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9.2: Fundamental Equations

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


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



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


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



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


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


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


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





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



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


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





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


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



    Thus, Equation (916) becomes


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



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


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



    We can write


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



    Thus, the integral equation (919) simplifies to


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


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



    The wavefunction is normalized such that


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


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



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


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


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


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



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


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


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





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


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



    Thus, the probability flux associated with the incident wavefunction,


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





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



    Likewise, the probability flux associated with the scattered wavefunction,


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





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





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





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