0
$\begingroup$

For an operator $T$ on a Hilbert space the residual spectrum is defined as the set of complex numbers $\lambda$ for which $(T-\lambda I)^{-1}$ exists but is not densely defined. The question is: could there be a complex number in the residual spectrum which is a generalised eigenvalue? By generalised eigenvalue I mean that there exists a sequence $\{x_n\}$ of unit vectors that satisfy: $$ (T-\lambda I)x_n \longrightarrow 0. $$ I was thinking about this from a physics point of view (discarding the fact that observables are assumed to be self adjoint) and wondering if one wanted to make an "observable" out of something which is hermitian but doesn't have an empty residual spectrum (like momentum on $L^2(0,\infty))$ would the problem be that for such operators the system could in principle collapse to some state with a complex spectral value associated to it? This spectral value would have to be in the residual spectrum because the continuous and point spectra of hermitian operators must be real.

$\endgroup$
4
  • $\begingroup$ Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue. $\endgroup$
    – amsmath
    Commented Mar 27, 2019 at 18:45
  • $\begingroup$ yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical. $\endgroup$ Commented Mar 27, 2019 at 18:58
  • $\begingroup$ What you want is an operator $T$ in a Hilbert space $\mathcal H$ (may be bounded) such that $\ker T = \{0\}$ and $\operatorname{ran}T$ is neither closed nor dense in $\mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now. $\endgroup$
    – amsmath
    Commented Mar 27, 2019 at 19:25
  • $\begingroup$ Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint) $\endgroup$ Commented Mar 27, 2019 at 19:27

1 Answer 1

1
$\begingroup$

Let $S : \mathcal H\to\mathcal H$ be any linear operator with non-closed range and $\ker S = \{0\}$ (for example $Sf = t\cdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $\mathcal H$ and let $V : \overline{\operatorname{ran}S}\to M$ be unitary. Then $T := VS$ satisfies the requirements.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .