Is this "position" operator $\hat{A}$ defined in the whole Hilbert space $L^2(\mathbb{R})$?

Question

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

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 $(A_4 f)(x) = x f(x)$

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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.