2
votes
2answers
53 views

Decomposition of a positive semidefinite self-adjoint operator?

If I have a positive semi-definite self-adjoint operator $H:D(H) \rightarrow L^2$, is it true that there is always a decomposition $H=A^* A$ available? If this is true, what can we say about the ...
3
votes
2answers
102 views

Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom.

I have the following two questions: The Fourier transform defines a unitary (provided that it is normalized properly) map $\hat{\cdot}:L^2(\mathbf{R})\rightarrow L^2(\mathbf{R})$. I figured out its ...
0
votes
2answers
49 views

Complete ONS and pure point spectrum

In all that follows all operators are taken to be densely defined on a Hilbert space $H$. Some textbooks state that an operator $A$ on $H$ has pure point spectrum if $H$ admits a complete ONS (Hilbert ...
3
votes
1answer
140 views

Why study Bergman Spaces?

I'm interested in Operator Algebras and mathematical physics; recently, a friend showed me Duren and Schuster's "Bergman Spaces". I've read about two chapters now and I see there is a nice play ...
0
votes
0answers
22 views

Hydrogenhamiltonian self-adjoint in one or two dimensions

let $d\in\{1,2\}$. I'd like to know if the operator $H=-\Delta - \frac{1}{|x|}$ is self-adjoint as an operator acting on a dense subset of $L^2(\mathbb R^d)$. In particular I'd like to know how its ...
2
votes
1answer
125 views

Relationship between adjoint operators, trace-class operators, compact operators and density operators in Quantum-Mechanics

I don't know much about Functional Analysis, but I was wondering about the following: In Banach spaces it is possible to define for every continuous opertor $T:X \rightarrow Y$ an adjoint Operator ...
0
votes
3answers
67 views

Positive unbounded operator with zero not as an eigenalue

I am currently doing Quantum Mechanics and I am supposed to show that zero is an eigenvalue of a positive operator. I have no knowledge of Functional Analysis at that kind of level, so I was wondering ...
2
votes
2answers
977 views

Commutator of $x$ and $p^2$

I have a question: If I have to find the commutator $[x, p^2]$ (with $p= {h\over i}{d \over dx} $) the right answer is: $[x,p^2]=x p^2 - p^2x = x p^2 -pxp + pxp - p^2x = [x,p]p + p[x,p] = 2hip$ But ...
4
votes
1answer
114 views

Can 0 be an eigenvalue?

Let $-\Delta $ be the positive Laplacian and consider the operator $$ -\Delta + V $$ on $L^2(\mathbb{R}^3)$ with domain the Sobolev space $W^{2,2}(\mathbb{R}^3)$. Here $V:\mathbb{R}^3\to \mathbb{R}$ ...
6
votes
1answer
122 views

How to find interesting operators for a quantum system?

How can we find "interesting" operators for a quantum mechanical system? I can think of the following method: Given some system with an associated Hilbert space $V$ and Hamiltonian $H:V\rightarrow ...
3
votes
1answer
155 views

A question about Moyal product

In Geometric quantization theory the Moyal product is one of main tools. We know Moyal product for the smooth functions $f$ and $g$ on $ℝ^{2n}$ takes the form $f\star g = fg + \sum_{n=1}^{\infty} ...
3
votes
1answer
655 views

Rayleigh-Ritz Theorem

Let $U$ be an $n$-dimensional subspace of $L:=L_2([-1,1])$. Let $F$ be an acting on $L$, given at $f \in L$ $$ (Ff)(x):=\int_{-1}^1 \frac{\sin a(x-y)}{(x-y)}f(y) dy, \quad x \in [-1,1], \quad a>0. ...
5
votes
0answers
351 views

Min Max Principle and Rayleigh-Ritz-Method for eigenvalues of unbounded operators?

Finding eigenvalues of matrices using the Rayleigh-Ritz quotient is well-known, c.f. http://en.wikipedia.org/wiki/Min-max_theorem Does the following generalization of that fact also hold? Theorem: ...
11
votes
1answer
386 views

Quantization of angular momentum: is Dirac's proof wrong?

I'm trying to understand the physicist's proof of the theorem on the spectral structure of angular momentum operators (I'm being told that this proof is due to Dirac). I will refer to Ballentine's ...
2
votes
2answers
177 views

Perturbation theorem of Weyl

Does anyone know where to find something about the perturbation theorem of Weyl, preferably on the internet. The theorem I'm talking about states: let $A$ be a self-adjoint operator on a Hilbert ...
20
votes
1answer
819 views

Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions. The first ...