Bounded Operators

A bounded operator is a linear operator that cannot stretch vectors by arbitrarily large factors. Bounded operators are the technically well-behaved operators on Hilbert space: they are continuous, defined on the whole Hilbert space, and have adjoints without domain surprises.

Finite-dimensional matrices are always bounded. Infinite-dimensional quantum mechanics is harder because important operators such as position on $\mathbb R$, momentum, and differential Hamiltonians are often unbounded.

Definition

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

$$A:\mathcal H\to\mathcal H$$

is bounded if there is a constant $C\ge0$ such that

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

for every $\psi\in\mathcal H$.

The smallest such $C$ is the operator norm:

$$\lVert A\rVert_{\mathrm{op}} = \sup_{\psi\ne0} \frac{\lVert A\psi\rVert}{\lVert\psi\rVert} = \sup_{\lVert\psi\rVert=1} \lVert A\psi\rVert.$$

When the norm is clear, physicists and mathematicians often write simply $\lVert A\rVert$.

Equivalent Continuity Statement

For linear maps between normed spaces, boundedness is equivalent to continuity. If $A$ is bounded and $\psi_n\to\psi$ in norm, then

$$\lVert A\psi_n-A\psi\rVert = \lVert A(\psi_n-\psi)\rVert \le \lVert A\rVert_{\mathrm{op}} \lVert\psi_n-\psi\rVert \to0.$$

So bounded operators preserve norm convergence. This is why they are easy to use with limits, approximations, and basis expansions.

Finite-Dimensional Case

Every linear operator on a finite-dimensional Hilbert space is bounded. In a finite orthonormal basis, an operator is represented by a matrix, and all matrix entries are finite. The matrix can stretch vectors, but only by a finite maximum factor on the unit sphere.

For a finite matrix $A$, the operator norm is the largest singular value:

$$\lVert A\rVert_{\mathrm{op}} = s_1(A).$$

This is one practical use of Singular Value Decomposition.

Standard Examples

The identity operator is bounded:

$$\lVert I\rVert_{\mathrm{op}}=1.$$

A unitary operator is bounded with norm one:

$$\lVert U\psi\rVert=\lVert\psi\rVert, \qquad \lVert U\rVert_{\mathrm{op}}=1.$$

An orthogonal projector $P$ is bounded with

$$\lVert P\rVert_{\mathrm{op}} = \begin{cases} 0, & P=0,\\ 1, & P\ne0. \end{cases}$$

A rank-one operator

$$A = \lvert u\rangle\langle v\rvert$$

is bounded, with

$$\lVert A\rVert_{\mathrm{op}} = \lVert u\rVert\,\lVert v\rVert.$$

These examples cover many finite-dimensional gates, projective measurements, and finite-rank approximations.

Multiplication Operators

On an $L^2$ space, multiplication by a bounded function is a bounded operator. If

$$(M_f\psi)(x) = f(x)\psi(x)$$

and $\lvert f(x)\rvert\le M$ almost everywhere, then

$$\lVert M_f\psi\rVert_2 \le M\lVert\psi\rVert_2.$$

Thus

$$\lVert M_f\rVert_{\mathrm{op}} \le M.$$

In fact the operator norm is the essential supremum of $\lvert f\rvert$.

This example also shows how boundedness depends on the space. Multiplication by $x$ is bounded on $L^2([0,1])$, but not on $L^2(\mathbb R)$.

Why Bounded Operators Are Easier

Bounded operators avoid several domain problems:

• they are defined on all of $\mathcal H$;
• their sums and products are bounded;
• their adjoints are bounded and defined on all of $\mathcal H$;
• norm-convergent input sequences have norm-convergent outputs;
• power series such as exponentials can be handled by operator-norm convergence.

For example, if $A$ is bounded, then

$$e^A = \sum_{n=0}^{\infty} \frac{A^n}{n!}$$

converges in operator norm. For unbounded operators, exponentials require spectral-theorem or semigroup machinery and domain hypotheses.

Quantum Interpretation

Bounded operators appear constantly:

• finite-dimensional observables and gates;
• unitary time-evolution operators;
• projectors and POVM effects;
• density operators and reduced density operators;
• finite-rank truncations and numerical approximations.

But many familiar observables are not bounded in their ideal infinite-dimensional form. Position on the real line, momentum, and Hamiltonians with unbounded spectra require domains. The focused Toolkit entry is Unbounded Operators, and the Core Formalism warning is Hermitian vs Self-Adjoint Operators.

Expectation-Value Stability

If $A$ is bounded and $\psi,\phi$ are normalized, then expectation values are stable under norm-small changes of state:

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

This is one reason bounded observables are numerically and conceptually easier. For unbounded observables, a small Hilbert-space norm difference alone may not control expectation values unless the states also satisfy domain and energy-type bounds.

Common Mistakes

• Assuming every operator on an infinite-dimensional Hilbert space is bounded.
• Forgetting that finite-dimensional matrices are automatically bounded.
• Confusing a bounded operator with an operator whose eigenvalues are all known.
• Treating the position operator on all of $L^2(\mathbb R)$ as if it were bounded.
• Applying operator-norm estimates to unbounded operators.
• Ignoring the phrase “almost everywhere” for multiplication operators on $L^2$ spaces.
• Assuming boundedness makes an operator physically observable; positivity, self-adjointness, and interpretation are separate questions.

Cross-Links

• Hilbert Spaces
• Norms and Metrics
• Adjoint Operators
• Unbounded Operators
• Unitary Operators
• Projectors
• Singular Value Decomposition
• Operators
• Hermitian vs Self-Adjoint Operators

References

• J. B. Conway, A Course in Functional Analysis, 2nd ed., Springer, 1990.
• 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.
• P. R. Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, 2nd ed., Chelsea, 1957.

Exercises

1. Show that a unitary operator has operator norm one.

Solution

For a unitary $U$,

$$\lVert U\psi\rVert=\lVert\psi\rVert$$

for every $\psi$. Therefore

$$\sup_{\lVert\psi\rVert=1}\lVert U\psi\rVert=1.$$

2. Let $P$ be a nonzero orthogonal projector. Show that $\lVert P\rVert_{\mathrm{op}}=1$.

Solution

For any $\psi$,

$$\lVert P\psi\rVert\le\lVert\psi\rVert,$$

so $\lVert P\rVert_{\mathrm{op}}\le1$. Since $P$ is nonzero, there is a unit vector $u$ in its range, and $Pu=u$. Therefore $\lVert P\rVert_{\mathrm{op}}\ge1$. Hence the norm is one.

3. On $L^2([0,1])$, find a bound for the multiplication operator $(M_x\psi)(x)=x\psi(x)$.

Solution

Since $0\le x\le1$ on the interval,

$$\lVert M_x\psi\rVert_2^2 = \int_0^1 x^2\lvert\psi(x)\rvert^2\,dx \le \int_0^1 \lvert\psi(x)\rvert^2\,dx = \lVert\psi\rVert_2^2.$$

Thus $\lVert M_x\rVert_{\mathrm{op}}\le1$. In fact the norm is one.

4. Why does the same multiplication rule by $x$ fail to be bounded on $L^2(\mathbb R)$?

Solution

On the real line, $x$ is not bounded. Wavefunctions can be concentrated far from the origin, making $\lVert x\psi\rVert_2$ arbitrarily large while $\lVert\psi\rVert_2=1$. Therefore there is no constant $C$ with $\lVert x\psi\rVert_2\le C\lVert\psi\rVert_2$ for all allowed $\psi$.