tag:blogger.com,1999:blog-3599722177679860131.post1626833539728391543..comments2022-11-23T11:24:15.248+01:00Comments on Chemical Quantum Images: Some more exciting mathFelixhttp://www.blogger.com/profile/05138335803929997277noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-3599722177679860131.post-39633180952969331972008-04-29T09:42:00.000+01:002008-04-29T09:42:00.000+01:00yes, thanks it's only for real scalars. i changed ...yes, thanks it's only for real scalars. i changed itFelixhttps://www.blogger.com/profile/05138335803929997277noreply@blogger.comtag:blogger.com,1999:blog-3599722177679860131.post-44306287004516307692008-04-28T23:23:00.000+01:002008-04-28T23:23:00.000+01:00Not a constructive comment (sorry), but the proof ...Not a constructive comment (sorry), but the proof is pretty; I like the integration by parts.<BR/><BR/>Actually, for the claim that the scalar multiple of a self-adjoint operator is self-adjoint ("You can do similar things with summation or scalar multiplication"), isn't it false for complex scalars? Or am I misunderstanding something? <BR/><BR/>(Using square brackets because I can't type the angle ones)<BR/>Given that [f, Ag] = [Af, g] for operator A and complex scalar &lambda,<BR/><BR/>[f, &lambda Ag] = &lambda [f, Ag]<BR/>= &lambda [Af, g]<BR/>= [&lambda *Af, g]<BR/>!= [&lambda Af, g] <BR/>if &lambda isn't realEchiralhttps://www.blogger.com/profile/02207127108260799588noreply@blogger.com