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Jul 23, 2021 at 19:20 comment added Quantum Mechanic @NorbertSchuch you don't think? The question asked about using $\rho_{SA}=\rho_S\otimes\mathbb{I}$... ahh, your answer is much more in the spirit of the question, this is explicitly just the letter
Jul 23, 2021 at 17:04 comment added Norbert Schuch @QuantumMechanic This is not at all in the spirit of the question.
Jul 15, 2021 at 2:06 vote accept Brown Hole
Jul 14, 2021 at 19:36 comment added Quantum Mechanic Notably, in the spirit of the question being posed, one can choose $\rho_A=\mathbb{I}/\mathrm{Tr}(\mathbb{I})$.
Jul 14, 2021 at 19:28 history answered Rammus CC BY-SA 4.0