Wave function of a scattered particle and cross section

In summary, Landau-Lifshitz's book on non-relativistic quantum mechanics discusses how to describe a particle with a plane wave function after a scattering event. The wave function is given by ##\psi \approx e^{ikz} + f(\theta) \frac{e^{ikr}}{r}##, with the first term representing the unscattered particle and the second term representing the scattered particle. The square value of the wave function is just ##|f(\theta)|^2##, neglecting the contribution of the first term. This is important in deriving the differential cross section, which is also given by ##|f(\theta)|^2##. The book goes on to explain how this approach
  • #1
dRic2
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In Landau-Lifsits's book about non relativistic QM it is said that if I have a particle described by a plane wave ##\phi = e^{ikz}## (I think he choses the ##z## direction for simplicity) the wave function after the scattering event is (far from the scattering event)
$$\psi \approx e^{ikz} + f(\theta) \frac{e^{ikr}} r$$
But then he said that the square value of ##\psi## is just ##|f(\theta)|^2## because we neglect the contribution of the first term (that is, we neglect some sort of interference phenomena). My first question is: then why bother says ##\psi = e^{ikz} + f(\theta) \frac{e^{ikr}} r## instead of simply putting ##\psi = f(\theta) \frac{e^{ikr}} r## ?

Then he says that the differential cross section is just ##|f(\theta)|^2## but I don't really follow the given explanation.

Can someone provide any hint?

Thanks
Ric
 
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In most scattering experiments, there is the main beam, often almost as strong as the incident beam, that passes through unscattered. This term, represented by ## \psi=e^{ikz} ## is discarded when computing ## \frac{d \sigma}{d \Omega} ##. In computing ## \psi^* \psi ##, there is a middle term consisting of products of ## e^{ikz} ## and ## \frac{f(\theta)e^{ikr}}{r} ## (a sum of two terms with the appropriate complex conjugates, etc.) . I think that term is being called the interference term, and it generally gets discarded, presumably because it is small. I'm no expert on this particular topic, but I believe my inputs here are accurate. Perhaps someone with expertise in this area can concur. ## \\ ## Edit: A text that covers this topic reasonably well is Squires' Introduction to the theory of Thermal Neutron Scattering. He shows how the differential cross section results from the scattered part of the wave function: ## \frac{d \sigma}{d \Omega}=\frac{I_{scattered}}{E_{incident}} ##, where ## E_{incident} ## is watts/m^2 and ##E_{scattered}=\frac{I_{scattered}}{r^2} ##. Thereby, the ## \frac{1}{r} ## of the scattered wave function isn't used in computing the cross section. (Note that irradiance ## E ## is proportional to ## \psi^* \psi ## for both the incident and the scattered beams). In Squires' text, he does the computation by using ## \psi_{scattered}=\frac{f(\theta)e^{ikr}}{r} ##, rather than using the complete wave function, with ## E_{scattered} ## being proportional to ## \psi^*_{scattered} \psi_{scattered} ##, with the result that ## I_{scattered} ## is proportional to ## |f(\theta)|^2 ##. ## \\ ## And I gave your question some additional thought: The incident beam is usually somewhat localized, so that ## \psi_{incident}=e^{ikz} ## can be a good description of the incident beam in the region where the incident beam has an approximately uniform flux, but it doesn't adequately describe the beam outside of this region. The question you ask is a good one=why not discard the ## e^{ikz} ## part in treating ## \psi_{scattered} ##? Squires basically discards that part, and works with ## \psi_{scattered} ##.
 
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This becomes much clearer, when you think about physical states rather than plane-wave representations. You have an asymptotic free incoming particle, described by a localized wave packet running towards the place where the potential is effective and you observe the scattered particle at a far distance from this place as an asymptotic outgoing state.

You get this description in terms of wave packets from the plane-wave scattering state by Fourier transformation. The plane-wave piece translates into a wave packet that moves as if it were unaffected by the potential. This is pretty much peaked in forward direction (i.e., in direction of the incoming particle). The spherical-wave part refers to the really scattered particles, and that's what's of main interest in terms of the cross section. Thus cutting out some small solid angle around the direction of the incoming particle leads to the conclusion of Landau Lifshitz, i.e., that the appropriate matrix element for the cross section is simply ##|f(\theta)|^2##.

For a good discussion of scattering along these arguments, see the textbook by Messiah.
 
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Sorry for my late reply. Basically they include in "scattered" wave function the non scattering event, right ? Sounds reasonable, but a bit counterintuitive to me... at least I still don't get the point of adding that contribution if it will be discarded lately (it represents a result I'm not interested in).

I'll check the books you both recommended.

Thank you very much.
 
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You need it to derive the "optical theorem", which is very important too!
 
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vanhees71 said:
You need it to derive the "optical theorem"
it -> the book, or was it a typo, meaning I should derive it as an exercise ?
 
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dRic2 said:
it -> the book, or was it a typo, meaning I should derive it as an exercise ?
I mean the complete asymptic wave function, including the incoming asymptotic free wave. As I said, a very thorough discussion of scattering theory (using the traditional wave-mechanics approach, which of course is particularly intuitive for scattering theory to begin with) is the famous textbook by Messiah:

Here's a Dover reprint, combining the two volumes of the original edition into one volume:

A. Messiah, Quantum Mechanics, Dover Publications, New York (1999).
 
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Thanks to both for the replies :smile:
 
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1. What is the wave function of a scattered particle?

The wave function of a scattered particle is a mathematical representation of the probability of finding the particle at a particular location after it has been scattered by an external force or potential. It describes the behavior of the particle in terms of its position, momentum, and energy.

2. How is the wave function of a scattered particle related to cross section?

The wave function of a scattered particle is used to calculate the cross section, which is a measure of the probability that the particle will interact with a target particle. The cross section is directly proportional to the square of the wave function, meaning that a larger wave function will result in a larger cross section and a higher likelihood of interaction.

3. What factors affect the wave function of a scattered particle?

The wave function of a scattered particle can be affected by the properties of the particle, such as its mass and charge, as well as the properties of the scattering potential. Additionally, the angle and energy of the incident particle can also impact the wave function and therefore the cross section.

4. How is the cross section experimentally determined?

The cross section can be experimentally determined by measuring the number of scattering events that occur between the incident particle and the target particle. By comparing this number to the total number of particles incident on the target, the cross section can be calculated.

5. What is the significance of the wave function and cross section in scattering experiments?

The wave function and cross section are important quantities in scattering experiments as they provide information about the interaction between particles. They can be used to understand the underlying physical processes and to make predictions about the behavior of particles in different scattering scenarios.

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