advanced35 min studyLesson 393 in Physics

Optical theorem in potential scattering

Derive the forward-amplitude constraint from elastic partial waves and diagnose perturbative apparent violations.

Before you begin

Partial-wave phase shifts

On this page

The question for today

What you’ll understand

Derive the 4π/k optical-theorem factor in the fixed amplitude convention.
Explain forward interference as the removal of incident flux.
Compare exact unitarity with order-by-order perturbative consistency.

The total number of particles scattered into all directions is encoded in one complex number: the forward scattering amplitude. This optical theorem can look surprising if scattered and unscattered beams are imagined as independent probabilities. They are coherent parts of one state. Their interference near the forward direction supplies the flux removed from the incident beam, and probability conservation links that removal to the total outgoing scattered flux.

Derive the relation in the elastic case

For a real short-range central potential, each partial wave has a real phase shift and unit-modulus S element. At zero scattering angle every Legendre polynomial equals one. The imaginary part of each partial amplitude is therefore its squared sine divided by wavenumber. Multiplying the forward imaginary amplitude by the appropriate kinematic factor reproduces exactly the integrated partial-wave cross section. This derivation fixes the numerical factor rather than borrowing it from a different scattering-amplitude normalization.

The real forward amplitude does not enter the total cross section by itself. It carries phase information that can be important in interference or propagation experiments, but it cannot substitute for the imaginary component in this theorem. Likewise, replacing the imaginary part by the squared forward magnitude is incorrect. The latter is a differential cross section at one angle, whereas the theorem relates an amplitude component to an integral over every accessible scattering direction.

\operatorname{Im}f(0)=\frac1k\sum_\ell(2\ell+1)\sin^2\delta_\ell,\qquad \sigma_{\rm el}=\frac{4\pi}{k}\operatorname{Im}f(0)

f has length units and k has inverse length, producing an area. This equality is for the unit incident plane wave plus f eⁱᵏʳ/r convention with only elastic scattering available.↔ Scroll wide equations horizontally.

Identify where the incident flux goes

The outgoing spherical current alone is positive in every direction, so adding it incoherently to an unchanged incident plane-wave current would create probability. The current of the full wavefunction includes cross terms. Their large-distance angular contribution is concentrated around forward propagation and removes the needed incident flux. A finite incident packet makes this redistribution physically explicit; the stationary plane-wave treatment is the limit of that beam description.

When inelastic channels are open, forward elastic scattering still records the total removal from the incident channel, including flux that exits in other internal states. The optical theorem then uses the total cross section, which equals elastic plus reaction contributions. Inelastic scattering need not change the number of particles; changing internal states while conserving total probability is sufficient. Unscreened Coulomb interactions complicate the standard forward statement because the amplitude and integrated cross section are singular and the short-range asymptotic assumptions fail.

\sigma_{\rm tot}=\sigma_{\rm el}+\sigma_{\rm reaction}=\frac{4\pi}{k}\operatorname{Im}f_{aa}(0)

The index a denotes the incident elastic channel. Other channels require their proper flux factors, and this standard finite total-cross-section statement assumes short-range asymptotics and a complete accounting of outgoing channels.↔ Scroll wide equations horizontally.

Check approximations at consistent order

A first-order weak-potential amplitude from a real potential is often real. Squaring it produces a nonzero second-order cross section, while taking its first-order imaginary forward part gives zero. This does not invalidate the optical theorem. It means that a second-order observable has been compared with an amplitude truncated one order too early. The missing second-order imaginary amplitude supplies exactly the contribution needed by unitarity at that order.

The same lesson appears in a small phase-shift expansion. The leading partial amplitude is real and linear in the phase, but its next term is imaginary and quadratic. Keeping the quadratic probability while dropping the quadratic imaginary term loses the flux balance. The wave lab can demonstrate why cross terms must be retained in coherent sums, but its same-frequency components are not a derivation of the spatial forward-interference integral and it has no probability-current or total-cross-section readout.

f_0=\frac{e^{i\delta_0}\sin\delta_0}{k}=\frac{\delta_0}{k}+i\frac{\delta_0^2}{k}+O(\delta_0^3),\qquad \sigma_0=\frac{4\pi\delta_0^2}{k^2}+O(\delta_0^4)

The expansion assumes a small real phase δ0 at fixed positive k. The imaginary term and the cross section first appear at the same second order; exact unitarity is recovered by keeping the full exponential.↔ Scroll wide equations horizontally.

Work it through

Forward amplitude from two phase shifts

Let $k=1.0\,\mathrm{nm^{-1}}$ , $\delta_0=\pi/4$ and $\delta_1=\pi/6$ , with higher waves absent. Compute the total elastic cross section and independently reconstruct it from the forward imaginary amplitude.

The integrated formula gives $\sigma=4\pi[\sin^2(\pi/4)+3\sin^2(\pi/6)]\,\mathrm{nm^2}=4\pi(1/2+3/4)=5\pi\,\mathrm{nm^2}$ .
At zero angle both angular polynomials equal one. The imaginary contribution is $(1/2+3/4)/k=1.25\,\mathrm{nm}$ . The real contributions do not enter this particular sum rule.
Multiplying the forward imaginary part by $4\pi/k$ gives $4\pi(1.25)=5\pi\,\mathrm{nm^2}$ , identical to the angularly integrated answer. Both routes use the same amplitude normalization and include both active angular momenta.

The result

The total elastic cross section is 5π nm², and Im f(0) = 1.25 nm gives precisely the same value through the optical theorem.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Why amplitudes cannot be added as probabilities

Reset and set BOTH Phase difference = 0° and Amplitude ratio = 1, predicting resultant amplitude 2. Reset and set Phase difference = 180° and Amplitude ratio = 1, predicting zero. Both constituent magnitudes remain one in these schematic same-frequency sums. The lab contains no radial scattering state, forward beam geometry, probability current or optical-theorem calculation.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

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 forCross terms can remove or reinforce a coherent signal even when separate component intensities are unchanged.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Squaring a first-order amplitude produces a second-order quantity. At that order the amplitude also acquires an imaginary forward contribution, so a first-order-only comparison is incomplete.

Your turn

Try it yourself

At fixed $k=2.0\,\mathrm{nm^{-1}}$ , an elastic s wave has a small phase $\delta_0=0.10$ . To second order in this phase, find the real and imaginary amplitudes and verify the optical theorem for the leading cross section.

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

Give me a hint

Use f0 ≈ δ0/k + iδ0²/k and retain probability terms only through second order.

Compare with the solution

The amplitude components are approximately $0.050\,\mathrm{nm}$ and $0.0050\,\mathrm{nm}$ , respectively. The leading cross section is $4\pi\delta_0^2/k^2=0.010\pi\,\mathrm{nm^2}$ . The optical theorem gives $(4\pi/2)(0.0050)=0.010\pi\,\mathrm{nm^2}$ as well. Squaring the approximate imaginary term would contribute fourth order and should not be included in a strictly second-order comparison. The exact amplitude would satisfy the theorem without any truncation if its sine and exponential were retained.

Keep the idea

Take it with you

The optical-theorem coefficient depends on the declared amplitude normalization.
Forward interference accounts for incident-channel flux removed into scattering.
Perturbative amplitudes and probabilities must be compared at compatible orders.

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.