# Tagged Questions

**2**

votes

**1**answer

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

**1**answer

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

**2**answers

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

**1**answer

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

**1**answer

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

**3**answers

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

**3**answers

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

**2**answers

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

**3**answers

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 ...