The Free-Particle Hamiltonian

The one-dimensional free Hamiltonian is the nonnegative self-adjoint Laplacian on its maximal Sobolev domain. Its continuum spectrum has no normalizable stationary eigenstates, but its resolvent and evolution are precisely defined. The Free Particle page develops the elementary stationary solutions; this page gives the operator realization and the convergence statements behind those solutions.

Required background. Momentum on the Real Line supplies the Fourier realization. Closed Quadratic Forms distinguishes finite energy from operator-domain membership.

The free operator and its form have different domains

Put $\alpha=\hbar^2/(2m)$, with $m>0$. Define

$$H_0=-\alpha\frac{d^2}{dx^2},\qquad D(H_0)=H^2(\mathbb R).$$

The unitary momentum transform makes $H_0$ multiplication by $p^2/(2m)$ on the domain where that product lies in $L^2(dp)$. This proves self-adjointness and nonnegativity. Smooth compactly supported functions form an operator core.

The associated closed form is

$$q_0[\psi]=\alpha\|\psi'\|_2^2,\qquad Q(H_0)=H^1(\mathbb R).$$

A normalized vector in $H^1$ has finite mean kinetic energy. Membership in $H^2$ requires the fourth momentum moment and permits the Hilbert-space vector $H_0\psi$. These are distinct requirements even for the free particle.

The expression $P^2/(2m)$ is exact because its spectral domain agrees with $H^2$. On an interval this identity can fail for a proposed momentum realization unless the squared-operator boundary conditions match those of the intended kinetic energy.

The resolvent has an exponentially decaying Green function

For $z\notin[0,\infty)$, choose $k$ satisfying $k^2=z/\alpha$ and $\operatorname{Im}k>0$. Then

$$G_z(x)=\frac{i}{2\alpha k}e^{ik|x|}$$

is the Green function of $H_0-z$. Away from zero it solves the homogeneous equation. Its derivative jump is

$$G_z'(0+)-G_z'(0-)=-\frac1\alpha,$$

so distributional differentiation gives $(-\alpha d^2/dx^2-z)G_z=\delta_0$. Thus

$$((H_0-z)^{-1}f)(x)=\int_{\mathbb R}G_z(x-y)f(y)\,dy.$$

Exponential decay makes this a bounded convolution operator for nonreal $z$ and for negative real $z$. The differential equation and Fourier characterization place its output in $H^2$.

The spectral theorem gives the exact norm

$$\|(H_0-z)^{-1}\| =\frac1{\operatorname{dist}(z,[0,\infty))}.$$

For $z=-\alpha\kappa^2$, $\kappa>0$, the kernel reduces to $e^{-\kappa|x|}/(2\alpha\kappa)$. This positive kernel is useful for localized perturbations, including the delta interaction.

The threshold belongs to a continuum, not to a bound state

The spectrum is purely absolutely continuous and equals $[0,\infty)$. For $E>0$, two momentum values $p=\pm\sqrt{2mE}$ have the same energy, giving multiplicity two. The explicit energy-transform Jacobian belongs to Generalized Eigenfunction Expansions.

At $E=0$ the branches meet, but the spectral projection onto that singleton is zero. A constant zero-energy solution is not square integrable on the line. The threshold therefore has no normalized eigenvector and contributes no extra discrete term to a state expansion.

The density of an energy measure can become singular as a function near the threshold while remaining integrable. An integrable continuum density is not a delta mass. Distinguishing those possibilities matters when estimating low-energy resolvents or scattering thresholds.

Unitary evolution preserves missing Sobolev regularity

Free evolution multiplies the momentum amplitude by $e^{-itp^2/(2m\hbar)}$. Its modulus is one, so every weighted momentum norm is preserved. Consequently a vector outside $H^2$ remains outside $H^2$ under free unitary evolution; dispersion does not create the missing operator-domain regularity.

The heat operator $e^{-sH_0}$, with $s>0$ measured in inverse-energy units, behaves differently. The decaying multiplier $e^{-sp^2/(2m)}$ overcomes any polynomial momentum weight and maps $L^2$ into every Sobolev space. It is a contraction, not a unitary time evolution.

For every initial vector, Schrödinger evolution still exists continuously in $L^2$. The strong differential equation $i\hbar\dot\psi=H_0\psi$ requires the operator domain. This qualification does not restrict the existence of the unitary solution.

Exercise: control a momentum cutoff with finite energy

Let a normalized state have finite free mean energy $\mathcal E$. Project onto $|p|\leq R$, leaving the result subnormalized. Bound the squared norm discarded, and explain why the bound remains valid at every free-evolution time.

Solution

On $|p|>R$, $p^2\geq R^2$, hence $\int_{|p|>R}|\widetilde\psi(p)|^2dp\leq2m\mathcal E/R^2$. The momentum cutoff commutes with free evolution, so its discarded norm is constant in time. This is a state-dependent approximation bound from a finite energy assumption, not operator-norm convergence of sharp cutoffs to the identity.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — free quantum evolution and Fourier analysis.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapter 7 — free domains, Green functions, and evolution.