Maaneli
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humanino said:Hey guys, instead of arguing, can you help me out and tell me something like "no Clifford valued operator can be hermitian, thus observable", so you'll pull me out of confusion
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true it has not been published, but true as well it has been posted on the arXiv by a decent physicist
Sorry man, I don't know anything about this yet. It looks interesting, but it'll take me some time to say anything useful.