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$\begingroup$ I am unable to see why $A'$ is self adjoint, or should the definition include $PBP$ instead of just $BP$ (or is $PB=BP=B$)? $\endgroup$
– Primo-uomo
Commented Jul 3, 2017 at 15:02
$\begingroup$ Right, I forgot a factor $P$... corrected. Actually, it was not necessary since $Ran(B) \subset \cal H_E$, it holds $PB=B$... $\endgroup$
– Valter Moretti
Commented Jul 3, 2017 at 15:24
$\begingroup$ As a clarification, this argument could be extended to a general Abelian group simply because the key point here, is that one can find a non-commuting operator if n>1, yes? In addition to this, could you please explain how the commutation relations between $A$ and $H$, and $P$ come to be? Thank you so much! $\endgroup$
– Primo-uomo
Commented Jul 3, 2017 at 15:29
$\begingroup$ Yes!... I added a comment regarding commutativity $\endgroup$
– Valter Moretti
Commented Jul 3, 2017 at 15:43

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