Domains of Operators

The domain of an operator is the set of vectors on which the operator is actually defined. In finite-dimensional quantum mechanics this is easy to miss because every matrix acts on every vector. In infinite-dimensional quantum mechanics it is central.

Position, momentum, and Hamiltonian operators are often written as simple formulas. Those formulas are incomplete until the domain is specified.

Definition

An operator on a Hilbert space $\mathcal H$ is written

A:D(A)\subseteq\mathcal H\to\mathcal H.

The domain $D(A)$ is the set of vectors $\psi$ for which $A\psi$ is defined and belongs to $\mathcal H$ . For a linear operator, $D(A)$ is usually required to be a linear subspace:

\psi,\phi\in D(A), \quad a,b\in\mathbb C \quad\Longrightarrow\quad a\psi+b\phi\in D(A).

Linearity then means

A(a\psi+b\phi) = aA\psi+bA\phi

for vectors in the domain.

The domain is not the range. The range is the set of possible outputs:

\operatorname{ran}A = \{A\psi:\psi\in D(A)\}.

The codomain in this setting is the ambient Hilbert space $\mathcal H$ . The domain, range, and codomain answer different questions.

Formula Plus Domain

Two operators can have the same formula but different domains. They are different operators.

For example, on $L^2([0,L])$ the formal derivative expression

-i\hbar\frac{d}{dx}

does not define a single operator until one says which wavefunctions are differentiable enough and which boundary behavior is allowed.

This is the operator-theoretic version of an elementary fact from functions: the formula $x^2$ on $\mathbb R$ and the formula $x^2$ on $[0,\infty)$ have different domains and different inverse behavior. For differential quantum operators, the consequences are much sharper: the spectrum, adjoint, self-adjointness, and time evolution can depend on the domain.

Bounded Versus Unbounded Domains

If $A$ is a bounded operator on a Hilbert space, it is normally taken to act on all of $\mathcal H$ :

D(A)=\mathcal H.

This is one reason Bounded Operators are technically easier.

For an Unbounded Operator, the domain is usually a proper subspace:

D(A)\subsetneq\mathcal H.

The domain can still be large in the important sense of being dense.

Dense Domains

A domain $D(A)$ is dense in $\mathcal H$ if every vector in $\mathcal H$ can be approximated arbitrarily well by vectors in $D(A)$ . Equivalently,

\overline{D(A)}=\mathcal H,

where the closure is taken in the Hilbert-space norm.

Dense domains matter for several reasons:

they allow the operator to act on a physically rich set of states;
they make the adjoint operation meaningful in the usual Hilbert-space sense;
they prevent the operator from ignoring an entire nonzero orthogonal subspace;
they let calculations on smooth or compactly supported test functions approximate more general states.

The dense-domain requirement does not mean $D(A)=\mathcal H$ . The space of smooth compactly supported functions $C_c^\infty(\mathbb R)$ is dense in $L^2(\mathbb R)$ , but it is much smaller than all of $L^2(\mathbb R)$ .

Position Operator Domain

On $L^2(\mathbb R)$ , the position operator is multiplication by $x$ :

(X\psi)(x)=x\psi(x).

Its natural domain is

D(X) = \left\{ \psi\in L^2(\mathbb R): x\psi(x)\in L^2(\mathbb R) \right\}.

This domain says exactly what is needed: the original wavefunction must be square-integrable, and the output after multiplication by $x$ must also be square-integrable.

Momentum on an Interval

Let $\mathcal H=L^2([0,L])$ and consider the formal momentum expression

P=-i\hbar\frac{d}{dx}.

The derivative already forces a domain such as a Sobolev space of square-integrable functions with square-integrable weak derivative. One commonly writes this as $H^1([0,L])$ .

Boundary conditions then choose the operator. If $\phi$ and $\psi$ are sufficiently regular, integration by parts gives

\langle\phi\vert P\psi\rangle - \langle P\phi\vert\psi\rangle = -i\hbar \left[ \phi(x)^*\psi(x) \right]_{0}^{L}.

Thus the boundary term controls whether the operator is symmetric on a proposed domain.

The domain

D_0(P) = \{\psi\in H^1([0,L]):\psi(0)=\psi(L)=0\}

makes the boundary term vanish. It gives a symmetric momentum operator, but it is not the standard self-adjoint momentum on an interval.

A one-parameter family of self-adjoint momentum operators is obtained from phase-twisted boundary conditions:

D_\theta(P) = \{\psi\in H^1([0,L]):\psi(L)=e^{i\theta}\psi(0)\}, \qquad \theta\in[0,2\pi).

If both $\phi$ and $\psi$ obey the same phase-twisted condition, then

\phi(L)^*\psi(L) = \phi(0)^*\psi(0),

so the boundary term vanishes. The resulting spectrum is

p_n = \hbar\frac{2\pi n+\theta}{L}, \qquad n\in\mathbb Z.

The differential expression did not change. The domain did.

Boundary Conditions as Domain Choices

Boundary conditions are not decorations added after solving an eigenvalue problem. They are part of the operator definition.

For a kinetic-energy Hamiltonian on an interval,

H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2},

the domain must say which functions are twice weakly differentiable enough and which endpoint conditions apply. Dirichlet, Neumann, periodic, and phase-twisted boundary conditions can define different self-adjoint operators and different spectra.

The analysis page Boundary Conditions explains the elementary differential-equation side. This page owns the operator-domain viewpoint.

Sums, Products, and Commutators

Domains also control algebraic operations. If $A$ and $B$ are unbounded operators, the natural domain of the sum is

D(A+B)=D(A)\cap D(B).

The product has a stricter domain:

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

Therefore the commutator

[A,B]\psi = AB\psi-BA\psi

is meaningful only on vectors for which both products exist. Formal operator algebra can be a useful guide, but for unbounded operators the domain is part of the statement.

Adjoint Preview

For a densely defined operator $A$ , the adjoint $A^\dagger$ is defined through the inner product. A vector $\phi$ lies in $D(A^\dagger)$ when there exists a vector $\eta\in\mathcal H$ such that

\langle\phi\vert A\psi\rangle = \langle\eta\vert\psi\rangle

for every $\psi\in D(A)$ . One then writes $A^\dagger\phi=\eta$ .

This definition shows why dense domains are not optional. Without density, the vector $\eta$ need not be uniquely determined. It also shows why self-adjointness is more than a formal integration-by-parts calculation: it requires equality of operators and equality of domains. The adjoint itself is developed in Adjoint Operators, and the domain equality distinction is developed in Symmetric versus Self-Adjoint Operators.

For the physics-facing warning, see Hermitian vs Self-Adjoint Operators.

Physical Interpretation

A normalizable vector can be a valid quantum state without lying in the domain of a particular unbounded observable. If $\psi\notin D(A)$ , then $A\psi$ is not defined as a Hilbert-space vector.

This matters for:

applying a differential operator to a wavefunction;
computing variances, which require control of $A^2\psi$ ;
using commutators involving unbounded observables;
specifying Hamiltonians and their boundary conditions;
proving that time evolution is unitary.

Some expectation values can be defined through quadratic forms on form domains, which are sometimes larger than the operator domain. That refinement is important in rigorous mathematical quantum mechanics, but the first rule remains: before applying an operator to a state, check that the state lies in the operator’s domain.

Common Mistakes

Treating an operator formula as complete without specifying $D(A)$ .
Confusing the domain with the range.
Assuming every $L^2$ wavefunction is differentiable.
Assuming a boundary condition is merely a solving trick rather than part of the operator.
Using $AB$ , $BA$ , or $[A,B]$ without checking product domains.
Calling a symmetric differential operator self-adjoint because one boundary term vanished.
Forgetting that the same differential expression can produce different spectra on different domains.

Cross-Links

References

M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.

Exercises

For the position operator on $L^2(\mathbb R)$ , explain why $D(X)$ is not all of $L^2(\mathbb R)$ .

Solution

A function can be square-integrable while $x\psi(x)$ is not square-integrable. For example, tails that decay just fast enough for $\psi$ to lie in $L^2(\mathbb R)$ may fail after multiplication by $x$ . The domain must require both $\psi\in L^2(\mathbb R)$ and $x\psi(x)\in L^2(\mathbb R)$ .

Verify that the phase-twisted momentum domain makes the boundary term vanish.

Solution

If $\psi(L)=e^{i\theta}\psi(0)$ and $\phi(L)=e^{i\theta}\phi(0)$ , then

\phi(L)^*\psi(L) = \left(e^{i\theta}\phi(0)\right)^* e^{i\theta}\psi(0) = e^{-i\theta}e^{i\theta} \phi(0)^*\psi(0) = \phi(0)^*\psi(0).

Therefore $\left[\phi(x)^*\psi(x)\right]_0^L=0$ .

Suppose $A$ and $B$ are unbounded. Why is $D(AB)$ generally not equal to $D(A)\cap D(B)$ ?

Solution

To form $AB\psi$ , one must first apply $B$ to $\psi$ , so $\psi\in D(B)$ . The result $B\psi$ must then lie in $D(A)$ . Thus

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

This can be smaller than $D(A)\cap D(B)$ .

Why is density useful when defining an adjoint?

Solution

The adjoint is defined by requiring

\langle\phi\vert A\psi\rangle = \langle\eta\vert\psi\rangle

for all $\psi\in D(A)$ . If $D(A)$ is dense, then $\eta$ is uniquely determined by its inner products against vectors in $D(A)$ . Without density, different vectors could agree on all of $D(A)$ while differing on an orthogonal complement.