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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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    $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Feb 17 '14 at 16:25
  • $\begingroup$ Comment to the question (v3): The momentum example proportional to $\frac{d}{dx}$ is mentioned on the Wikipedia page. $\endgroup$ – Qmechanic Feb 17 '14 at 19:44
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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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  • $\begingroup$ Could you list some examples of bounded operators used in quantum mechanics? $\endgroup$ – Seeker May 20 '17 at 8:37
  • $\begingroup$ Unitary evolutors, representations of symmetries, density matrices, $S$ matrices, spin operators, helicity operators, $\endgroup$ – Valter Moretti May 20 '17 at 9:27
  • $\begingroup$ Thank you so much. Could you please suggest some text which covers the bounded and unbounded operators in the context of quantum physics? $\endgroup$ – Seeker May 20 '17 at 14:53
  • $\begingroup$ Usually texts on quantum physics do not distinguish between bounded and unbounded operators, everything is treated as a big matrix. If you are a beginner on QM I do not suggest to enter into the details of these mathematical technicalities. If you are familiar with basic QM there are several good books on the mathematical formulation. See the list at the end of this page en.wikipedia.org/wiki/… $\endgroup$ – Valter Moretti May 20 '17 at 16:15

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