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Aug 19, 2020 at 9:31 comment added JSdJ You're welcome! Daftwullie's answer is indeed really nice :)
Aug 19, 2020 at 9:27 comment added Marco Fellous-Asiani Thanks for your answer/comment. It solved a part of my problem but DaftWullie gave me the exact step I was looking for so I accepted his answer instead. Thanks anyway for your instructive answer !
Aug 18, 2020 at 19:06 comment added JSdJ You might be able to use the first identity on this wikipedia page about the vectorization operator. (Coincidentally, this is also identity you can use to arrive at a 'natural representation' of a channel as a big matrix, which you asked about in one of your other questions today)
Aug 18, 2020 at 19:05 comment added JSdJ You're welcome! The $d$ is indeed because of normalization - I've been very sloppy with normalization in the above derivation (I also swept under the rug that the $a$ and $b$ vectors aren't properly normalized Hilbert-space vectors, but you can check that it works out in the end (if you still don't agree, I would be interested in a discussion :) )). I'm not $100 \%$ sure about the second question though - the partial trace is a vital part of the derivation so I don't think that the mapping works always. I'll have to think about it more.
Aug 18, 2020 at 18:54 comment added Marco Fellous-Asiani Thank you for your answer. Two questions: why do you have a $d$ that multiplies $tr_2$ in your definition of the isomorphism ? Is it because in your convention $| \Omega \rangle$ is normalized (whereas in mine it is not) ? Second question: does that mean in an indirect way that indeed any vector belonging in $H_1 \otimes H_0$ can be written under the form $A \otimes \mathcal{I} | \Omega \rangle$ ?
Aug 18, 2020 at 18:39 history answered JSdJ CC BY-SA 4.0