Talk:Hille–Yosida theorem

From Wikipedia, the free encyclopedia

[edit] Resolvent formalism

Hi CSTAR, fantastic article that I'm still trying to grok. I would like to sugges a low-brow section, or maybe a distinct article, called, for example, resolvent formalism, that uses the standard notation commonly seen in books on quantum mechanics. (even though its somewhat poorly defined) Viz:

The resolvent captures the spectral properties of an operator in the analytic structure of the resolvent. Given an operator A, the resolvent may be defined as

R(z;A)= \frac{1}{A-zI}

The residue may be understood to be a projection operator

\operatorname{res} R(z;A)\vert_{z=\lambda} =  \frac{-1}{2\pi i} \oint_{C_\lambda} R(\zeta;A) d\zeta =  P_\lambda = \vert \psi_\lambda \rangle \langle \psi_\lambda \vert

where λ corresponds to an eigenvalue of A

A\vert \psi_\lambda \rangle = \lambda \vert \psi_\lambda \rangle

and Cλ is a contour in the positive dirction around the eigenvalue λ.

The above is more-or-less a textbook defintion of the resolvant as used in QM. Unfortunately, I've never seen anything "better" than this, in particlar, don't know how to qualify \vert \psi_\lambda \rangle. Must this be an element of a Hilbert space? Something more general? Frechet space? Banach space? Does the operator A have to be Hermitian? Nuclear? Trace-class? or maybe not?

Since the above is occasionally seen in the lit. I'd like to get a good solid article for it, with the questions at least partly addressed (and it seems the Hille-Yosida theorem adresses these, in part). Let me know what you think. linas 21:30, 22 November 2005 (UTC)

[edit] Connection to the Laplace transform

By definitions of the infinitisimal operator and the semigroup we have that

\frac{d}{dt} T_t\phi=\lim_{h\to 0}\frac{T_{t+h}\phi-T_t\phi}{h}=AT_t \phi

If we formally do the Laplace transform of the semigroup

\mathcal{L} T_t\phi=\int_0^\infty e^{-\lambda t} T_t \phi dt

we get by integration by parts

\mathcal{L} \frac{d}{dt} T_t \phi = \lambda \mathcal{L} T_t\phi-\phi

so when applied to the differential equation above

\lambda \mathcal{L} T_t\phi-\phi=A\mathcal{L} T_t\phi

or

\mathcal{L} T_t\phi=(\lambda I-A)^{-1} \phi=R(\lambda,A)\phi


Is this discussion worthy of inclusion? (Igny 19:27, 12 March 2007 (UTC))

Re: Is this discussion worthy of inclusion? Yes.--CSTAR 19:39, 12 March 2007 (UTC)
You might be able to shorten it, since the Hille-Ypsida theorem already is stated (implicitly) in terms of the Laplacc transform.--CSTAR 00:05, 13 March 2007 (UTC)