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[mentor's note: This thread was forked from https://www.physicsforums.com/threads/is-the-time-derivative-hermitian.791879/ when it looked to to be raising issues beyond the original question. Refer back to that thread for any missing context]

The same argument holds true if time and Hamilton operator obey this commutation relation, and this they should, if you consider time as an observable in the sense of QT, because by definition the Hamiltonian is the generator for time translations.

Can you elaborate on this statement? For position and momentum you find the derivation in many textbooks that, if you have two operators with commutator ##[\hat{x},\hat{p}]= \mathrm{i} \hat{1}##, because momentum is the generator for spatial translations and position the generator for momentum translations. This implies that the spectrum of both operators is the entire real line. So, where is this argument flawed?vanhees71, your argument in the first paragraph is mathematically flawed. It's not necessary for both operators to have unbounded spectra.

The same argument holds true if time and Hamilton operator obey this commutation relation, and this they should, if you consider time as an observable in the sense of QT, because by definition the Hamiltonian is the generator for time translations.

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