Absolutely continuous spectrum invariant under unitary equivalence

I am doing an exercise calculating the absolutely continuous spectrum of some operator $A$ by calculating the spectrum of a different operator $B$ unitarily equivalent to $A$, i.e. $$A = U B U^*.$$ I know that $\sigma (A) = \sigma(B)$ as is easily seen by playing with the resolvent. But I cannot as easily conclude $\sigma_{ac}(A) = \sigma_{ac}(B)$. Am I missing something obvious? How can one show this? I first thought about something along $$\sigma_{ac}(A) = \sigma(A \mid_{\mathcal H_{ac}}) = \sigma(UBU^* \mid_{\mathcal H_{ac}}) \stackrel{?}{=} \sigma(U B \mid_{\stackrel{\sim}{\mathcal H_{ac}} }U^*) = \sigma_{ac}(B),$$ but I don't think the third equality is justifiable, especially we have different Hilbert spaces $\mathcal H, \stackrel{\sim}{\mathcal H}$.

Solution 1:

For each $h\in H$, you have the corresponding spectral measure $\mu_h^A$. Since $$ \langle f(UAU^*)h,h\rangle=\langle f(A)U^*h,U^*h\rangle, $$ it follows that $$\mu_h^{UAU^*}=\mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get $$\mu_{Uh}^{UAU^*}=\mu_{h}^A,\ \ h\in H.$$ From this you can deduce that $h\in H_{ac}^A $ if and only if $Uh\in H_{ac}^B$. In other words, $U:H_{ac}^A\to H_{ac}^B$ is an isomorphism. This shows exactly what you want $$ UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^* $$