Unbounded Operators

An unbounded operator is an operator whose action can amplify normalized vectors by arbitrarily large amounts. In quantum mechanics, the most important examples are position on the real line, momentum, angular-momentum-like differential operators on suitable domains, and Hamiltonians with spectra unbounded above.

The key warning is simple:

An unbounded operator is not just a formula. It is a formula together with a domain.

Finite-dimensional matrix notation hides this issue because every linear map on a finite-dimensional Hilbert space is bounded and defined everywhere. Wave mechanics does not have that luxury.

Basic Definition

Let $\mathcal H$ be a Hilbert space. A linear operator may be written as

$$A:D(A)\subseteq\mathcal H\to\mathcal H,$$

where $D(A)$ is the domain of vectors on which $A$ is defined. The operator is bounded on its domain if there is a constant $C\ge0$ such that

$$\lVert A\psi\rVert \le C\lVert\psi\rVert$$

for all $\psi\in D(A)$. It is unbounded if no such finite $C$ exists.

For most quantum-mechanical unbounded operators, $D(A)$ is a proper dense subspace of $\mathcal H$, not the whole Hilbert space. Density is important because it lets the operator act on enough states to be physically useful and mathematically connected to the ambient Hilbert space. The focused domain language is developed in Domains of Operators.

Why Unbounded Operators Are Not Defined Everywhere

A central theorem from functional analysis says that a linear operator defined on all of a Banach space and having a closed graph must be bounded. Self-adjoint operators are closed. Therefore a genuinely unbounded self-adjoint operator cannot be defined on all of $\mathcal H$.

For physics, the practical consequence is this: if $A$ is an unbounded observable, there are normalizable states $\psi\in\mathcal H$ for which $A\psi$ is not a Hilbert-space vector. The expression $A\psi$ is then not defined, even if the formal symbol looks meaningful.

Position on the Real Line

On $\mathcal H=L^2(\mathbb R)$, the position operator is formally

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

Its natural domain is

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

This condition is not automatic. A square-integrable wavefunction can have tails large enough that $x\psi(x)$ fails to be square-integrable.

The operator is unbounded. Choose a normalized packet $\psi$ and translate it far to the right:

$$\psi_n(x)=\psi(x-n).$$

Then $\lVert\psi_n\rVert_2=1$, while the position norm grows roughly like $n$:

$$\lVert X\psi_n\rVert_2^2 = \int_{\mathbb R} x^2\lvert\psi(x-n)\rvert^2\,dx.$$

After the change of variables $y=x-n$,

$$\lVert X\psi_n\rVert_2^2 = \int_{\mathbb R} (y+n)^2\lvert\psi(y)\rvert^2\,dy,$$

which becomes arbitrarily large as $n\to\infty$ for a packet with finite first and second moments. Thus no single constant $C$ can satisfy $\lVert X\psi\rVert\le C\lVert\psi\rVert$ on the natural domain.

Momentum as a Differential Operator

In position representation, momentum is formally

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

This formula does not act on every element of $L^2(\mathbb R)$. A general $L^2$ equivalence class need not have a classical derivative, and even when a weak derivative exists it may not be square-integrable.

A typical rigorous domain is a Sobolev-type space of square-integrable wavefunctions whose weak derivative is also square-integrable. On the real line, one often writes informally

$$D(P) = \left\{ \psi\in L^2(\mathbb R): \psi'\in L^2(\mathbb R) \right\},$$

with the derivative understood in the weak sense.

Momentum is unbounded. On a circle or interval with compatible boundary conditions, high-frequency modes have fixed norm but momentum proportional to their wave number. Schematically,

$$\psi_n(x) = \frac{1}{\sqrt L} e^{2\pi i n x/L}$$

has $\lVert\psi_n\rVert=1$, while

$$\lVert P\psi_n\rVert = \hbar\frac{2\pi\lvert n\rvert}{L}.$$

The output norm grows without bound as $\lvert n\rvert\to\infty$.

Hamiltonians

Hamiltonians in wave mechanics are often second-order differential operators:

$$H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x).$$

This expression is not a complete operator definition. The domain must encode:

• the Hilbert space, such as $L^2(\mathbb R)$ or $L^2([0,L])$;
• the differentiability or weak-differentiability required by the kinetic term;
• the integrability requirements imposed by $V(x)\psi(x)$;
• boundary conditions at walls, endpoints, singularities, or infinity.

The infinite square well is the standard elementary example. The differential expression $-\hbar^2 d^2/(2m\,dx^2)$ becomes a physical Hamiltonian only after specifying a domain such as twice weakly differentiable wavefunctions satisfying Dirichlet boundary conditions at the walls. Different boundary conditions correspond to different operators and can produce different spectra.

For the boundary-value side of this story, see Boundary Conditions. For the operator-language warning in the physics formalism, see Hermitian vs Self-Adjoint Operators.

Domains Are Part of the Operator

Two operators can have the same formula but different domains. They should be treated as different operators.

This matters for sums, products, and commutators. If $A$ and $B$ are unbounded, then

$$(A+B)\psi$$

is meaningful only on states where both $A\psi$ and $B\psi$ exist. Similarly,

$$AB\psi$$

requires $\psi\in D(B)$ and $B\psi\in D(A)$. A formal commutator

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

is meaningful only on the common domain where both products are defined.

This is why the canonical commutation relation

$$[X,P]=i\hbar I$$

is not just a matrix identity on all of $L^2(\mathbb R)$. It is a formal relation whose precise meaning involves a suitable common dense domain, or a more robust exponentiated form using unitary translation and multiplication operators.

Symmetric Is Not Automatically Self-Adjoint

For a densely defined operator $A$, being symmetric means

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

for all $\phi,\psi\in D(A)$. This is the condition often checked by integration by parts.

Self-adjointness is stronger. It requires equality with the adjoint operator, including equality of domains:

$$A=A^\dagger, \qquad D(A)=D(A^\dagger).$$

Boundary terms from integration by parts may vanish on a chosen domain and prove symmetry, but that alone does not prove self-adjointness. This distinction is developed in Symmetric versus Self-Adjoint Operators. It is physically important because self-adjoint Hamiltonians generate unitary time evolution and self-adjoint observables have the spectral structure needed for sharp measurements.

Why Quantum Mechanics Needs Them

Unbounded operators are not a technical nuisance added by mathematicians. They reflect physical idealizations:

• position on an unbounded space has no maximum possible value;
• momentum can be arbitrarily large in the nonrelativistic idealization;
• many Hamiltonians have spectra unbounded above;
• differential operators naturally increase high-frequency components.

If a self-adjoint operator is bounded, its spectrum is bounded. Thus an observable with an unbounded ideal spectrum cannot be represented by a bounded self-adjoint operator.

Bounded approximations are still useful. A finite-dimensional truncation of a Hamiltonian is a bounded matrix, and numerical work often uses such truncations. The approximation must then be interpreted as an approximation to an unbounded operator, not as proof that the original observable is bounded.

Common Mistakes

• Treating a differential expression as a complete operator definition.
• Writing $A:\mathcal H\to\mathcal H$ for an unbounded operator without specifying a proper domain.
• Assuming a normalizable $\psi$ automatically lies in $D(X)$, $D(P)$, or $D(H)$.
• Checking integration by parts and calling the operator self-adjoint without checking domains.
• Computing $AB-BA$ for unbounded operators without checking a common domain.
• Assuming finite-dimensional truncations preserve all domain and spectral facts of the infinite-dimensional operator.
• Forgetting that boundary conditions are part of the operator, not an afterthought.

Cross-Links

• Hilbert Spaces
• L2 Spaces
• Bounded Operators
• Domains of Operators
• Symmetric versus Self-Adjoint Operators
• Position and Momentum Representations
• Heisenberg Group
• Boundary Conditions
• Operators
• Hermitian vs Self-Adjoint Operators

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show directly that multiplication by $x$ is unbounded on $L^2(\mathbb R)$.

Solution

Let $\psi_n$ be normalized and supported in the interval $[n,n+1]$. Then

$$\lVert X\psi_n\rVert_2^2 = \int_n^{n+1} x^2\lvert\psi_n(x)\rvert^2\,dx \ge n^2 \int_n^{n+1} \lvert\psi_n(x)\rvert^2\,dx = n^2.$$

Thus $\lVert X\psi_n\rVert_2\ge n$ while $\lVert\psi_n\rVert_2=1$. No uniform bound exists.

2. Why is $P=-i\hbar d/dx$ not defined on every vector in $L^2(\mathbb R)$?

Solution

Elements of $L^2(\mathbb R)$ are equivalence classes of square-integrable functions. Such a function need not be differentiable, and even when it has a weak derivative, that derivative need not lie in $L^2(\mathbb R)$. The domain of $P$ must therefore restrict to wavefunctions for which the derivative exists in the appropriate sense and remains square-integrable.

3. In the infinite square well, why are boundary conditions part of the Hamiltonian rather than extra information added after solving?

Solution

The differential expression alone does not determine the operator. The domain tells us which wavefunctions are allowed and which boundary terms vanish under integration by parts. Dirichlet, Neumann, periodic, and phase-twisted boundary conditions can define different self-adjoint realizations with different spectra. Therefore the boundary conditions are part of the Hamiltonian’s definition.

4. Explain why norm closeness of states does not by itself control expectation values of an unbounded observable.

Solution

For a bounded operator $A$, one has estimates such as

$$\left\lvert \langle\psi\vert A\vert\psi\rangle - \langle\phi\vert A\vert\phi\rangle \right\rvert \le 2\lVert A\rVert_{\mathrm{op}} \lVert\psi-\phi\rVert$$

for normalized states. An unbounded operator has no finite operator norm, so this estimate is unavailable. Two states can be close in Hilbert-space norm while differing in high-energy or far-tail components that strongly affect an unbounded expectation value.