Quadratic Forms

A quadratic form assigns an energy-like quantity to vectors without requiring the associated operator to act on them. For a nonnegative Hamiltonian $H$, the form domain is $D(H^{1/2})$, usually larger than $D(H)$. This distinction allows finite-energy states and variational trial functions that need not have a strong Schrödinger derivative.

Helpful background. Unbounded Operators distinguishes first and second moments. Dense Domains explains the topology behind form cores.

Sesquilinear forms and lower bounds

Let $Q\subseteq\mathcal H$ be a dense linear subspace. A symmetric sesquilinear form $q(\phi,\psi)$ is conjugate-linear in its first argument, linear in its second, and satisfies $q(\phi,\psi)=\overline{q(\psi,\phi)}$. Its quadratic value is $q[\psi]=q(\psi,\psi)$, which is real.

Assume a lower bound

$$q[\psi]\geq c\|\psi\|^2.$$

After shifting by $c$, the form norm can be taken as

$$\|\psi\|_q^2=q[\psi]+(1-c)\|\psi\|^2.$$

Here the coefficient $1$ denotes a fixed positive reference scale if physical units are retained. The form is closed if $Q$ is complete in this norm. Equivalent positive shifts give equivalent norms.

A form is closable if it has a closed extension whose domain is obtained by completing the original form domain in the form norm and embedding that completion consistently in $\mathcal H$. For a nonnegative form, the sequence criterion is that $\psi_n\to0$ in Hilbert norm and $q[\psi_n-\psi_m]\to0$ imply $q[\psi_n]\to0$.

Closedness of the form is not closedness of an operator graph. The norms involve different numbers of derivatives in the common differential examples.

Representation by a self-adjoint operator

The representation theorem states that every densely defined closed semibounded symmetric form determines a unique self-adjoint operator $H$ with the same lower bound. Its operator domain is characterized by

$$\begin{aligned} D(H)=\{\psi\in Q:\ &\text{there exists }\eta\in\mathcal H\\ &\text{such that }q(\phi,\psi)=\langle\phi,\eta\rangle \text{ for all }\phi\in Q\}, \end{aligned}$$

and $H\psi=\eta$. Density makes $\eta$ unique. The condition is again continuity in the ambient Hilbert norm, now applied to the form pairing.

For $H\geq0$ the form is

$$Q=D(H^{1/2}), \qquad q(\phi,\psi)=\langle H^{1/2}\phi,H^{1/2}\psi\rangle.$$

A proof strategy shifts to a strictly positive form and solves the weak equation $q_c(\phi,\psi)=\langle\phi,f\rangle$ using the Riesz theorem in the complete form space. The resulting bounded solution operator on $\mathcal H$ is positive and self-adjoint; its inverse, with the appropriate range as domain, constructs the correspondingly shifted Hamiltonian. Completeness is the step that guarantees existence of the weak solution. This is a proof outline; the representation theorem also requires checking density, injectivity, and the inverse domain.

Kinetic energy requires fewer derivatives than the Hamiltonian

On $L^2(\mathbb R)$, set

$$q[\psi]=\frac{\hbar^2}{2m}\int_{\mathbb R}|\psi'(x)|^2\,dx, \qquad Q=H^1(\mathbb R).$$

This form is closed. Its associated operator is $H=-(\hbar^2/2m)d^2/dx^2$ on $H^2(\mathbb R)$.

The function $\psi(x)=e^{-|x|}$ belongs to $H^1(\mathbb R)$, so its kinetic form value is finite. Its first derivative jumps at zero, however, and its second distributional derivative contains a delta function. It is therefore outside $H^2$ and outside $D(H)$.

The expression $\int|\psi'|^2$ is valid for this state, but the Hilbert-space pairing $\langle\psi,H\psi\rangle$ is not literally defined. Writing them as equal without distinguishing the form extension hides the domain issue. Spectral integration still gives a well-defined mean kinetic energy.

Boundary conditions emerge from form domains

On $(0,L)$, the same derivative form on $Q=H_0^1(0,L)$ produces the Dirichlet Laplacian. The form domain imposes zero endpoint values, and the operator domain is $H^2\cap H_0^1$.

If instead $Q=H^1(0,L)$, the representation identity must hold against test vectors with arbitrary endpoint values. Integration by parts forces $\psi'(0)=\psi'(L)=0$, producing the Neumann Laplacian. The derivative action is unchanged; the form domain changes the boundary physics.

On a half-line, adding the boundary term

$$q_\alpha[\psi] =\int_0^\infty|\psi'|^2\,dx+\alpha|\psi(0)|^2, \qquad Q=H^1(0,\infty),$$

with real $\alpha$, produces the Robin condition $\psi'(0)=\alpha\psi(0)$. The trace term is controlled by the $H^1$ norm. A trace inequality with an arbitrarily small derivative coefficient shows semiboundedness even for negative $\alpha$.

These examples motivate Quadratic Form Methods, where forms are used to construct more complicated Hamiltonians.

Exercises

For $He_n=n e_n$ on $\ell^2(\mathbb N)$, compare its form and operator domains and find a vector in the difference.

Solution

The form domain requires $\sum n|c_n|^2<\infty$; the operator domain requires $\sum n^2|c_n|^2<\infty$. Taking $c_n=n^{-5/4}$ gives a convergent form sum $\sum n^{-3/2}$ but a divergent operator sum $\sum n^{-1/2}$. After normalization this is a finite-mean-energy state outside $D(H)$.

Why does minimizing the kinetic form on normalized $H^1(0,L)$ functions give a different lowest energy from minimizing on normalized $H_0^1(0,L)$ functions?

Solution

The first space contains constants, so the minimum is zero and corresponds to the Neumann ground state. Constants are excluded from $H_0^1$; the Poincaré inequality gives minimum $\hbar^2\pi^2/(2mL^2)$, attained by the first Dirichlet sine mode. A variational problem includes its admissible function space.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1, Chapter 10 — semibounded forms and self-adjoint operators.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — quadratic forms and the Friedrichs extension.