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This question is duplicate of the question Find projection-valued measure associated with parity operator.\

But in that question @Jacky Chong does not state how he found the operator \begin{align} P_\pi(\lambda) =P_{\pi}((-\infty,\lambda]) =\delta(\lambda-1)P_\text{even}+\delta(\lambda+1)P_\text{odd} \end{align}

Also we know that $P(R)=Id \enspace$ or we should have $P(\lambda_1)\leq P(\lambda_2)$ for $\lambda_1 < \lambda _2$ but if we put 2 and 1 into the $P_{\pi}$ we get $0:L^2\rightarrow L^2$ and $P_{even}$ resp.

My question is could someone explain me how to find $P_{\pi}$ explicitly using maybe Stieltjes inversion formula or something else

Notes:
(1) I need 50 reps to comment on the problem i linked
(2) I already found the spectrum and resolvent of parity operator.They're $$\sigma(\Pi)=\{1,-1\} \enspace R_{\Pi}(z)=\frac{\Pi+zI}{1-z^2}$$ (3) This is Problem 3.2 from Teschl G. - Mathematical methods in quantum mechanics

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This is simply the standard way of writing a finite sum as an integral relative to a corresponding atomic measure. In particular: if the operator in question can be written as a countable sum $$ A = \sum_i \lambda_i P_i $$ (with the $P_i$ equal to spectral projectors and hence mutually orthogonal and "complete"), then the measure associated with $A$ can be written as $$ P_A(\lambda) = \sum_i \delta(\lambda - \lambda_i)P_i. $$ To see that this works, it suffices to show that for a Borel function $f$, $f(A)$ is indeed equal to $\int_{\Bbb R} f(\lambda) \,dP_{A}(\lambda)$. In other words, this is simply a consequence of the fact that $$ f\left(\sum_i \lambda_i P_i\right) = \sum_i f(\lambda_i) P_i. $$

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  • $\begingroup$ First thank you for your answer.So by sayig "spectral projectors" what you mean is their sum is identity like if all the $\lambda_i=1$ and by making some of the $\lambda_i=0$ or something else we get $A$.In our case we have $P_{even}$ and $P_{odd}$ and their $\lambda$'s 1 and -1 resp. Also $\delta$ is dirac-delta right?It is not heaviside or whatever the name it is.Could you explain why it suffices to show that $f(A)=\int_{\mathbb{R}}f(\lambda)dP_A(\lambda)$ because it forms a $C*$ homomorphism ? but there could be some other things that might be $C*$ homo without being PVM. $\endgroup$ – Oğuzhan Kılıç Oct 28 at 8:45
  • $\begingroup$ and i wrote an example that if we had $\lambda_2=2$ we get $0$ map but we should have had a map that extends $P_{even}$.Maybe i'm understanding it wrong but im confused little bit :D.Also sorry for late return $\endgroup$ – Oğuzhan Kılıç Oct 28 at 8:45
  • $\begingroup$ For your first comment: all your statements are correct. The precise answer to why it suffices depends on what exactly it means for a projection valued measure to be "associated with" an operator. I'm rusty with all this and don't have a reference on hand; it would be helpful if you could consult your textbook or notes for how exactly this is defined. $\endgroup$ – Ben Grossmann Oct 29 at 1:22
  • $\begingroup$ I'm not exactly sure what you're saying about your example $\endgroup$ – Ben Grossmann Oct 29 at 1:23
  • $\begingroup$ i kind of get it now. I was confusing identity function and 1 function.Thanks for your answers anyway $\endgroup$ – Oğuzhan Kılıç Oct 29 at 9:08

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