Heisenberg Dynamics

Heisenberg conjugation is defined more generally than its formal differential equation. A self-adjoint Hamiltonian gives a well-defined evolution of every bounded observable, but that evolution need not be continuous in operator norm, let alone differentiable there. This page supplies the topology and domain qualifications; the physical picture comparison belongs to Heisenberg Picture.

Helpful background. Strongly Continuous Unitary Groups fixes the continuity of state evolution. Bounded Operators distinguishes operator norms from vector norms. Adjoints of Unbounded Operators explains transported domains.

Conjugation is strongly continuous on bounded observables

Let $H$ be self-adjoint and $U(t)=e^{-itH/\hbar}$. For $A\in\mathcal B(\mathcal H)$ define

$$\alpha_t(A)=U(t)^*AU(t).$$

Each $\alpha_t$ preserves products, adjoints, the identity, and the operator norm. It forms a group of *-automorphisms of $\mathcal B(\mathcal H)$.

For every fixed vector $\psi$,

$$\|(\alpha_t(A)-A)\psi\| \leq{}\|A\|\,\|(U(t)-I)\psi\| +\|(U(t)^*-I)A\psi\|\longrightarrow0.$$

Thus the orbit is continuous in the strong operator topology. Applying the same argument to $A^*$ gives strong-* continuity. Pairings with trace-class operators are continuous as well, by finite-rank approximation, so the orbit is ultraweakly continuous.

None of these assertions gives $\|\alpha_t(A)-A\|\to0$ for every bounded $A$. “Strong continuity” of an automorphism group in C*-algebra terminology usually means precisely that latter point-norm condition. The two uses of “strong” must be distinguished; see Derivations on Operator Algebras.

A translated sharp event is not norm-continuous

Use dimensionless units with $H=P=-i\,d/dx$ on $H^1(\mathbb R)$ and $U(t)\psi(x)=\psi(x-t)$. For the multiplication projection $A=M_{\mathbf1_{[0,\infty)}}$,

$$\alpha_t(A)=M_{\mathbf1_{[0,\infty)}(x+t)}.$$

For every nonzero $t$, the two indicator functions differ by one on an interval of positive measure. Hence

$$\|\alpha_t(A)-A\|=1.$$

Nevertheless their difference applied to a fixed $L^2$ wavefunction tends to zero: its squared norm is the probability in a shrinking interval. The projection moves continuously as tested on each fixed state while failing a uniform estimate over all states. States concentrated in the shrinking interval obstruct that estimate.

Bounded commutators yield strong differentiation

Suppose $A\in\mathcal B(\mathcal H)$ preserves $D(H)$ and the commutator

$$C\psi=(HA-AH)\psi,\qquad \psi\in D(H),$$

extends to a bounded operator $C$ on $\mathcal H$. Differentiating first on $D(H)$ and then integrating gives

$$\alpha_t(A)-A =\frac{i}{\hbar}\int_0^t\alpha_s(C)\,ds,$$

where the integral is understood strongly, on each vector. Density extends the identity to the entire Hilbert space. It follows that

$$\|\alpha_t(A)-A\|\leq\frac{|t|}{\hbar}\|C\|, \qquad \frac{d}{dt}\alpha_t(A)\psi =\frac{i}{\hbar}\alpha_t(C)\psi$$

for every $\psi\in\mathcal H$. This is a strong operator derivative. A sufficient additional condition for an operator-norm derivative is point-norm continuity of $s\mapsto\alpha_s(C)$ at zero. Boundedness of the commutator alone should not be used to skip that qualification.

A Lipschitz orbit need not have a norm derivative

In the translation example let $A=M_f$, where

$$f(x)=\begin{cases} 0,&x\leq0,\\ x,&0<x<1,\\ 1,&x\geq1. \end{cases}$$

This bounded Lipschitz multiplier preserves $H^1(\mathbb R)$. Its commutator with $P$ is the bounded multiplier $-iM_{f'}$, where $f'=\mathbf1_{(0,1)}$ almost everywhere. Therefore $\alpha_t(A)$ is norm-Lipschitz and strongly differentiable, with derivative $M_{f'}$ at zero.

For $0<t<1$, however, on $-t<x<0$ the quotient $[f(x+t)-f(x)]/t$ ranges up to one while $f'(x)=0$. The essential supremum of the derivative error is at least one. Thus the difference quotient does not converge in operator norm. This separates existence of a bounded commutator from membership in the norm-generator domain.

For a fixed $\psi\in D(H)$ and bounded $A$, differentiability of the scalar expectation requires less: the product rule gives

$$\frac{d}{dt}\langle\psi(t),A\psi(t)\rangle =\frac{i}{\hbar} \bigl(\langle H\psi(t),A\psi(t)\rangle -\langle\psi(t),AH\psi(t)\rangle\bigr).$$

This form identity remains meaningful when $A\psi(t)\notin D(H)$. It should not be rewritten as an operator product $HA\psi(t)$ without the missing domain condition.

Unbounded observables carry a transported domain

If $A$ is self-adjoint and unbounded, its Heisenberg representative is the self-adjoint operator

$$A_H(t)=U(t)^*AU(t),\qquad D(A_H(t))=U(t)^*D(A).$$

Its spectral projections are $U(t)^*E_A(B)U(t)$. These formulas define the observable and its probabilities without differentiating an unbounded product. Equality of the domains at different times requires an invariance argument. Writing $\dot A_H=(i/\hbar)[H,A_H]$ on an unspecified common space does not supply one.

The physical commutator examples are developed in Heisenberg Equations of Motion; Conserved Quantities gives a spectral criterion under which the transported observable is unchanged.

Exercises

1. Show that a rank-one operator has a norm-continuous Heisenberg orbit for every self-adjoint $H$.

Solution

For $A=|u\rangle\langle v|$, conjugation gives $|U(t)^*u\rangle\langle U(t)^*v|$. The rank-one norm formula and a two-term difference estimate reduce norm continuity to continuity of the two vectors. Approximation by finite-rank operators then proves the same statement for every compact operator.

2. For bounded self-adjoint $H$, explain why every bounded $A$ has a norm derivative.

Solution

The exponential $U(t)$ is norm-differentiable, so the operator-norm product rule gives $\dot\alpha_t(A)=(i/\hbar)\alpha_t([H,A])$. There are no proper operator domains in this bounded setting. The derivative bound is at most $2\|H\|\|A\|/\hbar$.

References

• O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987, Chapter 3, publisher edition — continuity topologies and generators of automorphism groups.
• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, publisher edition — unbounded observables and Hamiltonian dynamics.