# Bounded and Unbounded Operator

Can someone explain with a concrete example of how can I can check whether a quantum mechanical operator is bounded or unbounded?

EDIT: For example., I would like to check whether $\hat p=-i\hbar\frac{\partial}{\partial x}$ is bounded or not.

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If you have a specific operator you want to examine, you should mention it. Otherwise, this is a very broad question and there is little to do except refer you to the definition and that of continuity of a linear operator. The class of bounded operators is so broad that any method other than the definition can only apply to a small fraction of it. – Emilio Pisanty Feb 17 '14 at 16:25
Comment to the question (v3): The momentum example proportional to $\frac{d}{dx}$ is mentioned on the Wikipedia page. – Qmechanic Feb 17 '14 at 19:44

A linear operator $A: D(A) \to {\cal H}$ with $D(A) \subset {\cal H}$ a subspace and ${\cal H}$ a Hilbert space (a normed space could be enough), is said to be bounded if: $$\sup_{\psi \in D(A)\:, ||\psi|| \neq 0} \frac{||A\psi||}{||\psi||} < +\infty\:.$$ In this case the LHS is indicated by $||A||$ and it is called the norm of $A$.

Notice that, therefore, boundedness, is not referred to the set of values $A\psi$, which is always unbounded if $A\neq 0$, as $||A\lambda\psi|| = |\lambda|\: ||A\psi||$ for $\psi \in D(A)$ and $\lambda$ can be chosen arbitrarily large still satisfying $\lambda \psi \in D(A)$ since $D(A)$ is an subspace.

It is possible to prove that $A: D(A) \to {\cal H}$ is bounded if and only if, for every $\psi_0 \in D(A)$: $$\lim_{\psi \to \psi_0} A\psi = A \psi_0\:.$$

Another remarkable result is that a self-adjoint operator is bounded if and only if its domain is the whole Hilbert space.

Regarding $A= \frac{d}{dx}$, first of all you should define its domain to discuss boundedness. An important domain is the space ${\cal S}(\mathbb R)$ of Schwartz functions since, if $-id/dx$ is defined thereon, it turns out Hermitian and it admits only one self-adjoint extension that it is nothing but the momentum operator.

$d/dx$ on ${\cal S}(\mathbb R)$ is unbounded. The shortest way to prove it is passing to Fourier transform. Fourier transform is unitary so that it transforms (un)bounded operators into (un)bounded operators. ${\cal S}(\mathbb R)$ is invariant under Fourier transform, and $d/dx$ is transformed to the multiplicative operator $ik$ I henceforth denote by $\hat A$. So we end up with studying boundedness of the operator: $$(\hat A \hat{\psi})(k) = ik \hat{\psi}(k)\:,\quad \hat\psi \in {\cal S}(\mathbb R)\:.$$
Fix $\hat\psi_0 \in {\cal S}(\mathbb R)$ with $||\hat\psi_0||=1$ assuming that $\hat\psi_{0}$ vanishes outside $[0,1]$ (there is always such function as $C_0^\infty(\mathbb R) \subset {\cal S}(\mathbb R)$ and there is a function of the first space supported in every compact set in $\mathbb R$), and consider the class of functions $$\hat\psi_n(k):= \hat \psi_{0}(k- n)$$ Obviously, $\hat\psi_n \in {\cal S}(\mathbb R)$ and translational invariance of the integral implies $||\hat\psi_n||=||\hat\psi_0||=1$. Next, notice that: $$\frac{||\hat A\hat\psi_n||^2}{||\hat\psi_n||^2} = \int_{[n, n+1]} |x|^2 |\hat\psi_{0}(k-n)|^2 dk \geq \int_{[n, n+1]} n^2 |\hat\psi_{0}(k-n)|^2 dk$$ $$= n^2 \int_{[0,1]} |\hat\psi_{0}(k)|^2 dk = n^2\:.$$ We conclude that: $$\sup_{\hat{\psi} \in {\cal S}(\mathbb R)\:, ||\hat{\psi}||\neq 0} \frac{||\hat A\hat\psi||}{||\hat\psi||} \geq n \quad \forall n\in \mathbb N$$ So $\hat A$ is unbounded and $A$ is consequently.

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