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Since a QM operator is a linear map, it is useful to think about them as functions. An operator $\hat A$ on a finite $N$-dimensional Hilbert space $H_N$ is always such that $$\hat A:H_N\to H_N.$$

The domain and range of $\hat A$ are both $H_N$. $\hat A$ operates on states in Hilbert space and returns states in Hilbert space. My question regards the range of an operator $\hat B$ on an infinite dimensional Hilbert space $H_\infty$. If an operator always returns an eigenstate, we must write $$\hat B:H_\infty\not\to H_\infty,$$

because the eigenstate of an operator with a continuous spectrum cannot live in Hilbert space. The range cannot be the domain. So, what is the function notation $\hat B:H_\infty\to X$ appropriate for operators with continuous spectra?

Regarding the physics, I want to know about the mechanism by which the position operator can operate on a state to kick it out of the Hilbert space by returning a Dirac $\delta$ position eigenstate. I am aware that the rigged Hilbert space formalism offers a space for the position eigenstate to live in, but I am curious as to how this is described in the usual theory of linear operators on Hilbert space.

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    $\begingroup$ Note first that in an infinite dimensional Hilbert space the domain of an operator is usually not the whole space, but in QM one is interested in operators whose domain is dense in the Hilbert space. Then, the operators on the Hilbert space does not kick states out of the Hilbert space. $\endgroup$ Mar 6 at 9:29
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    $\begingroup$ I don't understand the question - the position operator never produces an "eigenstate" when acting on vectors inside the Hilbert space, precisely because its spectrum is purely continuous. The part of the question that starts at "If an operator always returns an eigenstate" makes no sense - why should an operator "always return an eigenstate"? What does that even mean? $\endgroup$
    – ACuriousMind
    Mar 8 at 20:38
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    $\begingroup$ @hodopsmith "Is it not a result in linear algebra that all linear operators return an eigenvalue?" No, certainly not, though its a bit hard to say what that even means. More to the point, operators with purely continuous spectra do not even have eigenvalues or eigenvectors in a strict sense. $\endgroup$
    – J. Murray
    Mar 8 at 20:57
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    $\begingroup$ @hodopsmith The Dirac delta is not an element of $L^2(\mathbb R)$, so it obviously can't be an eigenvector of an operator which is defined on a subset of $L^2(\mathbb R)$. It can be understood as a generalized eigenvector by invoking the rigged Hilbert space formalism, but you explicitly said you don't want to talk about that. And again, self-adjoint operators do not "return an eigenvalue." That terminology doesn't even make sense. Are you saying that the measurement of an observable represented by self-adjoint operator must return an eigenvalue? Because that's very different [...] $\endgroup$
    – J. Murray
    Mar 8 at 21:05
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    $\begingroup$ [...] from talking about the action of the operator on vectors in the Hilbert space. It's also strictly not true for operators with continuous spectra, precisely because they have no eigenvectors or eigenvalues. The machinery for dealing with purely continuous spectra is far more complex than what is necessary when dealing with operators with discrete spectra, which is why you don't learn about it in the first week of a typical QM class. $\endgroup$
    – J. Murray
    Mar 8 at 21:07

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The appropriate generalisation of eigenvalues of an operator $A$ on an infinite-dimensional Hilbert space $\mathcal{H}$ is the spectrum $\sigma(A)$ consisting of those $\lambda \in \mathbb{C}$ for which $\lambda - A$ does not admit a bounded inverse. The spectrum of a self-adjoint operator A with domain of self-adjointness $D(A)$ dense in $\mathcal{H}$ can be split into two parts: The discrete spectrum and the essential (or continuous) spectrum. The discrete spectrum consists of all isolated eigenvalues in $\sigma(A)$ with finite multiplicity, the essential spectrum of all other spectral values. If $\lambda$ lies in the spectrum, then, by Weyl's criterion, a sequence $(\psi_k)_{k\in\mathbb{N}}$ of normalised elements in $D(A)$ exists such that $\| A\psi_k - \lambda \psi_k\| \to 0$ as $k\to \infty$ (i.e. $A\psi_k$ approximates $\lambda \psi_k$). If $\lambda$ is an eigenvalue, simply take the constant sequence that consists of an eigenvector. If $\lambda$ lies in the essential spectrum, then $(\psi_k)_{k\in\mathbb{N}}$ can be chosen to have no convergent subsequence. Sometimes such a sequence is then called an approximate eigenvector.

Example: Let $X: D(X) \subset L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be the position operator defined by $X\psi(x) = x\psi(x)$ for $\psi \in D(X)$, where $D(X)$ is the domain of $X$ that makes $X$ self-adjoint. It is known that $\sigma(X) = \mathbb{R}$. Let $x_0 \in \sigma(X)$. Naively, $\delta_{x_0}$ is an eigenfunction of $X$ with eigenvalue $x_0$ because $$X\delta_{x_0}(x) = x \delta_{x_0}(x) = x_0 \delta_{x_0}(x).$$ However, $\delta_{x_0} \notin D(X)$ (and not even in $L^2(\mathbb{R})$). But $\delta_{x_0}$ can be approximated by normalised functions $f_\epsilon \in D(X)$, e.g. $$|f_\epsilon(x)|^2 = \frac{1}{\sqrt{2\pi\epsilon}} \mathrm{e}^{-\frac{(x-x_0)^2}{2\epsilon}}.$$ Indeed, Weyl's criterion is satisfied because $$\| Xf_\epsilon - x_0f_\epsilon\|^2_{L^2} = \frac{1}{\sqrt{2\pi\epsilon}} \int_{\mathbb{R}} (x-x_0)^2 \mathrm{e}^{-\frac{(x-x_0)^2}{2\epsilon}} \mathrm{d}x \stackrel{\epsilon \to 0}{\longrightarrow} 0. $$

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  • $\begingroup$ Thanks! Are you saying that the position operator takes a state in Hilbert space and simply returns a state in the spectrum of the position operator $\hat x:H_\infty\to\sigma(\hat x)$? Also, is your reference to Weyl's criterion meant to state that any position eigenstate can be approximated to arbitrary precision by wavepackets that do exist in the Hilbert space, and that we can then substitute those "approximate eigenvectors" into the spectrum of $\hat x$ to avoid $\hat x$ kicking states out of Hilbert space? $\endgroup$ Mar 8 at 19:56
  • $\begingroup$ The spectrum of an operator is a subset of $\mathbb{C}$, so it does not contain any states. $\endgroup$
    – Janik
    Mar 8 at 20:38
  • $\begingroup$ Yes, you are right. I meant it returns one of its eigenvectors whose eigenvalue is in the spectrum. How can we let $X: D(X) \subset L^2(\mathbb{R}) \to L^2(\mathbb{R})$ when the eigenstates of of $X$ are not in $ L^2(\mathbb{R})$, as you yourself say? That limit at the end of the your example is exactly the thing that is not in $ L^2(\mathbb{R})$. Do I have to settle for the convention being that we use an approximation and don't require that $X$ returns an eigenvector? $\endgroup$ Mar 8 at 20:48

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