The Lipmann-Schwinger equation

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

$$H= H_0 + H_1,$$ (891)

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

$$H_0 = \frac{p^2}{2 m},$$ (892)

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

$$H_0 \vert\phi\rangle = E \vert\phi\rangle,$$ (893)

whose wave-function $\langle {\bf r}'\vert\phi\rangle$ is $\phi({\bf r}')$. This state is a plane-wave state or, possibly, a spherical-wave state. Schrödinger's equation for the scattering problem is

$$(H_0 + H_1) \vert\psi\rangle = E \vert\psi\rangle,$$ (894)

where $\vert\psi\rangle$ is an energy eigenstate of the total Hamiltonian whose wave-function $\langle {\bf r}'\vert\psi\rangle$ is $\psi({\bf r}')$. 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 Eq. (894) which 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, (893), corresponding to the same energy eigenvalue.

Formally, the desired solution can be written

$$\vert\psi\rangle =\vert\phi\rangle +\frac{1}{E-H_0} H_1 \vert\psi\rangle.$$ (895)

Note that we can recover Eq. (894) by operating on the above equation with $E-H_0$, and making use of Eq. (893). Furthermore, the solution satisfies the boundary condition $\vert\psi\rangle \rightarrow \vert\phi\rangle$ as $H_1\rightarrow 0$. Unfortunately, the operator $(E-H_0)^{-1}$ is singular: i.e., it produces infinities when it operates on an eigenstate of $H_0$ corresponding to the eigenvalue $E$. We need a prescription for dealing with these infinities, otherwise the above solution is useless. The standard prescription is to make the energy eigenvalue $E$ slightly complex. Thus,

$$\vert\psi^\pm\rangle =\vert\phi\rangle +\frac{1}{E-H_0\pm {\rm i} \epsilon} H_1 \vert\psi^\pm\rangle,$$ (896)

where $\epsilon$ is real, positive, and small. Equation (896) is called the Lipmann-Schwinger equation, and is non-singular as long as $\epsilon >0$. The physical significance of the $\pm$ signs will become apparent later on.

The Lipmann-Schwinger equation can be converted into an integral equation via left multiplication by $\langle{\bf r}\vert$. Thus,

$$\psi^\pm({\bf r}) = \phi({\bf r}) +\int \left\langle{\bf r}\... ...le \langle {\bf r}'\vert H_1\vert\psi^\pm\rangle d^3{\bf r}'.$$ (897)

Adopting the Schrödinger representation, we can write the scattering problem (894) in the form

$$(\nabla^2 + k^2) \psi({\bf r}) = \frac{2 m}{\hbar^2} \langle {\bf r} \vert H_1\vert \psi\rangle,$$ (898)

where

$$E = \frac{\hbar^2 k^2}{2 m}.$$ (899)

This equation is called Helmholtz's equation, and can be inverted using standard Green's function techniques. Thus,

$$\psi({\bf r}) = \phi({\bf r}) + \frac{2 m}{\hbar^2} \int G(... ...r'}) \langle {\bf r}' \vert H_1\vert\psi\rangle d^3{\bf r}',$$ (900)

where

$$(\nabla^2 + k^2) G({\bf r}, {\bf r}') = \delta({\bf r} -{\bf r}').$$ (901)

Note that the solution (900) 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

$$G({\bf r}, {\bf r}') = -\frac{\exp(\pm {\rm i} k \vert{\bf r} - {\bf r}'\vert )}{4\pi \vert{\bf r} - {\bf r}'\vert}.$$ (902)

Thus, Eq. (900) becomes

$$\psi^\pm({\bf r}) = \phi({\bf r}) - \frac{2 m}{\hbar^2} \in... ...t} \langle {\bf r}' \vert H_1\vert\psi\rangle d^3{\bf r}'.$$ (903)

A comparison of Eqs. (897) and (903) suggests that the kernel to Eq. (897) takes the form

$$\left\langle{\bf r}\left\vert\frac{1}{E-H_0\pm {\rm i} \ep... ...f r} - {\bf r}'\vert )}{4\pi \vert{\bf r} - {\bf r}'\vert} .$$ (904)

It is not entirely clear that the $\pm$ signs correspond on both sides of this equation. In fact, they do, as is easily proved by a more rigorous derivation of this result.

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

$$\langle {\bf r}'\vert H_1\vert{\bf r}\rangle = V({\bf r}) \delta({\bf r} -{\bf r}').$$ (905)

We can write

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

$$= V({\bf r}') \psi^\pm ({\bf r}').$$ (906)

Thus, the integral equation (903) simplifies to

$$\psi^\pm({\bf r}) = \phi({\bf r}) - \frac{2 m}{\hbar^2} \in... ...\bf r}'\vert} V({\bf r}') \psi^\pm({\bf r'}) d^3{\bf r}'.$$ (907)

Suppose that the initial state $\vert\phi\rangle$ is a plane-wave with wave-vector ${\bf k}$ (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 wave-function takes the form

$$\langle {\bf r} \vert {\bf k}\rangle = \frac{ \exp( {\rm i} {\bf k}\!\cdot\!{\bf r}) }{(2\pi)^{3/2}}.$$ (908)

The wave-function is normalized such that

$$\langle {\bf k}\vert{\bf k}'\rangle = \int \langle {\bf k}\vert{\bf r}\rangle \langle {\bf r} \vert{\bf k}'\rangle d^3{\bf r}$$

$$= \int \frac{ \exp[-{\rm i} {\bf r}\!\cdot\!({\bf k} -{\bf k}')]} {(2\pi )^3} d^3{\bf r} = \delta ({\bf k} - {\bf k'}).$$ (909)

Suppose that the scattering potential $V({\bf r})$ is only non-zero in some relatively localized region centred on the origin (${\bf r} = 0$). Let us calculate the wave-function $\psi({\bf r})$ a long way from the scattering region. In other words, let us adopt the ordering $r\gg r'$. It is easily demonstrated that

$$\vert{\bf r} - {\bf r}'\vert \simeq r - \hat{{\bf r}}\!\cdot\!{\bf r}'$$ (910)

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

$$\hat{{\bf r}} = \frac{\bf r}{r}$$ (911)

is a unit vector which points from the scattering region to the observation point. Let us define

$${\bf k}' = k \hat{{\bf r}}.$$ (912)

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

$$\exp(\pm {\rm i} k \vert{\bf r} - {\bf r'} \vert ) \sime... ...rm i} k r) \exp(\mp {\rm i} {\bf k}'\!\cdot \!{\bf r}').$$ (913)

In the large-$r$ limit, Eq. (907) reduces to

$$\psi({\bf r})^\pm \simeq \frac{\exp( {\rm i} {\bf k}\!\cdo... ... \!{\bf r}') V({\bf r}') \psi^\pm ({\bf r}') d^3{\bf r}'.$$ (914)

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 wave-function a long way from the scattering region can be written

$$\psi({\bf r}) = \frac{1}{(2\pi)^{3/2}} \left[\exp( {\rm i}\... ... + \frac{\exp( {\rm i} kr)}{r} f({\bf k}', {\bf k}) \right],$$ (915)

where

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

$$= - \frac{(2\pi)^2 m}{\hbar^2} \langle {\bf k}'\vert H_1\vert\psi\rangle.$$ (916)

Let us define the differential cross-section $d\sigma/d\Omega$ as the number of particles per unit time scattered into an element of solid angle $d\Omega$, divided by the incident flux of particles. Recall, from Sect. 4, that the probability flux (i.e., the particle flux) associated with a wave-function $\psi$ is

$${\bf j} = \frac{\hbar}{m} {\rm Im}(\psi^\ast \nabla \psi).$$ (917)

Thus, the probability flux associated with the incident wave-function,

$$\frac{ \exp( {\rm i} {\bf k}\!\cdot\!{\bf r})}{(2\pi)^{3/2}},$$ (918)

is

$${\bf j}_{\rm inci} = \frac{\hbar}{(2\pi)^{3} m} {\bf k}.$$ (919)

Likewise, the probability flux associated with the scattered wave-function,

$$\frac{ \exp( {\rm i} k r)}{(2\pi)^{3/2}}\frac{ f({\bf k}', {\bf k})}{r},$$ (920)

is

$${\bf j}_{\rm scat}=\frac{\hbar}{(2\pi)^{3} m} \frac{\vert f( {\bf k}', {\bf k})\vert^2}{r^2} k \hat{\bf r}.$$ (921)

Now,

$$\frac{d\sigma}{d \Omega} d\Omega = \frac{ r^2 d\Omega \vert{\bf j}_{\rm scat}\vert}{\vert{\bf j}_{\rm inci}\vert},$$ (922)

giving

$$\frac{d\sigma}{d \Omega} = \vert f({\bf k}', {\bf k})\vert^2.$$ (923)

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 $d\Omega$ 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 shown in Eq. (905).