Optical Theorem

Purpose

For the asymptotic convention

$$\psi_{\mathbf k}^{(+)}(\mathbf r) \sim e^{i\mathbf k\cdot\mathbf r} + f(\theta,\phi) \frac{e^{ikr}}{r},$$

unitarity implies

$$\sigma_{\mathrm{tot}} = \frac{4\pi}{k} \operatorname{Im}f(0).$$

The amplitude on the right is the forward elastic amplitude for the incident channel. The cross section on the left is inclusive: it counts elastic scattering and every inelastic, reactive, or absorptive process that removes flux from that incident channel.

If elastic scattering is the only open process,

$$\sigma_{\mathrm{tot}} = \sigma_{\mathrm{el}} = \int d\Omega\, \lvert f(\theta,\phi)\rvert^2.$$

The derivation from $S^\dagger S=I$ belongs to Optical Theorem. This card collects the normalizations, partial-wave identities, perturbative order counting, and practical checks.

At a glance

Situation	Identity
One elastic channel	$(4\pi/k)\operatorname{Im}f(0)=\int d\Omega\,\lvert f\rvert^2$
Several open channels	$(4\pi/k_\alpha)\operatorname{Im}f_{\alpha\alpha}(0)=\sum_\beta(v_\beta/v_\alpha)\int d\Omega\,\lvert f_{\beta\alpha}\rvert^2$
Central partial waves	$\sigma_{\mathrm{tot}}=(4\pi/k^2)\sum_\ell(2\ell+1)\operatorname{Im}a_\ell$
Elastic partial wave	$\operatorname{Im}a_\ell=\lvert a_\ell\rvert^2=\sin^2\delta_\ell$
Inelastic diagonal element	$\operatorname{Im}a_\ell-\lvert a_\ell\rvert^2=(1-\eta_\ell^2)/4$
Perturbative second order	$\operatorname{Im}f^{(2)}(0)=(k/4\pi)\int d\Omega\,\lvert f^{(1)}\rvert^2$

The multichannel row assumes the displayed spherical-wave channel normalization. Other state normalizations move velocity and phase-space factors, but the underlying unitarity sum is unchanged.

Why the forward direction appears

The total scattered flux is distributed over all angles, so a relation to one forward amplitude can initially seem surprising. The incident plane wave and the outgoing spherical wave overlap coherently in the forward direction. The interference current there measures the depletion of the incident beam.

Probability conservation then requires

$$\text{forward extinction} = \text{flux into every allowed final state}.$$

The theorem is therefore an interference statement as well as a unitarity statement. It is not obtained by evaluating $\lvert f(0)\rvert^2$; the relevant quantity is the imaginary part of the complex forward amplitude.

A wave-packet or finite-beam treatment resolves the formal overlap of a plane wave with the narrow forward scattered contribution. The plane-wave result is the limiting cross-section identity.

Operator origin

In a schematic normalization, write

$$S=I+i\mathcal T.$$

Unitarity gives

$$S^\dagger S=I,$$

and therefore

$$i\left( \mathcal T^\dagger-\mathcal T \right) = \mathcal T^\dagger\mathcal T.$$

Taking a diagonal matrix element in an incoming state $\lvert i\rangle$ and inserting a complete set of final states gives

$$2\operatorname{Im}\mathcal T_{ii} = \sum_f \lvert\mathcal T_{fi}\rvert^2.$$

For continuum states, the sum includes energy-conserving delta functions and phase-space integrals. Converting this equation to the spherical-wave amplitude and dividing by incident flux produces the optical theorem.

This $\mathcal T$ is schematic. The chapter’s normalization-specific transition operator uses

$$S = I - 2\pi i\, \delta(E_f-E_i) T,$$

and the nonrelativistic $f$ is proportional to $-T$. Do not combine signs or $2\pi$ factors from the two conventions. The exact transition-operator identity is at Unitarity.

Elastic partial-wave check

For a short-range central potential,

$$f(\theta) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) a_\ell P_\ell(\cos\theta),$$

where

$$a_\ell = \frac{S_\ell-1}{2i}.$$

In a purely elastic channel,

$$S_\ell=e^{2i\delta_\ell}, \qquad a_\ell=e^{i\delta_\ell}\sin\delta_\ell.$$

Since $P_\ell(1)=1$,

$$f(0) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) e^{i\delta_\ell}\sin\delta_\ell.$$

The imaginary part is

$$\operatorname{Im}f(0) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) \sin^2\delta_\ell.$$

Multiplication by $4\pi/k$ gives

$$\frac{4\pi}{k} \operatorname{Im}f(0) = \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \sin^2\delta_\ell = \sigma_{\mathrm{el}}.$$

Equivalently, elastic unitarity holds channel by channel:

$$\operatorname{Im}a_\ell = \lvert a_\ell\rvert^2.$$

The identity is especially useful in a numerical phase-shift calculation: the independently integrated angular cross section and the forward partial sum must agree at the same angular-momentum cutoff.

Inelasticity and reaction cross sections

When other channels are open, parameterize the diagonal elastic partial wave as

$$S_\ell = \eta_\ell e^{2i\delta_\ell}, \qquad 0\le\eta_\ell\le1.$$

For

$$a_\ell = \frac{S_\ell-1}{2i},$$

one has

$$\operatorname{Im}a_\ell = \frac{1-\operatorname{Re}S_\ell}{2}$$

and

$$\operatorname{Im}a_\ell - \lvert a_\ell\rvert^2 = \frac{1-\eta_\ell^2}{4}.$$

The three partial-wave contributions are

$$\sigma_\ell^{\mathrm{el}} = \frac{\pi}{k^2} (2\ell+1) \lvert S_\ell-1\rvert^2,$$ $$\sigma_\ell^{\mathrm{reac}} = \frac{\pi}{k^2} (2\ell+1) \left(1-\lvert S_\ell\rvert^2\right),$$

and

$$\begin{aligned} \sigma_\ell^{\mathrm{tot}} &= \sigma_\ell^{\mathrm{el}} + \sigma_\ell^{\mathrm{reac}} \\ &= \frac{2\pi}{k^2} (2\ell+1) \left( 1-\operatorname{Re}S_\ell \right) \\ &= \frac{4\pi}{k^2} (2\ell+1) \operatorname{Im}a_\ell. \end{aligned}$$

Summing over $\ell$ proves

$$\sigma_{\mathrm{tot}} = \frac{4\pi}{k} \operatorname{Im}f_{\alpha\alpha}(0)$$

even though the observed elastic element has $\lvert S_\ell\rvert<1$. The full multichannel $S$ matrix remains unitary; the deficit in the elastic element is carried by the other channels.

Multichannel form

Let the component in outgoing channel $\beta$ have asymptotic form

$$\psi_\beta(\mathbf r) \sim \delta_{\beta\alpha} e^{i\mathbf k_\alpha\cdot\mathbf r} + f_{\beta\alpha}(\Omega) \frac{e^{ik_\beta r}}{r}.$$

With

$$v_\gamma = \frac{\hbar k_\gamma}{\mu_\gamma},$$

current conservation gives

$$\frac{4\pi}{k_\alpha} \operatorname{Im} f_{\alpha\alpha}(0) = \sum_{\beta\ \mathrm{open}} \frac{v_\beta}{v_\alpha} \int d\Omega\, \lvert f_{\beta\alpha}(\Omega)\rvert^2.$$

The $\beta=\alpha$ term is elastic. Every other open channel contributes to the reaction or inelastic total. Closed channels modify amplitudes virtually but carry no asymptotic flux and do not appear as final states in this sum.

If the channel states have been normalized to unit flux, the velocity factors are absorbed into the amplitudes. If additional final-state variables are continuous, the right-hand side also includes their phase-space integrals. Always compare formulas only after matching the channel normalization.

Low-energy s-wave example

Neglect effective-range corrections and write the unitary scattering-length amplitude as

$$f_0(k) = \frac{1}{-1/a-ik} = - \frac{a}{1+ika}.$$

Its real and imaginary parts are

$$f_0(k) = - \frac{a}{1+k^2a^2} + i \frac{ka^2}{1+k^2a^2}.$$

Therefore

$$\operatorname{Im}f_0 = \frac{ka^2}{1+k^2a^2} = k\lvert f_0\rvert^2.$$

The optical theorem gives

$$\sigma_{\mathrm{tot}} = \frac{4\pi}{k} \operatorname{Im}f_0 = \frac{4\pi a^2}{1+k^2a^2},$$

which equals $4\pi\lvert f_0\rvert^2$. At $\lvert a\rvert\to\infty$, the result reaches the $s$-wave unitarity limit $4\pi/k^2$.

This example shows why replacing the denominator by its leading real value, $f_0\approx-a$, gives the leading low-$k$ cross section but not exact unitarity.

Perturbative unitarity

Introduce a coupling parameter and expand

$$f = \lambda f^{(1)} + \lambda^2f^{(2)} + O(\lambda^3).$$

For one-channel elastic scattering, the optical theorem becomes

$$\operatorname{Im}f(0) = \frac{k}{4\pi} \int d\Omega\, \lvert f(\Omega)\rvert^2.$$

At order $\lambda$,

$$\operatorname{Im}f^{(1)}(0)=0$$

for a Hermitian first-order interaction. At order $\lambda^2$,

$$\operatorname{Im}f^{(2)}(0) = \frac{k}{4\pi} \int d\Omega\, \lvert f^{(1)}(\Omega)\rvert^2.$$

This is why a real first Born amplitude can produce the correct leading cross section even though it does not satisfy the exact optical theorem by itself. The second Born term supplies the required on-shell imaginary part.

For several channels, the second-order right-hand side is a sum over all first-order open-channel amplitudes with their flux factors. Omitting an intermediate open channel breaks perturbative unitarity.

Effective absorption

An optical potential often represents eliminated reaction channels by

$$V(\mathbf r) = U(\mathbf r) - iW(\mathbf r), \qquad W(\mathbf r)\ge0.$$

The reduced elastic $S$ matrix is then subunitary. For a unit-amplitude incident wave with speed $v$, the continuity equation gives the absorbed cross section

$$\sigma_{\mathrm{abs}} = \frac{2}{\hbar v} \int d^3r\, W(\mathbf r) \lvert\psi^{(+)}(\mathbf r)\rvert^2.$$

The generalized flux balance is

$$\sigma_{\mathrm{tot}} = \sigma_{\mathrm{el}} + \sigma_{\mathrm{abs}} = \frac{4\pi}{k} \operatorname{Im}f(0).$$

The missing elastic probability has not disappeared from the underlying closed system; it represents channels removed from the effective model.

Numerical diagnostic

Given an amplitude and independently summed cross sections, define

$$\Delta_{\mathrm{OT}} = \frac{4\pi}{k} \operatorname{Im}f(0) - \sigma_{\mathrm{tot}}.$$

A dimensionless residual is

$$\epsilon_{\mathrm{OT}} = \frac{ \lvert\Delta_{\mathrm{OT}}\rvert }{ \max\left( \sigma_{\mathrm{tot}}, \frac{4\pi}{k}\lvert\operatorname{Im}f(0)\rvert, \sigma_{\mathrm{scale}} \right) },$$

where $\sigma_{\mathrm{scale}}>0$ prevents an unstable relative error near a true zero. Report the chosen scale.

A nonzero residual can reveal:

• a mismatched amplitude or state normalization;
• omitted open channels or spin sums;
• an insufficient partial-wave cutoff;
• inaccurate forward-angle extrapolation;
• inconsistent quadrature between the angular and forward calculations;
• a lost $i0$ prescription or on-shell imaginary part;
• an approximation being tested beyond the order it retains.

Passing the optical-theorem test establishes consistency with unitarity in the represented space. It does not prove that the dynamics, potential, or channel model is physically accurate.

Scope limits

Long-range Coulomb scattering

For an unscreened Coulomb potential, the usual plane-wave plus spherical-wave asymptotic form is modified and the forward amplitude is singular. The ideal total cross section also diverges. The short-range formula should not be applied without screening, a Coulomb-distorted formulation, or a finite experimental angular resolution.

Identical particles

Unitarity applies to correctly symmetrized asymptotic states. Direct and exchange amplitudes, spin channels, and the one-count-per-final-pair rule must be incorporated before constructing the inclusive channel sum. Blindly inserting $f(\theta)\pm f(\pi-\theta)$ into a distinguishable-particle optical theorem can double count states.

Finite acceptance

The theorem returns the full inclusive total cross section. It does not equate the forward amplitude to a detector’s acceptance-limited integral. Detector cuts belong in a separate observable calculation.

Nonunitary approximations

A truncated or phenomenological amplitude can violate the theorem. The meaningful question is whether it satisfies unitarity through its claimed order or whether its explicit loss term accounts for the deficit.

Calculation workflow

1. Write the asymptotic amplitude, state normalization, and incident wave number.
2. Identify the forward elastic matrix element for the prepared incident channel.
3. List every open final channel and its velocity, spin, and phase-space factors.
4. Compute the inclusive total independently from the outgoing channels.
5. Compare it with $(4\pi/k)\operatorname{Im}f_{\alpha\alpha}(0)$.
6. In a partial-wave calculation, use the same $\ell_{\max}$ on both sides before taking a convergence limit.
7. For a perturbative result, compare equal orders in the coupling expansion.
8. Treat Coulomb tails, identical particles, absorption, and detector acceptance with their own conventions.

Common mistakes

• Using $\lvert f(0)\rvert^2$ instead of $\operatorname{Im}f(0)$.
• Equating the forward theorem with an angular average of $f$.
• Calling the elastic angular integral $\sigma_{\mathrm{tot}}$ when reaction channels are open.
• Omitting velocity or phase-space factors in a multichannel sum.
• Mixing $S=I+i\mathcal T$ with a differently normalized $T$ matrix.
• Demanding exact optical-theorem equality from a first-order Born amplitude.
• Comparing different partial-wave cutoffs on the two sides.
• Treating an acceptance-limited cross section as the inclusive total.
• Applying the short-range formula to unscreened Coulomb scattering.
• Ignoring identical-particle final-state counting.
• Assuming that a small unitarity residual proves the interaction model is correct.

Exercises

1. Use elastic partial waves to derive the optical theorem.

Solution

At $\theta=0$, $P_\ell(1)=1$, so

$$f(0) = \frac{1}{k} \sum_{\ell=0}^{\infty} (2\ell+1) e^{i\delta_\ell}\sin\delta_\ell.$$

Because

$$\operatorname{Im} \left( e^{i\delta_\ell}\sin\delta_\ell \right) = \sin^2\delta_\ell,$$

one obtains

$$\frac{4\pi}{k} \operatorname{Im}f(0) = \frac{4\pi}{k^2} \sum_{\ell=0}^{\infty} (2\ell+1) \sin^2\delta_\ell.$$

The right-hand side is the integrated elastic partial-wave cross section.

2. For $S_\ell=\eta_\ell e^{2i\delta_\ell}$, verify that the elastic and reaction pieces add to the forward total.

Solution

The sum of the two partial cross sections is

$$\begin{aligned} &\frac{\pi}{k^2}(2\ell+1) \left[ \lvert S_\ell-1\rvert^2 + 1-\lvert S_\ell\rvert^2 \right] \\ &\qquad= \frac{\pi}{k^2}(2\ell+1) \left[ 2-2\operatorname{Re}S_\ell \right] \\ &\qquad= \frac{2\pi}{k^2}(2\ell+1) \left( 1-\operatorname{Re}S_\ell \right). \end{aligned}$$

Since

$$\operatorname{Im}a_\ell = \frac{1-\operatorname{Re}S_\ell}{2},$$

this equals

$$\frac{4\pi}{k^2} (2\ell+1) \operatorname{Im}a_\ell,$$

the partial-wave contribution to $(4\pi/k)\operatorname{Im}f(0)$.

3. Verify the optical theorem for $f_0=-a/(1+ika)$ and identify the unitarity limit.

Solution

The magnitude and imaginary part are

$$\lvert f_0\rvert^2 = \frac{a^2}{1+k^2a^2}, \qquad \operatorname{Im}f_0 = \frac{ka^2}{1+k^2a^2}.$$

Thus

$$\frac{4\pi}{k} \operatorname{Im}f_0 = 4\pi\lvert f_0\rvert^2 = \frac{4\pi a^2}{1+k^2a^2}.$$

As $\lvert a\rvert\to\infty$ at fixed $k$,

$$\sigma\longrightarrow\frac{4\pi}{k^2},$$

the elastic $s$-wave unitarity limit.

4. A real first Born amplitude is order $\lambda$. At what order do the total cross section and the forward imaginary part first appear?

Solution

The leading cross section is

$$\sigma^{(2)} = \lambda^2 \int d\Omega\, \lvert f^{(1)}\rvert^2,$$

so it begins at order $\lambda^2$. The order-$\lambda$ forward imaginary part vanishes for a Hermitian first-order interaction. Unitarity requires

$$\operatorname{Im}f^{(2)}(0) = \frac{k}{4\pi} \int d\Omega\, \lvert f^{(1)}\rvert^2.$$

The first nonzero forward imaginary part is therefore also order $\lambda^2$ and comes from the second Born term or its equivalent on-shell intermediate-state contribution.

Canonical links

• Unitarity
• Optical Theorem
• Probability Current and Flux
• Partial-Wave Cross Sections
• Born Series
• Scattering Cross Section Formula Card

References

• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006, Chs. 3–4.
• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Springer, 1982, Chs. 5 and 10.
• M. L. Goldberger and K. M. Watson, Collision Theory, Wiley, 1964, Chs. 3–4.
• C. J. Joachain, Quantum Collision Theory, 3rd ed., North-Holland, 1983, Chs. 3 and 7.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, Ch. 7.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995, Sec. 7.3.