The #algebra of infinite dimensional matrices has no nonzero trace. #math #geometrypic.twitter.com/xVIgGFgUZ9
Also what if it's an |ℝ|- dimensional hilbert space? In QFT we do use the property that tr(H)=∫d³x〈x|H|x〉.. But then again, say, for the QM SHO we use tr(H)=∑〈n|H|n〉.. is this not well defined then?
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I think the theorem just means you don't have a "reasonable trace" if your family is too big (e.g. "all bounded operators"). If you shrink the family of operators you consider you are probably fine. (perhaps in your case it's the trace-class operators)
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