The Optical Theorem

Derivation

The Optical Theorem

D-363 Home PU-401 Threads waves · chance · energy Depends on Partial-Wave Analysis and Phase Shifts

Statement

For elastic potential scattering of a particle of wavenumber \(k\), the total cross section is fixed by the imaginary part of the scattering amplitude in the exact forward direction, \(\theta = 0\): \[ \sigma_{\text{tot}} = \frac{4\pi}{k}\,\operatorname{Im} f(0). \] This is the optical theorem. It is an exact identity, not an approximation, and expresses conservation of probability flux: whatever is removed from the forward beam (into all scattering angles and, more generally, into all open channels) must appear as a reduction of the forward wave, which is encoded in \(\operatorname{Im} f(0)\).

Why it matters

The optical theorem ties a bulk, angle-integrated quantity — the total cross section — to a single complex number, the forward amplitude. It is the scattering-theory face of unitarity: the statement that the scattering operator \(S\) conserves probability. Because it holds channel by channel, it survives even when the target absorbs the particle, so it is the workhorse relation linking measured total cross sections to the analytic structure of amplitudes throughout nuclear, particle, atomic, and condensed-matter physics.

Practically, it lets experimenters extract \(\operatorname{Im} f(0)\) from a total-cross-section measurement, and it constrains any candidate amplitude: a proposed \(f(\theta)\) whose forward imaginary part disagrees with \(\int|f|^2\,d\Omega\) is simply inconsistent with quantum mechanics. It also underlies dispersion relations and the shadow-scattering picture of diffraction.

Assumptions

Central, short-range potential \(V(r)\) falling faster than \(1/r\).If dropped, the partial-wave sum for \(f(0)\) diverges (as for the Coulomb field) and both the forward amplitude and \(\sigma_{\text{tot}}\) become ill-defined, so the theorem needs a screened/modified form.

Stationary (time-independent) scattering at fixed energy \(E=\hbar^2k^2/2m\).If dropped, there is no single sharp \(k\) and the interference of incident and scattered waves that produces the identity is smeared out.

Azimuthal symmetry, so the amplitude is a single scalar function \(f(\theta)\).If dropped, \(f\) becomes a matrix in spin/channel indices and the theorem must be written for the forward amplitude matrix rather than a scalar.

Purely elastic scattering: real phase shifts \(\delta_l\), unitary partial waves \(|S_l|=1\).If dropped (absorption present), \(\sigma_{\text{tot}}\) in the theorem still equals \((4\pi/k)\operatorname{Im}f(0)\), but it now counts elastic plus reaction cross sections; the intermediate elastic-only identity \(\sigma=\tfrac{4\pi}{k^2}\sum(2l+1)\sin^2\delta_l\) no longer holds.

Convergent partial-wave expansion and interchange of sum and angular integral.If dropped, the term-by-term use of Legendre orthogonality below is not justified and the closed forms fail.

Derivation

\[ f(\theta) = \frac{1}{k}\sum_{l=0}^{\infty}(2l+1)\,e^{i\delta_l}\sin\delta_l\,P_l(\cos\theta) \]

Prior result (partial-wave phase shifts): the asymptotic scattered wave organised into angular-momentum channels, each channel dephased by \(\delta_l\). A

\[ \sigma_{\text{tot}} = \int |f(\theta)|^2\, d\Omega \]

Definition: the total cross section is the differential cross section \(d\sigma/d\Omega=|f|^2\) integrated over all solid angle. A

\[ \sigma_{\text{tot}} = \frac{1}{k^2}\sum_{l,l'}(2l+1)(2l'+1)\,e^{i(\delta_l-\delta_{l'})}\sin\delta_l\sin\delta_{l'}\!\int P_l P_{l'}\, d\Omega \]

Insert the expansion for \(f\) and its conjugate \(f^*\); the Legendre polynomials are real. Symbols only, no numbers yet. B

\[ \int P_l(\cos\theta)\,P_{l'}(\cos\theta)\, d\Omega = \frac{4\pi}{2l+1}\,\delta_{ll'} \]

Orthogonality of Legendre polynomials on the sphere collapses the double sum to a single sum; the phase factors \(e^{i(\delta_l-\delta_{l'})}\to 1\) on the diagonal. B

\[ \sigma_{\text{tot}} = \frac{4\pi}{k^2}\sum_{l=0}^{\infty}(2l+1)\,\sin^2\delta_l \]

Carry out the \(l'\) sum against the Kronecker delta. This is the total elastic cross section in terms of phase shifts. A

\[ f(0) = \frac{1}{k}\sum_{l=0}^{\infty}(2l+1)\,e^{i\delta_l}\sin\delta_l, \qquad P_l(1)=1 \]

Evaluate the amplitude in the forward direction \(\theta=0\); every Legendre polynomial equals unity there. A

\[ e^{i\delta_l}\sin\delta_l = \sin\delta_l\cos\delta_l + i\sin^2\delta_l \;\Rightarrow\; \operatorname{Im} f(0) = \frac{1}{k}\sum_{l=0}^{\infty}(2l+1)\,\sin^2\delta_l \]

Take the imaginary part term by term (\(\delta_l\) real): the imaginary part of each channel is exactly \(\sin^2\delta_l\). B

\[ \sigma_{\text{tot}} = \frac{4\pi}{k^2}\sum(2l+1)\sin^2\delta_l = \frac{4\pi}{k}\left[\frac{1}{k}\sum(2l+1)\sin^2\delta_l\right] = \frac{4\pi}{k}\,\operatorname{Im} f(0) \]

Compare Step 5 with Step 7: the two sums are identical up to the factor \(k\). This is the optical theorem. A

\[ a_l \equiv \frac{e^{2i\delta_l}-1}{2ik}=\frac{e^{i\delta_l}\sin\delta_l}{k}\;\Rightarrow\; \operatorname{Im} a_l = \frac{\sin^2\delta_l}{k}=k\,|a_l|^2 \]

Deeper origin: with \(S_l=e^{2i\delta_l}\) and \(|S_l|=1\) (unitarity), each partial-wave amplitude obeys \(\operatorname{Im} a_l = k|a_l|^2\). Summing \(\sum(2l+1)\) gives \(\operatorname{Im}f(0)=\tfrac{k}{4\pi}\sigma_{\text{tot}}\) directly — the theorem is unitarity, term by term. For absorption, \(|S_l|<1\) and \(k|a_l|^2\) is replaced by the total (elastic + reaction) rate, yet the forward identity survives. C

Result

\[ \boxed{\;\sigma_{\text{tot}} = \dfrac{4\pi}{k}\,\operatorname{Im} f(0)\;}\qquad \operatorname{Im} f(0)=\frac{1}{k}\sum_{l}(2l+1)\sin^2\delta_l \]

Reading. The total probability scattered out of the incident beam is set entirely by how much the forward-going wave is attenuated — quantified by the imaginary (out-of-phase) part of \(f(0)\). A purely real forward amplitude would remove nothing from the beam and hence describe zero total cross section; scattering necessarily gives the forward amplitude a positive imaginary part. This is the shadow behind every scatterer.

Units check. The amplitude \(f\) carries dimension of length (\(\mathrm{m}\)), since the scattered wave is \(f(\theta)e^{ikr}/r\) with \(f/r\) dimensionless relative to the plane wave. Then \(\dfrac{4\pi}{k}\operatorname{Im}f(0)\) has dimension \(\dfrac{1}{\mathrm{m}^{-1}}\cdot\mathrm{m}=\mathrm{m}^2\), a cross-sectional area, as required. The prefactor \(4\pi\) is dimensionless.

Limiting cases

s-wave dominance (low energy, \(kR\ll1\)): only \(l=0\) survives, \(\sigma_{\text{tot}}=\dfrac{4\pi}{k^2}\sin^2\delta_0\) and \(\operatorname{Im}f(0)=\dfrac{\sin^2\delta_0}{k}\); the theorem reduces to a one-term identity.
Unitarity limit, \(\sin^2\delta_l=1\): channel \(l\) saturates at \(\dfrac{4\pi}{k^2}(2l+1)\); the forward amplitude of that channel becomes purely imaginary, \(f_l(0)=\dfrac{2l+1}{k}i\), the largest \(\operatorname{Im}f\) unitarity allows.
Resonance, \(\delta_l\to\pi/2\): that partial wave passes through its unitarity peak; the total cross section spikes and \(\operatorname{Im}f(0)\) is maximal there.
Weak potential (first Born): \(f^{(1)}(\theta)\) is real, so \(\operatorname{Im}f^{(1)}(0)=0\) at first order while \(\sigma\sim|f^{(1)}|^2=O(V^2)\); the missing \(\operatorname{Im}f\) appears only at second Born order, \(\operatorname{Im}f^{(2)}(0)=O(V^2)\) — a strict consistency check on perturbation theory.

Breaks when

Long-range (Coulomb) potentials, \(V\sim1/r\). The phase shifts do not vanish at large \(l\), the partial-wave sum for \(f(0)\) diverges, the forward amplitude has a non-integrable singularity, and \(\sigma_{\text{tot}}\) is infinite. The bare theorem must be replaced by its screened/Coulomb-modified version.
Absorptive targets with complex phase shifts, \(|S_l|=|\eta_l e^{2i\delta_l}|<1\). The elastic-only intermediate identity \(\sigma=\tfrac{4\pi}{k^2}\sum(2l+1)\sin^2\delta_l\) (Step 5) fails; the forward identity still holds but \(\sigma_{\text{tot}}\) now includes reaction/absorption cross sections, not elastic scattering alone.
Spin-dependent or coupled-channel scattering. When \(f\) is a matrix in spin/channel space, a single scalar \(\operatorname{Im}f(0)\) is insufficient; one must use \(\operatorname{Im}\big[\operatorname{tr}\,\mathbf{f}(0)\big]\) with the correct channel weighting.

Failure modes

Real-part slip: writing \(\sigma\propto\operatorname{Re}f(0)\) or \(|f(0)|\) instead of \(\operatorname{Im}f(0)\). Only the out-of-phase component depletes the beam.
Prefactor confusion: quoting \(\sigma=\tfrac{4\pi}{k^2}\operatorname{Im}f(0)\). Dimensional analysis exposes it — that expression has units of length, not area. The correct prefactor is \(4\pi/k\).
Elastic/total conflation: assuming \((4\pi/k)\operatorname{Im}f(0)\) equals the elastic cross section when the target absorbs. It equals the total (elastic + reaction).
Forgetting \(P_l(1)=1\): mishandling the forward direction and dropping the clean collapse of the sum in Step 6.
First-Born paradox: concluding the theorem is violated because \(\operatorname{Im}f^{(1)}(0)=0\). The imaginary part only enters at second order; the theorem is respected order by order once you keep it.
Sign-convention error: using an amplitude defined with the opposite sign (\(f\to-f\)) and then getting \(\operatorname{Im}f(0)<0\); the physical convention gives \(\operatorname{Im}f(0)>0\).

Discussion

The deepest reading of the optical theorem is as a conservation law for probability current. Superpose the incident plane wave \(e^{ikz}\) and the scattered wave \(f(\theta)e^{ikr}/r\), and compute the flux through a large sphere. The plane wave alone carries zero net flux (in equals out); the scattered wave carries outward flux proportional to \(\int|f|^2\,d\Omega=\sigma_{\text{tot}}\). For total probability to be conserved, this outgoing flux must be paid for by the interference cross term between incident and scattered waves. Stationary-phase evaluation of that cross term picks out precisely the forward direction and yields \(-\tfrac{4\pi}{k}\operatorname{Im}f(0)\) of removed forward flux. Equating the two gives the theorem. The scatterer casts a "shadow," and the shadow is nothing but forward interference.

Because the argument only used flux conservation, the theorem is far more robust than the partial-wave derivation might suggest. It holds for absorbing targets, for relativistic amplitudes, and for inelastic multichannel processes, always relating the total rate of removal from the forward beam to \(\operatorname{Im}f(0)\). In high-energy physics this is the basis for extracting total hadronic cross sections from forward elastic data, and, combined with analyticity, for the dispersion relations that connect real and imaginary parts of amplitudes.

Formally the theorem is the diagonal, forward matrix element of the unitarity relation \(S^\dagger S=\mathbb{1}\). Writing \(S=\mathbb{1}+iT\), unitarity gives \(-i(T-T^\dagger)=T^\dagger T\), i.e. \(2\operatorname{Im}T=T^\dagger T\). Sandwiching between identical forward states, the left side is proportional to \(\operatorname{Im}f(0)\) while the right side is a sum over all intermediate states — the total cross section including every open channel. The optical theorem is thus not a special property of potential scattering but a direct consequence of the probabilistic (unitary) structure of quantum mechanics itself; the partial-wave version is merely its angular-momentum-diagonal shadow.

Common misconceptions. The theorem does not say the forward amplitude is purely imaginary — it constrains only its imaginary part; the real part is fixed by the potential and dispersion relations. Nor does a large \(\operatorname{Im}f(0)\) mean strong forward scattering in the naive sense of a large \(|f(0)|^2\) peak; it means the beam is strongly depleted, which shows up as absorption and diffractive shadowing, not as brightness in the forward cone.

Worked examples

Example 1 — Pure s-wave scatterer. A low-energy particle scatters from a short-range target with wavenumber \(k=1.0\times10^{14}\,\mathrm{m^{-1}}\); only the \(l=0\) phase shift is appreciable, \(\delta_0=30^\circ=\pi/6\). Verify the optical theorem.

\[ \sigma_{\text{tot}}=\frac{4\pi}{k^2}\sin^2\delta_0 \]

Keep only \(l=0\) in Step 5. Symbols first. A

\[ \sin^2\!\left(\tfrac{\pi}{6}\right)=\left(\tfrac12\right)^2=0.250,\qquad \sigma_{\text{tot}}=\frac{4\pi(0.250)}{(1.0\times10^{14})^2}=\frac{\pi}{1.0\times10^{28}} \]

Insert numbers and units. A

\[ \operatorname{Im}f(0)=\frac{\sin^2\delta_0}{k}=\frac{0.250}{1.0\times10^{14}\,\mathrm{m^{-1}}}=2.5\times10^{-15}\,\mathrm{m} \]

Forward imaginary part from Step 7 with one term. A

\[ \frac{4\pi}{k}\operatorname{Im}f(0)=\frac{4\pi(2.5\times10^{-15}\,\mathrm{m})}{1.0\times10^{14}\,\mathrm{m^{-1}}}=3.14\times10^{-28}\,\mathrm{m^2} \]

Independent route via the theorem. A

\[ \sigma_{\text{tot}}=3.14\times10^{-28}\,\mathrm{m^2}=3.14\ \text{barn} \]

Reading. Both routes agree: the direct \(\int|f|^2\) and the forward-imaginary-part route give the identical cross section, confirming the theorem for a single unitary channel. Units: \(\mathrm{m^{-1}}\cdot\mathrm{m}=\) dimensionless in \(4\pi\operatorname{Im}f(0)\)? No — \(\tfrac{1}{k}\,\mathrm{Im}f=\mathrm{m}\cdot\mathrm{m}=\mathrm{m^2}\); \(1\ \text{barn}=10^{-28}\,\mathrm{m^2}\).

Example 2 — Two partial waves. At \(k=5.0\times10^{14}\,\mathrm{m^{-1}}\) a target has \(\delta_0=50^\circ\) and \(\delta_1=20^\circ\), higher waves negligible. Compute \(\sigma_{\text{tot}}\) both ways.

\[ \sigma_{\text{tot}}=\frac{4\pi}{k^2}\Big[(1)\sin^2\delta_0+(3)\sin^2\delta_1\Big] \]

Step 5 with \(l=0\) (weight 1) and \(l=1\) (weight \(2l+1=3\)). A

\[ \sin^2 50^\circ=0.587,\quad \sin^2 20^\circ=0.117 \;\Rightarrow\; (1)(0.587)+(3)(0.117)=0.938 \]

Evaluate the bracket (the same sum appears in \(\operatorname{Im}f(0)\)). A

\[ \sigma_{\text{tot}}=\frac{4\pi(0.938)}{(5.0\times10^{14})^2}=\frac{11.79}{2.5\times10^{29}}=4.7\times10^{-29}\,\mathrm{m^2} \]

Direct route. A

\[ \operatorname{Im}f(0)=\frac{0.938}{k}=\frac{0.938}{5.0\times10^{14}}=1.876\times10^{-15}\,\mathrm{m},\quad \frac{4\pi}{k}\operatorname{Im}f(0)=\frac{4\pi(1.876\times10^{-15})}{5.0\times10^{14}} \]

Theorem route uses the identical bracket, so agreement is guaranteed. B

\[ \sigma_{\text{tot}}=4.7\times10^{-29}\,\mathrm{m^2}=0.47\ \text{barn} \]

Reading. The \(p\)-wave, though it has a smaller phase shift, contributes comparably (\(3\times0.117=0.351\) vs \(0.587\)) because of its degeneracy weight \(2l+1=3\). The optical theorem bundles both channels into the single number \(\operatorname{Im}f(0)\). Units: area in \(\mathrm{m^2}\), \(0.47\) barn.

Problems

(A) One-term theorem. A neutron scatters with \(k=2.0\times10^{14}\,\mathrm{m^{-1}}\) and only \(\delta_0=90^\circ\). Find \(\sigma_{\text{tot}}\) and \(\operatorname{Im}f(0)\), and comment.
Solution
Unitarity limit: \(\sin^2\delta_0=1\). \(\sigma_{\text{tot}}=\dfrac{4\pi}{k^2}=\dfrac{4\pi}{4.0\times10^{28}}=3.14\times10^{-28}\,\mathrm{m^2}=3.14\) barn. \(\operatorname{Im}f(0)=\dfrac{1}{k}=5.0\times10^{-15}\,\mathrm{m}\); check \(\dfrac{4\pi}{k}(5.0\times10^{-15})=3.14\times10^{-28}\,\mathrm{m^2}\). At \(\delta_0=\pi/2\) the s-wave saturates the unitarity bound and \(f(0)=i/k\) is purely imaginary.
(B) Dimensional trap. A student writes \(\sigma=\dfrac{4\pi}{k^2}\operatorname{Im}f(0)\). With \(k=1.0\times10^{14}\,\mathrm{m^{-1}}\) and \(\operatorname{Im}f(0)=2.5\times10^{-15}\,\mathrm{m}\), show by units that this is wrong and give the correct value.
Solution
The wrong form: \(\dfrac{4\pi}{(10^{14})^2}(2.5\times10^{-15})=\dfrac{4\pi(2.5\times10^{-15})}{10^{28}}\) has units \(\mathrm{m^{-2}}\cdot\mathrm{m}=\mathrm{m^{-1}}\) — not an area, so it cannot be a cross section. Correct form uses \(4\pi/k\): \(\dfrac{4\pi}{10^{14}}(2.5\times10^{-15})=3.14\times10^{-28}\,\mathrm{m^2}\), units \(\mathrm{m}\cdot\mathrm{m}=\mathrm{m^2}\). ✓
(C) Absorption. A partial wave has \(S_0=\eta_0 e^{2i\delta_0}\) with inelasticity \(\eta_0=0.6\) and \(\delta_0=45^\circ\), \(k=4.0\times10^{14}\,\mathrm{m^{-1}}\). Find the elastic and reaction cross sections and \(\operatorname{Im}f(0)\), and verify the optical theorem for the total.
Solution
With \(a_0=\dfrac{S_0-1}{2ik}\): \(S_0=0.6\,e^{i\pi/2}=0.6i\). \(\sigma_{\text{el}}=\dfrac{\pi}{k^2}|S_0-1|^2=\dfrac{\pi}{k^2}|{-1+0.6i}|^2=\dfrac{\pi}{k^2}(1.36)\). \(\sigma_{\text{reac}}=\dfrac{\pi}{k^2}(1-|S_0|^2)=\dfrac{\pi}{k^2}(1-0.36)=\dfrac{\pi}{k^2}(0.64)\). \(\sigma_{\text{tot}}=\sigma_{\text{el}}+\sigma_{\text{reac}}=\dfrac{\pi}{k^2}(2.00)\). Forward: \(\operatorname{Im}f(0)=\operatorname{Im}a_0=\dfrac{1-\operatorname{Re}S_0}{2k}=\dfrac{1-0}{2k}=\dfrac{1}{2k}\). Then \(\dfrac{4\pi}{k}\operatorname{Im}f(0)=\dfrac{4\pi}{k}\cdot\dfrac{1}{2k}=\dfrac{2\pi}{k^2}=\sigma_{\text{tot}}\). ✓ Numerically \(k^2=1.6\times10^{29}\), \(\sigma_{\text{tot}}=\dfrac{2\pi}{1.6\times10^{29}}=3.9\times10^{-29}\,\mathrm{m^2}=0.39\) barn. The theorem gives the total, not the elastic part.
(C) Born-order consistency. Explain quantitatively why the first Born amplitude appears to violate the optical theorem, and at what order it is restored.
Solution
The first Born amplitude \(f^{(1)}(\theta)=-\dfrac{2m}{\hbar^2}\dfrac{1}{4\pi}\int e^{i\mathbf{q}\cdot\mathbf r}V(r)\,d^3r\) is real for a real potential, so \(\operatorname{Im}f^{(1)}(0)=0\). Yet \(\sigma_{\text{tot}}=\int|f^{(1)}|^2\,d\Omega=O(V^2)\neq0\). The resolution: \(f=f^{(1)}+f^{(2)}+\cdots\) with \(f^{(1)}=O(V)\), \(f^{(2)}=O(V^2)\). The theorem at order \(V^2\) reads \(\dfrac{4\pi}{k}\operatorname{Im}f^{(2)}(0)=\int|f^{(1)}|^2\,d\Omega\); the second Born term supplies exactly the imaginary forward part needed. So there is no violation — the two sides simply enter at different orders, and matching them is a standard check on \(f^{(2)}\).
(B) Hard sphere, low energy. For a hard sphere of radius \(R\) with \(kR\ll1\), the s-wave phase shift is \(\delta_0=-kR\). Find \(\sigma_{\text{tot}}\) and \(\operatorname{Im}f(0)\) to leading order and confirm the theorem; take \(R=3.0\times10^{-15}\,\mathrm{m}\), \(k=1.0\times10^{13}\,\mathrm{m^{-1}}\).
Solution
\(\delta_0=-kR\) small, \(\sin^2\delta_0\approx(kR)^2\). \(\sigma_{\text{tot}}=\dfrac{4\pi}{k^2}(kR)^2=4\pi R^2\) — the familiar low-energy result (four times the geometric area). \(\operatorname{Im}f(0)=\dfrac{\sin^2\delta_0}{k}\approx\dfrac{(kR)^2}{k}=kR^2\). Check: \(\dfrac{4\pi}{k}(kR^2)=4\pi R^2\). ✓ Numbers: \(kR=(10^{13})(3.0\times10^{-15})=0.030\ll1\), \(\sigma_{\text{tot}}=4\pi R^2=4\pi(9.0\times10^{-30})=1.13\times10^{-28}\,\mathrm{m^2}=1.13\) barn; \(\operatorname{Im}f(0)=kR^2=(10^{13})(9.0\times10^{-30})=9.0\times10^{-17}\,\mathrm{m}\). Note \(\operatorname{Im}f(0)\propto(kR)^2\) is second order in the small parameter, while \(\operatorname{Re}f(0)\approx-R\) is first order — consistent with the Born-order lesson of Problem 4.