Sharp Lieb–Thirring bounds: one state, matrix channels, and equality

Source and verification

This article explains family 262 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.

🏷️ Bound states and the cost of a potential

A one-dimensional Schrödinger operator describes the competition between kinetic energy and an attractive potential. For a scalar potential $W\ge 0$, its formal expression is $H_W=-d^2/d{x}^2-W$. The negative eigenvalues describe bound states. A deeper or wider potential can bind more states, and the Lieb–Thirring problem asks how their total binding energy is controlled by an integral of $W$.

The matrix version acts on $L^2(\mathbb{R};{\mathbb{C}}^m)$, where $m$ is a finite number of internal components. The potential $W(x)$ is Hermitian positive semidefinite almost everywhere. Fix

$$\frac{1}{2}<\gamma <\frac{3}{2},p=\gamma +\frac{1}{2},$$

and assume ${\int }_{\mathbb{R}}\mathrm{tr}(W^p)<\infty$. The trace is the ordinary unnormalized matrix trace. Powers of a positive matrix are defined by applying the scalar power to its eigenvalues.

For rough potentials, define the operator through its quadratic form on $H^1(\mathbb{R};{\mathbb{C}}^m)$:

$$h_W[\psi ]=\int \Vert {\psi }^{\prime }{\Vert }^2-\int ⟨\psi ,W\psi ⟩.$$

This form is closed and bounded below. One useful estimate is

$$\int ⟨\psi ,W\psi ⟩\le \epsilon \Vert {\psi }^{\prime }{\Vert }_2^2+C_{\epsilon }\Vert \psi {\Vert }_2^2.$$

It follows from Hölder, the one-dimensional Sobolev estimate, and Young’s inequality, since $p>1$. Truncating the potential in both space and size also shows compactness of multiplication by $W^{1/2}$ from $H^1$ to $L^2$. Therefore the negative spectrum consists of isolated eigenvalues of finite multiplicity, with possible accumulation only at zero. Smoothness or boundedness of $W$ is not assumed.

The spectral moment is

$$\mathrm{Tr}(H_W{)}_{-}^{\gamma }=\sum _{{\lambda }_j(H_W)<0}|{\lambda }_j(H_W){|}^{\gamma },$$

counting multiplicity. Before a bound is established, this sum may be infinite. The capital trace refers to the operator on the function space; the lowercase trace refers to a finite matrix.

🏷️ The family theorem and its three parts

Sharp bound, as claimed in the family

For every finite $m$ and every measurable positive semidefinite Hermitian potential in the class above,

$$\mathrm{Tr}(H_W{)}_{-}^{\gamma }\le C_{\gamma }{\int }_{\mathbb{R}}\mathrm{tr}(W^p).$$

The optimal constant is

$$\begin{array}{c}{C}_{\gamma }={\left(\frac{\gamma -1/2}{\gamma +1/2}\right)}^{\gamma -1/2}\\ \times \frac{\Gamma (\gamma +1)}{\sqrt{\pi }\Gamma (\gamma +3/2)}.\end{array}$$

The same constant works in every finite matrix dimension. (OpenAI, 2026)

Here $\Gamma$ is Euler’s gamma function, satisfying $\Gamma (z+1)=z\Gamma (z)$. The conclusion concerns the full spectral moment, not a fixed number of eigenvalues. In particular, it proves finiteness of that moment under the stated potential assumption.

Manuscript	Main contribution	Scope
September scalar paper	Sharp constant throughout the open exponent interval	Scalar potentials; gives an attaining profile
October matrix paper	The same sharp constant for arbitrary finite matrix size	Potentials need not commute at different positions
October equality paper	Classification of every equality case	One fixed internal basis; independent scalar profiles

The scalar paper uses matrices to organize a finite family of scalar eigenfunctions. That does not make its potential matrix-valued. The October inequality paper supplies the additional comparison needed for genuine matrix potentials. The equality paper then strengthens the argument to retain the defects that must vanish at an optimizer. (OpenAI, 2026a) (OpenAI, 2026b) (OpenAI, 2026a)

The exponent range is open. The family does not extend the equality classification below to $\gamma =1/2$ or $\gamma =3/2$; the upper endpoint permits different scalar equality configurations.

🏷️ Scaling and the one-state profile

The integral exponent is dictated by scaling. If $W_a(x)=a^{2}W(ax)$, a unitary rescaling multiplies each eigenvalue by $a^2$, while

$$\int \mathrm{tr}(W_a^p)=a^{2p-1}\int \mathrm{tr}(W^p).$$

Matching the spectral factor $a^{2\gamma }$ requires $p=\gamma +1/2$.

Set $r=(\gamma -1/2{)}^{-1}>1$. The scalar profile at scale $a>0$ and center $x_0$ is

$$\begin{array}{c}V(x)\\ =(r+1)a^2{\mathrm{sech}}^2(ra(x-x_0)).\end{array}$$

Here $\mathrm{sech}t=1/\cosh t$. Its bound-state function is proportional to $\psi (x)={\mathrm{sech}}^{1/r}(ra(x-x_0))$. Indeed,

$$\begin{array}{c}\frac{{\psi }^{\prime }}{\psi }=-a\tanh (ra(x-x_0)),\\ \frac{{\psi }^{\prime \prime }}{\psi }=a^2-V(x),\end{array}$$

so $(-d^2/d{x}^2-V)\psi =-a^2\psi$. The function decays exponentially and lies in $L^2$.

It is useful to check that this is the only negative eigenvalue. Put $w=a\tanh (ra(x-x_0))$ and $Q={\partial }_x+w$. Then

$$\begin{array}{c}{Q}^{\ast }Q=H_V+a^2,\\ Q{Q}^{\ast }=-{\partial }_x^2+a^2\\ +(r-1)a^2{\mathrm{sech}}^2(ra(x-x_0))\\ \ge a^2.\end{array}$$

The first factorization excludes eigenvalues below $-a^2$. Its kernel is the one-dimensional span of $\psi$. If another eigenvalue $\lambda$ lay between $-a^2$ and zero, applying $Q$ to its eigenfunction would give a nonzero eigenvector of $Q{Q}^{\ast }$ with eigenvalue $\lambda +a^2<a^2$, contradicting the second bound. The strict range $r>1$ supplies the favorable partner potential.

For the potential cost, substitute $y=ra(x-x_0)$ and then $s=\tanh y$:

$$\begin{array}{c}\int V^p=\frac{(r+1{)}^p}{r}{a}^{2\gamma }\\ \times {\int }_{-1}^1(1-s^2{)}^{p-1}ds,\\ {\int }_{-1}^1(1-s^2{)}^{p-1}ds\\ =\frac{\sqrt{\pi }\Gamma (p)}{\Gamma (p+1/2)}.\end{array}$$

Since the spectral moment is exactly $a^{2\gamma }$, the ratio of moment to cost equals the displayed $C_{\gamma }$. This proves that no smaller constant can work. For example, $\gamma =1$ gives $r=2$, $V(x)=3{\mathrm{sech}}^2(2x)$ at unit scale, and $C_1=4/(3\sqrt{3}\pi )$.

The remaining difficulty is the universal upper bound: showing that many bound states, or noncommuting internal channels, cannot improve this ratio. An explicit optimizer alone does not settle that direction. (OpenAI, 2026b) (OpenAI, 2026a)

🏷️ A finite-interval action inequality

Write $\sigma =\gamma -1/2$, so $p=1+\sigma$, $r=1/\sigma$, and $q=r+1$ is conjugate to $p$. Start with $N$ orthonormal real vector-valued functions on a bounded interval $I$, vanishing at its endpoints. Arrange them as the rows of $U(x)\in {\mathbb{R}}^{N\times m}$; orthonormality means ${\int }_{I}U{U}^{\mathsf{T}}=I_N$.

Let $K=\mathrm{diag}(k_1,\dots ,k_N)$ with $k_i>0$. For $A=CU$, where $C$ is lower triangular, put $\rho =A{A}^{\mathsf{T}}$ and consider the action

$$\begin{array}{c}{\mathcal{E}}_K(A)\\ ={\int }_I\left(\Vert A^{\prime }{\Vert }_{\mathrm{HS}}^2+\mathrm{tr}(K^2\rho )-\mathrm{tr}({\rho }^q)\right).\end{array}$$

The Hilbert–Schmidt norm satisfies $\Vert A{\Vert }_{\mathrm{HS}}^2=\mathrm{tr}(A^{\mathsf{T}}A)$. The action theorem claims

$$\begin{array}{c}\underset{C\text{ lower triangular}}{\sup }{\mathcal{E}}_K(CU)\\ \ge \sum _{i=1}^N{\int }_{-k_i}^{k_i}(k_i^2-s^2{)}^{\sigma }ds.\end{array}$$

Each integral equals $b_{\sigma }{k}_i^{2\gamma }$, where $b_{\sigma }=\mathrm{B}(1/2,1+\sigma )$ and $\mathrm{B}$ is Euler’s beta function. The action lower bound thus retains every individual eigenvalue scale. (OpenAI, 2026c)

To connect this to the spectral problem, order trial energies as $k_1\ge \cdots \ge k_N$. Lower triangularity makes row $i$ a combination only of the first $i$ trial functions. For an orthonormal family with diagonal form matrix $-K^2$,

$$\begin{array}{c}{h}_W[a_i]+k_i^2\Vert a_i{\Vert }_2^2\\ =\sum _{j\le i}{C}_{ij}^2(k_i^2-k_j^2)\le 0.\end{array}$$

The sign would not follow for arbitrary row mixing. Summing and using the common nonzero eigenvalues of $A{A}^{\mathsf{T}}$ and $A^{\mathsf{T}}A$ gives, with $Z=A^{\mathsf{T}}A$,

$${\mathcal{E}}_K(A)\le {\int }_I(\mathrm{tr}(WZ)-\mathrm{tr}(Z^q)).$$

Scalar maximization yields

$$\begin{array}{c}yz-z^q\le D_{\sigma }{y}^p(y,z\ge 0),\\ D_{\sigma }=\frac{{\sigma }^{\sigma }}{(1+\sigma {)}^{1+\sigma }}.\end{array}$$

The same constant works in the matrix trace inequality even if $W$ and $Z$ do not commute. Diagonalize $Z$ with eigenvectors $v_j$ and eigenvalues $z_j$, and put $y_j=⟨v_j,W{v}_j⟩$. Convexity gives $y_j^p\le ⟨v_j,W^p{v}_j⟩$. Summing the scalar inequality therefore bounds the action by $D_{\sigma }\int \mathrm{tr}(W^p)$.

Combining the lower and upper action bounds gives the spectral constant $D_{\sigma }/b_{\sigma }=C_{\gamma }$. Smooth compactly supported approximations to each finite eigenfunction family are corrected by their Gram matrices and their form matrices. Passing to the approximation limit proves every finite partial-sum bound; taking the supremum over partial sums gives the full negative moment. The complex Hermitian case follows by realification: both the spectral moment and the potential trace integral double, so the ratio is unchanged.

🏷️ The noncommuting matrix comparison

The action lower bound is the main analytic input. The source constructs a symmetric path $B(x)$ joining $K$ to $-K$, with a regularized matrix field imposing a density constraint. A penalty enforces the constraint, and continuation supplies both the path and the lower triangular coefficients. The penalty limit is taken at fixed regularization, then the regularization is removed. This construction is not supplied by the elementary Young inequality above.

For scalar potentials, the density $A{A}^{\mathsf{T}}$ has rank at most one. Matrix potentials require a trace comparison for arbitrary rank. The October paper isolates it as follows. For Hermitian matrices $B,V$, assume

$$\begin{array}{c}{X}_s=(sI-B{)}^2+V\ge bI\\ (s\in \mathbb{R})\end{array}$$

for some $b>0$. Define a weighted integrated resolvent difference

$$\begin{array}{c}R(t)=\frac{1}{\pi }{\int }_{\mathbb{R}}(X_s+tI{)}^{-1}ds,\\ N\\ =c_{\sigma }{\int }_0^{\infty }{t}^{\sigma -1/2}(t^{-1/2}I-R(t))dt.\end{array}$$

The positive normalization $c_{\sigma }$ is chosen so that in the scalar case $V=v>0$, one gets $N=v^{\sigma }$. The claim is

$$N\ge 0⟹\mathrm{tr}(NV)\ge \mathrm{tr}(N^q).$$

The perturbation $V$ itself may be indefinite: positivity is required of the full symbol $X_s$, and $N\ge 0$ is an additional hypothesis. Replacing either condition by an informal positivity assumption would alter the statement. (OpenAI, 2026c)

The proof represents the integrated resolvent using decaying solutions of a constant-coefficient differential equation on two half-lines. Their boundary matrices satisfy Riccati identities with commutator terms. A logarithmic-determinant identity expresses the comparison deficit through nonnegative matrix terms. These terms control the noncommutative error while preserving the scalar constant. The full identity and global connecting-path argument remain source inputs in these notes.

🏷️ Equality and fixed internal channels

The equality companion claims a complete classification. Equality holds precisely when, for some constant unitary matrix $U$,

$$\begin{array}{c}W(x)=UD(x)U^{\ast },\\ D(x)\\ =\mathrm{diag}(w_1(x),\dots ,w_k(x))\oplus 0_{m-k},\\ w_j(x)\\ =(r+1)a_j^2{\mathrm{sech}}^2(r{a}_j(x-x_j)),\end{array}$$

almost everywhere, with $0\le k\le m$, $a_j>0$, and arbitrary real centers $x_j$. Scales and centers are independent; repeated profiles are permitted. The case $k=0$ is the zero potential. Every nonzero channel contributes exactly one negative eigenvalue $-a_j^2$, so every extremizer has at most $m$ negative eigenvalues counting multiplicity. (OpenAI, 2026a)

The fixed basis is stronger than pointwise diagonalizability. A position-dependent unitary change of basis also changes the kinetic energy through derivatives of that basis; it cannot simply split the problem into scalar copies.

The proof retains quantitative defects in the action estimate. Besides $\Vert A^{\prime }-BA{\Vert }_{\mathrm{HS}}^2$, it controls failure of

$$\begin{array}{c}[B,K^2]=0,\\ K^2-B^2=(A{A}^{\mathsf{T}}{)}^r.\end{array}$$

The coefficient is independent of the number of trial rows and of the smallest binding scale. This is essential because an equality potential is not initially known to have finitely many negative eigenvalues. The number of columns remains $m$ as the trial list grows; the source uses this rank bound and the small tail binding energies to obtain compactness and exact limiting relations.

Group the limiting rows by their common binding scale $\alpha$. Their matrices $G_{\alpha }$ satisfy $G_{\alpha }^{\prime }=B_{\alpha }{G}_{\alpha }$ with uniformly bounded $B_{\alpha }$. For a fixed vector $v$,

$$\Vert (G_{\alpha }v{)}^{\prime }\Vert \le \kappa \Vert G_{\alpha }v\Vert .$$

Gronwall’s inequality implies that if $G_{\alpha }v$ vanishes at one point, it vanishes everywhere. Thus its kernel, and hence its row space in the physical internal space, is constant in position. Different energy blocks have orthogonal row spaces, so their total dimension is at most $m$.

After the corresponding range spaces are also fixed, the source obtains a finite matrix Riccati equation

$$B_0^{\prime }=r(B_0^2-{\alpha }^{2}I),|B_0|<\alpha I.$$

Diagonalize $B_0$ at one point. The differential equation is polynomial in $B_0$, and uniqueness preserves that basis. Each diagonal entry is $-\alpha \tanh (r\alpha (x-x_j))$, producing the displayed scalar profiles. A separate realification argument recovers a constant complex unitary basis. Conversely, the one-state factorization and integral computation above verify equality for every listed finite direct sum.

🏷️ Key steps and proof outlines

The scalar paper removes a noncommuting average by a trace argument

Even the scalar-potential action proof contains noncommuting auxiliary matrices. In its rank-one comparison, write $M(B)=a{a}^T$, $m=|a{|}^2$, and $D=K^2-B^2$. The regularized tangent inequality produces an averaged matrix $\mathcal{P}(D)$ rather than $D$ itself. In an eigenbasis of $B$, the averaging multiplier is

$$P_{ij}={\int }_0^1{\left(1+\frac{\tau (1-\tau )(z_i-z_j{)}^2}{d+\delta }\right)}^{\sigma -1}d\tau ,$$

where $\mu (d)=m$ for the scalar regularized field $\mu$.

Proof outline. A gamma-integral representation expresses this multiplier as a mixture of Gaussian characteristic functions. Thus $\mathcal{P}(Z)=\int e^{itB}Z{e}^{-itB}d\nu (t)$ for an even probability measure $\nu$. For any increasing scalar function $f$ and Hermitian $X,Y$,

$$\mathrm{tr}((f(X)-f(Y))(X-Y))\ge 0;$$

expanding in their two eigenbases gives a sum of scalar monotonicity terms with nonnegative overlap weights. Apply this to a matrix and its conjugate by $e^{itB}$, integrate the field parameter, then average in $t$. The result is $\mathrm{tr}(MD)\ge \mathrm{tr}(M\mathcal{P}(D))$. Pairing the tangent inequality with rank-one $M$, where $M^2=mM$, now gives $a^{T}Da\ge dm\ge m^{1+1/\sigma }$. The rank-one relation is essential to this cancellation; the matrix-potential companion needs the stronger mechanism below. (OpenAI, 2026b)

Half-line boundary maps expose the full matrix deficit

Use $R(t),N,V,B$ from the integrated comparison above and set $\beta =\sigma -\frac{1}{2}$. For the constant-coefficient equation $(-({\partial }_x-iB{)}^2+V+tI)\psi =0$, let the decaying right and left solutions with boundary value $v$ have covariant derivatives $-Sv$ and $Tv$. Coercivity of the full symbol gives their existence and uniqueness.

Gluing the two solutions makes a point source $(S+T)v$. Fourier inversion at zero therefore gives $(S+T)/2=R(t{)}^{-1}$. Put $L=(S+T)/2$ and $Z=(S-T)/2$. Their Riccati identities imply

$$V+tI=L^2+Z^2+i[B,Z],LZ+ZL+i[B,L]=0.$$

Multiplying the second relation by $R=L^{-1}$ and taking traces turns the commutator in the first into a signed square: $\mathrm{tr}(R(V+tI))=\mathrm{tr}L-\mathrm{tr}(R{Z}^2)$.

A logarithmic-determinant integral and integration by parts in $t$ then give the exact deficit formula

$$\begin{array}{rl}\mathrm{tr}(NV)-\mathrm{tr}(N^q)=& c_{\sigma }{\int }_0^{\infty }{t}^{\beta }[\mathrm{tr}(R{Z}^2)+q\mathcal{D}(t)]dt,\\ \mathcal{D}(t)=& {\Vert R^{-1/2}-R^{1/2}(tI+N^{1/\sigma }{)}^{1/2}\Vert }_{\mathrm{HS}}^2.\end{array}$$

Both terms are nonnegative. In particular, $\mathrm{tr}(R{Z}^2)=\mathrm{tr}(ZRZ)\ge 0$. The second square works without commuting $R$ and $N$: cyclicity of trace handles the quadratic term, and adjacent inverse square roots cancel in the cross term. This identifies where noncommutativity is paid for while keeping the scalar constant. (OpenAI, 2026c)

Fixing the range completes the equality argument

The constant kernel argument above fixes the physical channel, but one must also fix the range in eigenfunction-index space. Restrict a nonzero energy block to its constant physical row space, obtaining a full-column-rank matrix $E(x)$. Its range projection is

$$P(x)=E(E^{T}E{)}^{-1}{E}^T.$$

The limiting equality relations say that $E^{\prime }=B_{\alpha }E$ and that $B_{\alpha }$ preserves this range. Differentiation gives

$$P^{\prime }=(I-P)E^{\prime }(E^{T}E{)}^{-1}{E}^T+E(E^{T}E{)}^{-1}(E^{\prime }{)}^T(I-P)=0.$$

Thus both spaces admit constant bases. In those bases the block is an invertible square matrix $D$, and $B_0=D^{\prime }{D}^{-1}$ is uniquely defined. Substituting the row eigenfunction equation and the equality relation for the potential gives $B_0^{\prime }=r(B_0^2-{\alpha }^{2}I)$. This justifies the constant-basis Riccati reduction and excludes an unaccounted position-dependent channel rotation. It also avoids choosing the earlier auxiliary matrix measurably at every point: the genuine $B_0$ is recovered from the differentiable block itself. (OpenAI, 2026a)

🏷️ Connections and verification scope

Anderson spectral types explains spectral measures and the distinction between spectral sets and eigenfunction decompositions. The present problem concerns a decaying attractive potential and the sum of its discrete negative energies, rather than an iid potential spread throughout the lattice. Complete Crouzeix inequalities illustrates another setting where a scalar sharp constant must survive finite matrix amplification, although its numerical-range mechanism is different.

The source review covered the three theorem statements, the action and trace-comparison interfaces, the spectral deduction, selected equality compactness and fixed-channel arguments, and the explicit optimizer calculation. The full regularized field, continuation, integrated trace identity, and compactness proofs were not independently verified. The factorization, beta-integral normalization, scaling, and triangular spectral sign are elementary deductions checked here; no numerical experiment is presented as a certificate for the universal upper bound.

🏷️ Manuscripts and verification scope

• Equality cases in the sharp one-dimensional matrix Lieb–Thirring inequality — pinned manuscript and companion source.
• Sharp one-dimensional Lieb–Thirring inequalities for matrix potentials — pinned manuscript and companion source.
• Sharp one-dimensional Lieb–Thirring constants — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. Only the scalar September manuscript is listed. The selected statement covers the full negative-eigenvalue moment for every nonnegative potential in the stated Lebesgue class and every exponent strictly between one half and three halves, with the optimal one-state constant and its explicit attaining profile. No finite spectral cutoff is imposed. The two matrix-potential companions and their full equality classification are not listed. 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 2026a. Equality cases in the sharp one-dimensional matrix Lieb–Thirring inequality.

🐻  OpenAI 2026b. Sharp one-dimensional Lieb–Thirring constants.

🐻  OpenAI 2026c. Sharp one-dimensional Lieb–Thirring inequalities for matrix potentials.