Assumptions made in stationary-state scattering

In summary, the stationary-state scattering model is a theoretical framework used to study the interaction between particles and potential fields. Its main assumptions include the particles being in a steady state, the absence of external forces, and the particles being non-interacting. This model simplifies the analysis of particle scattering by reducing the complex dynamics into a simpler, steady-state system. However, it has limitations such as only being applicable to non-relativistic particles and assuming point-like particles without internal structure. The stationary-state scattering model has practical applications in various fields, including electronics, medical imaging, and nuclear reactions.
  • #1
WWCY
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Hi all, I recently started learning about quantum scattering in school and came across a few things I find confusing. Thanks in advance for any assistance!

1. Plane wave approximation to incident waves.

In past QM courses, I kept reading that plane waves were not "physical" since they do not normalise to unity. As such, would using them as a mathematical description for incident waves affect results? I have read that such approximations are valid if the incident waveform is much larger than the scatterer, which i can picture. However, the positional spread of the wavefunction is equal at all points in space, under what conditions would an actual scatterer encounter such a wavefunction?

2. Stationary-state solutions

A stationary-state solution to a scattering Hamiltonian would mean that we are solving a problem in which "everything" (wavefunction of incident and scattered particle) looks the same everywhere, at all points in time. Again, what sort of scattering set-up would allow for such an approximation?
 
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  • #2
Could anyone assist? Many thanks
 
  • #3
WWCY said:
1. Plane wave approximation to incident waves.

In past QM courses, I kept reading that plane waves were not "physical" since they do not normalise to unity. As such, would using them as a mathematical description for incident waves affect results? I have read that such approximations are valid if the incident waveform is much larger than the scatterer, which i can picture. However, the positional spread of the wavefunction is equal at all points in space, under what conditions would an actual scatterer encounter such a wavefunction?
You can always consider that the incoming state is a wave packet. If you decompose it in terms of plane waves, then the result of the scattering will be a superposition of the solutions found for each plane wave, weighted by the proper coefficient. For most well-behaved potentials, the scattering result varies slowly as a function of the incoming wave vector, such that for a narrow enough wave packet in momentum space, there is no significant difference between the full solution and that obtained from a single plane wave corresponding to the peak of the wave packet.

WWCY said:
2. Stationary-state solutions

A stationary-state solution to a scattering Hamiltonian would mean that we are solving a problem in which "everything" (wavefunction of incident and scattered particle) looks the same everywhere, at all points in time. Again, what sort of scattering set-up would allow for such an approximation?
The scattering potential has to be independent of time. Otherwise, I don't think that there are any other considerations, as you can write the wave packet as a superposition of these stationary states.

You can also model the scattering process using time-dependent perturbation theory (see for instance the textbook by Sakurai).
 
  • #4
Thank you for replying

DrClaude said:
You can always consider that the incoming state is a wave packet. If you decompose it in terms of plane waves, then the result of the scattering will be a superposition of the solutions found for each plane wave, weighted by the proper coefficient. For most well-behaved potentials, the scattering result varies slowly as a function of the incoming wave vector, such that for a narrow enough wave packet in momentum space, there is no significant difference between the full solution and that obtained from a single plane wave corresponding to the peak of the wave packet.

Apologies, but I'm not sure I follow here. When you mention "scattering result", what do you mean by this? Also, what does it mean to "vary slowly" as a function of an incoming wave-vector?
 
  • #5
WWCY said:
Apologies, but I'm not sure I follow here. When you mention "scattering result", what do you mean by this? Also, what does it mean to "vary slowly" as a function of an incoming wave-vector?
There are different ways to represent the outcome of a scattering event, but I was thinking specifically about the scattering amplitude ##f(\theta, \phi)##, which gives the angular distribution of the scattered particle and can be used to calculate the scattering cross section. In many cases, ##f(\theta, \phi)## is a function of the wave vector ##k##, but if that function is slowly varying with respect to ##k##, then the difference between the scattering of a plane wave of wave vector ##k## and a wave packet of width ##\delta k## centered on ##k## will be minimal.
 
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1. What is the concept of stationary-state scattering?

Stationary-state scattering refers to the process of a particle or wave interacting with a stationary potential, resulting in a change in its direction or energy. This phenomenon is commonly studied in quantum mechanics and is used to understand the behavior of particles in various physical systems.

2. What are the assumptions made in stationary-state scattering?

The main assumptions made in stationary-state scattering include the assumption that the potential is stationary, meaning it does not change with time, and that the scattering process is elastic, meaning there is no energy loss or gain during the interaction. Additionally, it is assumed that the particles are non-relativistic and that the scattering occurs in a vacuum.

3. How do these assumptions affect the accuracy of the results?

The assumptions made in stationary-state scattering can affect the accuracy of the results in different ways. For example, if the potential is not truly stationary, the results may not accurately represent the behavior of the particles. Similarly, if the scattering process is not elastic, the energy of the particles may change, leading to incorrect predictions. However, in most cases, these assumptions provide a good approximation of the behavior of particles in a given system.

4. Can these assumptions be relaxed in certain situations?

Yes, these assumptions can be relaxed in certain situations. For example, if the potential is not stationary, more advanced techniques such as time-dependent perturbation theory can be used to analyze the scattering process. Similarly, if the scattering is not elastic, the results can still be useful in understanding the overall behavior of the particles, but may not accurately predict their exact energy levels.

5. How is stationary-state scattering used in practical applications?

Stationary-state scattering has various practical applications, such as in the study of nuclear reactions, particle accelerators, and electron microscopy. It is also used in the development of new materials and technologies, such as in the design of semiconductors and solar cells. Additionally, stationary-state scattering is used in medical imaging techniques, such as X-ray and MRI, to understand the scattering of radiation through different tissues in the body.

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