I was reading about bounded operators, and I was wondering that if we assume a "position" operator $A$ that acts on $\psi(x)$ and here $x$ is bounded in the interval $[-1,1]$ for example. What can we say about the operator $A$ in this case? Is it defined in the whole Hilbert space $L^2(\mathbb{R})$, or just $L^2[-1,1]$ because of the restriction on $x$? Is it considered a bounded operator if it is defined on the whole Hilbert space $L^2(\mathbb{R})$?

  • $\begingroup$ That operator would be unbounded on (and thus well-defined only on a dense subspace of) $L^2(\mathbb R)$ but is bounded and globally well-defined on $L^2[-1,1]$. $\endgroup$ Jun 15, 2020 at 9:36
  • $\begingroup$ @BenceRacskó Can I then say that A is a bounded operator because we restrict 𝑥 to the bounded interval [−1,1]. But A, in general, is a self-adjoint defined on the whole Hilbert space 𝐿^2 (ℝ) ??? $\endgroup$
    – Quantally
    Jun 15, 2020 at 10:05
  • $\begingroup$ Are you saying that the domain of $A$ is only those elements of $L^2(\mathbb R)$ which are zero outside the interval $[-1,1]$? $\endgroup$
    – J. Murray
    Jun 15, 2020 at 13:46
  • $\begingroup$ @J.Murray I am not saying anything about $A$. If you read my question again, I want to know what can we say about $A$ if $x$ is restricted in the interval [-1,1] ?? $\endgroup$
    – Quantally
    Jun 15, 2020 at 14:19
  • $\begingroup$ @Sally I don't know what that means. The elements of $L^2(\mathbb R)$ are functions which are defined on the entire real line. You can restrict your attention to functions which are zero outside of $[-1,1]$, or you could choose a different space such as $L^2([-1,1])$ on which to define your operator. But an operator on $L^2(\mathbb R)$ acts on functions as a whole. $\endgroup$
    – J. Murray
    Jun 15, 2020 at 14:24

1 Answer 1


There seems to be some underlying confusion here. Let the Hilbert space be $L^2(\mathbb R)$ - roughly, the space of functions $f$ such that

$$\int_{-\infty}^\infty |f(x)|^2 dx < \infty$$

A linear operator $A$ is a linear map from some domain $D_A \subseteq L^2(\mathbb R)$ to $L^2(\mathbb R)$. Consider the following four operators:

  1. $A_1$ acts on those functions contained within $L^2(\mathbb R)$ which vanish outside the interval $[-1,1]$. For all functions $f$ in this domain, $(A_1 f)(x) = xf(x)$
  2. $A_2$ acts on all functions in $L^2(\mathbb R)$. For all functions $f\in L^2(\mathbb R)$, $$(A_2 f)(x) = \cases{x f(x) & $x\in[-1,1]$\\f(x) & else}$$
  3. $A_3$ also acts on all functions in $L^2(\mathbb R)$, and its action is defined by $$(A_3 f)(x) = \cases{x f(x) & $x\in[-1,1]$\\ 0 & else}$$
  4. $A_4$ acts on those functions in $L^2(\mathbb R)$ such that $$\int_{-\infty}^\infty x^2|f(x)|^2 dx < \infty$$ and its action is defined by $(A_4 f)(x) = x f(x)$

Operator $A_1$ is bounded, but is not defined on all of $L^2(\mathbb R)$. $A_2$ and $A_3$ are bounded, and they are defined on all of $L^2(\mathbb R)$. $A_4$ is the standard position operator and is not bounded, and therefore is not defined on all of $L^2(\mathbb R)$ by necessity.

As a side note (though you didn't ask about this), all four operators are Hermitian, but only the last three are self-adjoint. $A_1$ is not self adjoint because its domain is not a dense subspace of $L^2(\mathbb R)$, which means its adjoint operator is not even well-defined.


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