Infinite-Dimensional Caveats

Infinite-dimensional quantum mechanics retains the Hilbert-space probability rules, but loses several automatic properties of finite matrices. Bounded sets need not be norm compact, spectra need not consist of eigenvalues, weak limits can lose normalization, and accurate state vectors can have inaccurate energy expectations. The examples below identify the failed implication in each case and the stronger hypothesis that repairs it.

Required background. Hilbert Space as State Space supplies orthonormal expansions and state convergence. Expectation Functionals separates bounded-observable values from moment and domain conditions.

Weak convergence can lose normalization

Let $\{e_n\}_{n\geq1}$ be an orthonormal basis of $\ell^2(\mathbb N)$. For every fixed $\phi\in\ell^2$, its coordinates tend to zero, and therefore $\langle\phi,e_n\rangle\to0$. Thus $e_n$ converges weakly to zero, although $\|e_n\|=1$ for every $n$.

The sequence does not converge in norm, and no subsequence does: distinct elements remain a distance $\sqrt2$ apart. This proves that the infinite-dimensional unit sphere is not norm compact. A sequence of normalized trial states can escape to successively different modes without approaching any normalized state.

For the rank-one density operators $\rho_n=P_{e_n}$, every fixed matrix element tends to zero:

$$\langle\phi,\rho_n\psi\rangle =\langle\phi,e_n\rangle\langle e_n,\psi\rangle \longrightarrow0.$$

Yet $\operatorname{Tr}\rho_n=1$ throughout. Weak-operator convergence does not preserve the trace. It is also weaker than convergence of expectations against every bounded observable: the identity always has expectation one. No trace-norm-convergent subsequence exists, because distinct $\rho_n$ have trace distance one.

A useful repair for vectors is weak convergence plus convergence of norms. If $\psi_n\rightharpoonup\psi$ and $\|\psi_n\|\to\|\psi\|$, then expanding $\|\psi_n-\psi\|^2$ shows norm convergence.

Small state error need not mean small energy error

Use a basis $\{e_k\}_{k\geq0}$ with $He_k=E_0k e_k$, where $E_0>0$. For $n\geq2$, define

$$\psi_n=\sqrt{1-1/n}\,e_0+\frac1{\sqrt n}e_{n^2}.$$

These are unit vectors in $D(H)$, and

$$\|\psi_n-e_0\|^2 =2-2\sqrt{1-1/n}\longrightarrow0.$$

Their pure-state trace distance from $P_{e_0}$ is $1/\sqrt n$, so every bounded measurement probability converges. Nevertheless

$$\langle\psi_n,H\psi_n\rangle=E_0n\longrightarrow\infty,$$

while the limiting state has zero energy. A vanishing probability tail carries an increasing energy. Thus even trace-norm convergence of states is insufficient for convergence of an unbounded expectation.

One sufficient repair is uniform integrability of the energy distributions. For example, suppose $H\geq0$, $\rho_n\to\rho$ in trace norm, and for some $\delta>0$ all states, including $\rho$, satisfy

$$\operatorname{Tr}(\rho_n H^{1+\delta})\leq M, \qquad \operatorname{Tr}(\rho H^{1+\delta})\leq M.$$

Interpret these as spectral moments. The bounded spectral cutoff $H_R=H\mathbf1_{[0,R]}(H)$ has norm at most $R$, while each omitted first-moment tail is at most $MR^{-\delta}$. Hence

$$|\mathbb E_{\rho_n}[H]-\mathbb E_\rho[H]| \leq R\|\rho_n-\rho\|_1+2MR^{-\delta}.$$

First make the tails small by choosing $R$, then take $n$ large. This separates state convergence from the additional energy-tail control it needs.

Spectral values need not have eigenvectors

Consider the bounded self-adjoint operator $Ae_n=n^{-1}e_n$ on $\ell^2(\mathbb N)$. It is injective, so zero is not an eigenvalue. But its inverse on its range is unbounded because $\|Ae_n\|=1/n$ while $\|e_n\|=1$.

Zero belongs to the spectrum. The range is dense, since it contains every finite-support sequence, but is not all of $\ell^2$: the sequence $y_n=1/n$ belongs to $\ell^2$, whereas its candidate preimage has every coordinate equal to one. Under the usual disjoint point/continuous/residual classification, zero is a continuous-spectrum point.

Finite truncations of $A$ are invertible matrices. Their inverse norms grow with the truncation size, revealing the missing uniform bound. A finite matrix calculation therefore cannot establish bounded invertibility of the limiting operator merely by finding no exact zero eigenvalue.

For an observable with no eigenvectors at all, the position operator on $L^2(\mathbb R)$ has spectrum $\mathbb R$ and zero singleton spectral projections. The Unbounded Spectral Theorem supplies the appropriate measure-based replacement for matrix diagonalization.

Operator formulas must carry their domains

A densely defined symmetric operator need not be self-adjoint. On an interval, the momentum expression with two zero endpoint values is symmetric, but its adjoint has a larger domain. The same differential expression with a suitable quasiperiodic endpoint condition is self-adjoint. The Self-Adjoint Operators page gives the boundary-form calculation.

Products also carry extra information:

$$D(AB)=\{\psi\in D(B):B\psi\in D(A)\}.$$

Knowing that both $A\psi$ and $B\psi$ exist does not guarantee that either product exists. A formal commutator identity must specify a common product domain. An identity on a dense subspace also need not imply commutation of the associated spectral measures.

Likewise, cyclicity of the trace requires hypotheses. Products of a trace-class operator and bounded operators obey the familiar trace identities; moving arbitrary unbounded factors around a formal trace can create undefined products. The correct repair is to state which products are trace class and justify the relevant limiting operation.

Exercise: a limit that cannot be a density operator

Show that the equal mixture $\sigma_N=N^{-1}\sum_{n=1}^{N}P_{e_n}$ converges to zero in operator norm but cannot converge to zero in trace norm. What does this say about normalization?

Solution

The nonzero eigenvalues of $\sigma_N$ are all $1/N$, so $\|\sigma_N\|=1/N$. Positivity gives $\|\sigma_N\|_1=\operatorname{Tr}\sigma_N=1$. Operator-norm convergence is therefore too weak to preserve trace-one normalization in an infinite-dimensional space. The rank grows with $N$, exactly compensating the decreasing eigenvalues in the trace.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — spectral and domain qualifications beyond finite matrices.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapters 1–3 and Section 6.3 — weak convergence, spectra, self-adjointness, and trace ideals.