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Sandbox Physics
Nuclear physics

L3 · Chapter 5

Where the wave goes after a neutron meets a nucleus

Sandbox Physics ·

Some neutrons leave a collision with the same internal states as they entered. Others excite the nucleus or enter reaction channels. A potential with a negative imaginary part represents the loss from the elastic channel. The size of its outgoing wave carries information about processes we have chosen not to follow individually.

Specify which neutrons are counted

Let a neutron beam travel to the right and strike calcium-40. Elastic scattering preserves the internal states of the projectile and target. It changes their relative direction of motion, and recoil still redistributes laboratory kinetic energy. We therefore define both the incident energy and the scattering angle in the centre-of-mass frame. Zero angle is forward.

Set E=1.2 MeVE=1.2\,\mathrm{MeV}. Both objects move, so the wave equation contains the reduced mass μ=mnM/(mn+M)\mu=m_nM/(m_n+M). Approximating the calcium mass as forty atomic mass units gives k=0.23767 fm−1k=0.23767\,\mathrm{fm}^{-1}, or a wavelength of about 26.4 fm. This exceeds the 4.5 fm well radius. Wave interference is essential at this scale.

The calcium-40 ground state has spin zero. We also omit neutron spin to isolate a scalar central-potential problem. Spin–orbit scattering and polarization observables require a more complete model.

[−ℏ22μ∇2+U(r)]ψ=Eψ,k=2μEℏ\left[-\frac{\hbar^2}{2\mu}\nabla^2+U(r)\right]\psi=E\psi,\qquad k=\frac{\sqrt{2\mu E}}\hbar
The potential depends on the neutron–target separation. Energy and wave number use the same centre-of-mass convention.

Source: A. M. Moro

Represent omitted channels by a smaller wave

Inside a sphere of radius RR, take the real potential to be negative, so it attracts the neutron. Add a negative imaginary part. With the usual time factor e−iEt/ℏe^{-iEt/\hbar}, that imaginary part attenuates the wave. The probability continuity equation gives the same sign: it acts as a sink within the elastic channel.

The neutron has not vanished from the complete physical system. Excitation, capture and other channels carry the missing probability. The optical potential keeps the elastic channel and summarizes their combined effect. Its imaginary depth combines their loss; resolving individual reactions requires their channel dynamics.

U(r)={−V−iW,r<R,0,r≥R,V,W≥0U(r)=\begin{cases}-V-iW,&r<R,\\0,&r\geq R,\end{cases}\qquad V,W\geq0
Depths are in MeV and radius in fm. The discontinuous surface is a deliberate square-well approximation.
∂∣ψ∣2∂t+∇⋅j=2Im⁡Uℏ∣ψ∣2=−2Wℏ∣ψ∣2(r<R)\frac{\partial|\psi|^2}{\partial t}+\nabla\cdot\mathbf j=\frac{2\operatorname{Im}U}{\hbar}|\psi|^2=-\frac{2W}{\hbar}|\psi|^2\quad(r<R)
Multiply the time-dependent Schrödinger equation by the conjugate wave and subtract the conjugate equation. The resulting negative source fixes the absorption convention.

Source: A. M. Moro

Match each partial wave at the surface

A central potential preserves orbital angular momentum. Expand the wave in terms labeled by ℓ\ell, called partial waves. Each has an incoming and an outgoing spherical component. One complex number, SℓS_\ell, describes the extra phase and amplitude of the outgoing component.

For a real potential, conservation of flux gives ∣Sℓ∣=1|S_\ell|=1. With absorption, the outgoing-to-incoming elastic flux ratio is ∣Sℓ∣2<1|S_\ell|^2<1. Match the regular interior spherical Bessel solution to the exterior wave and its derivative at the boundary to obtain this number.

The complex amplitudes from all partial waves must be added before their squared modulus is taken. Their interference produces peaks and dips as the detector angle changes.

q2=2μ(E+V+iW)ℏ2,ψℓ(r>R)∝12[hℓ−(kr)+Sℓhℓ+(kr)]q^2=\frac{2\mu(E+V+iW)}{\hbar^2},\qquad\psi_\ell(r>R)\propto\frac12\left[h_\ell^-(kr)+S_\ell h_\ell^+(kr)\right]
Choose the interior wave number with nonnegative imaginary part. The Hankel functions label incoming and outgoing radial waves.
f(θ)=12ik∑ℓ(2ℓ+1)(Sℓ−1)Pℓ(cos⁡θ),dσdΩ=∣f(θ)∣2f(\theta)=\frac{1}{2ik}\sum_\ell(2\ell+1)(S_\ell-1)P_\ell(\cos\theta),\qquad\frac{d\sigma}{d\Omega}=|f(\theta)|^2
The amplitude has units of length. Differential cross section has units of area per solid angle; the numerical example uses fm²/sr.

Source: David Tong · A. M. Moro

Worked example: scattering and lost flux

Choose V=42 MeVV=42\,\mathrm{MeV}, W=4 MeVW=4\,\mathrm{MeV} and R=4.5 fmR=4.5\,\mathrm{fm}. Boundary matching gives S0=−0.452−0.788iS_0=-0.452-0.788i. Its squared modulus is about 0.826, so roughly 17.4 percent of the incoming flux in this partial wave leaves the elastic channel.

Integrating over all angles removes cross terms between different Legendre polynomials. Summing through ℓ=12\ell=12 gives an elastic cross section of 232.477 fm2232.477\,\mathrm{fm}^2 and a reaction-loss cross section of 109.947 fm2109.947\,\mathrm{fm}^2. Their sum is 342.424 fm2342.424\,\mathrm{fm}^2. Since one barn is one hundred square femtometres, these are about 2.32, 1.10 and 3.42 barns.

These are calculated values for the specified well. The next chapter reverses the direction of reasoning: given only several angular counts, which wells could have produced them?

σel=πk2∑ℓ(2ℓ+1)∣Sℓ−1∣2,σreac=πk2∑ℓ(2ℓ+1)(1−∣Sℓ∣2)\begin{aligned}\sigma_{\rm el}&=\frac\pi{k^2}\sum_\ell(2\ell+1)|S_\ell-1|^2,\\\sigma_{\rm reac}&=\frac\pi{k^2}\sum_\ell(2\ell+1)(1-|S_\ell|^2)\end{aligned}
Elastic scattering measures the change from free propagation. Reaction loss measures probability carried by omitted channels. Together they give the total cross section.

Source: David Tong

Try the reasoning in an experiment

Follow the wave into the cross section

  1. Predict the reaction-loss cross section when the imaginary depth reaches zero. Lower that depth continuously and inspect both wave and readout.
  2. Keep the well fixed and drag the detector around the circle. Find a large forward-to-backward change and locate those angles in the distribution below.
  3. Restore an imaginary depth of four. Explain why the wave display and angular prediction must use the same scattering matrices.

Exercises and explanations

  1. A partial wave has Sℓ=0.6eiπ/3S_\ell=0.6e^{i\pi/3}. What fraction of its incoming elastic flux is lost?

    Read the explanation

    The elastic outgoing-to-incoming flux ratio is ∣Sℓ∣2=0.36|S_\ell|^2=0.36, so the lost fraction is 0.64. The phase affects interference with other partial waves, but not this flux ratio.

  2. Does Sℓ=0S_\ell=0 imply a zero elastic scattering contribution?

    Read the explanation

    No. The scattering amplitude contains Sℓ−1S_\ell-1, whose squared modulus is one here. A missing outgoing elastic radial wave still produces diffraction relative to the free wave that would otherwise emerge.

Sources and reproduced scope

Evaluated level energies and published reaction measurements retain their source versions and uncertainties. Model curves, schematic figures and synthetic scattering records are identified separately. The offline benchmark files state which calculations have been reproduced.

  1. A. M. Moro, FRESCO notes, section 3.1, Optical Model

    Equations 1–8: elastic-channel wave, partial-wave cross sections and complex-potential absorption. The local square-well implementation fixes a neutral projectile and centre-of-mass energy.

  2. David Tong, Applications of Quantum Mechanics, scattering theory

    Partial-wave amplitude, optical theorem, phase shifts and matching. Neutron scattering here fixes a centre-of-mass energy and a spinless central potential.

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