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I have a conceptual question about the resolution of the identity in quantum mechanics.

For a continuous spectrum, a self-adjoint operator $F$ can be written in the following way $$ F = \int \lambda\,dE_F(\lambda) = \int \lambda\frac{dE}{d\lambda}d\lambda = \int \lambda |\lambda\rangle\langle\lambda|d\lambda $$ where $E_F(\lambda)$ is the projection operator.

Now suppose we want of find out the resolution of the identity $E_{F^2}(\lambda)$ for $F^2$.

By definition we have $$ F^2 = \int \lambda^2\,dE_F(\lambda) = \int \lambda\,dE_{F^2}(\lambda) $$ And I don't know how to proceed to write out $E_{F^2}$ in terms of $E_F$. I also wonder if there is a way to find the resolution of the identity for any operator $g(F)$ that is a function of $F$?

LarryC
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  • What is $E_F$ and why are we integrating over it? – DanielSank Oct 25 '18 at 04:16
  • @DanielSank The Stiltjes measure on the parameter space. – DanielC Oct 25 '18 at 04:40
  • Ping @ValterMoreti – DanielC Oct 25 '18 at 04:42
  • That's the projection-valued measure. – Max Lein Oct 25 '18 at 04:52
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    I mean, why wouldn't it just be $g(F)=\int d\lambda~g(\lambda)~|\lambda\rangle\langle\lambda|?$ I mean you can renumber your basis vectors if you really want but it seems like a painful choice and that construction would appear to give the result for any $|\lambda'\rangle$ that $g(F)~|\lambda'\rangle=g(\lambda') ~|\lambda'\rangle,$ no? – CR Drost Oct 25 '18 at 04:58

1 Answers1

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The two projection-valued measures (PVMs) are related in a straight-forward fashion. For better legibility, let me call the spectral parameter for $F^2$ with the letter $\mu$ instead of $\lambda$. The spectrum of $F^2$ really equals $\bigl ( \sigma(F) \bigr )^2 = \sigma(F^2)$. Moreover, the equality you give tells you that the level set \begin{align*} U_{\mu} := \bigl \{ \lambda \in \sigma(F) \; \; \vert \; \; \mu = \lambda^2 \bigr \} \end{align*} is crucial in relating the two PVMs, so that morally speaking you have \begin{align*} \mathrm{d} E_{F^2}(\mu) = \sum_{\lambda \in U_{\mu}} \mathrm{d} E_{F}(\lambda) . \end{align*} To make this rigorous, you have to consider subsets of the spectra which are related by the function $q(\lambda) = \lambda^2$, considered as a function from $\mathbb{R}$ to $\mathbb{R}$. If $\Omega$ is any Borel set, then the associated PVMs are related by \begin{align*} E_{F^2}(\Omega) = E_{F} \bigl ( q^{-1}(\Omega) \bigr ) . \end{align*} This generalizes the “differential” version since the level set $U_{\mu} = q^{-1}(\{ \mu \})$ is just the preimage of the square function $q$.

Max Lein
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  • Thanks for the detailed explanation. I roughly understand what you're trying to say here. Just to confirm, can I rewrite the equality in the following way: $F^2 = \int \lambda^2 E_F(\lambda) = \int \mu E_{F^2}(\mu) = \int (\sqrt{\mu})^2 E_F(\sqrt{\mu}) $ so that $E_{F^2}(\lambda) = E_{F}(\sqrt{\lambda})$? – LarryC Oct 25 '18 at 17:44
  • The first two equalities are correct, the last one is not. Consider the momentum operator $- \mathrm{i} \partial$ on $L^2(\mathbb{R})$ endowed with the usual domain. The spectrum is all of $\mathbb{R}$. Compare that to $- \Delta$ (again equipped with the usual domain), whose spectrum is $[0,\infty)$. The PVM of $-\Delta$ at $\mu$ gets a contribution from $- \mathrm{i} \partial$'s PVM at $+ \sqrt{\mu}$ and $- \sqrt{\mu}$. Your last equality picks up only the positive part of that. – Max Lein Oct 31 '18 at 06:14