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  What is the physical interpretation/application of the residual spectrum of a linear operator?

+ 4 like - 0 dislike

Assume that $X$ is an infinite dimensional compact Banach space, $D(A) \subseteq X $ is a linear subspace, and $A: D(A) \rightarrow X$ is a linear operator. The spectrum $\sigma(A)$ can be separated into three disjoint components:

  • the point spectrum $\sigma_p(A)$
  • the continuous spectrum $\sigma_c(A)$
  • the residual spectrum $\sigma_r(A)$

When applied to a physics context, what is the interpretation of these three different components of the spectrum of an operator?

In my naive (and maybe not correct) understanding, I always thought that for example in quantum mechanics, the discrete energy values of the harmonic oscillator would correspond to the point spectrum (of the Hamilton operator), and the continuous values of momentum are an example for a continuous spectrum (of the momentum operator).

But I have no clue how to physically interpret the residual spectrum, in the context of quantum mechanics for example.

Could anybody explain to me what kind of observables would be described by the residual spectrum of an operator, and give some examples?

asked Feb 27, 2015 in Mathematics by Dilaton (6,240 points) [ revision history ]
edited Mar 3, 2015 by Arnold Neumaier

What is a compact Banach space?

@Yiyang in my functional analysis course, a space $X$ was defined to be compact, if each open cover of $X$ has a finit partial cover.

i know. What puzzles me is that, since the topology of a banach space is a normed metric topology, there seems to be no way for the space to be compact.

3 Answers

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In quantum physics, we consider an Hilbert space $H$, i.e. we have an hermitian scalar product $(.,.)$ on $H$, and we consider self adjoint operators $T$, i.e. which satisfy $T=T^*$, where $T^*$ is the adjoint of $T$, defined by the property that $(Tx,y)=(x,T^*y)$ for every $x$ and $y$ in $H$.

Let us assume $T$ self adjoint. By definition, a complex number $\lambda$ is in the residual spectrum of $T$ if and only if $Ker(T-\lambda)=0$ (i.e. $\lambda$ is not an eigenvalue, i.e. not in the point spectrum) and $Im(T-\lambda)$ is not dense in $H$. But it is an easy computation to show that the closure of $Im(T-\lambda)$ in $H$ is $Ker(T^*-\bar{\lambda})=Ker(T-\bar{\lambda})$. So if $\lambda$ is in the residual spectrum of $T$ then $\bar{\lambda}$ is an eigenvalue of $T$. But an eingenvalue of a self-adjoint operator is real (if $Tx=\mu x$ with $x$ non-zero then $(Tx,x)=(\mu x,x)=\bar{\mu}(x,x)$ but also $(Tx,x)=(x, T^*x)=(x,Tx)=(x,\mu x)= \mu (x,x)$, thus $\bar{\mu}=\mu$ i.e. $\mu$ is real) so $\lambda$ is real and so $\lambda$ is an eigenvalue of $T$: contradiction.

Conclusion: the residual spectrum of a self-adjoint operator is empty, which explains why we do not often hear about residual spectrum in quantum mechanics courses...

Remark: in general, the operators relevant in quantum mechanics are unbounded, i.e. not continous for the norm operator, and so not defined everywhere. The preceding argument applies without assuming $T$ bounded.

answered Feb 28, 2015 by 40227 (5,140 points) [ revision history ]
Thanks for these very clear and patient explanations :-)
+ 4 like - 0 dislike

First of all, notice that a Banach space cannot be compact: If it is finite-dimensional, compactness is forbidden by the Heine-Borel-Plancherel theorem (compact sets are closed and bounded), if infinite-dimensional it is impossible as well because the the closed unit sphere would be compact (a closed subset of a compact set) and, as well known, this is false. In Hilbert spaces, (densely defined closed) normal operators, i.e. the ones commuting with their adjoint operators, do not have residual spectrum. Therefore self-adjoint operators (observables) and unitary operators (symmetries) do not have a residual spectrum. The only relevant operators in QM which might have a residual spectrum are partial isometries. Moeller operators in scattering theory may be of this type, but, as far as I know, there is no interest in the residual spectrum (if any).

answered Mar 2, 2015 by Valter Moretti (2,085 points) [ revision history ]
edited Mar 3, 2015 by Arnold Neumaier
+ 3 like - 0 dislike

Given a few-particle quantum system, the point spectrum of the Hamiltonian describes the bound states of the system, while the continuous spectrum describes its scattering states. Since the Hamiltonian is self-adjoint, there is no singular (=residual) spectrum. 

However, one may describe resonances of unstable systems as discrete eigenvalues of non-Hermitian Hamiltonians obtained by analytically continuation. The analytically continuation is done by complex scaling, which, intuitively rotates the continuous spectrum into the second sheet of a double cover of the complex plane, thereby uncovering resonances. For these, a singular spectrum might exist, and proving that in fact it doesn't (which is expected on physical grounds) is nontrivial, and constitutes a main step in proving asymptotic completeness. See Section 4.4. in Thirring's Course in Mathematical Physics 3, Quantum mechanics of atoms and molecules.

answered Mar 3, 2015 by Arnold Neumaier (15,787 points) [ revision history ]
edited Mar 3, 2015 by Arnold Neumaier


It seems to me that you are speaking about the singular spectrum which, as far as I know, has nothing to do with the residual spectrum. E.g., the singular spectrum, in general exists for self-adjoint operators, the residual spectrum does not.


@Walter: You are possibly right. I never saw the concept of a residual spectrum before and thought it was the same as the singular spectrum, since then there should be four parts of the spectrum - discrete, continuous, singular, and residual. But the Wikipedia page and the OP both only list three.

So could you please clarify the ''position'' of the singular spectrum in the general classification statement? Is it there subsumed under the continuous spectrum?

The decomposition into absolutely continuous,  pure point spectrum and singular one is referred to the natural measure of $\mathbb R$, that is the Lebesgue measure. Take a self-adjoint (generally unbounded) operator $T$ and consider its spectral measure $P$. If $\psi$ is a vector the map $E \mapsto \langle \psi | P(E) \psi \rangle = \mu_\psi(E)$, where $E \subset \mathbb R$ is Borel measurable, is a positive measure. That measure, in view of a celebrated theorem can always be decomposed into three parts $\mu_\psi = \mu_\psi^{ac} + \mu_\psi^{pp} + \mu_\psi^{s}$. The first measure is nothing but  a regular (measurable) desity fuctions times the standard Lebesgue measure, the second one is the sum of measures concentrated on a discrete set of points of $\mathbb R$, the third one is not regular with respect to the Lebesgue measure but it is not concentrated on a discrete set of points, it is a "strange" measure. Now we can define three subspaces:  The first one is spanned by the vectors $\psi$ whose measure $\mu_\psi$ is of the first class, the second subspace is spanned by the vectors whose associated measures belong to the second class and so on. It is possible to prove that the closures of these spaces are mutually orthogonal and their sum is the whole Hilbert space. Moreover they are separately invariant under the action of $T$. The restrictions of $T$ to the three subspaces define respective self-adjoint operators. The spectra of these operators are disjoint sets whose union is the spectrum of the initial  overall $T$. These disjoint parts of the spectrum of $T$ by definition are, obviously, the absolutely continuous part, the pure point part and the singular part of the spectrum of $T$. The physical relevance of this decomposition is due to various reasons in particular, assuming  certain hypotheses, in a sense the decomposition turns out to be "invariant" under perturbation of the initial self-adjoint operator (typically a Hamiltonian operator).

This decomposition of the spectrum is different from the standard one into continuous spectrum, point spectrum and residual spectrum which is instead based on properties of $(T-\lambda I)^{-1}$. In particular, for self-adjoint operators there is no residual spectrum. However, if I correctly remember(*) it turns out that the absolutely continuous spectrum is a subset of the continuous spectrum, the pure point spectrum is a subset of the point spectrum. The singular spectrum, in general has intersection with both the continuous and the point spectrum.

I think you are right in relating resonances  with the singular spectrum (not the residual one which is empty)  of the (self-adjont) Hamiltonian operator (though I am not an expert on these issues).

(*) There are further parts of the spectrum called essential spectrum, discrete spectrum, purely continuous spectrum, approximate point spectrum and and purely residual spectrum, due to various similar but inequivalent decompositions. I (very superficially) handled these non-standard decompositions several years ago when I wrote my book on spectral theory... and  I forgot almost everything!

@Valter: Thanks. This still leaves a number of details unclear that I'd prefer to discuss separately. I therefore asked this question. Please move the main part of the above answer there, for further discussion.

But please leave the part about resonances here, as this should be discussed here: Resonances don't belong to the singular spectrum of a self-adjoint Hamiltonian operator but to the discrete spectrum of its non-self-adjoint deformation.

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