Operators in Quantum Mechanics

An operator in quantum mechanics can describe a transformation of vectors, a measurable quantity, an event, or an infinitesimal generator. Those roles impose different conditions. A unitary transformation is bounded and acts on every state vector; its Hamiltonian generator can be unbounded and need not act on every vector. Keeping the role separate from the symbol prevents many apparently contradictory statements about quantum mechanics.

Helpful background. State Vectors supplies the interpretation of vectors and inner products. Self-Adjoint Operators gives the domain-sensitive observable criterion used here.

Operators as actions with specified inputs

A linear operator is a map $A:D(A)\subseteq\mathcal H\to\mathcal H$ on a linear subspace. Specifying a differential expression without specifying $D(A)$ does not yet specify the operator. Equality $A=B$ means equality of domains and equality of actions on that domain.

The same Hilbert-space vector may be admissible for one calculation and inadmissible for another. For example, a normalized $\psi\in L^2(\mathbb R)$ always defines position probabilities

$$\mathbb P_\psi(Q\in\Delta)=\int_\Delta|\psi(x)|^2\,dx.$$

It need not satisfy $x\psi\in L^2$, the condition for applying the position operator $Q\psi=x\psi$. The probability law exists even when its second moment diverges. This separates the existence of a state from the existence of particular moments.

An operator acting on a vector need not return a normalized vector, or even a nonzero vector. Applying a general operator is therefore not automatically a physical deterministic evolution. The transformation law must also say how probabilities and normalization are treated.

Transformations, events, and observables

A unitary $U$ is an everywhere-defined linear operator satisfying $U^*U=UU^*=I$. It preserves inner products and maps normalized vectors to normalized vectors. Time evolution of a closed system and continuously connected symmetry transformations have this form. Antiunitary symmetries are conjugate-linear and require a separate treatment; they are not members of the complex-linear operator algebra.

A projection $P=P^*=P^2$ describes a sharp yes/no event. It is bounded, with norm one unless $P=0$. More generally an effect satisfies $0\leq E\leq I$, so

$$0\leq\langle\psi,E\psi\rangle\leq1$$

for every normalized vector. Effects express probabilities without specifying a state update. Measurement operations require additional data; see Instruments.

A real-valued sharp observable is represented by a self-adjoint operator $A$ together with its spectral measure. The bounded projections $E_A(\Delta)$ determine events even when $A$ itself is unbounded. The distinction matters experimentally: asking whether energy lies in a finite interval makes sense in more states than applying the Hamiltonian or computing an energy variance.

For a bounded observable, expectations obey

$$|\langle\psi,A\psi\rangle|\leq\|A\|$$

on normalized states. An unbounded observable has no such uniform bound. This is a mathematical statement about all admissible states, not a claim that a particular experiment can prepare arbitrarily large energies.

Generators and finite transformations

For a self-adjoint Hamiltonian $H$, spectral calculus defines

$$U(t)=e^{-itH/\hbar}$$

on all of $\mathcal H$. Its norm is one. The exponential is defined by a bounded function of the spectrum, so it does not require a convergent power series on an arbitrary state.

The strong derivative has a narrower domain:

$$i\hbar\lim_{t\to0}\frac{U(t)\psi-\psi}{t}=H\psi, \qquad \psi\in D(H).$$

In fact, for this group the derivative exists exactly on $D(H)$. Thus every initial vector has a continuous unitary trajectory, but only vectors in $D(H)$ satisfy the Schrödinger equation as a differentiable Hilbert-space equation. Stone’s theorem supplies the precise equivalence between strongly continuous unitary groups and self-adjoint generators.

For the free particle, $H=P^2/(2m)$ is unbounded, while each $e^{-itH/\hbar}$ remains bounded. In momentum space the latter multiplies by a phase of modulus one. This is why the regularity required to differentiate time evolution is stronger than the regularity needed to evolve.

Equations between operators need domains

The sum and product have natural domains

$$\begin{aligned} D(A+B)&=D(A)\cap D(B),\\ D(AB)&=\{\psi\in D(B):B\psi\in D(A)\}. \end{aligned}$$

Consequently $AB\psi$ and $BA\psi$ may not both exist. A commutator identity must specify a common domain, often a dense invariant test space. An identity there need not automatically extend to the full domains of the closed operators.

For position and momentum on the line, Schwartz functions form a useful common invariant space: differentiation and multiplication by $x$ preserve that space. The relation $[Q,P]\psi=i\hbar\psi$ can be checked there. Extending such a statement to the Weyl relations or to strong commutation is a separate theorem, as explained in Domain Subtleties of Canonical Commutation Relations.

These distinctions are the practical reason for writing an operator’s domain next to its action. The domain records which question the formula actually answers.

Exercises

For a projection $P$, explain why $\psi\mapsto P\psi$ is not generally a deterministic normalized state transformation, despite being bounded.

Solution

Its output has squared norm $\|P\psi\|^2=\langle\psi,P\psi\rangle$, which is the yes-event probability and can lie anywhere between zero and one. Conditional on that outcome, the vector becomes $P\psi/\|P\psi\|$ when the denominator is nonzero. This normalization makes the conditional transformation nonlinear in $\psi$; the underlying measurement operation and its outcome probability must be described together.

Let $H e_n=n e_n$ on an orthonormal basis indexed by positive integers. Construct a state with a well-defined $e^{-itH}\psi$ but without a strong time derivative at zero.

Solution

Take $\psi=C\sum_{n\geq1}n^{-1}e_n$, where $C^{-2}=\sum n^{-2}$. The phase-multiplied coefficients remain square-summable for every $t$, so unitary evolution exists. But $\sum n^2|C/n|^2$ diverges, so $\psi\notin D(H)$ and the strong derivative does not exist.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — operator roles and unitary dynamics.
• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — domains, spectral measures, and generators.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s book page — mathematical foundations and quantum dynamics.