2
votes
1answer
83 views

Eigenvalues and Compact Resolvent

For $A$ an unbounded (densely defined) operator on a separable Hilbert space, what conditions on its eigenvalues will show that, for $\lambda \notin $spec$(A)$, we have that $(A-\lambda)^{-1}$ is a ...
1
vote
1answer
104 views

Special form of unbounded operators on $L_2(\mathbb{R}_+, \mathcal{H})$

I have the following problem; Fix a Hilbert space $\mathcal{H}$. Let $S \colon \mathrm{Dom}S \subset L_2(\mathbb{R}_+, \mathcal{H}) \rightarrow L_2(\mathbb{R}_+, \mathcal{H}) $ be a closed densely ...
8
votes
2answers
152 views

why is this a sufficient condition for a domain to be a core of an unbounded operator?

Let $\alpha:\mathbb R\to U(H)$ be a strongly continuous action of the reals on some Hilbert space, and let $A=-i\frac d{dt}\alpha(t)|_{t=0}$ be its infinitesimal generator, so that ...
2
votes
1answer
118 views

Self-adjointness of a perturbed quantum mechanical Hamiltonian specified in an infinite matrix form

Consider an operator $H$ on the Hilbert space $\ell_2$ given as an infinite matrix with two pieces, one diagonal and one arbitrary: $H_{ij}=E_i\delta_{ij}+V_{ij}$. This has a physical meaning in ...
7
votes
1answer
208 views

Product of commuting nonnegative operators

Let $V$ be a real vector space with an inner product and $A,B : V \to V$ linear maps which are self-adjoint nonnegative-definite, i.e. $\langle Ax,y \rangle = \langle x,Ay \rangle$ and $\langle Ax,x ...
6
votes
3answers
344 views

ordered exponential of unbounded operators

Let $H$ be a Hilbert space, and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation $$ ...
1
vote
3answers
548 views

Countability of eigenvalues of a linear operator

Is it true that every closed operator on a separable Hilbert H space only has countably many eigenvalues? Or put the other way around, if I want to ensure that a (not necessarily bounded) linear ...
5
votes
2answers
681 views

Nice Classes of Non-Closable Operators

The only thing I know about non-closable operators can be summarised as "they exist, but they're nasty, so let's not talk about them!" This seems to be the case with everyone else I've talked to. I'd ...
2
votes
3answers
1k views

Infinite hermitian matrix

Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric: $H^\dagger = H$ and $H^T = -H$. (T denotes transpose, $\dagger$ denote conjugate transpose. I ...