Polarization observables in R-matrix theory

The idea, before any formulas#

Fire an unpolarized beam at a target and count what comes out at angle $\theta$. You measure a cross section, and you learn how much scattering happens.

Now spin-polarize the beam, so every projectile enters with its spin pointing the same way. Count again — and count separately on the left and on the right of the beam. In general the two counts differ. The asymmetry between them is the analyzing power:

$$A_y = \frac{N_{\mathrm{left}} - N_{\mathrm{right}}}{N_{\mathrm{left}} + N_{\mathrm{right}}} \Big/ P_{\mathrm{beam}}$$

where the division by the beam polarization normalizes to a perfectly polarized beam. It runs from $-1$ to $+1$.

Why should there be an asymmetry at all? Because the nuclear force depends on spin. A projectile whose spin points "up" relative to its orbital motion feels a different potential than one pointing "down" — this is the spin–orbit force, the same one that splits nuclear shells. Left and right correspond to opposite relative orientations, so they scatter differently.

The reason this is worth measuring is subtler and more valuable. A cross section is a sum of squared amplitudes. Squaring destroys phase information. An analyzing power is a ratio, whose numerator is an interference between two amplitudes and whose denominator is the cross section — so the overall size and the overall phase both cancel, and what survives is the relative phase between amplitudes. Where a cross section shows a smooth bump, an analyzing power can swing from $+1$ to $-1$ in a few tens of keV. That is why a single analyzing-power measurement can settle a spin-parity assignment that a great deal of cross-section data leaves open.

What the calculation must keep track of#

With spin, a single complex amplitude is no longer enough. The scattering must say what it does to the spins as well as where it sends the particle, so the amplitude becomes a matrix in spin space. Everything below is bookkeeping for that matrix.

Symbol	Name	What it means, plainly	Values it takes
$\theta$	scattering angle	where the detector sits, measured from the beam	$0$ to $180$ degrees
$I_1$	projectile spin	intrinsic spin of the incoming particle	$1/2$ for a proton
$I_2$	target spin	intrinsic spin of the target nucleus	$0$ for carbon-12, $1/2$ for nitrogen-15
$m_1, m_2$	spin projections	which way each spin points along the chosen axis	$-I$ to $+I$ in steps of 1
$s$	channel spin	the two intrinsic spins added together as vectors	$\lvert I_1-I_2\rvert$ to $I_1+I_2$
$\nu$	its projection	which way the combined spin points	$-s$ to $+s$
$l$	orbital angular momentum	how much the pair is "swinging around" each other on the way in	$0, 1, 2, \ldots$
$l'$	the same, on the way out
$\mu = \nu - \nu'$	orbital projection, outgoing	the angular momentum the motion must absorb if the spins flipped	$-l'$ to $+l'$
$J$	total angular momentum	orbital plus channel spin; conserved, so it labels resonances	$\lvert l-s\rvert$ to $l+s$
$\pi$	parity	conserved; decides which $l$ can appear	$+$ or $-$
$U$	collision matrix	what the R-matrix calculation produces: how much of each entrance channel turns into each exit channel	complex, one per $(J, \text{in}, \text{out})$
$M$	amplitude matrix	the amplitude to come in with one spin state and leave with another	complex, one per (in state, out state)
$C(\theta)$	Coulomb amplitude	pure electrostatic (Rutherford) scattering	complex
$\omega_l$	Coulomb phase	the phase the long-range electric field adds	real
$Y_{l'}^{\mu}$	spherical harmonic	the angular pattern of the outgoing motion	function of $\theta$
$k$	wave number	momentum of relative motion	$1/\mathrm{fm}$
$\hat{n}$	normal to the scattering plane	the only direction a polarization can point	unit vector

A prime always means "on the way out". So $M_{s'\nu' s\nu}$ reads: the amplitude to enter with combined spin $s$ pointing $\nu$, and leave with combined spin $s'$ pointing $\nu'$.

Why the spins are added together first#

It would seem simpler to track the projectile and target spins separately. The reason not to is that the nuclear force conserves total angular momentum $J$ and parity, and $J$ is built from the orbital motion plus the combined spin:

$\mathbf{J} = \mathbf{l} + \mathbf{s}, \qquad \mathbf{s} = \mathbf{I}_1 + \mathbf{I}_2 .$

Resonances have definite $J$ and parity. So if one organizes the calculation by $(l, s)$, each resonance couples to a small, definite set of channels, and the whole problem block-diagonalizes. Organized by $(m_1, m_2)$ instead, every resonance would smear across everything. The channel spin is the basis in which the physics is simple — and, not coincidentally, the basis in which the model parameters, the level energies and reduced widths, are defined.

This choice has one consequence that dominates the second worked example: the channel spin is not the projectile's spin, except in the special case where the target has none.

The master formula#

The amplitude matrix follows from the collision matrix as

$$\begin{aligned} M_{s'\nu' s\nu}(\theta) = \frac{\sqrt{\pi}}{k}\Big[ &-C(\theta)\,\delta_{ss'}\delta_{\nu\nu'} \\[4pt] &+ i\sum_{J,l,l'} \sqrt{2l+1}\; \langle s\,\nu\,l\,0 \mid J\,\nu \rangle\, \langle s'\,\nu'\,l'\,\mu \mid J\,\nu \rangle \\[4pt] &\qquad\times\; e^{i(\omega_l+\omega_{l'})} \left(\delta_{ss'}\delta_{ll'}-U^J_{s'l' sl}\right) Y_{l'}^{\mu}(\theta) \Big] \end{aligned}$$

Written this way it takes the collision matrix $U$ — the object an R-matrix calculation already produces — directly, rather than going through phase shifts. It looks forbidding, but every piece is forced by something physical.

The Coulomb term $C(\theta)$ carries the deltas $\delta_{ss'}\delta_{\nu\nu'}$ because an electric field does not touch spin: whatever spin state goes in comes out unchanged. It appears only in elastic scattering, since Rutherford scattering cannot transmute one nucleus into another.

The first Clebsch–Gordan coefficient $\langle s\,\nu\,l\,0 \mid J\,\nu \rangle$ says how the entrance channel spin and orbital motion combine into $J$. Its orbital projection is fixed at zero — that is not an approximation but a choice of axis. Point the $z$-axis along the beam; a plane wave carries no angular momentum about its own direction of travel, so the incoming orbital projection vanishes and the total projection is $\nu$ alone.

The second coefficient does the same on the way out, and enforces conservation: whatever the exit spins take as $\nu'$, the orbital motion must carry the remainder $\mu = \nu - \nu'$.

The collision-matrix term $\delta_{ss'}\delta_{ll'} - U$ is the part that actually scatters: subtracting the identity removes the piece of the wave that went straight through. The spherical harmonic $Y_{l'}^{\mu}(\theta)$ is the angular pattern of the outgoing motion.

The one line that explains why polarization needs new machinery#

Look at $\mu = \nu - \nu'$ in that spherical harmonic.

If the spins do not flip, $\nu = \nu'$, so $\mu = 0$, and $Y_l^{0}$ is just a Legendre polynomial $P_l(\cos\theta)$ — the familiar angular distributions of ordinary cross sections.

If the spins do flip, $\mu \neq 0$, and one needs the associated Legendre functions, which describe angular patterns that are not symmetric about the beam axis in the same way.

So a formalism equipped only with $P_l(\cos\theta)$ cannot produce an analyzing power at all — not for want of an option, but because the amplitudes whose interference is the analyzing power cannot be written down in that angular basis. Spin flip is inseparable from sideways angular momentum.

From amplitudes to what is measured#

Describe the beam by a spin density matrix $\rho_{\mathrm{in}}$ — a compact way of saying "this fraction of the beam has its spin here, that fraction there". Then the outgoing spin state is

$\rho_{\mathrm{out}} = M\,\rho_{\mathrm{in}}\,M^{\dagger},$

and every observable is a trace of $\rho_{\mathrm{out}}$ against whatever operator the apparatus is sensitive to. Two cases:

$$\frac{d\sigma}{d\Omega} = \frac{\mathrm{Tr}\left(M M^{\dagger}\right)}{(2I_1+1)(2I_2+1)}, \qquad A_y = \frac{\mathrm{Tr}\left(M\,\sigma_y\,M^{\dagger}\right)}{\mathrm{Tr}\left(M M^{\dagger}\right)} .$$

The denominator in the cross section counts the spin states an unpolarized beam populates equally. In $A_y$ that factor cancels top and bottom — which is the formal reason an overall normalization cannot change an analyzing power. It is already a ratio.

Why the polarization can only point one way#

$M$ is a matrix in spin space, so it can be written as a piece proportional to the identity plus a piece proportional to the Pauli matrices, with coefficients built from the only two vectors in the problem, the incoming and outgoing momenta.

Now apply parity. The Pauli matrices form a pseudovector (they do not change sign under reflection), while momenta are ordinary vectors (they do). For the whole expression to have definite parity, the coefficient multiplying the Pauli matrices must itself be a pseudovector. Out of two ordinary vectors there is exactly one pseudovector available: their cross product,

$$\hat{n} = \frac{\mathbf{k}_{\mathrm{in}} \times \mathbf{k}_{\mathrm{out}}} {\lvert \mathbf{k}_{\mathrm{in}} \times \mathbf{k}_{\mathrm{out}}\rvert}$$

So a vector polarization can only point along the normal to the scattering plane. The components in the plane vanish identically, which is why there is one vector analyzing power and not three. Writing $\sigma_y$ above is shorthand for "the Pauli matrix along $\hat n$".

What the numerator is really doing#

Write $u$ for the amplitude when the beam spin is up, $d$ for spin down, at the same outgoing configuration. The trace works out to

$$A_y = \frac{2\,\sum \mathrm{Im}\left[\, u\, d^{*} \,\right]} {\sum \left( \lvert u \rvert^{2} + \lvert d \rvert^{2} \right)} ,$$

summed over everything not measured. Read it slowly, because it contains the whole phenomenology:

• if spin-up and spin-down scatter identically, then $u = d$, the product $u d^{*}$ is real, and $A_y = 0$;
• if they scatter differently but in phase, the product is still real, and $A_y = 0$ again;
• a non-zero analyzing power needs both a spin dependence and a relative phase between the two.

That is why analyzing powers are large near resonances — a resonance sweeps its phase through 180 degrees — and why they are such sharp probes of interference between overlapping levels.

Four places it must vanish#

Where	Why
exactly forward or backward	there is no scattering plane, so $\hat n$ is undefined; the sideways angular functions vanish there
far below any resonance	scattering is pure Coulomb, which is spin-blind, so up and down are identical
if there were no spin–orbit force	nothing would distinguish the two orientations
a single isolated resonance with no background	one common phase, which cancels in the ratio

The second is a trap when checking a calculation: at low energy, zero is the correct answer, so agreement there proves nothing.

Example A — protons on carbon-12#

Carbon-12 has spin zero. This is the case where the bookkeeping collapses to nothing.

The channel spin. With $I_1 = 1/2$ and $I_2 = 0$, the only possible value is $s = 1/2$. Its projection $\nu$ is then simply the proton's own spin projection: with a spinless target there is nothing else to combine with. Channel spin and beam spin coincide.

Size of the problem. Two entrance spin states, two exit states, so the amplitude matrix has $2 \times 2 = 4$ entries.

Its structure. From the parity argument above, a $2\times2$ amplitude matrix for a spin-1/2 particle on a spinless target must have the form

$$M = g(\theta)\,\mathbf{1} + h(\theta)\,\boldsymbol{\sigma}\cdot\hat{n} = \begin{pmatrix} g & -ih \\ ih & g \end{pmatrix},$$

with $g$ the non-flip amplitude and $h$ the spin-flip one. So the two diagonal entries must be equal, and the two off-diagonal entries equal and opposite. Calculating the four amplitudes independently, at 1.75 MeV and 80 degrees, gives

	leave spin up	leave spin down
enter spin up	$-1951 - 1955\,i$	$+0.0027 - 0.0080\,i$
enter spin down	$-0.0027 + 0.0080\,i$	$-1951 - 1955\,i$

exactly the required symmetry, recovered from a sum over many different $J$, $l$ and $l'$ pathways that had no reason to conspire unless the couplings and phases are right. Substituting into the trace gives the classical result

$A_y = \frac{2\,\mathrm{Re}\left(g\,h^{*}\right)}{\lvert g\rvert^{2}+\lvert h\rvert^{2}} .$

Notice how small the flip amplitude is — around $10^{-2}$ against $2\times10^{3}$, because the non-flip amplitude is dominated by Rutherford scattering. And yet, at 1.75 MeV, between the $3/2^-$ level at 3.503 MeV excitation and the $5/2^+$ at 3.545 MeV:

angle	20	40	60	80	100	120	140	160
$A_y$	+0.18	+0.21	−0.25	−0.99	−0.05	+0.79	+0.91	+0.40

At 80 degrees the analyzing power reaches $-0.99$: essentially every scattered proton goes to one side. A spin-flip amplitude four orders of magnitude smaller than the non-flip one produces a nearly complete asymmetry — because $A_y$ measures an interference, not a magnitude, and interference is first order in the small amplitude while the cross section is second order. Two overlapping resonances of different $J$ and opposite parity supply the phase difference that makes it possible. This is one of four points at which Baumann and co-workers measured $\lvert A_y\rvert$ reaching unity in this system, at the same energy and angle.

Example B — protons on nitrogen-15#

Nitrogen-15 has spin 1/2. Now the bookkeeping matters.

The channel spin. With $I_1 = I_2 = 1/2$,

$s = 0 \quad \text{or} \quad s = 1, \qquad \text{but never } 1/2 .$

Two spin-1/2 particles combine into a singlet and a triplet. The channel spin is no longer the proton's spin, and a polarized proton is not a state of definite channel spin at all.

Size of the problem. Entrance states: one from $s=0$, three from $s=1$, so four — as it must be, since two spin-1/2 particles have $2\times2$ orientations. Same on the way out, so the amplitude matrix has $4 \times 4 = 16$ entries, against 4 for carbon.

The translation between the two languages. The singlet and triplet states, in terms of the individual spins (first arrow the proton, second the nitrogen), are

$$\lvert 1,+1\rangle = \lvert \uparrow\uparrow \rangle, \qquad \lvert 1,-1\rangle = \lvert \downarrow\downarrow \rangle,$$ $$\lvert 1,0\rangle = \frac{1}{\sqrt{2}}\left( \lvert \uparrow\downarrow \rangle + \lvert \downarrow\uparrow \rangle \right), \qquad \lvert 0,0\rangle = \frac{1}{\sqrt{2}}\left( \lvert \uparrow\downarrow \rangle - \lvert \downarrow\uparrow \rangle \right).$$

Turn that around. A proton with spin up, on a target nucleus with spin down, is

$$\lvert \uparrow\downarrow \rangle = \frac{1}{\sqrt{2}}\left( \lvert 1,0\rangle + \lvert 0,0\rangle \right).$$

This is the physical heart of the general case. The polarized beam prepares a coherent superposition of singlet and triplet channels, and the analyzing power is sensitive to the relative phase between them. To compute it one must first translate the amplitudes out of the channel-spin language and back into "which way is the proton pointing", which for each outgoing configuration reads

$$\begin{aligned} M_{\uparrow\uparrow} &= M_{s=1,\,\nu=+1}, & M_{\uparrow\downarrow} &= \frac{1}{\sqrt{2}}\left(M_{1,0} + M_{0,0}\right), \\[2pt] M_{\downarrow\uparrow} &= \frac{1}{\sqrt{2}}\left(M_{1,0} - M_{0,0}\right), & M_{\downarrow\downarrow} &= M_{s=1,\,\nu=-1}. \end{aligned}$$

The two beam orientations are then compared at fixed target orientation, and the target orientations are summed incoherently — nobody prepared or measured them, so they are averaged over rather than added as amplitudes.

Why a cross section could never substitute. In an unpolarized cross section the entrance orientations are averaged, and that average destroys precisely the cross terms between $s=0$ and $s=1$. Their relative phase is invisible to every cross section one can measure on this system. It becomes observable only with a polarized beam. That is why the measurement exists.

Result at 3.0 MeV, elastic:

angle	20	40	60	80	100	120	140	160
$A_y$	−0.00	+0.00	−0.01	−0.08	−0.16	−0.20	−0.17	−0.09

Smaller than the carbon example — though the comparison is not fair, since the carbon numbers sit deliberately on top of two overlapping resonances. There is nonetheless a real systematic effect: with a spin-carrying target the denominator collects amplitudes from every target orientation, while the numerator only collects the beam-spin interference at each fixed orientation. Averaging over something one did not measure dilutes the asymmetry.

How the treatment changes with other particles#

Situation	What happens	Why
spin-1/2 beam, target of any spin	works exactly as in Example B	the translation between channel spin and individual spins is general
any parity, any resonance spin	no change at all	parity and $J$ only decide which $(l,s)$ channels exist; the spin algebra is untouched
inelastic or rearrangement, $(\vec p, p')$ or $(\vec p, \alpha)$	works	entrance and exit are independent throughout; only the Coulomb term is restricted to elastic scattering
spin-1 beam (a polarized deuteron)	needs different observables	a spin-1 particle has a $3\times3$ density matrix, so besides a vector polarization it can be aligned — stretched along an axis without pointing. That carries tensor moments, which need rank-2 operators, not a Pauli matrix
capture, $(\vec p, \gamma)$	needs a separate treatment	the outgoing photon is described by multipole radiation rather than by orbital angular momentum in a channel
identical particles, such as $p+p$	needs symmetrization	one cannot tell "beam scattered by $\theta$" from "target recoiled at $180^\circ-\theta$", so the two possibilities must be added as amplitudes before squaring

Two practical consequences#

A thick target destroys an analyzing power. A cross section measured on a thick target is an integral: the beam loses energy as it goes, and every depth contributes. An analyzing power is a ratio, so it cannot be integrated that way. What the experiment returns is the ratio of the two integrated yields,

$$\langle A_y \rangle = \frac{\int A_y(E)\,\sigma(E)\,dE}{\int \sigma(E)\,dE},$$

weighted by the cross section, so the depths where the reaction is likely count for more. For charged-particle scattering this is severe. Rutherford scattering makes the cross section blow up as the beam slows down, exactly where the analyzing power goes to zero — so the low-energy tail of the target dominates the weighting and drowns the asymmetry. A resonant $A_y$ of $0.8$ can average down to $10^{-6}$ through a thick gas target. This is why analyzing powers are measured on thin foils: Baumann's carbon targets were 85 nm, about 3 keV of energy loss.

Uncertainties should be absolute, not a percentage. An analyzing power passes through zero at many angles and energies. A point sitting near a zero crossing does not have a correspondingly small uncertainty — the measurement is a difference of two counts, and its error depends on the counts, not on how close their difference happens to be to zero. Quoting a fixed percentage of the value gives those points an enormous and entirely artificial statistical weight, and any fit will chase them at the expense of everything else.