5.2. Quantum scattering

In quantum scattering, we know parts of the wave function describing the particles that are scattered off some other particles. Just as for classical scattering, we consider a two-particle collision, which we can reduce to a single particle problem in which an incident particle scatters off a static potential which is nonzero only near the origin -- see problem 5.1. Assuming that the potential vanishes indeed outside of a sphere with radius $r_{\text{max}}$ centered at the origin, we know the wave functions for the incident beam along the $z$-axis and the scattered waves. So, outside of the sphere, we have $$\psi \propto e^{{\rm i}kz} + f(\vartheta, \varphi) \frac{e^{{\rm i}kr}}{r}.$$ We shall justify the form of the second term below. The main message for us at this stage is that we have an incident wave and a scattered wave. The amplitude of the scattered wave (the second term) depends on the polar angles $\vartheta, \varphi$ that are defined with respect to the $z$-axis along which the incident particles approach the scattering centre.

The detection of particles far from the scattering centre provides information about the amplitude $f(\vartheta,\varphi)$. Knowing that the outgoing wave $e^{ikr}/r$ represents a unit flux per solid angle $d\Omega = \sin\vartheta d\vartheta d\varphi$, we conclude that $$\frac{d\sigma}{d\Omega} = \left| f(\theta,\varphi)\right|^2.$$ Therefore, we need to find $f(\vartheta,\varphi)$ in order to determine the experimentally relevant quantity $d\sigma/d\Omega$.

In order to get a handle on this quantity, we start from the Schrödinger equation: $$\left[ - \frac{\hslash^2}{2m} \nabla^2 + V({\bf r}) \right] \psi({\bf r}) = E \psi({\bf r}).$$ For $V({\bf r})\equiv 0$, the solution to this equation would be an incoming plane wave. It turns out possible to write the solution to the Schrödinger equation with a potential formally as an integral expression. This is done using the Green's function formalism, discussed in Appendix B. The Green's function is an operator -- in this case that means that it depends on two positions ${\bf r}$ and ${\bf r}'$ -- it is defined by $$\left[ E + \frac{\hslash^2}{2m} \nabla^2 - V({\bf r}) \right] G({\bf r}, {\bf r}') = \delta ({\bf r}- {\bf r}').$$ In fact, these are the matrix elements of the Green's function operator: $$G({\bf r}, {\bf r}') = \left\langle {\bf r} \left| \hat{G} \right| {\bf r}' \right\rangle.$$ You may view the delta function on the right hand side as a unit operator, so that $G$ may be called the inverse of the operator $E \unicode{120793}-\hat{H}$, where $\unicode{120793}$ is the unit operator, all in line with the theory of chapter Appendix B. Now we want to calculate the unperturbed Green's function, i.e. the one for $V({\bf r}) \equiv 0$, which we denote as usual as $G_0$: $$\left[ E + \frac{\hslash^2}{2m} \nabla^2\right] G_0({\bf r}, {\bf r}') = \delta ({\bf r}- {\bf r}').$$

Before calculating $G_0$ let us assume we have it at our disposal. We then may write the solution to the full Schrödinger equation, i.e. including the potential $V$, in terms of a solution $\phi({\bf r})$ to the 'bare' Schrödinger equation, that is, the Schrödinger equation with potential $V\equiv 0$: $$\tag{5.14}\label{Eq:Dyson} \psi({\bf r}) = \phi({\bf r}) + \int G_0({\bf r}, {\bf r}') V({\bf r}') \psi({\bf r}')\; d^3 r',$$ which is an explicit form of the equation we met in chapter Appendix B: $$\left| \psi \right\rangle= \left| \phi\right\rangle+ G_0 V \left| \psi \right\rangle.$$ For this case, Eq. \eqref{Eq:Dyson} can easily be checked by substituting the solution into the full Schrödinger equation and using the fact that $E \unicode{120793}-\hat{H}_0$, acting on the Green's function $\hat{G}_0$, gives a delta-function: $$\left( E\unicode{120793}-H_0 \right)\psi({\bf r}) = 0 + \int \delta({\bf r}-{\bf r}') V({\bf r}') \psi({\bf r}') d^3 r' = V({\bf r}) \psi({\bf r}),$$ showing that $\psi({\bf r})$ satisfies the Schrödinger equation for the full Hamiltonina $H_0 + V$.

Now we consider the scattering problem with an incoming beam of the form $\phi({\bf r}) = \exp({\rm i}{\bf k}_{\text{i}} \cdot {\bf r})$ (the subscript 'i' denotes the incoming wave vector; do not confuse it with ${\rm i}=\sqrt{-1}$!). We see from Eq. \eqref{Eq:Dyson} that this wave persists but that it is accompanied by a scattering term which is the integral on the right hand side. Now the wavefunction $\psi({\bf r})$ is still very difficult to find, as it occurs in Eq. \eqref{Eq:Dyson} in an implicit form. We can make the equation explicit if we assume that the potential $V({\bf r})$ is small, so that the scattered part of the wave is much smaller than the wavefunction of the incoming beam. In a first approximation we might then replace $\psi({\bf r}')$ on the right hand side of Eq. \eqref{Eq:Dyson} by $\phi({\bf r}')$ which is a plane wave: $$\psi({\bf r}) = \phi({\bf r}) + \int G_0({\bf r}, {\bf r}') V({\bf r}') \phi({\bf r}') \; d^3 r'= e^{{\rm i}{\bf k}_{\text{i}} \cdot {\bf r}} +\int G_0({\bf r}, {\bf r}') V({\bf r}') e^{{\rm i}{\bf k}_{\text{i}} \cdot {\bf r}'} \; d^3 r' .$$ The key to the scattering amplitude is given by the notion that it must always be possible to write the solution \eqref{Eq:Dyson} in the form: $$\psi({\bf r}) = e^{{\rm i}{\bf k}_{\text{i}} \cdot {\bf r}} + f(\vartheta, \varphi) \frac{e^{{\rm i}kr}}{r}.$$ At this moment we hardly recognise this form in the expression obtained for the wavefunction. We first must find the explicit expression for the Green's function $G_0$. Without going through the derivation (see for example Griffiths, section 11.4¹ ) we give it here: $$G_0({\bf r}, {\bf r}') = - \frac{2m}{\hslash^2} \frac{e^{{\rm i}k \left| {\bf r}- {\bf r}'\right|}}{4\pi \left| {\bf r}- {\bf r}'\right|}$$ with $k = \sqrt{2m E/\hslash^2}$.

We are interested in the wave function at the detector -- this position is ${\bf r}$. Therefore we can take ${\bf r}$ far from the origin. As the range of the potential is finite, we know that only contributions with $r'\ll r$ have to be taken into account. Taylor expanding the exponent occuring in the Green's function: $$\left| {\bf r}- {\bf r}' \right| = \sqrt{ r^2 -2{\bf r}\cdot {\bf r}'+ r'^2} \approx r - \frac{{\bf r}\cdot {\bf r}'}{r}$$ leads to $$G({\bf r}, {\bf r}') = - \frac{2m}{\hslash^2} \frac{e^{{\rm i}kr}}{4\pi r} e^{-{\rm i}k {\bf r} \cdot {\bf r}'/r}.$$ The denominator does not have to be taken into account as it gives a much smaller contribution to the result for $r\gg 1/k$. Now we define ${\bf k}_{\text{f}}= k{\bf r}/r$, i.e. ${\bf k}_{\text{f}}$ is a wave vector corresponding to an outgoing wave from the scattering centre to the point ${\bf r}$ at the detector. We have $$\psi({\bf r}) = \phi({\bf r}) - \frac{2m}{\hslash^2} \frac{e^{{\rm i}kr}}{4\pi r} \int V({\bf r}') e^{-{\rm i}{\bf k}_{\text{f}} \cdot {\bf r}'} e^{{\rm i}{\bf k}_{\text{i}} \cdot {\bf r}'} d^3 r'.$$ This is precisely of the required form provided we set $$f(\vartheta, \varphi) = -\frac{m}{2\pi \hslash^2} \int V({\bf r}') e^{{\rm i}({\bf k}_{\text{i}}-{\bf k}_{\text{f}})\cdot {\bf r}'} \; d^3 r'.$$ This is the so-called first Born approximation. It is valid for weak scattering -- higher order approximations can be made by iterative substitution for $\psi({\bf r}')$ in the integral occurring in Eq. \eqref{Eq:Dyson}. In the first order Born approximation, the scattering amplitude $f(\vartheta, \varphi)$ is in fact a Fourier transform of the scattering potential.

For future reference, we note that the exact expression for the scattering amplitude is given by $$f(\vartheta, \varphi) = -\frac{m}{2\pi \hslash^2} \int V({\bf r}') \tag{5.15}\label{ExactF} e^{{\rm i}{\bf k}\cdot {\bf r}} \psi({\bf r}) \; d^3 r'.$$ where ${\bf k}$ is the outgoing wave with polar angles $\vartheta,\varphi$. This is the equation one would obtain from \eqref{Eq:Dyson} without the Born approximation.

As an example, we consider the Coulomb potential which is strictly speaking not weak, but we pretend that we can use the Born approximation for this. The Coulomb potential has the form $$V(r) = \frac{q_1 q_2}{4\pi \epsilon_0} \frac{1}{r}.$$ The Fourier transform of this potential reads $$V(k) = \frac{q_1 q_2}{\epsilon_0} \frac{1}{k^2}.$$ Therefore, we immediately find for $f(\vartheta)$: $$f(\vartheta) = -\frac{m q_1 q_2}{2\pi \epsilon_0 \hslash^2 ({\bf k}_{\text{i}}-{\bf k}_{\text{f}})^2}.$$ The angle $\vartheta$ is hidden in ${\bf k}_{\text{i}}-{\bf k}_{\text{f}}$, the norm of which is equal to $2k\sin(\vartheta/2)$. The result therefore is, using $E=\hslash^2 k^2/(2m)$: $$\frac{d\sigma}{d\Omega} = \left[ \frac{q_1 q_2}{16 \pi \epsilon_0 E \sin^2 (\vartheta/2)}\right]^2.$$ This is precisely the classical Rutherford formula, which also turns out to be the correct quantum mechanical result. This could not possibly be anticipated beforehand, but it is a happy coincidence.

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1. The free Schrödinger equation is also known as the Helmholtz equation. You will probably derive the Green's function for this in your math course under that name. ↩