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I have read many times the term "optical potential" in nuclear physics papers. Here is an example paper where it is mentioned a few times. It seems to be explained in these lecture slides as a phenomenological approximation of an effectively non-local potential arising from a nucleon-target scattering equation. The "phenomenological" part seems to be, in general, Woods-Saxon and Coulomb nucleon-interaction terms and a source/drain term to account for transfer/capture reactions. But since they were meant to be accompanied by a speaker it is not super easy to follow.

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    $\begingroup$ Note: not a question about neutron optics, despite the similarity of names. $\endgroup$
    – rob
    Commented Dec 26, 2016 at 7:23
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    $\begingroup$ You might benefit from reading my answer to this question: physics.stackexchange.com/q/231618 $\endgroup$ Commented Dec 26, 2016 at 20:01

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The "optical potential" is just a model to include in your potential not only the elastic scattering (such as nuclear scattering or Coulomb scattering), but also inelastic scattering.

First of all since the nuclear physics is characterized by low energies (usually in the scale of 10-100MeV) the Schroedinger quantum mechanics is good enough to describe the physics involved. This means that you can describe every system (nucleon, nucleus, etc.) using a wavefunction $\psi(\vec{x},t)$ which is a solution of the well-known Schroedinger equation:

$$i\hbar\frac{\partial}{\partial t}\psi(\vec{x},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\vec{x},t)+V(\vec{x})\psi(\vec{x},t)$$

whee $V(\vec{x})$ is the potential. Since you can describe your system as a wavefunction you can compute, you can begin to understand why in nuclear physics it's pretty useful an analogy with optics: light is described using wavefunction as nucleons or nuclei.

In this particular case, also known as the "optical model", what you do is taking from optics the concept of complex refractive index. As you know in optics given $n\in\mathbb{C}$ the real part describes is the refractive index while the imaginary part is the extinction factor: the first one describes how the light is transmitted and the second one how it is absorbed by the medium. In the same way you introduce in nuclear physics an imaginary potential describing the absorption of your wavefunction (it means this imaginary part describes the inelastic scattering removing particles from the flux). In general the optical model is used when you know the real part of the potential and you can measure the fraction of the beam subject to it.

Making the calculation suppose you have a nucleon interacting with a nucleus (radius $R$), then you can approximate the potential (I will consider a one dimensional case) as $$V(x)=-U(x)-iW(x)\quad x\leq R$$ and null otherwise. Solving the static Schroedinger equation considering constant potential (this is just as an example, the real physics is way harder) and for "high" energy (high for nuclear physics, it means $O(100\text{MeV})$) to have $E>>U,W$ $$-\frac{\hbar^2}{2m}\nabla^2\psi(x)-\left[U+iW\right]\psi(x)=E\,\psi(x)\\ \left[\frac{\hbar^2}{2m}\nabla^2+\left(E+U+iW\right)\right]\psi(x)=0$$ you get the solution (inside the target nucleus $x\leq R$) in the form: $$\psi(x)\sim e^{\frac{i}{\hbar}\sqrt{2m(E+U+iW)}x}$$ where the wavenumber can be seen as $$k+k_R+\frac{i}{2}k_I=\frac{1}{\hbar}\sqrt{2m(E+U+iW)}$$ where $k=\frac{\sqrt{2mE}}{\hbar}$.

Now to have the optical analogy let's introduce the "complex refractive index" defined as $$n=\sqrt{\frac{E+U+iW}{E}}=1+\frac{k_R}{k}+\frac{i}{2}\frac{k_I}{k}$$ where this is possible because $n\approx 1$ so $n^2-1\approx 2(n-1)$. This means we can write the wavefunction as $$\psi(x)\sim e^{iknx}=e^{i(k+k_R)x}e^{-\frac{1}{2}k_Ix}.$$ Since we are working with Schoredinger quantum mechanics the observable is the squared module and this means that the probability density (the nucleon projectile beam flux is proportional to it) goes as $$|\psi(x)|^2\sim e^{-k_Ix}.$$

You can see now why the imaginary part of the potential is related to the absorption of the flux and how this really works as the complex refractive index known from optics. As the imaginary part of the refractive index describes the attenuation of the intensity of a light beam, in the same way the imaginary part of the "refractive index" for the nucleon-nucleus scattering describes the attenuation of the projectile beam. Also the exponential law for attenuation is the same both in optics and in nuclear physics.

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The term optical potential emphasizes the fact that one can treat the nucleus as some sort of a medium which can give rise to diffraction effects. These effects are absent in the so called black nucleus for which the cross section does not show any oscillatory behavior with energy. For optical potentials the transmitted waves interferes with the incident waves giving rise to oscillatory pattern with energy.

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  • $\begingroup$ The optical potential consists of the real part which leads to no absorption of incident particles and of an imaginary part which leads to some but not all (as in the black nucleus) absorption of incident particles. $\endgroup$
    – SAKhan
    Commented Dec 26, 2016 at 17:51

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