Domain Pitfalls Index

Most domain errors occur when a valid first operation is followed by an unjustified second one. A bounded projection can act on every vector while sending a finite-energy state outside the Hamiltonian’s form domain. A derivative can exist on a dense set without extending to every state. This index identifies the missing condition and links its canonical treatment.

Missing assumptions in common manipulations

Tempting step	Condition that must be checked	Full explanation
Apply $AB$ whenever $A\psi$ and $B\psi$ exist	$\psi\in D(B)$ and $B\psi\in D(A)$	Unbounded operators
Replace a form commutator by $\langle\psi,[A,B]\psi\rangle$	Both product domains, beyond the first-action domains	Commutation relations on domains
Identify a formal adjoint with the Hilbert adjoint	The maximal pairing domain and its boundary conditions	Adjoints of unbounded operators
Extend an identity from a dense set	A graph-norm core or an appropriate bounded-continuity argument	Dense domains
Differentiate $e^{-itH/\hbar}\psi$ in Hilbert norm	$\psi\in D(H)$	Schrödinger equation in Hilbert space
Infer $H\psi$ from a finite energy mean	A second spectral moment, not merely an integrable first moment	Quadratic forms
Commute an observable through every time evolution	Spectral commutation or a justified invariant-domain theorem	Conserved quantities
Cancel the trace of an unbounded dissipator term by term	Trace-class products and the actual generator domain, plus conservativity	Bounded and unbounded generators

The repair is usually local: write the input domain, the intermediate output space, and the next operator’s domain. If a needed inclusion fails, one may still have a meaningful quadratic form, weak equation, or mild solution. Those are different statements and should be named explicitly.

Bounded operations need not preserve finite energy

On $L^2(\mathbb R)$, let $P=-i\hbar\,d/dx$ with self-adjoint domain $H^1(\mathbb R)$, and let

$$(B\psi)(x)=\mathbf1_{(0,\infty)}(x)\psi(x).$$

$B$ is a bounded orthogonal projection and is defined on every Hilbert vector. Yet it does not preserve $D(P)$. For a smooth rapidly decaying $\psi$ with $\psi(0)\ne0$, the weak derivative is

$$\frac{d}{dx}(B\psi) =\mathbf1_{(0,\infty)}\psi'+\psi(0)\delta_0.$$

The delta term is not an $L^2$ function, so $B\psi\notin H^1(\mathbb R)$. Therefore $PB\psi$ is not a Hilbert-space operator action, even though $P\psi$, $B\psi$, and $BP\psi$ all exist. This gives a direct counterexample to the first tempting step in the table.

The free kinetic Hamiltonian $H=P^2/(2m)$ has form domain $H^1(\mathbb R)$. Its mean kinetic energy is finite exactly when the weak derivative is square integrable. Thus the normalized state after an ideal selective Lüders operation,

$$\psi_{\mathrm{out}}= \frac{B\psi}{\|B\psi\|},$$

has infinite kinetic-energy form value for this input. The outcome probability $\|B\psi\|^2$ is nevertheless perfectly finite and the postselected state is a normalized Hilbert vector.

The conclusion does not invalidate the PVM or its probabilities. It says that an ideal infinitely sharp spatial cutoff can leave the finite-energy class. The projective-measurement article supplies the measurement rule; the domain test determines its energy consequence.

A smooth edge exposes the singular limit

Let $\chi$ be a smooth function with $0\le\chi\le1$, equal to zero for $x\le-1$ and one for $x\ge1$. Replace the projector branch by the bounded Kraus operator

$$(K_\varepsilon\psi)(x)=\chi(x/\varepsilon)\psi(x),\qquad \varepsilon>0.$$

For each fixed $\varepsilon$, this preserves $H^1$ because the multiplier and its derivative are bounded. The output derivative contains $\varepsilon^{-1}\chi'(x/\varepsilon)\psi(x)$. For smooth $\psi$,

$$\left\|\varepsilon^{-1}\chi'(x/\varepsilon)\psi(x)\right\|_2^2 =\frac1\varepsilon \int|\chi'(u)|^2|\psi(\varepsilon u)|^2\,du.$$

If $\psi(0)\ne0$, this grows as $|\psi(0)|^2\|\chi'\|_2^2/\varepsilon$. The remaining derivative contribution is bounded and its cross term is at most lower order, so the kinetic energy diverges as the edge sharpens. Meanwhile $K_\varepsilon\psi\to B\psi$ in $L^2$ by dominated convergence.

This is another useful distinction: Hilbert-norm convergence of states does not imply convergence of an unbounded energy expectation. To pass the energy form to a limit, one needs control in the form norm or an appropriate uniform integrability estimate.

Exercises

When does the sharp edge preserve the domain? For $\psi\in H^1(\mathbb R)$, identify the additional condition that makes $B\psi\in H^1(\mathbb R)$.

Solution

The one-dimensional Sobolev trace $\psi(0)$ is well defined. The half-line pieces already have square-integrable derivatives; gluing them to zero on the negative half-line gives an $H^1$ function exactly when their traces agree, namely $\psi(0)=0$. This removes the delta term.

Completing the smooth operation. Give a second Kraus operator that makes the smooth branch part of a two-outcome instrument.

Solution

Take the bounded multiplication operator $L_\varepsilon=\sqrt{I-K_\varepsilon^\dagger K_\varepsilon}$. Then $K_\varepsilon^\dagger K_\varepsilon+L_\varepsilon^\dagger L_\varepsilon=I$, so the two branches define a trace-preserving total operation. This completeness identity does not by itself provide a uniform energy bound for either branch as $\varepsilon\downarrow0$.

References

• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Momentum domains, forms, and distributional derivatives.
• Teschl, Gerald. Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed. Graduate Studies in Mathematics 157, American Mathematical Society, 2014. Author’s edition. Sobolev domains and Schrödinger quadratic forms.