0
$\begingroup$

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 $L^2(\mathbb{R})$?

$\endgroup$
5
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.