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.

Helpful background. Scattering amplitudes and cross sections connect waves to flux. Phase shifts supplies the first proof; the T-matrix supplies the operator identity used to remove the central-potential restriction.

Forward Amplitude and Total Scattered Flux

Use time dependence $e^{-iEt/\hbar}$ and the asymptotic convention $\psi\sim e^{ikz}+f(\Omega)e^{ikr}/r$ with unit incident amplitude. Here $k>0$ is a wave number, so $f$ has length units. 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:

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

Here $k$ is the incident wave number and $f_{\alpha\alpha}$ multiplies $e^{ik r}/r$ relative to a unit incident wave. The formula assumes distinguishable-particle channel counting and short-range asymptotics; unscreened Coulomb scattering has a singular forward limit and cannot be substituted directly into it.

For open two-body channels with coordinate amplitudes $f_{\beta\alpha}$ and outgoing speeds $v_\beta$, the total on the left is

$$\sigma_{\mathrm{tot},\alpha} =\sum_{\beta\ \mathrm{open}}\frac{v_\beta}{v_\alpha} \int d\Omega\,|f_{\beta\alpha}(\Omega)|^2.$$

The elastic term has $v_\beta=v_\alpha$; other terms need their velocity ratios. An effective absorptive description replaces the explicitly summed missing channels by their loss cross section. The forward theorem determines the sum, not its partition among final channels.

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 same phase-shift sum therefore determines the forward imaginary part and the integrated outgoing flux.

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.

From operator unitarity to the factor 4π/k

The partial-wave proof used rotational symmetry. The operator proof needs only a complete conservative scattering problem with the stated short-range asymptotics. For one channel, use $\langle\mathbf r|\mathbf k\rangle=(2\pi)^{-3/2}e^{i\mathbf k\cdot\mathbf r}$ and $E_k=\hbar^2k^2/(2\mu)$. Write $T_{\mathbf p\mathbf k}=\langle\mathbf p|T(E_k+i0)|\mathbf k\rangle$. The transition-operator discontinuity gives

$$\operatorname{Im}T_{\mathbf k\mathbf k} =-\pi\int d^3p\, \delta\!\left(E_k-\frac{\hbar^2p^2}{2\mu}\right) |T_{\mathbf p\mathbf k}|^2.$$

The minus sign follows from $S=I-2\pi i\,\delta(E_f-E_i)T$. The radial density of states on the energy shell is

$$\int_0^\infty p^2dp\, \delta\!\left(E_k-\frac{\hbar^2p^2}{2\mu}\right) =\frac{\mu k}{\hbar^2}.$$

Now convert to the measured amplitude with $C=4\pi^2\mu/\hbar^2$ and $f=-CT$. Both the sign and the continuum-normalization factors matter:

$$\begin{aligned} \operatorname{Im}f(0) &=C\pi\frac{\mu k}{\hbar^2} \int d\Omega\,|T_{\mathbf k'\mathbf k}|^2\\ &=\frac{\pi\mu k}{C\hbar^2} \int d\Omega\,|f(\Omega)|^2 =\frac{k}{4\pi}\sigma_{\mathrm{el}}. \end{aligned}$$

No central potential was assumed. Including all open channels in the same operator identity gives their separate energy-shell densities and flux factors, hence the total cross section above. Signer’s scattering exercise presents this route in a different plane-wave normalization; its $T$ must be converted before comparing coefficients.

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 elastic flux fraction is $\eta_\ell^2$, not $\eta_\ell$. With $f_\ell=(S_\ell-1)/(2ik)$, the identity

$$\operatorname{Im}f_\ell-k|f_\ell|^2 =\frac{1-|S_\ell|^2}{4k}$$

is the missing-flux contribution in that partial wave. Partial-wave cross sections derives the corresponding elastic and reaction sums. The optical theorem counts both terms.

A two-channel calculation

At a fixed energy, consider two open spinless channels and a unit-flux $s$-wave matrix

$$S_0=i\begin{pmatrix} \eta&\sqrt{1-\eta^2}\\ \sqrt{1-\eta^2}&-\eta \end{pmatrix}, \qquad 0\le\eta\le1.$$

Take $S_\ell=I$ for every $\ell>0$. This is an on-shell model for checking flux balance, not a proposed potential or an energy-dependent threshold law. Direct multiplication gives $S_0^\dagger S_0=I$. For incidence in channel 1 with wave number $k$, the elastic amplitude is isotropic:

$$f_{11}=\frac{S_{11}-1}{2ik} =\frac{\eta+i}{2k}.$$

The outgoing coefficient in channel 2 must convert unit-flux normalization back to coordinate amplitude. Since its speed is $v_2$,

$$f_{21}=\sqrt{\frac{v_1}{v_2}}\frac{S_{21}}{2ik} =\sqrt{\frac{v_1}{v_2}}\frac{\sqrt{1-\eta^2}}{2k}.$$

Integrating these constant amplitudes over $4\pi$ gives

$$\begin{aligned} \sigma_{\mathrm{el}}&=4\pi|f_{11}|^2 =\frac{\pi}{k^2}(1+\eta^2),\\ \sigma_{2\leftarrow1}&=4\pi\frac{v_2}{v_1}|f_{21}|^2 =\frac{\pi}{k^2}(1-\eta^2),\\ \sigma_{\mathrm{tot}}&=\frac{2\pi}{k^2} =\frac{4\pi}{k}\operatorname{Im}f_{11}. \end{aligned}$$

At $\eta=1$, scattering is entirely elastic with phase shift $\delta_0=\pi/4$. At $\eta=1/2$, the elastic and reaction cross sections are $5\pi/(4k^2)$ and $3\pi/(4k^2)$; the forward imaginary part and total cross section are unchanged. Thus the theorem alone cannot distinguish those two partitions. Keeping only channel 1 makes the second process appear absorptive, while the full two-channel model conserves flux exactly. Even at $\eta=0$, elastic diffraction remains: $\sigma_{\mathrm{el}}=\sigma_{2\leftarrow1}=\pi/k^2$.

Use as a Check

Before using a mismatch as an error diagnostic, match the channel set, amplitude normalization, angular coverage, and perturbative order. A residual can then reveal omitted channels, numerical error, or an uncontrolled approximation. Agreement is necessary for an exact conservative solution, but cannot by itself establish that its dynamics is accurate.

For $V\mapsto\lambda V$ with real $\lambda$ and Hermitian $V$, write

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

The forward matrix element of $V$ is real, so $\operatorname{Im}f^{(1)}(0)=0$, even when the nonforward amplitude is complex. The leading elastic cross section is order $\lambda^2$, and the matching identity is

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

The second Born integral obtains this imaginary part from the on-shell delta term in $(E-H_0+i0)^{-1}$. Keeping $f^{(1)}$ for the leading differential cross section is consistent; using it alone to infer that the total cross section vanishes is not. For a complex effective optical potential the Hermitian assumption fails, and loss can already enter at first order.

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. At a fixed energy, an elastic scatterer has $\delta_0=\pi/2$, $\delta_1=\pi/4$, and all higher phase shifts zero. Calculate the forward amplitude, full cross section, and cross section in each hemisphere. Does doubling the forward-hemisphere result provide a valid optical-theorem test?

Solution

With $x=\cos\theta$, the amplitude is

$$f(x)=\frac1k\left[i+\frac32(1+i)x\right], \qquad f(\theta=0)=\frac{3+5i}{2k}.$$

Therefore the optical theorem predicts $\sigma_{\mathrm{tot}}=10\pi/k^2$. An independent angular integration uses

$$|f(x)|^2=\frac1{k^2}\left(1+3x+\frac92x^2\right).$$

Integrating $2\pi\,dx$ over $[0,1]$ and $[-1,0]$ gives

$$\sigma_{\mathrm F}=\frac{8\pi}{k^2}, \qquad \sigma_{\mathrm B}=\frac{2\pi}{k^2}.$$

Their sum agrees with the theorem. Doubling the forward result would instead give $16\pi/k^2$: full angular coverage cannot be inferred from hemisphere data when partial-wave interference produces a forward–backward asymmetry.

2. For the weak real Gaussian potential $V(r)=\lambda V_0e^{-r^2/a^2}$, the first Born result is

$$f^{(1)}(\theta)=-A\exp\!\left[-\frac{(ka)^2}{2}(1-\cos\theta)\right], \qquad A=\frac{\mu V_0\sqrt\pi a^3}{2\hbar^2}.$$

Find the coefficient of $\lambda^2$ in the total cross section and the imaginary forward second Born coefficient required by unitarity. Check the $ka\to0$ limit.

Solution

Let $b=(ka)^2$. Squaring the first-order amplitude and integrating gives

$$\sigma^{(2)}=2\pi A^2\int_{-1}^{1}e^{-b(1-x)}dx =\frac{2\pi A^2}{b}(1-e^{-2b}).$$

The matching forward coefficient is consequently

$$\operatorname{Im}f^{(2)}(0) =\frac{k}{4\pi}\sigma^{(2)} =\frac{A^2}{2ka^2}(1-e^{-2k^2a^2}).$$

As $ka\to0$, $1-e^{-2b}=2b+O(b^2)$, so

$$\sigma^{(2)}\to4\pi A^2, \qquad \operatorname{Im}f^{(2)}(0)=kA^2+O(k^3a^2A^2).$$

The imaginary amplitude vanishes linearly at threshold while the cross section remains finite because the theorem contains $1/k$. Here $\sigma(\lambda)=\lambda^2\sigma^{(2)}+O(\lambda^3)$. This determines only the imaginary forward part of $f^{(2)}$; its dispersive real part requires the principal-value Born integral.

3. In the two-channel model above, set $\eta=1/2$ and $v_2=v_1/4$. Calculate $f_{21}$ and the reaction cross section. What error results from omitting the velocity ratio?

Solution

The slower channel has coordinate amplitude

$$f_{21}=2\frac{\sqrt3/2}{2k}=\frac{\sqrt3}{2k}.$$

Its reaction cross section is

$$\sigma_{2\leftarrow1}=4\pi\left(\frac14\right)|f_{21}|^2 =\frac{3\pi}{4k^2}.$$

Using $4\pi|f_{21}|^2$ would overestimate this by a factor of four. Adding that erroneous result to $\sigma_{\mathrm{el}}=5\pi/(4k^2)$ would give $17\pi/(4k^2)$ instead of $2\pi/k^2$, falsely suggesting a failure of the optical theorem. Unit-flux matrix elements and coordinate-wave amplitudes must not be interchanged.

References

• R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Springer, 1982.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. Signer, “Quantum Mechanics II, Series 10”, ETH Zürich, Spring 2013, Exercise 1 — operator unitarity, on-shell density of states, and the forward cross-section relation.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• B. Zwiebach, “Chapter 7: Scattering”, Quantum Physics III, MIT OpenCourseWare 8.06, 2018, §7.2.1 — partial-wave normalization and the elastic optical theorem.