An exactly solvable quantum four-body problem associated with the symmetries of an octacube

The relationship between exactly solvable quantum one-dimensional multi-particle problems and kaleidoscopes has been long appreciated [1]. A kaleidoscope is a system of mirrors where none of the multiple images of the original object is broken at the mirror junctions. A viewer situated inside a kaleidoscopic cavity has no means of distinguishing the images of objects from the original. For a broad class of boundary conditions, the eigenmodes of a kaleidoscopic cavity are represented by finite superpositions of plane waves; they can be found exactly, using the method of images [2]. When a many-body problem reduces to an exactly solvable kaleidoscope, the resulting solution is known as the (coordinate) Bethe Ansatz solution [3, 4]. The list of particle systems solvable using Bethe Ansatz is so far exhausted by: equal mass hard-core bosons, on an open line, on a circle or in between walls [5]; δ-interacting, equal-mass, generally distinguishable particles on a line and on a circle [6–8]; this includes bosons [9], which could also be in the presence of one or two walls [1]; systems of two hard-core particles with masses m and $3m$ interacting with a wall, for both wall-m-$3m$ and wall-$3m$-m spatial orders were briefly commented on in [6], but discarded as inferior to the problems with finite strength interactions.

It has been long conjectured that no exceptional—specific to a given number of spatial dimensions—kaleidoscopes lead to exactly solvable particle problems with realistic interactions [1, 3]. The search for physical realizations was limited to systems of particles of the same mass, with a possible addition of immobile walls. In this Letter, we show that the exceptional closed kaleidoscope ${\tilde{F}}_{4}$ induces a novel quantum integrable system: four one-dimensional quantum hard-core particles with different masses, $6m,$ $2m,$ m, and $3m,$ in a hard-wall box. The solution utilizes the symmetries of an octacube, a Platonic solid unique to four dimensions.

Consider six hard-core (i.e. impenetrable) particles on a line, with masses m₀, m₁, m₂, m₃, m₄, and m₅. Their coordinates will be denoted as x₀, x₁, x₂, x₃, x₄, and x₅ respectively. The natural coordinate transformation

$$\begin{eqnarray*}&&{x}_{i}=\sqrt{\displaystyle \frac{{\mathcal{M}}}{{m}_{i}}}\;{y}_{i},\;{\rm{for}}\;{\rm{}}i=0,1,\ldots ,5\end{eqnarray*}$$

reduces the system to a single six-dimensional particle of mass ${\mathcal{M}}$ (to be fixed later) whose motion is constrained by the following set of five inequalities: ${y}_{i+1}/\sqrt{{m}_{i+1}}\gt {y}_{i}/\sqrt{{m}_{i}},$ for $i=0,1,\ldots ,4.$

Consider now the affine Coxeter diagram ${\tilde{F}}_{4}$ [10] depicted in the top line of figure 1. It encodes the geometry of a particular simplex-shaped four-dimensional kaleidoscope. The vertices label its five mirrors: the three-faces (represented by three-dimensional simplexes) of the four-dimensional simplex. The edges—the links—of the diagram (and their absences) encode the angles between the corresponding mirrors, i.e. the angles between their normals. When two vertices are not linked by an edge, the respective mirrors are orthogonal to each other. An edge with no numbers above it corresponds to an angle of $\pi /3.$ A number n above an edge would produce an angle of $\pi /n$ between the corresponding mirrors. The meaning of the mark inside the leftmost vertex will be explained later. In our case, the non-right angles between the mirrors are $\pi /3,$ $\pi /3,$ $\pi /4,$ and $\pi /3,$ in the order of their appearance on the Coxeter diagram, left to right.

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Let us try to identify the five consecutive vertices of the Coxeter diagram with the five hyperplanes of contact between neighboring particles: $\displaystyle \frac{{y}_{i+1}}{\sqrt{{m}_{i+1}}}=\displaystyle \frac{{y}_{i}}{\sqrt{{m}_{i}}}$ for $i=0,1,\ldots ,4$ (see the particle diagrams in figure 1). This choice is natural: contact hyperplanes for two non-overlapping pairs of neighboring particles are indeed orthogonal to each other. Then the four edges of the diagram correspond to the four triplets of consecutive particles.

For three consecutive particles, i, $i+1,$ and $i+2,$ the angle between the i versus $i+1$ and $i+1$ versus $i+2$ contact hyperplanes is given by

$$\begin{eqnarray*}&&{\theta }_{i(i+1)(i+2)}=\mathrm{arctan}\sqrt{\displaystyle \frac{{m}_{i+1}({m}_{i}+{m}_{i+1}+{m}_{i+2})}{{m}_{i}{m}_{i+2}}}\end{eqnarray*}$$

(see e.g. [6]). Notice that these angles do not depend on the overall mass scale. Therefore, any constraints on these angles can only lead to relationships between the five mass ratios, ${m}_{i+1}/{m}_{i}$ for $i=0,1,2,3,4$ alone, with no overall mass scale involved. On the other hand, for our six-body system to lead to the ${\tilde{F}}_{4}$ kaleidoscope, the angles between the hyperplanes of contact must satisfy four equations only,

$$\begin{eqnarray*}&&{\theta }_{012}=\displaystyle \frac{\pi }{3};\quad {\theta }_{123}=\displaystyle \frac{\pi }{3};\quad {\theta }_{234}=\displaystyle \frac{\pi }{4};\quad {\theta }_{345}=\displaystyle \frac{\pi }{3},\end{eqnarray*}$$

thus being seemingly underdetermined. Nevertheless, once the non-negativity of the masses (${m}_{i}\geqslant 0$ for $i=0,1,2,3,4,5$) is invoked, a single solution (up to an overall scale m) survives:

$$\begin{eqnarray*}{m}_{0} & = & +\infty ;{m}_{1}=6m;\;{m}_{1}=2m;\\ {m}_{3} & = & m;{m}_{4}=3m;{m}_{5}=+\infty .\end{eqnarray*}$$

In retrospect, that both the leftmost and the rightmost mass should be infinite could have been predicted from the onset: this is probably the only way to map six particles on an open line to a single four-dimensional particle in a closed cavity.

Without loss of generality, we can place the infinitely massive particles at

$$\begin{eqnarray*}&&{x}_{0}=-L;{x}_{5}=0,\end{eqnarray*}$$

where L is the size of the resulting hard-wall box (see the hyperplanes of contact in terms of the original x-coordinates in figure 1). Now observe that ${m}_{3}=m$ and L appear to be convenient mass and length scales, and ℏ a natural scale of action. From now on, we will be using the system of units where

$$\begin{eqnarray*}&&{m}_{3}=L={\hslash }=1.\end{eqnarray*}$$

Also, it will become clear from what follows that the choice ${\mathcal{M}}=2{m}_{1}$ allows us to respect the existing conventions on the size of the octacube.

Figure 1 also shows the corresponding formulae for the contact hyperplanes expressed through the transformed coordinates y₁, y₂, y₃, and y₄. These hyperplanes form the boundaries of the ${\tilde{F}}_{4}$ kaleidoscope we have in mind: $-1\lt \sqrt{2}\;{y}_{1};\;$ ${y}_{1}\lt \sqrt{3}\;{y}_{2};\;$ ${y}_{2}\lt \sqrt{2}\;{y}_{3};\;$ $\sqrt{3}\;{y}_{3}\lt {y}_{4};\;$ $\;{y}_{4}\lt 0.$

The coordinate transformation ${\boldsymbol{z}}={\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{y}}}\cdot {\boldsymbol{y}},$ with the rotation-inversion matrix ${\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{y}}}$ given by

$$\begin{eqnarray*}{\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{y}}}=\displaystyle \frac{1}{2\sqrt{3}}\left(\begin{array}{rrrr}-\sqrt{6} & -\sqrt{2} & -1 & -\sqrt{3}\\ -\sqrt{6} & \sqrt{2} & 1 & \sqrt{3}\\ 0 & -2\sqrt{2} & 1 & \sqrt{3}\\ 0 & 0 & -3 & \sqrt{3}\end{array}\right),\end{eqnarray*}$$

brings the domain of our ${\tilde{F}}_{4}$ kaleidoscope to the conventional form [2, 3, 11–15]:

$$\begin{eqnarray}&&{{\mathcal{D}}}_{\hat{e}}\equiv \left\{{\boldsymbol{z}}{\rm{}}\;{\rm{such}}\;{\rm{that}}\;{\rm{}}\;{{\boldsymbol{\alpha }}}_{0}\cdot {\boldsymbol{z}}\gt -\displaystyle \frac{1}{2}{{\boldsymbol{\alpha }}}_{0}\cdot {{\boldsymbol{\alpha }}}_{0}=-1\quad {\rm{}}\;{\rm{and}}\;{\rm{}}{{\boldsymbol{\alpha }}}_{\mathrm{1,2},\mathrm{3,4}}\cdot {\boldsymbol{z}}\gt 0\right\},\end{eqnarray} \tag{ 1 }$$

where ${\boldsymbol{y}}\equiv ({y}_{1},{y}_{2},{y}_{3},{y}_{4}),$ and ${\boldsymbol{z}}\equiv ({z}_{1},{z}_{2},{z}_{3},{z}_{4}).$ According to this convention, the kaleidoscope (1) is a four-dimensional simplex with five 3-faces (mirrors) defined by five inward normals ${{\boldsymbol{\eta }}}_{j}={{\boldsymbol{\alpha }}}_{j}/| {{\boldsymbol{\alpha }}}_{j}|$ for $j=0,1,2,3,4,$ with ${{\boldsymbol{\alpha }}}_{0}=(-1,-1,0,0),$ ${{\boldsymbol{\alpha }}}_{1}=(0,+1,-1,0),$ ${{\boldsymbol{\alpha }}}_{2}=(0,0,+1,-1),$ ${{\boldsymbol{\alpha }}}_{3}=(0,0,0,+1),$ ${{\boldsymbol{\alpha }}}_{4}=\left(+\displaystyle \frac{1}{2},-\displaystyle \frac{1}{2},-\displaystyle \frac{1}{2},-\displaystyle \frac{1}{2}\right)$ being the so-called minimal (or negative of maximal) root (j = 0) and the simple roots ($j=1,2,3,4$) of the corresponding finite reflection group F₄[16]. The mirrors from 1 to 4 pass through the origin; the 0th mirror (marked by a dot in the Coxeter diagram of figure 1) passes through the point $(1/2,1/2,0,0).$ The hyperplanes that define the inequalities (1) are also identified in figure 1. In equation (1), the subscript $\hat{e}$ in ${{\mathcal{D}}}_{\hat{e}}$ has the meaning of the identity element of the finite group F₄, a convention whose meaning will become clear from what follows. Note that the 'natural coordinates' ${\boldsymbol{z}}$ can be expressed through the particle coordinates ${\boldsymbol{x}}$ as ${\boldsymbol{z}}={\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}}\cdot {\boldsymbol{x}},$ with

$$\begin{eqnarray*}{\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}}=\displaystyle \frac{1}{12}\left(\begin{array}{rrrr}-6 & -2 & -1 & -3\\ -6 & 2 & 1 & 3\\ 0 & -4 & 1 & 3\\ 0 & 0 & -3 & 3\end{array}\right).\end{eqnarray*}$$

The key to Bethe Ansatz solvability of the closed kaleidoscopic billiards lies in their ability to periodically tile the full space—with neither holes nor overlaps—via consecutive reflections about their faces³ [2]. Consider the ${\tilde{F}}_{4}$ kaleidoscope (1). A union of the figures produced by sequential reflections of (1) about its three-faces from the 1st to the 4th ('simple root' mirrors),

$$\begin{eqnarray}{\mathcal{C}} & \equiv & {\cup }_{\hat{g}\in {F}_{4}}{{\mathcal{D}}}_{\hat{g}},\\ & = & \left\{{\boldsymbol{z}}{\rm{}}\;{\rm{such}}\;{\rm{that}}\;{\rm{}}\;(\hat{g}{{\boldsymbol{\alpha }}}_{0})\cdot {\boldsymbol{z}}\gt -\displaystyle \frac{1}{2}(\hat{g}{{\boldsymbol{\alpha }}}_{0})\cdot (\hat{g}{{\boldsymbol{\alpha }}}_{0})=-1,\quad {\rm{}}\;{\rm{for}}\;{\rm{all}}\;{\rm{}}\hat{g}\in {F}_{4}\right\},\end{eqnarray} \tag{ 2 }$$

with

$$\begin{eqnarray}{{\mathcal{D}}}_{\hat{g}} & \equiv & \left\{{\boldsymbol{z}}{\rm{}}\;{\rm{such}}\;{\rm{that}}\;{\rm{}}\;(\hat{g}{{\boldsymbol{\alpha }}}_{0})\cdot {\boldsymbol{z}}\gt -\displaystyle \frac{1}{2}(\hat{g}{{\boldsymbol{\alpha }}}_{0})\cdot (\hat{g}{{\boldsymbol{\alpha }}}_{0})=-1\quad {\rm{}}\;{\rm{and}}\;{\rm{}}(\hat{g}{{\boldsymbol{\alpha }}}_{\mathrm{1,2},\mathrm{3,4}})\cdot {\boldsymbol{z}}\gt 0\right\},\end{eqnarray} \tag{ 3 }$$

leads to the so-called octacube, otherwise known as the 24-cell [10]; it is the only four-dimensional Platonic solid that does not have any three-dimensional analogues. Its 24 vertices lie at points given by all coordinate permutations and sign choices applied to $(\pm 1,0,0,0)$ and $\left(\pm \displaystyle \frac{1}{2},\pm \displaystyle \frac{1}{2},\pm \displaystyle \frac{1}{2},\pm \displaystyle \frac{1}{2}\right).$ Its 24 octahedron-shaped three-faces are centered at all coordinate permutations and sign choices of $\left(\pm \displaystyle \frac{1}{2},\pm \displaystyle \frac{1}{2},0,0\right).$ Sequential reflections about the 'simple root' mirrors of the ${\tilde{F}}_{4}$ kaleidoscope form the symmetry group of the octacube, the reflection group F₄. (The same label is used to mark both an open kaleidoscope, where the 0th, 'minimal root' mirror is absent and the corresponding simple, 'non-affine' Coxeter diagram.) The union in (2) runs over all elements $\hat{g}$ of the group F₄. The 'physical' domain (1) is the identity member of the set (3).

The four-dimensional space can be periodically tiled by identical octacubes; the centers of the octacubical cells lie on all points with integer coordinates whose coordinates sum to an even number: $\{({z}_{1},{z}_{2},{z}_{3},{z}_{4})\;:\;{z}_{1}+{z}_{2}+{z}_{3}+{z}_{4}={\rm{even}}\}.$ The very same tiling can be obtained by a consecutive reflection of the (1) simplex about all five of its mirrors, including the 'minimal root' one. Transformations generated by these reflections form the symmetry group of the octacubical tiling— the affine reflection group ${\tilde{F}}_{4},$ homonymous to the corresponding closed kaleidoscope.

We choose the lattice vectors ${{\boldsymbol{a}}}_{\mathrm{1,2},\mathrm{3,4}}$ to be proportional to the four simple roots (in an unconventional order) of the reflection group D₄[16], closely associated with⁴ F₄ : ${{\boldsymbol{a}}}_{1}=(0,1,-1,0),$ ${{\boldsymbol{a}}}_{2}=(0,0,1,-1),$ ${{\boldsymbol{a}}}_{3}=(0,0,1,1),$ and ${{\boldsymbol{a}}}_{4}=(1,-1,0,0).$ The reciprocal lattice vectors are thus ${{\boldsymbol{\kappa }}}_{1}=(2\pi ,2\pi ,0,0),$ ${{\boldsymbol{\kappa }}}_{2}=(\pi ,\pi ,\pi ,-\pi ),$ ${{\boldsymbol{\kappa }}}_{3}=(\pi ,\pi ,\pi ,\pi ),$ and ${{\boldsymbol{\kappa }}}_{4}=(2\pi ,0,0,0).$ Any plane wave whose wavevector is equal to an integer-coefficient linear superposition of the reciprocal lattice vectors will be periodic on the original spatial lattice.

Let us briefly describe the parallel structures in the case of a much simpler two-dimensional kaleidoscope, ${\tilde{G}}_{2}$—a right triangle with angles $\pi /2,$ $\pi /3,$ and $\pi /6,$ which allows one to solve the two-body problem with particles of mass m and $3\;m$ in between two hard walls [6]. In this case, the minimal root wall will be represented by the shortest of the two catheti (sides enclosing the right angle). The other cathetus and the hypothenuse would give the simple-root walls. Consecutive reflections about the latter produce a hexagonal periodicity cell, an analogue of the octacubical cell (2). The lattice vectors are given by the simple roots of the reflection group A₂: two equal-length vectors at an angle $\pi /3$ between them.

In general, finding the eigenenergies and eigenstates of a D-dimensional kaleidoscopic billiard amounts to solving a system of D nonlinear algebraic equations—the so-called Bethe Ansatz equations [2, 3, 11–15]. However, in the case of hard-core interactions, the problem greatly simplifies (see corollary 3.2 in [13]). The eigenstates acquire the form

$$\begin{eqnarray}&&\psi ({\boldsymbol{z}})=\displaystyle \frac{1}{\sqrt{{V}_{{\mathcal{D}}}| G| }}\displaystyle \sum _{\hat{g}\in G}{(-1)}^{{\mathcal{P}}(\hat{g})}\mathrm{exp}[{\rm{i}}(\hat{g}{\boldsymbol{k}}){\boldsymbol{z}}],\end{eqnarray} \tag{ 4 }$$

where $\hat{g}$ are the members of the finite reflection group G($={F}_{4}$ in our case) generated by sequential reflections about the 'simple root' mirrors of the kaleidscope in question; ${\mathcal{P}}(\hat{g})$ is the parity of the number of the elementary reflections needed to reach $\hat{g}$ from the identity transformation; ${V}_{{\mathcal{D}}}={V}_{{\mathcal{C}}}/| G|$($=1/576$ in our case) is the kaleidoscope volume; ${V}_{{\mathcal{C}}}=| \mathrm{det}[({{\boldsymbol{a}}}_{1},{{\boldsymbol{a}}}_{2},{{\boldsymbol{a}}}_{3},\ldots )]|$ is the unit cell volume; $| G|$($=1152$ for F₄) is the order, i.e. the number of elements, of the group G. The conjecture contained in theorem 1 in [15] indicates that the wavefunction (4) is normalized to unity: ${\displaystyle \int }_{{{\mathcal{D}}}_{\hat{e}}}{\rm{d}}{z}_{1}{\rm{d}}{z}_{2}{\rm{d}}{z}_{3}\ldots | \psi ({\boldsymbol{z}}){| }^{2}=1$ when integrated over the corresponding kaleidoscope ${{\mathcal{D}}}_{\hat{e}}$ ((1) in the case of ${\tilde{F}}_{4}$). This conjecture covers not only the case of the hard-wall boundary conditions but also a much more general class of boundary conditions, associated with finite strength interactions. The wavevectors k are simply drawn from the reciprocal lattice of the corresponding tiling⁵ . In order to prevent both double counting and the formal appearance of the eigenstates that are identically zero, the wavevectors k must be further restricted to those lying inside the open kaleidoscope associated with the kaleidoscope in question, obtained by removing the 'minimal root' mirror: ${{\boldsymbol{\eta }}}_{j}\cdot {\boldsymbol{k}}\gt 0$ for $j=1,2,3,\ldots$ (see the end of section 3 of [13]).

For the case of the hard-wall boundary conditions, the normalization formula used above can also be proven directly without resorting to the conjecture [15]. Observe that the octacube (2) is a unit cell of the lattice spanned by the lattice vectors ${{\boldsymbol{a}}}_{\mathrm{1,2},\mathrm{3,4}}$ and that the 'seed' wavevector ${\boldsymbol{k}},$ along with all its images $\hat{g}{\boldsymbol{k}}$ belong to the respective reciprocal lattice. In that case, any two distinct plane waves with wave vectors $\hat{g}{\boldsymbol{k}}$ and $\hat{g}^{\prime} {\boldsymbol{k}}$ will be orthogonal to each other if integrated over the whole cell:

$$\begin{eqnarray*}&&\langle \hat{g}^{\prime} {\boldsymbol{k}}| \hat{g}{\boldsymbol{k}}{\rangle }_{{\mathcal{C}}}={\delta }_{\hat{g}^{\prime} ,\;\hat{g}}{V}_{{\mathcal{C}}},\end{eqnarray*}$$

with

$$\begin{eqnarray*}&&\langle {{\boldsymbol{k}}}_{1}| {{\boldsymbol{k}}}_{2}{\rangle }_{{\mathcal{A}}}\equiv {\displaystyle \int }_{{\mathcal{A}}}{{\rm{d}}}^{d}{\boldsymbol{z}}\mathrm{exp}[{\rm{i}}({{\boldsymbol{k}}}_{2}-{{\boldsymbol{k}}}_{1}){\boldsymbol{z}}],\end{eqnarray*}$$

where ${\mathcal{A}}$ is an area of space. The normalization integral in question can now be evaluated as

$$\begin{eqnarray*}&&{\displaystyle \int }_{{{\mathcal{D}}}_{\hat{e}}}{{\rm{d}}}^{d}{\boldsymbol{z}}| \displaystyle \sum _{\hat{g}\in G}{(-1)}^{{\mathcal{P}}(\hat{g})}\mathrm{exp}[{\rm{i}}(\hat{g}{\boldsymbol{k}}){\boldsymbol{z}}]{| }^{2}\end{eqnarray*}$$

$$\begin{eqnarray*}= & \displaystyle \sum _{\hat{g}^{\prime} }\displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime} )+{\mathcal{P}}(\hat{g})}\langle \hat{g}^{\prime} {\boldsymbol{k}}| \hat{g}{\boldsymbol{k}}{\rangle }_{{{\mathcal{D}}}_{\hat{e}}}\overset{\hat{g}^{\prime\prime} \equiv {\hat{g}}^{-1}\hat{g}^{\prime} }{=}\displaystyle \sum _{\hat{g}^{\prime\prime} }\displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g}\hat{g}^{\prime\prime} )+{\mathcal{P}}(\hat{g})}\langle \hat{g}\hat{g}^{\prime\prime} {\boldsymbol{k}}| \hat{g}{\boldsymbol{k}}{\rangle }_{{{\mathcal{D}}}_{\hat{e}}},\\ = & \displaystyle \sum _{\hat{g}^{\prime\prime} }{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime\prime} )}\displaystyle \sum _{\hat{g}}\langle \hat{g}\hat{g}^{\prime\prime} {\boldsymbol{k}}| \hat{g}{\boldsymbol{k}}{\rangle }_{{{\mathcal{D}}}_{\hat{e}}}\overset{\hat{g}\prime \prime \prime \equiv {\hat{g}}^{-1}}{=}\displaystyle \sum _{\hat{g}^{\prime\prime} }{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime\prime} )}\displaystyle \sum _{\hat{g}\prime \prime \prime }\langle \hat{g}^{\prime\prime} {\boldsymbol{k}}| {\boldsymbol{k}}{\rangle }_{{{\mathcal{D}}}_{\hat{g}\prime \prime \prime }},\\ = & \displaystyle \sum _{\hat{g}^{\prime\prime} }{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime\prime} )}\langle \hat{g}^{\prime\prime} {\boldsymbol{k}}| {\boldsymbol{k}}{\rangle }_{{\mathcal{C}}},\\ = & \displaystyle \sum _{\hat{g}^{\prime\prime} }{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime\prime} )}{\delta }_{\hat{g}^{\prime} ,\hat{e}}{V}_{{\mathcal{C}}},\\ = & {V}_{{\mathcal{C}}}={V}_{{\mathcal{D}}}| G| ;\end{eqnarray*}$$

this result justifies the choice of the normalization factor in the expression (4). Above, we used ${(-1)}^{{\mathcal{P}}(\hat{g}\hat{g}^{\prime\prime} )}={(-1)}^{{\mathcal{P}}(\hat{g})}{(-1)}^{{\mathcal{P}}(\hat{g}^{\prime\prime} )}$ and ${(-1)}^{2{\mathcal{P}}(\hat{g})}=1.$

We are now in the position to write down the spectrum of our system explicitly. Starting from this point we abandon the ${m}_{3}=L={\hslash }=1$ system of units. The eigenenergies, $E=({{\hslash }}^{2}/2{\mathcal{M}}){k}^{2},$ and the eigenstates (4), with ${\boldsymbol{k}}={\displaystyle \sum }_{j=\mathrm{1,2},\mathrm{3,4}}{n}_{j}{{\boldsymbol{\kappa }}}_{j},$ ${n}_{j=\mathrm{1,2},\mathrm{3,4}}\in {\mathbb{Z}},$ and ${{\boldsymbol{\eta }}}_{j=\mathrm{1,2},\mathrm{3,4}}\cdot {\boldsymbol{k}}\gt 0,$ are then given by

$$\begin{eqnarray}&&{E}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}=\displaystyle \frac{{\pi }^{2}{{\hslash }}^{2}}{6{m}_{3}{L}^{2}}\left(2{n}_{1}({n}_{1}+{n}_{2}+{n}_{3}+{n}_{4})+{n}_{2}^{2}+{n}_{3}^{2}+{n}_{4}^{2}+{n}_{2}{n}_{3}+{n}_{2}{n}_{4}+{n}_{3}{n}_{4},\right)\end{eqnarray} \tag{ 5 }$$

$$\begin{eqnarray}&&{{\rm{\Psi }}}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}({\boldsymbol{x}})=\displaystyle \frac{1}{\sqrt{48{L}^{4}}}\displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g})}\mathrm{exp}[{\rm{i}}(\hat{g}{{\boldsymbol{k}}}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}})\cdot {\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}}\cdot {\boldsymbol{x}}]\end{eqnarray} \tag{ 6 }$$

for ${n}_{1}\geqslant 1$ and ${n}_{4}\gt {n}_{3}\gt {n}_{2}\geqslant 1$, where

$$\begin{eqnarray*}&&{{\boldsymbol{k}}}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}=\left(\begin{array}{c}2{n}_{1}+{n}_{2}+{n}_{3}+2{n}_{4}\\ 2{n}_{1}+{n}_{2}+{n}_{3}\\ {n}_{2}+{n}_{3}\\ -{n}_{2}+{n}_{3}\end{array}\right)\;\displaystyle \frac{\pi }{L};\end{eqnarray*}$$

the sum in (6) runs over all 1152 members $\hat{g}$ of the reflection group F₄ [17] (transformations generated by consecutive reflections about the four 'simple root' mirrors of the ${\tilde{F}}_{4}$ kaleidoscope); ${{\rm{\Psi }}}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}({\boldsymbol{x}})\equiv \sqrt{| \mathrm{det}[{\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}}]| }{\psi }_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}({\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}}\cdot {\boldsymbol{x}})$ are the eigenstates of the system expressed through the particle coordinates ${x}_{1},{x}_{2},{x}_{3},{x}_{4},$ normalized as ${\displaystyle \int }_{{x}_{1}=-L}^{0}{\displaystyle \int }_{{x}_{2}={x}_{1}}^{0}{\displaystyle \int }_{{x}_{3}={x}_{2}}^{0}{\displaystyle \int }_{{x}_{4}={x}_{3}}^{0}{{\rm{d}}}^{4}{\boldsymbol{x}}| {{\rm{\Psi }}}_{{n}_{1}{n}_{2}{n}_{3}{n}_{4}}({\boldsymbol{x}}){| }^{2}=1;\;$ $\left(\displaystyle \frac{{\mathcal{D}}{\boldsymbol{z}}}{{\mathcal{D}}{\boldsymbol{x}}}\right)$ is the Jacobian matrix of the ${\boldsymbol{z}}\leftarrow {\boldsymbol{x}}$ transformation.

At high energies, the spectrum converges to the Weyl law prediction for the number of states with energies below a given energy E: ${\mathcal{N}}(E)=\displaystyle \frac{{\mathcal{W}}(E)}{{(2\pi {\hslash })}^{D=4}}=\displaystyle \frac{{m}_{3}^{2}{L}^{4}{E}^{2}}{32{\pi }^{2}{{\hslash }}^{4}},$ where ${\mathcal{W}}(E)$ is the classical phase-space volume occupied by points with energies below E, and D is number of spatial dimensions (see figure 2).

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The ground state energy is

$$\begin{eqnarray}&&{E}_{{\rm{ground}}\;{\rm{state}}}={E}_{{n}_{1}=1,{n}_{2}=1,{n}_{3}=2,{n}_{4}=3}=\displaystyle \frac{13{\pi }^{2}{{\hslash }}^{2}}{2{m}_{3}{L}^{2}}.\end{eqnarray} \tag{ 7 }$$

The ground state wave function can be obtained from equation (6), using

$$\begin{eqnarray}&&{{\boldsymbol{k}}}_{{\rm{ground}}\;{\rm{state}}}={{\boldsymbol{k}}}_{{n}_{1}=1,{n}_{2}=1,{n}_{3}=2,{n}_{4}=3}=\left(\begin{array}{c}11\\ 5\\ 3\\ 1\end{array}\right)\;\displaystyle \frac{\pi }{L}.\end{eqnarray} \tag{ 8 }$$

figure 3 shows a particular section of the ground state density distribution within the ${\tilde{F}}_{4}$ simplex, along with its space-tiling mirror images.

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The first excitation, ${n}_{1}=1,{n}_{2}=1,{n}_{3}=2,{n}_{4}=4,$ has an energy ${E}_{{\rm{1st}}\;{\rm{excited}}\;{\rm{state}}}={E}_{{n}_{1}=1,{n}_{2}=1,{n}_{3}=2,{n}_{4}=4}=\displaystyle \frac{17{\pi }^{2}{{\hslash }}^{2}}{2{m}_{3}{L}^{2}}.$ The first degeneracy unites the 14th and the 15th excited states. All degeneracies appear to be purely 'number-theoretical,' i.e. not associated with the symmetries of the system. While the rich symmetry and the tiling properties of the octacubical periodicity cell constitute the very reason for the existence of the exact solutions, the actual kaleidoscope—of whose mirror images the cell consists—does not possess any relevant symmetry.

The Bethe Ansatz integrability of our system can be also reinterpreted in terms of the Liouville integrability. To construct the three additional integrals of motion in involution with the Hamiltonian and each other, we suggest invoking the invariant polynomials of the group in question [18]: finite polynomials of coordinates $w({z}_{1},{z}_{2},{z}_{3},\ldots )$ that remain invariant under the group action,

$$\begin{eqnarray}&&w(\hat{g}{\boldsymbol{z}})=w({\boldsymbol{z}}).\end{eqnarray} \tag{ 9 }$$

Consider now an operator $\hat{I}$ that is constructed by taking an invariant polynomial w as a function of the momenta associated with the ${\boldsymbol{z}}$ coordinates:

$$\begin{eqnarray*}&&\hat{I}\equiv w(-{\rm{i}}{{\boldsymbol{\nabla }}}_{{\boldsymbol{z}}}).\end{eqnarray*}$$

Let us show that any eigenstate (4) of the kaleidoscope associated with the group in question is at the same time an eigenstate of $\hat{I}.$ Indeed,

$$\begin{eqnarray*}\hat{I}\psi ({\boldsymbol{z}}) & = & w(-{\rm{i}}{{\boldsymbol{\nabla }}}_{{\boldsymbol{z}}})\displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g})}\mathrm{exp}[{\rm{i}}(\hat{g}{\boldsymbol{k}}){\boldsymbol{z}}],\\ & = & \displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g})}w(\hat{g}{\boldsymbol{k}})\mathrm{exp}[{\rm{i}}(\hat{g}{\boldsymbol{k}}){\boldsymbol{z}}],\\ & = & w({\boldsymbol{k}})\displaystyle \sum _{\hat{g}}{(-1)}^{{\mathcal{P}}(\hat{g})}\mathrm{exp}[{\rm{i}}(\hat{g}{\boldsymbol{k}}){\boldsymbol{z}}].\end{eqnarray*}$$

Q.E.D. for the reflection group F₄, the four lowest power functionally independent invariant polynomials read [19]:

$$\begin{eqnarray}{w}_{M}^{({F}_{4})}({\boldsymbol{z}}) & = & {({z}_{1}-{z}_{2})}^{{l}_{M}}+{({z}_{1}+{z}_{2})}^{{l}_{M}}+{({z}_{1}-{z}_{3})}^{{l}_{M}}\\ & & +{({z}_{1}+{z}_{3})}^{{l}_{M}}+{({z}_{1}-{z}_{4})}^{{l}_{M}}+{({z}_{1}+{z}_{4})}^{{l}_{M}}\\ & & +{({z}_{2}-{z}_{3})}^{{l}_{M}}+{({z}_{2}+{z}_{3})}^{{l}_{M}}+{({z}_{2}-{z}_{4})}^{{l}_{M}}\\ & & +{({z}_{2}+{z}_{4})}^{{l}_{M}}+{({z}_{3}-{z}_{4})}^{{l}_{M}}+{({z}_{3}+{z}_{4})}^{{l}_{M}}\\ & & {l}_{M=1}=2;{l}_{M=2}=6;{l}_{M=3}=8;{l}_{M=4}=12.\end{eqnarray} \tag{ 10 }$$

Accordingly, the four fundamental integrals of motion for our system read

$$\begin{eqnarray}&&{\hat{I}}_{M}^{({F}_{4})}={w}_{M}^{({F}_{4})}({({({\hat{T}}_{{\boldsymbol{z}}\leftarrow {\boldsymbol{x}}})}^{-1})}^{\top }\cdot {\hat{{\boldsymbol{p}}}}_{{\boldsymbol{x}}}),\quad \quad M=1,2,3,4,\end{eqnarray} \tag{ 11 }$$

with

$$\begin{eqnarray}&&{\hat{{\boldsymbol{p}}}}_{{\boldsymbol{x}}}\equiv (-{\rm{i}}{\hslash }\displaystyle \frac{\partial }{\partial {x}_{1}},-{\rm{i}}{\hslash }\displaystyle \frac{\partial }{\partial {x}_{2}},-{\rm{i}}{\hslash }\displaystyle \frac{\partial }{\partial {x}_{3}},-{\rm{i}}{\hslash }\displaystyle \frac{\partial }{\partial {x}_{4}})\end{eqnarray} \tag{ 12 }$$

being the particle momenta, corresponding to the particle coordinates x₁, x₂, x₃, x₄. Above, ${(\ldots )}^{\top }$ stands for the transpose of a matrix. Note that the first integral of motion is proportional to the Hamiltonian of the system: ${\hat{I}}_{1}^{({F}_{4})}/m=144\hat{H}.$

In this Letter, we obtain—using Bethe Ansatz—an exact expression for the eigenenergies and eigenfunctions of four hard-core particles with mass ratios $6\ :2\ :1\ :3$ in a hard-wall box. The Ansatz is induced by a hidden symmetry of the system related to the symmetries of the tiling of a four-dimensional space by octacubes. The exact spectrum stands in good agreement with the approximate Weyl's law prediction.

The following observation may serve as a seed for a longer research program. The procedure, outlined in this Letter, for identifying a few-body problem relevant to a particular Coxeter diagram [17] is not unique to ${\tilde{F}}_{4}.$ Any diagram, affine or not, that does not have bifurcations can potentially be used to generate a solvable few-body hard-core problem. The ${\tilde{A}}_{N-1}$ (N identical hard-cores on a circle) and ${\tilde{C}}_{N}$ (N identical hard-cores in a hard-wall box) diagrams have been already successfully introduced in the first years of many-body Bethe Ansatz, in [5] and [1], respectively. The ${I}_{2}(n)$ diagrams were explored in [20], albeit classically, still awaiting a quantum treatment: for each integer n a continuous one-parametric family of solvable three-body mass ratios (on a line with no boundary conditions) was obtained. A system of two hard-core particles with mass ratio 3:1 in a box (whose one-wall version was explored in [6]) is expected to exhibit an exact solution associated with the ${\tilde{G}}_{2}$ Coxeter diagram. It can be conjectured that H₃ and H₄ Coxeter diagrams lead to solvable four- and five-body problems on a line, respectively.

The true challenge is posed by the Coxeter diagrams (affine or otherwise) with bifurcations, most notably the ${\tilde{E}}_{\mathrm{6,7,8}}$ series. At the moment, it is not clear if there are any respective realistic 6-, 7-, or 8-body problems that can be solved using these symmetries. (Some examples of 'unphysical' few-body realizations of Coxeter diagrams—e.g. those where a given particle can interact with an empty point in between the other two—are given in [3].)

The potential empirical context for this and further planned explorations is one-dimensional cold gas mixtures in optical lattices where the dynamics is governed by an effective (rather than physical) mass, controlled at will [21].

We are profoundly indebted to late Marvin Girardeau for guidance and inspiration. We are thanking Jean-Sébastien Caux for reading the manuscript and for the comments that followed. The impact of the numerous in-depth discussions of the subject with Dominik Schneble can not be overestimated. This work was supported by grants from National Science Foundation (PHY-1402249) and the Office of Naval Research (N00014-12-1-0400).