2
$\begingroup$

This question is a spin-off from this related question: Why does the Born approximation for the scattering amplitude depend on the potential $V$ everywhere in space, unlike classical scattering? This question deals with a broadly similar topic, which is solving scattering problems "locally", i.e., without knowing the potential $V$ everywhere in space, which is possible in the case of classical scattering.

In the textbook derivation of the scattering cross section (e.g., Griffiths QM) we look for solutions to the Schrödinger equation which have the form

$$ \psi \propto e^{ikz}+f( \theta) \frac{e^{ikr}}{r}$$

where $f$ the scattering amplitude has the interpretation that $\lvert f \rvert^2 = \frac{d \sigma}{d \Omega} $. The way I understand cross sections is as they relate to the attenuation of a particle beam, but a particle beam is a localized in a way that a plane wave is not. If wave packets in a particle beam are written as a linear combination of plane waves $\psi_{\text{beam}} = \sum_{k \in K} A_k e^{ikz},$ wouldn't we find the appropriate scattering amplitude to be some $\tilde f = \sum_{k \in K} f_k \neq f$ from solving the Schrödinger equation for each plane wave? Is there some identity like $\frac{d\sigma}{d \Omega} = \lvert \tilde f \rvert^2 = \lvert f \rvert^2$ or $\sigma_{\text{tot}}=\int_0^{\pi} \lvert \tilde f \rvert^2 \sin \theta\, d\theta = \int_0^{\pi} \lvert f \rvert^2 \sin \theta\, d\theta$?

$\endgroup$
1
$\begingroup$

Quantum mechanical scattering theory is a difficult compromise between wave phenomena on the one hand, and the classical view of particle scattering on the other. Just like an electromagnetic wave, the scattering solution exists everywhere in space - what really makes it different from an eigenvalue problem is that it is a propagating (rather than a standing) wave, which is correctly incorporated via the appropriate boundary conditions (i.e. incoming and outgoing waves). This is also consistent with the classical view of scattering in the sense that the incident particle has a well-defined momentum.

There are two main reasons why one speaks about wave packets in this context. One is pedagogical - since it helps to establish the connection with more intuitive picture of scattering as particles hitting an obstacle. I personally find such explanations confusing, but I suppose it helps some people to think this way. The price to pay is somewhat confusing math.

The deeper reason for talking about wave packets is because a scattering wave couldn't realistically extend through the whole space and exist forever. This means that it is actually a wave packet, whose size is controlled by the geometry and the duration of our experiment.

Regarding the formulas: they relate the quantities on the same energy shell (the magnitudes of the momenta can really vary in some problems), and better understood/interpreted in terms of the wave picture. The general relation between the cross-section and the scattering amplitude (to all orders in the scattering potential) is known as the Optical theorem.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If I'm understanding your question correctly, the answer is yes - assuming a plane wave solution would yield a wavefunction of the form

$$\psi_k(\mathbf r,t) = e^{i(kz-\omega_k t)} + f_k(\theta) \frac{e^{i(kr-\omega_k t)}}{r}$$

so a wavepacket solution would take the form

$$\Psi(\mathbf r,t) = \underbrace{\int d^3\mathbf k \ A(\mathbf k) e^{i(\mathbf k \cdot \mathbf r - \omega_k t)}}_{\text{incoming beam}} + \int d^3\mathbf k A(\mathbf k) f_k(\theta) \frac{e^{i(kr-\omega_k t)}}{r}$$

in the asymptotic region in which the potential is (effectively) zero.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.