\pm\rangle = \int d^{\,3} {\bf x}''\,\langle {\bf x}'\vert\,H_1\,\vert{\bf x}''\rangle \langle {\bf x}'' \vert\psi^\pm\rangle$$

$$=\int d^{\,3}{\bf x}'' \,V({\bf x}')\,\delta^{\,3}({\bf x}'-{\bf x}'')\,\psi({\bf x}'') = V({\bf x}') \,\psi^\pm ({\bf x}'),$$ (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

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

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

The preceding wavefunction is conveniently normalized such that

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

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

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

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

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

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

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

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

where

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

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

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

Thus, the particle flux associated with the incident wavefunction,

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

is proportional to

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

Likewise, the particle flux associated with the scattered wavefunction,

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

is proportional to

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

Now, by definition,

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

giving

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