Stability of Matter

Stability of matter is a lower-bound theorem uniform in the number and arrangement of particles. For nonrelativistic Coulomb electrons with fermionic statistics and bounded nuclear charges, the ground energy is bounded below by a constant times the total particle count. The distinction from mere existence of a ground-energy lower bound is crucial: an estimate growing quadratically or faster with particle number does not prevent macroscopic collapse.

Required background. Quadratic form methods supplies the semibounded Hamiltonian realization. Many-body Hilbert spaces fixes the fermionic sector and spin multiplicity.

Coulomb stability has a uniform quantifier

Use units $\hbar^2/(2m_e)=e^2=1$. Let $N$ electrons have $q<\infty$ spin states, with wave functions in

$$\bigwedge^N\!\left(L^2(\mathbb R^3)\otimes\mathbb C^q\right).$$

Fix $K$ distinct nuclear positions $R_k$ and charges $0<Z_k\leq Z_{\max}<\infty$. The electronic Hamiltonian, including nuclear repulsion, is

$$H_{N,K}=\sum_{i=1}^N(-\Delta_i)+V_{\mathrm C},$$ $$V_{\mathrm C} =\sum_{i<j}\frac1{|x_i-x_j|} -\sum_{i,k}\frac{Z_k}{|x_i-R_k|} +\sum_{k<\ell} \frac{Z_kZ_\ell}{|R_k-R_\ell|}.$$

For each finite system its Coulomb quadratic form defines a semibounded self-adjoint operator. Stability of the first kind means that its infimum is finite for the specified particle numbers and nuclear data.

Stability of the second kind is the stronger theorem

$$\inf\sigma(H_{N,K}) \geq-C(q,Z_{\max})(N+K),$$

uniformly in $N,K$, positions, and charges in the stated range. It does not require exact charge neutrality. The constant depends on the fixed physical units and spin degeneracy, but not on system size. Including nuclear kinetic energy preserves the lower bound because that additional term is nonnegative.

The theorem is for the stated nonrelativistic, nonmagnetic Coulomb model. Relativistic kinetic energy and dynamical electromagnetic fields require separate hypotheses and sometimes coupling restrictions.

Pauli exclusion supplies a density lower bound

For a normalized fermionic state with finite kinetic energy, let $\rho(x)$ be its spin-summed density, normalized by $\int\rho=N$. The Lieb–Thirring kinetic inequality gives

$$T=\left\langle\sum_i-\Delta_i\right\rangle \geq a_q\int_{\mathbb R^3}\rho^{5/3}, \qquad a_q=K_{\mathrm{LT}}q^{-2/3}>0.$$

This is a universal lower bound, not an assertion that a local-density approximation is exact. Its origin is the Pauli restriction on the one-particle density matrix: occupation of each spin orbital is at most one. Packing a fixed density into a small region therefore forces occupation of larger momenta.

For a roughly uniform density $N/L^3$ in a region of size $L$, the right side scales as $N^{5/3}/L^2$. This scaling identifies the fermionic pressure, but is not alone a proof of stability: the attraction, electron repulsion, and nuclear repulsion must be handled together.

A proof architecture through electrostatic inequalities

The following argument proves the linear bound using three established inequalities, whose full proofs are given by Lieb and Seiringer. It displays where each input enters.

First, the Lieb–Oxford inequality controls the indirect electron repulsion:

$$\left\langle\sum_{i<j}|x_i-x_j|^{-1}\right\rangle \geq D(\rho,\rho)-C_{\mathrm{LO}}\int\rho^{4/3},$$

where

$$D(\rho,\rho)=\frac12 \iint\frac{\rho(x)\rho(y)}{|x-y|}\,dx\,dy.$$

Second, elementary Young-type absorption gives, for every $r\geq0$,

$$C_{\mathrm{LO}}r^{4/3} \leq\frac{a_q}{2}r^{5/3}+C_qr.$$

Indeed, divide by $r$ when $r>0$ and maximize a quadratic polynomial in $r^{1/3}$. Integrating costs only $C_qN$.

Combining these estimates with the exact nuclear terms yields

$$\langle H_{N,K}\rangle \geq\mathcal E_{\mathrm{TF},a_q/2}(\rho)-C_qN,$$

where $\mathcal E_{\mathrm{TF},a}$ contains $a\int\rho^{5/3}$, the classical density self-repulsion, electron–nuclear attraction, and nuclear repulsion.

Third, the Thomas–Fermi no-binding inequality bounds this functional below by the sum of neutral atomic Thomas–Fermi energies. Atomic scaling gives

$$\mathcal E_{\mathrm{TF},a}(\rho) \geq-C_a\sum_{k=1}^K Z_k^{7/3}.$$

This comparison remains valid when the density has any fixed total mass: allowing arbitrary electronic mass only lowers the atomic reference minima. Since $Z_k\leq Z_{\max}$, the right side is at least $-C_a Z_{\max}^{7/3}K$. Together with the absorbed $-C_qN$ term, this proves the desired linear estimate.

The kinetic inequality, indirect-energy inequality, and Thomas–Fermi no-binding theorem are the substantive imported results in this proof. Discarding positive Coulomb terms before applying the electrostatic comparison would lose the cancellation that makes the bound uniform.

What the theorem does and does not establish

Stability prevents energy per particle from tending to minus infinity. It does not prove that every finite molecule binds, that the system has a spectral gap, or that a unique thermodynamic phase exists. Existence of a thermodynamic free-energy limit also requires control of boundaries and long-range interactions; stability is one necessary structural ingredient.

A bound such as $E_N\geq-CN^3$ proves only first-kind stability at each fixed $N$. It is insufficient for ordinary extensive thermodynamics. Conversely, a linear lower bound allows positive surface energies, fragmentation, and nontrivial phase diagrams.

The fermionic hypothesis matters. Bosonic Coulomb systems do not satisfy this same uniform second-kind conclusion in general. The kinetic density inequality above is the point at which the fermionic proof would fail; bosons can populate one spatial orbital without a Pauli occupation ceiling.

Exercises

Absorb the indirect-energy term. Produce an explicit admissible $C_q$ in the displayed Young-type inequality.

Solution

Set $y=r^{1/3}$. The required inequality is $C_{\mathrm{LO}}y\leq(a_q/2)y^2+C_q$. The maximum of $C_{\mathrm{LO}}y-(a_q/2)y^2$ is $C_{\mathrm{LO}}^2/(2a_q)$. This value is an admissible $C_q$.

Why bound the nuclear charge? Explain why fixed $K$ with arbitrarily large $Z_k$ cannot share the same constant in a linear $N+K$ estimate.

Solution

Already one electron and one nucleus have energy $-Z^2/4$ in these units. Taking $Z\to\infty$ with $N=K=1$ defeats any charge-independent constant. A charge bound is genuine theorem data, not merely a convention for notation.

References

• E. H. Lieb and S. Oxford, “Improved lower bound on the indirect Coulomb energy,” International Journal of Quantum Chemistry 19, 427–439, 1981, doi:10.1002/qua.560190306.
• E. H. Lieb and R. Seiringer, The Stability of Matter in Quantum Mechanics, Cambridge University Press, 2009, Chapters 4–7, publisher.
• E. H. Lieb and W. E. Thirring, “Bound for the kinetic energy of fermions which proves the stability of matter,” Physical Review Letters 35, 687–689, 1975; erratum 35, 1116, 1975, doi:10.1103/PhysRevLett.35.687.