Optical Theorem

The optical theorem relates the total scattering cross section to the imaginary part of the forward scattering amplitude. With the standard nonrelativistic convention,

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

Here $f(0)$ means the scattering amplitude in the forward direction $\theta=0$. The theorem is a direct consequence of unitarity: probability removed from the incident beam must appear in all outgoing channels.

Its operator origin is $S^\dagger S=I$. S-Matrix constructs that operator, and Unitarity develops its channel and partial-wave constraints. This page derives the corresponding forward-amplitude identity in the stated normalization.

Statement

For elastic scattering from a real short-range potential with only one open channel,

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

If inelastic or absorptive channels are open, the same structure relates the forward elastic amplitude to the total cross section into all channels:

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

The precise channel labels and normalization factors depend on convention, but the unitarity content is invariant.

Partial-Wave Proof

For a central potential with elastic phase shifts,

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

At forward angle, $P_\ell(1)=1$, so

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

Since

$$e^{i\delta_\ell}\sin\delta_\ell = \sin\delta_\ell\cos\delta_\ell + i\sin^2\delta_\ell,$$

the imaginary part is

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

Multiplying by $4\pi/k$ gives

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

But the right-hand side is exactly the elastic sum derived at Partial-Wave Cross Sections. The present page owns its identification with the forward imaginary part.

Physical Meaning

The theorem can seem surprising because it connects all scattering angles to the amplitude at $\theta=0$. The reason is interference. The incident plane wave and the forward scattered wave overlap in the forward direction. Their interference accounts for the reduction of flux in the incident beam.

The word “optical” comes from the analogous extinction theorem in wave optics: forward scattering encodes the total attenuation of an incident wave. Hard-Sphere Scattering makes this concrete: a forward diffraction peak supplies one geometric area in the high-energy integrated cross section even though it narrows away from every fixed nonzero angle.

Unitarity View

For a schematic convention used only in this subsection, write the scattering operator as

$$S=1+iT$$

The $T$ in this subsection includes convention-dependent normalization and energy-conservation factors. T-Matrix fixes the volume convention $S=I-2\pi i\,\delta(E_f-E_i)T$ and derives the corresponding discontinuity relation. In the schematic convention above, unitarity,

$$S^\dagger S=1,$$

implies

$$i(T-T^\dagger) = - T^\dagger T.$$

Matrix elements of this relation between the same incoming state connect the imaginary part of the forward transition amplitude to a sum over probabilities for all possible final states. The optical theorem is the cross-section form of this statement.

Different books distribute signs and factors of $2\pi$ differently by defining $S=1-2\pi i\,\delta(E_f-E_i)T$ or related conventions. The observable content is the same.

Absorption and Inelasticity

When inelastic channels are open, a single elastic phase shift is not enough. A common partial-wave parametrization is

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

The parameter $\eta_\ell$ measures how much flux remains in the elastic channel. The optical theorem then counts both elastic scattering and loss into inelastic channels. This is why the forward elastic amplitude knows about reactions that do not end in the elastic final state.

Use as a Check

The optical theorem is an excellent diagnostic for approximations. If an approximate amplitude gives a total cross section that badly violates the optical theorem in a regime where unitarity should be visible, the approximation is being pushed beyond its domain.

The first Born approximation often gives a real amplitude for real weak potentials at leading order. Then $\operatorname{Im} f_{\mathrm B}(0)$ may vanish while $|f_{\mathrm B}|^2$ gives a nonzero cross section. This is not a paradox: unitarity is restored order by order only when the appropriate higher Born terms are included.

QFT Bridge

In quantum field theory, the optical theorem becomes a relation between the imaginary part of a forward scattering amplitude and a sum over intermediate on-shell states. Diagrammatically, this is the origin of cutting rules. The nonrelativistic theorem here is the fixed-particle ancestor of that unitarity logic, but relativistic normalizations and phase space change the formula.

Common Mistakes

• Forgetting that the theorem uses the forward amplitude, not an angular average of the amplitude.
• Applying the simple elastic formula when inelastic channels are open without modifying the interpretation of $\sigma_{\mathrm{tot}}$.
• Treating a leading real Born amplitude as if it satisfied unitarity by itself.
• Losing convention factors when comparing $T$-matrix and $f$-amplitude versions.
• Assuming the theorem is special to central potentials; partial waves give a clean proof, but unitarity is more general.

Exercises

1. Use the partial-wave amplitude to prove the elastic optical theorem.

Solution

At $\theta=0$,

$$P_\ell(1)=1.$$

Thus

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

Taking the imaginary part gives

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

Therefore

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

2. If only the $s$-wave contributes, verify the optical theorem explicitly.

Solution

For $\ell=0$,

$$f(0) = \frac{1}{k}e^{i\delta_0}\sin\delta_0.$$

Hence

$$\operatorname{Im}f(0) = \frac{1}{k}\sin^2\delta_0.$$

The optical theorem gives

$$\sigma = \frac{4\pi}{k^2} \sin^2\delta_0,$$

which is the $s$-wave total elastic cross section.

3. Why can a leading Born amplitude fail the optical theorem even when it is useful?

Solution

For a real potential, the leading Born amplitude is often real. Then its forward imaginary part vanishes, while $|f_{\mathrm B}|^2$ gives a nonzero cross section. The optical theorem relates terms of different perturbative order: the imaginary forward contribution needed for unitarity appears at higher order. The leading Born approximation can still give the leading cross section, but it does not by itself satisfy exact unitarity.

References

• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Springer, 1982.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.