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I'm going to ask 2 questions but they're (I believe) related.

1) Let $z_0$ be an eigenvalue and $\psi$ a corresponding normalized eigenvector. Compute $μ_{\psi}$.
This is the whole question but we could assume operator $A$ which we know has eigenvalue $z_0$ is self-adjoint and $\mu_{\psi}(\Omega)$ is given by $\langle\psi,P_A(\Omega)\psi\rangle$ where $P_A$ is associated PVM to $A$.

2) Show that $z_0$ is an eigenvalue if and only if P({$z_0$})$\neq0$. Show that $Ran(P(\{z_0\}))$is the corresponding eigenspace in this case. Again we could assume $A$ is self-adjoint.

I've already (kind of) answered both of the questions but reason for why i'm asking these questions is both questions seems very intuitive like the moment i saw the first question i immediately said if $\Omega$ contains $z_0$ it is $||\psi||^2$ if not it is $0$. And i would like to answer them more formally \

3) Under the shadow of these 2 questions i would like to ask another question.We know spectrum $$\sigma(A)=\{z\in\mathbb{R} : P_A(z-\epsilon,z+\epsilon)\neq0 \enspace \forall\epsilon>0\}$$ but this does not differentiate between different types of spectrums pure point,point embedded in continuum and purely continuousSchullers lectures definition of spectrum for self adjoint operator.Clearly eigenvalues are in point part of the spectrum and above results shows us $P(\{z_0\})$ is an eigenspace but what happens if $z_0\in$ purely cont.?Could someone give me an example(maybe example from physics like quantum harmonic oscillator etc.)?

Related links Show that eigen-vectors belong in range of projection valued measure. Spectral measure associated to eigenvector of self-adjoint operator
Any hint and solution is appreciated THANKS!

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  • $\begingroup$ Quick note: for angled brackets, expressions are better formatted if you use \langle, \rangle instead of <,>. $\endgroup$ Commented Oct 29, 2020 at 14:45
  • $\begingroup$ Also, I'm not sure what you're trying to do with the \ before each new line. If you're trying to get a new line without a paragraph break, then the way to do this is by putting four spaces at the end of your line before going to the next line $\endgroup$ Commented Oct 29, 2020 at 14:46
  • $\begingroup$ @Ben Grossmann Thanks :D i'm fixing them now $\endgroup$ Commented Oct 29, 2020 at 14:50
  • $\begingroup$ Have you seen this wiki page? I think this answers most of your question $\endgroup$ Commented Oct 29, 2020 at 15:13
  • $\begingroup$ Most of your third question, that is. $\endgroup$ Commented Oct 29, 2020 at 15:20

2 Answers 2

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  1. I'm not sure if there's a way to see this directly from the definition of $P_A$, but here is a proof using the resolvent, which is quite natural in view of the proof of the spectral theorem. With $R_A(z)=(A-zI)^{-1}$, we know that $$\langle\psi,R_A(z)\psi\rangle=\langle \psi,\frac{1}{z_0-z}\psi\rangle=\frac{1}{z_0-z}||\psi||^2$$ This is in turn, by the spectral theorem, equal to $F(z):=\int_{\mathbb{R}}\frac{1}{\lambda-z}\,d\mu_{\psi}(\lambda)$, which is the Borel transform of $\mu_\psi$. We may recover a measure from its Borel transform via the Stieljes inversion formula: $$\mu_\psi(\lambda)=\lim_{\delta\downarrow0}\lim_{\epsilon\downarrow 0}\frac{1}{\pi}\int_{-\infty}^{\lambda+\delta}\text{Im}(F(t+i\epsilon))\,dt$$ I suggest your try this computation, but if you cannot get it I am happy to include details. The key point is that the integrand has a singularity as $\epsilon\rightarrow 0$ precisely at $t = z_0$.

  2. If $\psi$ is an eigenvector corresponding to $z_0$ then $$0\neq\langle \psi,\psi\rangle=\int_{\mathbb{R}}\,d\mu_\psi=\int_{\{z_0\}}\,d\mu_\psi=\langle \psi,P(\{z_0\})\psi\rangle$$ where we have used the fact that $\mu_\psi$ is a point-mass from $1.$ Conversely, if $P(\{z_0\})\neq 0$, we may find $\psi$ such that $P(\{z_0\})\psi=\psi$ because $P(\{z_0\})$ is a projection. This also means that $P(\mathbb{R}\setminus\{z_0\})x=0$. Now, the result follows from DisintegratingByParts answer here. They are proving the backwards implication (more or less), but if you look at the argument it works to prove what we want by tracing it in reverse.

  3. The canonical example of an operator with purely ac spectrum is the free energy $-\Delta$, where $\Delta$ is the Laplacian. For this fact, see Theorem 7.8 in Mathematical Methods in Quantum Mechanics by Teschl. Teschl's book is an excellent reference for this material and you may find it free online. While extremely important, this example is a little hard to make sense of because $-\Delta$ is an unbounded operator. For a simpler example, you may try to show that $f(x)\mapsto xf(x)$ on $L^2([0,1])$ has purely ac spectrum.

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  • $\begingroup$ actually these are problems from Teschl book 3.9 and 3.10.I'm combining Teschl book with Schuller's lectures (They're on youtube and has a pdf version).Also if you found the $\mu_{\psi}$ using inversion formula then ok i would try my best to actually find it because this formula is the first one that come to my mind but it seemed too tricky to actually come up with something.and thanks :) $\endgroup$ Commented Oct 29, 2020 at 16:08
  • $\begingroup$ I figured they might be from Teschl but I wasn't sure. From looking at a lot of sources I think it's the easiest to read on this topic. Anyway, you shouldn't be afraid of this integral computation. It's not that bad if you actually write it out and similar computations come up a lot. I will try to think of a simpler proof though. $\endgroup$
    – user293794
    Commented Oct 29, 2020 at 18:02
  • $\begingroup$ Essentially, it is the computation in the proof of Stone's formula, which you should learn about. $\endgroup$
    – user293794
    Commented Oct 29, 2020 at 18:06
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Claim: $A\psi=\lambda\psi$ for some $\psi\ne 0$ iff $$ P(\{\lambda\})\psi = \psi. $$ Proof: First assume that $A\psi=\lambda\psi$ for some $\psi\ne 0$. Then $$ 0=\|(A-\lambda I)\psi\|^2=\int_{-\infty}^{\infty}|\mu-\lambda|^2d\rho(\mu), $$ where $\rho(S)=\langle P(S)\psi,\psi\rangle= \|P(S)\psi\|^2$ is the measure associated with $\psi$. It follows that $\rho$ is concentrated only at $\lambda$; indeed, if $\rho$ had positive mass on $\mathbb{R}\setminus\{\lambda\}$, then the above equality could not hold. Therefore, $$ \psi=P(\{\lambda\})\psi. $$ Conversely, if $\psi=P(\{\lambda\})\psi$ for some $\psi\ne 0$, then $P(\mathbb{R}\setminus\{\lambda\})=0$, which gives $$ A\psi=\int_{\mathbb{R}}\mu dP(\mu)\psi=\int_{\{\lambda\}}\mu dP(\mu)\psi=\lambda P(\{\lambda\})\psi=\lambda \psi. $$ Q.E.D.

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