A Coulomb density without Kohn–Sham ensemble representation

Source and verification

This article explains family 278 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Densities and noninteracting representation

Kohn–Sham theory seeks a noninteracting electron system whose ground-state density equals that of an interacting system. A density can come from a perfectly legitimate state without coming from a ground state of any potential in a prescribed class. Family 278 constructs a finite Coulomb molecule for which even allowing arbitrary ground-state ensembles does not produce such a representation. (OpenAI, 2026)

The prerequisites are Sobolev spaces, quadratic forms, the spectral theorem, fermionic antisymmetry, and basic elliptic estimates. One electron has Hilbert space $\mathfrak{h}=L^2({\mathbb{R}}^3;{\mathbb{C}}^2)$, where the two components describe spin. Three electrons occupy the antisymmetric tensor space ${\bigwedge }^3\mathfrak{h}$. For a normalized state $\Psi$, its spin-summed density $n_{\Psi }(x)$ is three times the squared wave function integrated over the other two positions and summed over all spins; hence $\int n_{\Psi }=3$.

An ensemble is a positive trace-class operator $\Gamma$ of trace one on this three-electron space. Its density is obtained by averaging the densities in a spectral decomposition. A ground-state ensemble has range in the eigenspace at the bottom of the Hamiltonian’s spectrum. This includes arbitrary degeneracy and mixtures, and requires an actual ground eigenspace.

The candidate local potentials belong to

$$\mathcal{V}=L^{3/2}({\mathbb{R}}^3;\mathbb{R})+L^{\infty }({\mathbb{R}}^3;\mathbb{R}).$$

The sum means $v=v_1+v_{\infty }$ with one summand in each space. Such potentials define semibounded closed Schrödinger forms on $H^1$. The same real scalar potential acts on both spins. The noninteracting three-electron Hamiltonian is $H_s(v)={\sum }_{i=1}^3(-{\Delta }_i/2+v(x_i))$.

🏷️ The finite Coulomb counterexample

Manuscript's nonrepresentation theorem

Fix $D=10^{200}$. The manuscript specifies a finite positive integer $Z$ such that the three-electron Coulomb system with two nuclear charges $Z,Z$ at $\pm (D/Z)e_z$ has a normalizable absolute ground state $\Psi$ in the spin-projection sector $S_z=1/2$, whose density is not the density of any ground-state ensemble of $H_s(v)$ for $v\in \mathcal{V}$. (OpenAI, 2026)

Here $e_z=(0,0,1)$. Electron repulsion has its usual coefficient one, and nuclear attraction is $-Z/|x-(D/Z)e_z|-Z/|x+(D/Z)e_z|$. The integer is defined by exact spectral and variational data, with no evaluated numerical value or effective upper bound claimed. The nuclei are at distinct finite positions. The system need not be neutral.

The conclusion allows complex spinors, spin coherences, and unrestricted ground-space degeneracy in a hypothetical representing ensemble. It concerns a single real, spin-independent local potential in $\mathcal{V}$; it does not settle representation by spin-dependent, nonlocal, or more singular distributional potentials.

Scale positions by $x=Zr$ and put $\lambda =1/Z$. The physical Hamiltonian becomes $Z^2{H}_{3,\lambda }$, where

$$\begin{array}{rl}{H}_{m,\lambda }& =\sum _{i=1}^m{h}_D(x_i)\\ & +\lambda \sum _{i<j}|x_i-x_j{|}^{-1},\\ h_D& =-\Delta /2+u_D,\\ u_D(x)& =-|x-D{e}_z{|}^{-1}\\ & -|x+D{e}_z{|}^{-1}.\end{array}$$

The potential class is preserved: a physical $v$ becomes $Z^{-2}v(x/Z)$, whose $L^{3/2}$ norm is unchanged for the integrable summand. Thus it suffices to rule out representation in the scaled problem.

For large fixed $D$, the two lowest spatial orbitals of $h_D$ are a positive even orbital $b$ and an odd orbital $e$, positive on $z>0$. They have a small positive energy splitting. At $\lambda =0$, the three-electron determinant occupies $b\uparrow ,b\downarrow ,e\uparrow$. The manuscript follows its simple continuation in the $S_z=1/2$ sector and establishes that it remains an absolute bound ground state for sufficiently small positive coupling.

🏷️ A slow tail on the median plane

Let $E_2,E_3$ be the ionic and three-electron ground energies in scaled coordinates. Define

$$\begin{array}{c}k=\sqrt{2(E_2-E_3)},\\ \beta =\frac{2-2\lambda }{k}-1,\\ A(r)=e^{-kr}{r}^{\beta }.\end{array}$$

The leading electron-removal amplitude is odd in the median plane $z=0$, so it vanishes there. That does not force the full density to have a faster exponential decay on the plane. Coupling to excited ionic channels supplies a smaller polynomial prefactor with the same exponential rate.

To express this, view $F(x)=\sqrt{3}\Psi (x,\cdot )$ as a vector in the remaining ionic and spin variables, so $n(x)=\Vert F(x){\Vert }^2$. Let $P$ project onto the ionic singlet ground channel, $Q=I-P$, and $B_Q$ be the positive ionic energy operator on the complementary space. In the exterior channel expansion, the order-$r^{-2}$ and order-$r^{-3}$ complementary coefficients vanish on the plane. The angular Laplacian produces the first nonzero planar coefficient at order $r^{-4}$.

Its vector is proportional to $B_Q^{-2}Q(y_{1z}+y_{2z})\varphi$, where $\varphi$ is the even ionic ground vector. Odd parity makes the projection $Q$ redundant on this vector. It is nonzero: a nonzero $L^2$ vector cannot be supported on the hyperplane $y_{1z}+y_{2z}=0$, and the reduced inverse is injective. The nonzero slope of the direct odd amplitude supplies the remaining factor.

The manuscript controls the full channel remainder, not merely a truncated formal expansion. For each selected fixed $\lambda >0$, it obtains

$$\begin{array}{c}\sqrt{n(p,0)}\ge c_{\lambda }{e}^{-kp}{p}^{\beta -4}\\ \text{for sufficiently large }p,\end{array}$$

where $p=\sqrt{x_1^2+x_2^2}$ and $c_{\lambda }>0$. It also proves a translation limit. For distant centers $y_j=p_j{\omega }_j$ on the plane with horizontal unit vectors ${\omega }_j\to \omega$,

$$\frac{n(y_j+x)}{(A(p_j)/p_j{)}^2}⟶a_{\ast }^2{z}^2{e}^{-2k\omega \cdot x}$$

locally uniformly, with $a_{\ast }>0$. For centers whose heights tend to infinity in absolute value, the normalized limit is $n(y_j+x)/n(y_j)\to e^{-2k\omega \cdot x}$ after selecting a direction.

These assertions are compatible: the planar tail has an extra factor $p^{-3}$ in amplitude relative to the translation normalization, so it disappears in the limit. The exact density remains positive on the plane. The uniform channel estimates and their crossover with the faster unperturbed core are substantial manuscript inputs; their complete proof was not independently verified here.

🏷️ Two orbitals from an angle minimization

For the interacting density $n$, minimize $\int n|\nabla \theta {|}^2$ among odd angles with $\int n{\sin }^2\theta =1$, folded into $[0,\pi /2]$ on the upper half-space. Compactness, integrability of $n$, and the Euler equation give a minimizer with multiplier $d>0$:

$$\mathrm{div}(n\nabla \theta )=-dn\sin (2\theta ).$$

Define $g=\sqrt{n}\cos \theta$ and $h=\sqrt{n}\sin \theta$. Then $g$ is positive and even, $h$ is odd, $\Vert g{\Vert }_2^2=2$, $\Vert h{\Vert }_2^2=1$, and $g^2+h^2=n$. Oddness gives orthogonality. Their common formal potential is

$$\begin{array}{rl}V=\frac{\Delta g}{2g}& =\frac{\Delta \sqrt{n}}{2\sqrt{n}}-\frac{|\nabla \theta {|}^2}{2}\\ & +d{\sin }^2\theta .\end{array}$$

The distributional orbital equations are $(-\Delta /2+V)g=0$ and $(-\Delta /2+V)h=dh$. At this stage $V$ is locally square integrable and $\int n|V|<\infty$; membership in $\mathcal{V}$ is not assumed.

Two orbital equations alone do not prove that these orbitals minimize kinetic energy. The additional ingredient is the weighted spectral gap

$$\begin{array}{c}\frac{1}{2}\int g^2|\nabla \xi {|}^2\ge d{\mathrm{Var}}_{g^2}(\xi ),\\ {\mathrm{Var}}_{g^2}(\xi )=\int g^2|\xi {|}^2-\frac{|\int g^2\xi {|}^2}{\int g^2}.\end{array}$$

For odd tests this follows from the positive function $h/g$ on the upper half-space and the zero boundary trace. For even tests the proof combines the unperturbed even spectral gap with uniform exterior coercivity. A quadratic partition localizes a potentially escaping sequence: the interior limit pays the even gap, while any escaped weighted mass pays the exterior gap. Both exceed the small target level. This prevents an invalid inference from local convergence alone. The detailed exterior strip estimates remain manuscript inputs.

🏷️ Why arbitrary ensembles are excluded

Normalize $\hat{g}=g/\sqrt{2}$. The determinant $\Phi =(\hat{g}\uparrow )\wedge (\hat{g}\downarrow )\wedge (h\uparrow )$ has density $n$. The weighted gap and the positive-orbital identity give, for scalar orbitals $f$ in the relevant form class,

$$\begin{array}{rl}& \frac{1}{2}\int |\nabla f{|}^2+\int V|f{|}^2\\ & \ge d(\Vert f{\Vert }_2^2-|⟨\hat{g},f⟩{|}^2).\end{array}$$

The identity starts on compact smooth tests and extends by local approximation and cutoffs using absolute potential integrability.

For any three-electron ensemble, the one-particle density matrix $\gamma$ satisfies $0\le \gamma \le I$ and $\mathrm{Tr}\gamma =3$: any normalized spin orbital can be occupied at most once. The projection onto $\hat{g}$ with both spins has rank two. For an ensemble with density $n$ and finite kinetic energy, applying the scalar inequality to its natural spin orbitals and summing yields

$$\begin{array}{rl}T[\Gamma ]+\int Vn& \ge d\left(3-\mathrm{Tr}(\gamma P)\right)\\ & \ge d.\end{array}$$

The determinant $\Phi$ attains equality. This proves a kinetic minimum over all ensembles with density $n$, including arbitrary complex spinors and mixtures. No rank bound on $\Gamma$ is needed.

Suppose a representing potential $v\in \mathcal{V}$ existed. Its form bound gives finite kinetic energy to a ground-state ensemble, even if the ground eigenspace is infinite dimensional. Replacing that ensemble by $\Phi$ preserves the potential energy and cannot raise kinetic energy. Equality at the spectral bottom then makes $\Phi$ itself a ground eigenvector.

To extract a one-particle equation, vary only its down-spin orbital. Spin orthogonality shows that the form pairing of $\hat{g}$ with every orthogonal spatial test vanishes. Hence $(-\Delta /2+v)g=e_{g}g$ for a real constant $e_g$. Since $g>0$, comparison with the formal equation gives $v=V+e_g$ almost everywhere. Therefore every hypothetical representing potential would force $V\in \mathcal{V}$. This argument is what extends the obstruction beyond a chosen orbital ansatz.

🏷️ Translated potentials and the trace contradiction

Assume $V=V_c+V_b$ with $V_c\in L^{3/2}$ and $V_b\in L^{\infty }$. Along escaping translations, $V_c$ tends to zero locally in $L^{3/2}$, while subsequences of $V_b$ converge weak star to bounded potentials $W$. Local form estimates and Rellich compactness allow the normalized orbital equations to pass to such limits.

For centers on the median plane, the density limit implies that the nonnegative limit of $g$ vanishes on the plane. Harnack’s inequality forces that limit to vanish everywhere. The limit of $h$ is consequently $a_{\ast }z{e}^{-k\omega \cdot x}$. Its eigenfunction equation identifies $W=d+k^2/2$ almost everywhere. The manuscript controls the angle above the plane to treat centers whose heights diverge; bounded-height sequences reduce to plane translations. Thus every escaping potential limit is the same constant.

This is a weak-limit statement, not an assumed pointwise asymptotic for $V$. Uniform local form compactness and a quadratic partition turn it into an exterior lower bound with potential level $c_{\infty }=d+k^2/2$. Testing the zero-energy equation for $g$ with cutoff exponential weights then gives

$$\begin{array}{c}{e}^{a|x|}g\in H^1({\mathbb{R}}^3)\\ \text{for }k<a<\sqrt{k^2+2d}.\end{array}$$

The trace of an $H^1({\mathbb{R}}^3)$ function on a plane lies in $L^2({\mathbb{R}}^2)$. But $\theta =0$ on the plane, so $g=\sqrt{n}$ there. The planar lower bound would force its squared weighted trace integral to dominate a positive multiple of

$${\int }^{\infty }{e}^{2(a-k)p}{p}^{2\beta -7}dp=\infty .$$

The exponent includes the planar area element $pdp$. Since $a>k$, no polynomial factor can restore integrability. This contradiction excludes $V\in \mathcal{V}$, and the ensemble certificate excludes every representing $v$.

🏷️ Exact parameters and a functional consequence

The construction first fixes $D$, then chooses a sufficiently small positive coupling, and finally takes the far-field limit with that coupling fixed. Setting $\lambda =0$ would remove the required nonzero planar coefficient.

For the exact charge prescription, the paper defines a bad set $\mathcal{B}\subset (0,1]$ using auxiliary spectral isolation, energy and density-tail tests, and tests on every folded angle minimizer. The tests include a fixed exterior radius and strict strip-comparison margin. They do not ask whether another potential represents the density. Uniform small-coupling estimates prove that $\mathcal{B}$ avoids some interval next to zero. Consequently

$$Z=2+\lceil \underset{0<t\le 1}{\sup }{t}^{-1}{1}_{\mathcal{B}}(t)\rceil$$

is finite, and $1/Z$ passes every test: otherwise the supremum would be at least $Z$. Fixed-parameter channel and gap criteria then apply directly at this coupling. This is an exact noneffective specification, not a computed molecular example.

There is also a variational consequence. Let $T_{\mathrm{ens}}[n]$ be the constrained kinetic minimum and $F_{\mathrm{ens}}[n]$ the constrained interacting internal-energy minimum. On their common finite-energy density domain, set $E_{\mathrm{Hxc}}=F_{\mathrm{ens}}-T_{\mathrm{ens}}$. The physical density has no subgradient of $T_{\mathrm{ens}}$ represented by a potential in $\mathcal{V}$: such a subgradient would make its attaining determinant a ground state for the negative subgradient potential.

Nor can one $w\in \mathcal{V}$ represent the one-sided variation of $E_{\mathrm{Hxc}}$ along every segment from this density to another finite-energy density. Convexity of $T_{\mathrm{ens}}$ and the interacting ground-state supporting inequality would then put $-v_{\mathrm{ext}}-w$ in that forbidden subdifferential. The claim concerns all such segments, not every possible restricted notion of derivative.

The selected review covers the ensemble certificate and potential identification, the full translation and trace obstruction, the exact parameter prescription, and the functional consequence. The channel and weighted-gap sections were read selectively; their full uniform estimates have not been independently verified.

Coulomb ionization provides context for interpreting the removal-energy decay rate. Continuum Coulomb energy gives another example where the full fermionic variational class and a finite-dimensional or orbital construction must be connected by an explicit comparison theorem.

🏷️ Key steps and proof outlines

Why the first nonzero planar tail occurs at order four

The channel expansion is more informative than a general statement that electron interaction creates a tail. Let $\varphi$ be the ionic ground state, $Q$ its orthogonal complement, and $B_Q$ the positive ionic operator shifted by the relevant threshold. In the notation of the channel analysis, the slowly decaying part of $QF$ has the form

$$\lambda A(r)\left(\frac{T_2(\omega )}{r^2}+\frac{T_3(\omega )}{r^3}+\frac{T_4(\omega )}{r^4}\right),$$

with an error bounded by $C_D\lambda A(r)r^{-5}$, including the derivatives needed later. The other displayed channel component decays faster. Here $A(r)$ is the radial exponential factor already introduced, and the $T_j$ take values in the ionic Hilbert space.

To obtain these coefficients, expand the interaction into angular multipoles $W_j$ and the scalar channel into coefficients $a_j$. Matching successive inverse powers of $r$ gives

$$\begin{array}{rl}{B}_Q{T}_2& =-a_{0}Q{W}_2\varphi =:S_2,\\ B_Q{T}_3& =S_3+2k{T}_2,\\ B_Q{T}_4& =S_4+3k{T}_3-(K_2+C_2)T_2.\end{array}$$

The operators $K_2$ and $C_2$ collect the angular/radial and potential terms at the indicated order. These equations are interpreted on the localized domains established in the source; the multipole operators are not asserted to be bounded on the whole ionic space.

Reflection symmetry makes the $a_j$, and hence the $S_j$, vanish on the equator. It also makes $T_2$ and $T_3$ vanish there. However, angular differentiation can detect how $T_2$ crosses that equator. The surviving term is

$$T_4{|}_{\mathrm{eq}}=\frac{1}{2}{B}_Q^{-1}{\Delta }_{{\mathbb{S}}^2}{T}_2{|}_{\mathrm{eq}}=-a_0^{\prime }(0)B_Q^{-2}Q{Y}_z\varphi ,Y_z=y_{1z}+y_{2z}.$$

The product rule supplies the factor two canceled by $1/2$: the normal angular derivative hits both $a_0$ and $W_2$. The vector $Y_z\varphi$ is odd and nonzero, while the ionic ground state is even; consequently it survives $Q$. Injectivity of the positive inverse then prevents cancellation. This is the mechanism behind the nonzero planar channel-amplitude tail $\lambda A(r)r^{-4}$, hence a density contribution of order ${\lambda }^{2}A(r{)}^2{r}^{-8}$. The remainder estimate is essential: a formal nonzero coefficient alone would not establish the lower bound. (OpenAI, 2026, p.c hannel expansion)

A weighted spectral gap must also charge mass escaping to infinity

The angle construction uses a weighted energy

$${\mathcal{E}}_w(\xi )=\frac{1}{2}\int w|\nabla \xi {|}^2.$$

Its compactness argument does not assume that normalized weighted eigenfunctions remain spatially tight. For the even sector, suppose a sequence of real even tests with weights $w_j$ had mean zero, norm one, and energies at most $d_{\ast }<0.03$. Local compactness produces a limit $\xi$, but its limiting weighted mass

$$s=\int w_0|\xi {|}^2$$

may be strictly less than one. Here $w_0=2b^2$. Tightness of the weights suffices to pass the mean-zero condition, by Cauchy–Schwarz, even though the squared eigenfunction mass might escape.

Choose a quadratic partition ${\chi }_R^2+{\rho }_R^2=1$, with derivatives $O(R^{-1})$. The localization identity is

$${\mathcal{E}}_{w_j}({\xi }_j)={\mathcal{E}}_{w_j}({\chi }_R{\xi }_j)+{\mathcal{E}}_{w_j}({\rho }_R{\xi }_j)-O(R^{-2}).$$

The independently established exterior estimate charges the escaping mass at rate $0.03$. The unperturbed even spectral gap charges the retained mean-zero mass at rate ${\Delta }_{\mathrm{ev}}>0.2$. Passing first along the sequence and then letting $R$ grow yields

$$d_{\ast }\ge {\Delta }_{\mathrm{ev}}s+0.03(1-s)\ge 0.03,$$

a contradiction even when $s=0$. The odd sector is treated separately before combining the two parity sectors and real and imaginary parts. Thus the local spectral calculation and the exterior estimate cover complementary failure modes. This weighted gap is an actual input to the ensemble obstruction, not a consequence of local convergence alone. (OpenAI, 2026)

🏷️ Manuscripts and verification scope

• A Coulomb ground-state density without Kohn–Sham ensemble representation — pinned manuscript and companion source.

No family scope page was present at lean/docs/278.md in the pinned snapshot. This does not establish that all related formalization code is absent. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. A Coulomb ground-state density without Kohn–Sham ensemble representation.