A Particle on the Line

A spinless particle on the line uses the configuration Hilbert space $L^2(\mathbb R,dx)$, but that choice does not yet specify a Hamiltonian or its domain. The physical model also needs observables, dynamics, and any interaction or boundary data. This page organizes those inputs and distinguishes normalization from the regularity needed for momentum or energy.

Required background. The Schrödinger Representation constructs the canonical operators and their domains. Measure Theory Minimum explains equivalence classes and spectral moments.

The configuration measure determines probability normalization

A state vector is an almost-everywhere equivalence class $\psi$ with

$$\int_{\mathbb R}|\psi(x)|^2\,dx=1.$$

For a Borel region $\Delta$, its position probability is $\int_\Delta|\psi(x)|^2dx$. Changing a representative at a point changes no such probability. In particular, normalization alone gives no well-defined number $\psi(x_0)$.

The wavefunction carries units of inverse square root of length when $dx$ has length units. This matters when changing coordinates or discretizing the model: an amplitude and a probability density transform differently.

If $x=f(y)$ is a smooth orientation-preserving diffeomorphism from $\mathbb R$ onto $\mathbb R$, with $f'(y)>0$, the unitary map to $L^2(dy)$ is

$$(U_f\psi)(y)=\sqrt{f'(y)}\,\psi(f(y)).$$

Omitting the square root changes the inner product. The transformed momentum also includes the corresponding Jacobian term; it cannot be assigned the expression $-i\hbar\,d/dy$ merely by renaming a nonlinear coordinate.

Normalization, momentum, and kinetic energy require different domains

The maximal position and momentum domains on the line are

$$D(Q)=\{\psi:x\psi\in L^2\},\qquad D(P)=H^1(\mathbb R).$$

Here $H^1$ means that the weak first derivative belongs to $L^2$. The free Hamiltonian is $H_0=P^2/(2m)$ with operator domain $H^2(\mathbb R)$ and form domain $H^1(\mathbb R)$.

Thus a normalized vector always gives probabilities, but it need not have a momentum operator vector $P\psi$ or a strong Schrödinger derivative $H_0\psi$. A finite mean kinetic energy requires the form domain; membership in $D(H_0)$ is stronger and gives a finite second energy moment.

For example, $\psi(x)=\sqrt a\,e^{-a|x|}$ with $a>0$ is normalized and belongs to $H^1$. Its weak first derivative is square integrable, but its second distributional derivative contains a delta contribution at zero. Hence $\psi\notin H^2$.

This state has finite free mean energy $\hbar^2a^2/(2m)$, although the literal Hilbert vector $H_0\psi$ does not exist. The point-interaction model will turn this same cusp into an eigenvector of a different self-adjoint Hamiltonian.

A length scale gives a unitary dimensionless model

Choose $\ell>0$, put $y=x/\ell$, and define

$$(U_\ell\psi)(y)=\sqrt\ell\,\psi(\ell y).$$

Then

$$U_\ell Q U_\ell^*=\ell Y,\qquad U_\ell P U_\ell^*=\frac{\hbar}{\ell}(-i\partial_y).$$

For a sufficiently regular Schrödinger realization,

$$U_\ell H U_\ell^* =-\frac{\hbar^2}{2m\ell^2}\partial_y^2+V(\ell y),$$

with the domain transported by $U_\ell$. The unitary transformation preserves all probabilities while exposing the dimensionless coupling ratios of the model.

Setting a coefficient to one is therefore a choice of units accompanied by a rescaling of state normalization and domains. It is not permission to compare couplings of different dimensions without restoring the relevant scale.

The potential expression does not determine every realization

For real bounded $V$, multiplication by $V$ is bounded self-adjoint, so $H_0+V$ is self-adjoint on $H^2$ by bounded perturbation theory. Singular or growing potentials require additional arguments, such as quadratic-form methods.

A point interaction, for instance, is specified by matching conditions rather than ordinary multiplication by a function. Removing the point $0$ does not change the measure-space Hilbert space up to canonical identification, but starting from test functions supported away from zero changes the minimal differential operator and its possible extensions.

Similarly, passing from the line to an interval or circle changes the configuration and boundary specification. These Hilbert spaces may be abstractly isomorphic, but their labeled position and momentum observables need not be unitarily equivalent. A Hilbert-space isomorphism alone does not identify quantum models.

Exercise: check the nonlinear-coordinate momentum

For the map $U_f$ above, differentiate $U_f^*\chi$ to determine $U_f P U_f^*$ on transformed smooth compactly supported test functions.

Solution

Since $\psi(f(y))=\chi(y)/\sqrt{f'(y)}$, the chain rule gives

$$(U_f P U_f^*\chi)(y) =-i\hbar\left[ \frac{\chi'(y)}{f'(y)} -\frac{f''(y)}{2f'(y)^2}\chi(y) \right].$$

The domain is the unitary image of $D(P)$, not an independently guessed Sobolev domain. For $f(y)=\ell y$ the Jacobian derivative vanishes and the dimensionless scaling formula is recovered.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — configuration representations and canonical operators.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 7–9 — Sobolev domains and Schrödinger realizations.