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Dec 28, 2014 at 21:38 comment added Red Pencil Dunno why I was struggling with this thing. I think the main problem was that I did not think of pulling out $\langle ψ_1|ψ_2\rangle$ out of the trace... :x Cheers!
Dec 28, 2014 at 21:35 comment added Phoenix87 yes, provided $|\phi_i\rangle$ form an orthonormal basis of the Hilbert space and you allow the sum to be an integral when needed
Dec 28, 2014 at 21:31 comment added Red Pencil Thank you! Is this justified by: $tr(|ψ_1\rangle \langleψ_2|) = \sum_i \langle φ_i|ψ_1\rangle \langle ψ_2|φ_i\rangle = \sum_i \langle ψ_2|φ_i\rangle \langle φ_i|ψ_1\rangle = \langle ψ_2|ψ_1\rangle$?
Dec 28, 2014 at 21:05 vote accept Red Pencil
Dec 28, 2014 at 20:44 history answered Phoenix87 CC BY-SA 3.0