The plane in which three bodies are flat

A particle at rest that breaks into two gives each product a definite energy, fixed by the three masses alone. The cone a decay cannot leave built its measurements on that fact: boosted, the definite energy spreads into a rectangle with sharp edges, and the edges weigh the parent. The slope a spin leaves in a spectrum kept the edges and let the parent’s spin tilt the top.

Break the parent into three and the definite energy is gone. Energy and momentum conservation are four equations, and three products in the parent’s rest frame have nine momentum components; fixing the masses and discarding the three angles that merely orient the whole event still leaves two quantities free. Each product can have a range of energies, depending on how the other two share what is left. The energy that did not all arrive met the consequence in beta decay, where the electron’s continuous spectrum was the clue that a third, unseen particle was taking a share.

The two free quantities are not an inconvenience to be integrated away. Chosen correctly, they form a plane on which a decay with no forces between its products is perfectly uniform, and that makes the plane the most sensitive place to look for forces.

The two numbers a three-body decay leaves free #

For a parent of mass $M$ decaying to particles 1, 2 and 3, the natural choice is two invariant masses squared: $m_{12}^2$, the mass of the pair 1 and 2 taken together, and $m_{13}^2$. Each is an invariant — the combination that survives a boost — so it is the same in every frame, and each is also an energy in disguise. In the parent’s rest frame, particle 3’s energy is

$$E_3 = \frac{M^2 + m_3^2 - m_{12}^2}{2M},$$

a straight line in $m_{12}^2$, and particle 2’s energy is the same straight line in $m_{13}^2$. A plot of the two invariant masses is a plot of two products’ energies, rescaled. Richard Dalitz introduced it in 1953 in exactly that form, as a plot of energies.

The third pair’s mass is not free: the three masses squared always add to $M^2 + m_1^2 + m_2^2 + m_3^2$. And not every point of the plane is reachable. The boundary is traced by the events in which the three products fly along a single line, since those are the extremes of how the energy can be shared, and inside it lies everything that energy and momentum allow.

The plane that phase space fills evenly #

The decays in the figure contain no dynamics at all. Each was made as two successive two-body decays in random directions — the D into a $K^-\pi^+$ pair and a $\pi^0$, the pair into its two particles — with the only weighting the one kinematics requires, the momenta available at each step. Each D was then set moving at nine tenths of the speed of light in a random direction, and the two invariant masses were computed from the laboratory four-momenta, where the energies and angles are scrambled beyond recognition.

The result is flat. The 219 cells lying wholly inside the boundary hold their counts at a $\chi^2$ of 216, as a uniform density predicts, and no cell is further off than chance allows. The rate of a three-body decay is the squared amplitude times a constant, times the area element in this plane:

$$d\Gamma = \frac{|\mathcal{M}|^2}{32\,(2\pi)^3 M^3}\, dm_{12}^2\, dm_{13}^2.$$

Everything kinematic has been absorbed into the choice of coordinates. The reason is the same one that made a two-body spectrum flat. For a fixed $m_{12}^2$, the other mass squared is a straight-line function of the angle at which the pair decays in its own rest frame, so a decay isotropic in that frame fills its line evenly; and the momentum factors that weight each value of $m_{12}^2$ are exactly cancelled by the length of that line. What is left, if $|\mathcal{M}|^2$ is constant, is an even shade.

What a single spectrum cannot tell apart #

The same force-free decays, projected onto one product’s energy, look nothing like flat.

The spectrum has a hump and two soft edges, and all of it is geometry. Projecting a flat region onto an axis gives a curve proportional to the region’s width along that axis, and the Dalitz boundary is narrow at both ends of the $\pi^0$’s range — where the $\pi^0$ is at rest, and where the kaon and the other pion leave as a single lump — and widest in between. A spectrum with this shape says three bodies and nothing else. It cannot say whether the products attract one another, because a force that crowds decays into one part of the plane moves a spectrum’s hump by amounts no larger than the hump the boundary makes by itself.

That is why beta-decay spectra could be analysed only after Fermi wrote down the phase-space factor: every departure from phase space had to be measured against a curve that was already strongly shaped. The plane removes the shaping. Against a flat background a band of a few per cent of the decays is visible to the eye.

A resonance is a band #

Real three-body decays are rarely force-free. Often two of the three products are the decay products of something short-lived that the parent made first — a $D^0$ decaying to a $\rho^+$ and a $K^-$, with the $\rho^+$ then decaying to $\pi^+\pi^0$ in about $4\times10^{-24}$ seconds. The $\rho^+$ never reaches a detector. It lives only as a preference in the plane.

A short-lived intermediate state fixes the invariant mass of the pair it decays into, up to its width, and leaves the rest of the event free. So it appears as a stripe at constant $m^2$ for its pair: a $K^{*0}$ decaying to $K^-\pi^+$ as a vertical band, a $K^{*-}$ decaying to $K^-\pi^0$ as a horizontal one, and a $\rho^+$ decaying to $\pi^+\pi^0$ as a diagonal one, because $m^2(\pi^+\pi^0)$ is fixed by the sum rule once the other two are known. The band’s position is the state’s mass, and its thickness is its width — the quantity the width that is a lifetime relates to how long it lived, which for the ρ is a width of about 150 MeV and for the K* about 50.

The three bands in the figure were imposed by accepting force-free decays in proportion to the squared amplitude, so what is drawn is sampled decays and not the model’s contours. Each band holds markedly more of the decays than phase space would put in the same strip. But the three rates on their own — 67.0, 25.5 and 33.2 per cent of the total — add to 126 per cent. The missing 26 per cent is not an error. It is interference, which returns below.

Counting dark points counts the spin #

A band is not uniformly bright along its length, and the pattern along it is the intermediate state’s spin.

Along a vertical band, $m_{12}^2$ is fixed, and the position along the band — the value of $m_{13}^2$ — is a straight-line function of the angle at which the pair’s first particle leaves, measured in the pair’s rest frame from the direction of the third. A state of spin $J$ decaying to two spinless particles has an amplitude proportional to the Legendre polynomial $P_J$ of the cosine of that angle, so the band’s brightness follows $|P_J|^2$.

A spin-0 band is evenly bright. A spin-1 band is dark in the middle, where the angle is ninety degrees, and bright at both ends. A spin-2 band goes dark at two places, where the cosine is $\pm 1/\sqrt 3$. A band of spin $J$ goes dark at $J$ places, because $P_J$ has $J$ zeros, and the zeros can be counted by eye. Nothing about any particle’s spin was detected; the spin is read from where the particles went. The two computations of the angle in the figure — one boosting laboratory four-vectors into the pair’s frame, one from the invariant masses alone — agree to a part in $10^{14}$, which is the statement that the plane already contains the angle.

The first use of the plane was exactly this. In the early 1950s two charged kaons were known, the “tau”, decaying to three pions, and the “theta”, decaying to two. Dalitz plotted the handful of tau decays then available and found them spread with no dark points and no bands — the pattern of a particle of spin zero and negative parity. Two pions from a spin-zero particle have positive parity. So if parity were conserved in these decays, the tau and theta had to be different particles, yet they had the same mass and the same lifetime. Lee and Yang’s resolution in 1956 was that they were one particle and that the weak interaction does not conserve parity, which is the fact whose consequence for polarised decays the slope a spin leaves in a spectrum measured.

A phase that only a crossing can show #

Where two bands cross, a single set of three products can have come through either intermediate state, and quantum mechanics adds the two amplitudes before squaring.

The figure holds everything about the two resonances fixed — masses, widths, strengths, spins — and turns only the phase between their amplitudes. The share of decays in the crossing moves by a factor of 2.6. Away from the crossing, where only one band contributes, nothing changes at all: a phase multiplies an amplitude by a number of size one and is invisible in its square. It becomes visible only where a second amplitude gives it something to interfere with, which is the same logic by which a resonance on a smooth background acquires a zero and by which a magnetic flux leaves a phase on a path it never touches that only an interference pattern can report.

This is why the plane is used to measure phases that are otherwise inaccessible. The most important is an angle of the matrix that describes how quarks of different generations mix, which is extracted from the Dalitz plots of D mesons produced in B meson decays: the interference between two routes to the same D decay, one through a $D^0$ and one through its antiparticle, is read off the plane’s crossings. The same interference is why fractions do not add to 100 per cent, and why a model of a Dalitz plot must be fitted as amplitudes, with phases, and never as a sum of rates.

A band’s shadow on the other axes #

A band is a line of constant mass for one pair, and along it the masses of the other two pairs are not constant at all. Walking along the $K^{*-}$ band, which is horizontal, $m^2(K^-\pi^+)$ runs across the whole width of the plane, and because the three masses squared add to a constant, $m^2(\pi^+\pi^0)$ runs across a wide range in the opposite direction. Projected onto the $\pi^+\pi^0$ mass, the $K^{*-}$ therefore appears not as a peak at its own mass — it has no mass in that pair — but as a broad hump spread over everything the band crosses.

That hump is called a reflection, and it is not always broad. The band’s brightness along its length is its spin pattern, so a spin-1 band, dark in the middle and bright at both ends, projects to two humps, one near each end of its range; and where a band meets the boundary obliquely, a long stretch of it maps onto a short range of the other mass, and the projection piles up into something narrow enough to be mistaken for a resonance of its own. A projected spectrum of $m^2(\pi^+\pi^0)$ alone would show a structure there that no particle decaying to two pions produced.

The plane is what makes the distinction possible. A genuine state in the $\pi^+\pi^0$ pair lies along a diagonal line of constant $m^2(\pi^+\pi^0)$; a reflection lies along a horizontal one and only looks diagonal after projection. Every claim of a new state made from a three-body decay has to show that its peak is not the shadow of a known band, and the way to show it is to put every known band into the amplitude, reflections included, and find that the plane still disagrees along a line of the new state’s mass.

What the plane has found #

The flatness of phase space makes the plane a search instrument as much as a measuring one. A structure that no known state explains is a claim that something unknown was made in between.

In 2014 the plane of $B^0$ decays to $\psi(2S)K^+\pi^-$ was shown to require a band in $m(\psi(2S)\pi^-)$ at 4430 MeV — a charged state containing a charm quark and its antiquark, which therefore had to contain at least two more quarks. In 2015 the plane of $\Lambda_b$ decays to $J/\psi\,K^-p$ required a band in $m(J/\psi\,p)$ near 4450 MeV that none of the many known states decaying to $K^-p$ could produce by their reflections across the plane, and the state was reported as a pentaquark. Both claims rested on the argument of this essay: kinematics and every known resonance were put in, the plane was not flat where they predicted, and the excess sat along a line of constant invariant mass.

Where the plane stops being flat #

The products were taken as spinless. When the final particles have spin, the rate is still the squared amplitude times the flat area element, but the squared amplitude is summed over unobserved spin states and the angular patterns along bands are no longer single Legendre polynomials. The plane remains flat for phase space; reading a spin off it needs more careful bookkeeping.

Two identical particles fold the plane. In a decay to $\pi^+\pi^+\pi^-$, exchanging the two identical pions is not a different event, so the plane is symmetric about a diagonal and each band has a mirror image; the amplitude must be symmetrised, and its reflections can be mistaken for structure of their own.

The bands were built as isolated two-body resonances. The model treats each intermediate state as decaying independently of the third particle — the isobar model. In reality the bachelor particle can rescatter off the pair, and such three-body effects can produce shapes, including peaks near thresholds, that no single band accounts for. Most of the model dependence in modern amplitude analyses lives here.

The detector is assumed to see every decay equally. A real experiment’s efficiency varies across the plane, lowest near the boundary, where one product is slow. That variation multiplies the density before anything is fitted, and an efficiency map measured wrongly paints bands that are not there.

And the widths were constant. A Breit–Wigner with a fixed width is a good description of a narrow state far from thresholds and a poor one for a broad state like the ρ, whose width depends on the pair’s mass. The model in the figures uses the simple form throughout.

What the figures leave out #

The resonance figure’s strengths and phases were chosen so that all three bands show, and no real $D^0$ decay is being reproduced; the $\rho^+$ band in the measured decay is far stronger relative to the K* bands than drawn. The figure is an argument about what a band looks like, not a measurement of these branching fractions.

The crossing figure uses a fixed window of one width in each mass and reports a share of decays; a real analysis fits the entire plane at once, and extracts the phase from the whole pattern of constructive and destructive interference rather than from a count in one box.

Still open: whether a band is a particle #

A band at constant invariant mass says that the pair’s amplitude varies rapidly near that mass. A short-lived particle does that. So can other things: a threshold, where a new pair of particles becomes energetically available; or a triangle singularity, a rescattering loop whose kinematics happens to put all three internal particles nearly on their mass shells at once, which produces a logarithmic peak with no particle behind it.

The narrow states reported near 4312, 4440 and 4457 MeV in $J/\psi\,p$ sit close to the thresholds for producing a charmed baryon and a charmed meson together, and whether they are compact five-quark states, loosely bound molecules of a baryon and a meson, or partly kinematic effects of that kind has not been settled. Distinguishing them requires the phase of the amplitude as the mass crosses the band — a genuine resonance’s phase turns by 180 degrees — which the plane can measure through interference with other bands, and which so far has been measured precisely for very few such states.

The habit worth carrying away is to find the coordinates in which ignorance is uniform. In a plane where nothing kinematic prefers any point, every preference is physics, and a structure there is a claim about forces, spins or phases that no single projected spectrum could have made.