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I read some answers about this and the wikipedia page that basically always say that a spectrum can be decomposed into:

$$\mu = \mu_{ac} + \mu_{sc} + \mu_{pp}, $$

where $\mu_{ac}$ is absolutely continuous, $\mu_{pp}$ is a pure point spectrum, and $\mu_{sc}$ is "singular continuous".

What does it mean, physically, to have a singular continuous spectrum? Are there examples in physics?

Can I say that all free states have absolutely continuous spectra, and all bound states have pure point spectra? So what lies in between? Localised states?

DanielSank
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SuperCiocia
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1 Answers1

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I know it's an old question, but it's a topic I love. ;) I always recommend this paper which goes into details in a very clear way. Let's try to sum up the most important parts for your question.

Step 1: momentum is hermitian but not self-adjoint

Let's take the momentum operator $P=-i\hslash\frac{d}{dx}$. It's an operator acting on functions (usually wavefunctions), so its domain of definition is all functions of $L^2$ which have a derivative also in $L^2$ (page 11).

$P$ is a hermitian operator, so $P$ and $P^\dagger$ act in the same way on functions in the domain of definition of $P$ (page 13).

But $P^\dagger$ has a larger domain of definition than $P$ (page 12). Because of this, $P\neq P^\dagger$ for every function, so $P$ is hermitian but not self-adjoint (page 13).

Step 2: spectral theorem

A self-adjoint $A$ operator has a spectrum in two parts (page 14):

  • a discrete part, for eigenfunctions that are its domain of definition,
  • a continuous part, for eigenfunctions that are generalized eigenfunctions (page 16).

But an operator $A$ that is hermitian but not self-adjoint has a third part to its spectrum:

  • a residual part, for functions that are eigenfunctions of $A^\dagger$ but not of $A$ (page 44).

We saw above that this is possible, that something as simple and well-known as the momentum operator is in this situation.

The paper studies a very common example: a wavefunction for a bound state in $x\in[0,1]$, so that it vanishes at the boundaries. It yields the surprising result that both the discrete and continuous spectra are empty, and that all eigenvalues are in the residual spectrum (page 44).

Miyase
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  • How does this answer the question? It asks about the meaning of the singular continuous part of the spectrum. Your answer never uses the words "singular continuous", even if one might be led to assume that you mean to imply that the "third part" at the end is that spectrum (I'm not convinced that that is true), and it never assigns a physical meaning to this type of spectrum. (Presumably there isn't one, cf. this quote from Simon and Reed). – ACuriousMind Aug 08 '22 at 16:03
  • (You might simply be confusing two different decompositions here: Your answer talks about the decomposition of the spectrum of a general operator's spectrum into discrete, continuous and residual, but the question is about the decomposition of a self-adjoint operator's spectrum into discrete, absolutely continuous and singularly continuous.) – ACuriousMind Aug 08 '22 at 16:06
  • @ACuriousMind As far as I know, "singular continuous" and "residual" are the same thing, just different words by different authors. The question was about an example of a physical situation where an operator has a residual part, I gave one: the momemtum operator for a simple bounded system. – Miyase Aug 08 '22 at 16:18
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    Do you have any reference for this belief? Note that the residual spectrum of any self-adjoint operator is empty, but e.g. Wiki explicitly puts the discussion of absolutely vs. singularly continuous in the section on self-adjoint operators, which would be pretty pointless if the singularly continuous spectrum was always empty in that case. – ACuriousMind Aug 08 '22 at 16:23
  • It isn't a belief, but a memory, which might not be better, I'll grant you that. I'll do a bit of research first. Thanks for pointing this out. – Miyase Aug 08 '22 at 16:25
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    Here's an MO question also using the two terms with different meanings, for example, and Simon and Reed also give two distinct definitions for the two terms. – ACuriousMind Aug 08 '22 at 16:28