advanced34 min studyLesson 587 in Physics

Direct reactions and optical potentials

Interpret complex elastic-channel potentials and partial-wave absorption without confusing flux loss with particle destruction.

On this page

What you’ll understand

  • Separate direct, compound, and elastic-channel descriptions.
  • Determine the sign of an absorptive optical potential.
  • Calculate elastic and reaction partial-wave cross sections from S-matrix elements.

A projectile need not equilibrate with every degree of freedom in a nucleus before emerging. Direct excitation or transfer can retain a strong connection to the entrance motion. An optical potential instead summarizes the elastic channel after other channels are eliminated. Its imaginary part represents the loss of elastic flux into those omitted possibilities. This effective loss is compatible with probability conservation in a larger coupled-channel description.

A reaction mechanism is not an energy-balance label

Direct processes typically involve a relatively brief interaction and a limited change of nuclear configuration, such as transferring a nucleon or exciting a collective mode. Compound processes redistribute excitation more extensively before decay. The distinction concerns dynamics and timescales; both still obey the same conservation laws and may lead to the same final particle species. Angular distributions, correlations, and energy dependence help test which amplitudes contribute rather than assigning a mechanism from Q alone.

A distorted-wave treatment uses entrance and exit waves already modified by nuclear and Coulomb interactions, then computes a transition between specified channels. Its extracted strength depends on the wave functions, optical parameters, and transition operator used. A fitted transfer normalization is therefore model dependent and should not be identified uncritically with an exact occupation probability. Competing direct and compound contributions can also require coherent amplitude treatment when the same resolved final state is involved.

Uopt(r)=V(r)−iW(r),W(r)≥0,∂∣ψ∣2∂t+∇⋅j=−2Wℏ∣ψ∣2U_{\rm opt}(r)=V(r)-iW(r),\quad W(r)\ge0,\qquad\frac{\partial |\psi|^2}{\partial t}+\nabla\cdot\mathbf j=-\frac{2W}{\hbar}|\psi|^2
V and W are energies, and this sign convention uses Schrödinger evolution exp(−iEt/ℏ). The right side is an elastic-channel probability sink; it does not claim total nuclear probability is lost. A local energy-dependent optical model is an effective approximation to eliminated channels.↔ Scroll wide equations horizontally.

The sign of absorption can be checked directly

In a region with a constant positive W under the declared convention, an amplitude acquires a decaying exponential in time and its norm decays twice as fast in the exponent. Integrating the continuity equation shows that the lost elastic flux supplies reactions into the omitted channels. Reversing the imaginary sign would create elastic norm instead, describing gain rather than the intended absorptive model. Different references write the imaginary part with different signs, so read the convention rather than memorizing the letter W.

Realistic optical potentials may be nonlocal, energy dependent, spin dependent, and surface dominated. A local Woods–Saxon parametrization is a practical representation, not a unique microscopic force. Distinct parameter sets can fit an elastic angular distribution similarly, particularly over a restricted energy and angular range. Measurements of reaction cross sections and additional channels constrain the ambiguity, but extrapolation outside the fitted regime still needs a physically justified model and uncertainty assessment.

σreac=πk2∑ℓ=0∞(2ℓ+1)(1−∣Sℓ∣2),σel=πk2∑ℓ=0∞(2ℓ+1)∣1−Sℓ∣2\sigma_{\rm reac}=\frac{\pi}{k^2}\sum_{\ell=0}^\infty(2\ell+1)(1-|S_\ell|^2),\qquad\sigma_{\rm el}=\frac{\pi}{k^2}\sum_{\ell=0}^\infty(2\ell+1)|1-S_\ell|^2
These integrated formulas assume spinless distinguishable particles with no long-range Coulomb amplitude, k in inverse length, and dimensionless partial-wave Sℓ. Cross sections are areas. Absorption requires |Sℓ| ≤ 1. Charged-particle elastic scattering needs separate Coulomb treatment and interference.↔ Scroll wide equations horizontally.

A complex elastic S matrix identifies omitted flux

A real single-channel potential gives unit-modulus partial-wave S elements and zero reaction cross section, even when elastic deflection is large. Absorption reduces that modulus below one. The phase still determines elastic interference, so an elastic cross section is not simply one minus the reaction probability. In particular, full absorption of a set of waves also produces diffraction, which contributes to elastic scattering even when those waves have no outgoing elastic S amplitude.

The reaction cross section sums every process outside the chosen elastic channel. It is not automatically equal to one named transfer, capture, or fusion cross section. That identification requires showing which other channels are negligible. Partial-wave formulas are valuable consistency tests for signs and unitarity, but a teaching model with a prescribed S element does not calculate a nuclear optical potential from first principles. The measured differential pattern supplies more information than one integrated area alone.

Elastic deflection and missing flux in one partial wave

Use a spinless neutral toy model with only ℓ=0\ell=0 different from free propagation, S0=0.6S_0=0.6, and π/k2=2 b\pi/k^2=2\,\mathrm b. Compute elastic and reaction cross sections, their sum, and check the optical-theorem expression σtot=(2π/k2)(1−Re⁡S0)\sigma_{\rm tot}=(2\pi/k^2)(1-\operatorname{Re}S_0).

  1. The outgoing elastic probability in this wave is ∣S0∣2=0.36|S_0|^2=0.36, leaving reaction probability 0.64. Multiplying by the s-wave area gives σreac=2(0.64)=1.28 b\sigma_{\rm reac}=2(0.64)=1.28\,\mathrm b.

  2. Elastic scattering depends on the amplitude difference from free propagation: ∣1−S0∣2=0.42=0.16|1-S_0|^2=0.4^2=0.16. Thus σel=2(0.16)=0.32 b\sigma_{\rm el}=2(0.16)=0.32\,\mathrm b, not 2(0.36)2(0.36).

  3. The sum is 1.28+0.32=1.60 b1.28+0.32=1.60\,\mathrm b. Independently the optical-theorem form gives 2(2)(1−0.6)=1.60 b2(2)(1-0.6)=1.60\,\mathrm b, confirming the normalization and distinguishing elastic amplitude from surviving flux.

The result

The reaction cross section is 1.28 b, elastic cross section 0.32 b, and total 1.60 b. Elastic-channel absorption conserves probability when omitted reaction channels are restored.

Explore the model

Interactive lab

Interactive lab · waves

Interference belongs to amplitudes

Reset and set both controls to phase difference 180° and amplitude ratio 0.60. Predict resultant amplitude 0.40 in the readout, mirroring the algebraic difference 1 − 0.6. Reset and set both to phase 0° and ratio 0.60; predict amplitude 1.60. These are equal-frequency classical sine waves, not incoming/outgoing nuclear radial waves or an optical-potential solver.

Interference belongs to amplitudesTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAmplitude differences and phases matter before a cross section can be formed by squaring.

Check your understanding

Under U = V − iW with W ≥ 0 and exp(−iEt/ℏ), what does W represent?

Try it yourself

For the same neutral spinless normalization π/k2=2 b\pi/k^2=2\,\mathrm b, take S0=−1S_0=-1 with all other waves free. Find reaction, elastic, and total cross sections. Explain why a large elastic area need not imply absorption.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Its modulus is one but its phase differs maximally from free propagation S = 1.

Compare with the solution

Since ∣S0∣2=1|S_0|^2=1, the reaction cross section is zero. The elastic factor is ∣1−(−1)∣2=4|1-(-1)|^2=4, so the elastic and total cross sections are both 8 b. The optical-theorem expression gives 2(2)(1−(−1))=8 b2(2)(1-(-1))=8\,\mathrm b as a check. The large area is caused by the elastic phase shift, while no probability leaves the elastic channel.

Take it with you

  • Direct and compound mechanisms need dynamical evidence.
  • An absorptive potential removes elastic-channel norm under a stated sign convention.
  • Reaction and elastic cross sections probe different parts of a partial-wave S element.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.