16 Jul 2026
Partial-Wave Analysis and Phase Shifts
Construction of the elastic scattering amplitude from angular-momentum channels, partial cross sections, unitarity, and the optical theorem.
Take the incident direction as the $z$-axis. Because $z=r\cos\theta$, expand the plane wave in Legendre polynomials:
\[e^{ikr\cos\theta} =\sum_{l=0}^{\infty}a_l(kr)P_l(\cos\theta).\]Multiplication by $P_l(x)$ and integration over $x=\cos\theta$ use
\[\int_{-1}^{1}P_l(x)P_{l'}(x)\,dx =\frac{2}{2l+1}\delta_{ll'}.\]The coefficient is
\[a_l(kr) =\frac{2l+1}{2} \int_{-1}^{1}e^{ikrx}P_l(x)\,dx =(2l+1)i^l j_l(kr).\]Thus
\[\boxed{ e^{ikz} =\sum_{l=0}^{\infty} i^l(2l+1)j_l(kr)P_l(\cos\theta). }\]Incoming and outgoing parts
For a central potential, distinct values of $l$ do not mix. Outside the potential, the $l$th radial wave is a superposition of incoming and outgoing spherical waves. If their amplitudes are $A_l^{(-)}$ and $A_l^{(+)}$, define
\[S_l=\frac{A_l^{(+)}}{A_l^{(-)}}.\]Elastic probability conservation requires $|A_l^{(+)}|=|A_l^{(-)}|$, hence $|S_l|=1$. A unit-modulus complex number can be written
\[\boxed{ S_l=e^{2i\delta_l}. }\]
The incoming part must match the incident plane wave. The free outgoing part corresponds to $S_l=1$; the extra outgoing coefficient is therefore $S_l-1$. Matching it to $f(\theta)e^{ikr}/r$ gives
\[\boxed{ f(\theta) =\frac{1}{2ik} \sum_{l=0}^{\infty} (2l+1)(S_l-1)P_l(\cos\theta). }\]Now
\[\begin{aligned} \frac{e^{2i\delta_l}-1}{2i} &=\frac{e^{i\delta_l} \left(e^{i\delta_l}-e^{-i\delta_l}\right)}{2i}\\ &=e^{i\delta_l}\sin\delta_l. \end{aligned}\]The phase-shift form of the amplitude is
\[\boxed{ f(\theta) =\frac1k\sum_{l=0}^{\infty} (2l+1)e^{i\delta_l}\sin\delta_l P_l(\cos\theta). }\]Partial and total cross sections
Insert this amplitude into $\sigma_{\mathrm{tot}}=\int|f|^2d\Omega$. The angular integral contains
\[\int P_l(\cos\theta)P_{l'}(\cos\theta)\,d\Omega =2\pi\int_{-1}^{1}P_l(x)P_{l'}(x)\,dx =\frac{4\pi}{2l+1}\delta_{ll'}.\]All terms with $l\neq l’$ vanish, and $|e^{i\delta_l}\sin\delta_l|^2=\sin^2\delta_l$. Therefore
\[\boxed{ \sigma_{\mathrm{tot}} =\frac{4\pi}{k^2} \sum_{l=0}^{\infty} (2l+1)\sin^2\delta_l. }\]The contribution of one channel is
\[\boxed{ \sigma_l =\frac{4\pi}{k^2}(2l+1)\sin^2\delta_l. }\]Since $\sin^2\delta_l\leq1$, elastic unitarity imposes
\[\sigma_l\leq\frac{4\pi}{k^2}(2l+1).\]Forward amplitude and total scattering
At $\theta=0$, $P_l(1)=1$. Expanding $e^{i\delta_l}\sin\delta_l$,
\[f(0) =\frac1k\sum_l(2l+1) \left[ \sin\delta_l\cos\delta_l+i\sin^2\delta_l \right].\]Its imaginary part is
\[\operatorname{Im}f(0) =\frac1k\sum_l(2l+1)\sin^2\delta_l.\]Comparison with the integrated cross section produces the optical theorem:
\[\boxed{ \sigma_{\mathrm{tot}} =\frac{4\pi}{k}\operatorname{Im}f(0). }\]The coefficient identity, Legendre orthogonality, partial cross sections, and optical-theorem equality are verified in the Maxima worksheet.
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