6

Let $A$ be a bounded or unbounded selfadjoint linear operator on a complex Hilbert space $H$ with spectral representation $A=\int_{\sigma}\lambda \, dE(\lambda)$ given by the Spectral Theorem for Selfadjoint Operators. Stone's Formula gives a way to constructively obtain the spectral measure $E[a,b]$ of a finite interval as a strong (vector) limit: $$ \frac{1}{2}\{E[a,b]+E(a,b)\}x=\lim_{\varepsilon\downarrow 0}\frac{1}{2\pi i}\int_a^b \{R(u+i\varepsilon)-R(u-i\varepsilon)\}x\,du,\;\;\; x \in H, $$ where $R$ is the resolvent operator $R(\lambda)=(A-\lambda I)^{-1}$.

Question: What is the easiest way to prove that Stone's Formula is valid?

Application: This may be viewed as a generalized residue of the resolvent function. If $A$ has isolated spectrum, then such an integral may be evaluated using residues of the resolvent.

Michael Hardy
  • 1
Disintegrating By Parts
  • 92,626
  • 6
  • 78
  • 173
  • Great question @T.A.E., actually I thought about it recently and wanted to ask it here. – m_gnacik Nov 06 '14 at 19:55
  • @mgn : Thanks. It's interesting to me that Spectral Theorem methods--at least for separable spaces--are constructive in nature. Methods used in Quantum Mechanics must be constructive, or they cannot be taken seriously. – Disintegrating By Parts Nov 06 '14 at 21:50

1 Answers1

4

I think that an easy method is delivered by functional calculus.

Consider the right-hand side of your equation as $$ \lim_{ \varepsilon \downarrow 0} \frac{1}{2\pi i}f_{\varepsilon, a, b}(A)x, $$ where $$f_{\varepsilon, a, b}(t) = \int_{a}^b \left(\frac{1}{t-(u+i\varepsilon)} - \frac{1}{t-(u-i\varepsilon)}\right) \mbox{d}u.$$

We can show that (by doing calculations) $$\lim_{ \varepsilon \downarrow 0} \frac{1}{2\pi i}\int_{a}^b \left(\frac{1}{t-(u+i\varepsilon)} - \frac{1}{t-(u-i\varepsilon)}\right) \mbox{d}u = \begin{cases} 1 & \mbox{if } t \in (a,b) \\ \frac{1}{2} &\mbox{if } t = a \\ \frac{1}{2} &\mbox{if } t = b \\ 0 &\mbox{if }t\notin [a, b] \end{cases}$$

If now you apply functional calculus to the right-hand side of above experssion you will obtain excatly $$ \frac{1}{2}\left(E[a,b] + E(a,b)\right)x.$$

m_gnacik
  • 3,403
  • 1
    +1. Minor typo: I think you meant "$+$" in your last equation instead of "$-$". Also, to be completely correct, we will need some estimate like $|f_{\varepsilon, a,b}| \leq C$ for all $\varepsilon > 0$, I think. – PhoemueX Nov 07 '14 at 14:44
  • 2
    Yes, an estimate would be needed. $\int_{a}^{b}\frac{\varepsilon}{(t-u)^{2}+\varepsilon^{2}},du$ has an explicit form in terms of arctan, which is uniformly bounded. Thank you. – Disintegrating By Parts Nov 07 '14 at 15:25
  • 1
    Yes, indeed I actually converted these two integrals to the above one (in @T.A.E comment, since it is easier to notice that it is $\mbox{arctan}$ and then calculate a limit). I didn't have time to include all the things I was rushing a bit, so I'm glad that I didn't make some more mistakes and functional calculus methods worked:). – m_gnacik Nov 07 '14 at 19:47
  • 2
    @mgn : Amazing how well the functional calculus works for this case. It's really made to order, isn't it? – Disintegrating By Parts Nov 07 '14 at 21:30
  • Won't you need interchange of limits somewhere? The way it is proven up to now is sort of: $\frac12{\ldots}=\frac12{\ldots}$ – freishahiri Feb 24 '15 at 20:03
  • Sorry, but I don't understand your comment. Could you please be a bit more precise? – m_gnacik Feb 25 '15 at 10:05