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Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}\to\mathcal{H}:\quad N^*N=NN^*$$ and its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Regard a unitary transformation: $$U:\mathcal{H}\to\mathcal{K}:\quad U^*=U^{-1}$$

How to check that the transformed becomes: $$M:=UNU^{-1}=\int_\mathbb{C}\lambda\mathrm{d}UEU^{-1}(\lambda)=:\int_\mathbb{C}\lambda\mathrm{d}F(\lambda)$$


marked as duplicate by C-Star-W-Star, wythagoras, user91500, A.D, Tom-Tom Jun 24 '15 at 11:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


First show that if $E$ is a spectral measure then so is $UE(\cdot)U^*$.

Now note that for a spectral measure $UE(\cdot)U^*$ we have $$ \left<f, \int \limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^* g\right>= \int\limits_{\sigma(N)} z\ F_{f, g}(dz) \ \quad (f \in \mathcal{H}, g \in U \mathcal{D}),$$ where $F_{f, g}(\cdot):=\left<f, UE(\cdot)U^*g\right>$ is a countably additive measure on a Borel sigma-algebra over $\sigma(N)$ with total variation less or equal $\|f\|\|g\|$.

For any $f \in \mathcal{H}$ and $g \in U\mathcal{D}$ we have

$$\begin{align}\left<f, UNU^* g \right>&=\left<U^*f, NU^*g\right> = \left<U^*f, \left( \ \int\limits_{\sigma(N)}z \ E(\mathrm{d}z)\right)U^*g\right> \\&= \int\limits_{\sigma(N)} z \ \left<U^*f, E(\mathrm{d}z)U^*g\right>=\int\limits_{\sigma(N)} z\left<f, UE(\mathrm{d}z)U^*g\right>\\&=\left<f, \int\limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^*g \right>. \end{align}$$

Hence, $UNU^*= \int_{\sigma(N)}z \ UE(\mathrm{d}z)U^*$.

  • $\begingroup$ I was gonna post an answer anyway. Sorry if this was a misunderstanding. Anyway: (+1) $\endgroup$ – C-Star-W-Star Jan 20 '15 at 22:05

It remains a spectral measure: $$F(A):=UE(A)U^{-1}:\quad F(\varnothing)=0\quad F(\Omega)=1$$

There are no domain issues since: $$\|UE(A)U^*\psi\|=\|E(A)U^*\psi\|$$

But a formal calculation shows: $$U^*\psi\in\mathcal{D}(N):\quad\langle\varphi,UNU^*\psi\rangle=\int\lambda\mathrm{d}\langle U^*\varphi,E(\lambda)U^*\psi\rangle=\int\lambda\mathrm{d}\langle\varphi,UE(\lambda)U^*\psi\rangle\quad(\varphi\in\mathcal{H})$$

Concluding that the identity holds!

  • $\begingroup$ I don't understand why you asked this question if you knew the answer. It is a standard question not a tricky/interesting one. What is the point to write almost identical second answer ? If you want to be precise change $A$ to $\lambda$ in your longer sequence of equalities, also after second equality symbol (in a long sequence of equalities) you should have integral symbol inside the inner product and then take it outside (as you did after writing $\left<U^*\varphi, NU^*\psi\right>$). $\endgroup$ – m_gnacik Jan 20 '15 at 22:48
  • $\begingroup$ @mgn: Thanks for the typo. Having integral symbol inside the scalar product would be just a pointless symbolic step. I needed these intermediate results for the thread on Moeller operators. I found it worth it to have it recorded on MSE. For the Q&A issue see this blog and this meta $\endgroup$ – C-Star-W-Star Jan 20 '15 at 22:58
  • $\begingroup$ ok that's a good point. $\endgroup$ – m_gnacik Jan 20 '15 at 23:00

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