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Welcome Ggf4t!

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Sincerely, – Paine EllsworthCLIMAX! 11:28, 6 February 2015 (UTC)   (Leave me a message)[reply]

Because you thanked me

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Ggf4t, you thanked me for one of my recent edits, so here is a heart-felt...
 YOU'RE WELCOME!
It's a pleasure, and I hope you have a lot of fun while you edit this inspiring encyclopedia phenomenon! – Paine EllsworthCLIMAX!

11:28, 6 February 2015 (UTC)

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Split exact sequence

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Hi Luca, just letting you know that there is a small mistake in the commutative diagram you added to the split exact sequence article – the map in the bottom row should be the inclusion , not the identity (thank you for adding a commutative diagram though!). Joel Brennan (talk) 13:54, 6 December 2021 (UTC)[reply]

Thanks for pointing that out! Should be fixed now Luca (talk) 14:02, 6 December 2021 (UTC)[reply]

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Fidelity of quantum states

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Could I interest you in making some edits to Fidelity of quantum states where I have a major conflict of interest? I see you have expertise in this area.

In the subsection Consistency with fidelity between probability distributions it is noted that the fidelity formula can be simplified for the case that and commute to give , and that this then simplifies further in terms of the eigenvalues of and . However we have shown in Efficiently computing the Uhlmann fidelity for density matrices (arxiv version here) that the form can be used in the general case, for any and (the next step, in terms of eigenvalues, does require commutation).

Would it be possible to add this result? Thanks, Jonathan A Jones (talk) 11:10, 25 September 2023 (UTC)[reply]

Hello! Sorry, yes, I agree it would be useful to add that to the page, and I understand the conflict of interest situation. I started doing it but then ended up modifying a bunch of other things because I was a bit annoyed by the overall structuring of the page. I'm planning to add the expression, but I wanted to understand the paper a bit better first. I'll get to it, just give me a few days.
Interesting paper by the way. I assume you're familiar with this quantumcomputing.SE post? It's where I first heard about that simplified expression for the fidelity a few years ago. It might be nice to link your paper there as well. Either as a comment, or even better as an answer if you feel like going into a more detailed explanation of the main arguments. Luca (talk) 16:20, 26 September 2023 (UTC)[reply]
That would be great, many thanks. As you can see I also was unhappy with some of the structure of the page, and have tweaked some of your improvements. Our result was independently rediscovered by Adrian Muller, and he put it on Scirate where Bartosz Regula pointed him at our paper, an earlier equivalent result from Sarah Plosker's group, and that stackexchange post. It looks to me like this result keeps on being rediscovered, and so it would be good to have it on the Wikipedia page so it doesn't get forgotten yet again. Jonathan A Jones (talk) 19:24, 26 September 2023 (UTC)[reply]
Armed with these other versions it's possible to come up with a much simpler proof. From the fact that the Uhlmann form is normal by construction you can use the spectral decomposition to work out the fidelity in the eigenbasis, leading to the square of the sum of the sqare-roots of the eigenvalues form. Then you can use the fact that the the eigenvalues of a matrix product are invariant uder cyclic permutation to get the eigenvalues from those of . Note that this doesn't have to be normal, because you apply the spectral decomposition argument to a different matrix with the same eigenvalues which is normal. The invariance of the eigenvalues under cyclic permutation was new to me, but was described to me by Plosker as "well known", and Muller has a simple sketch proof in his paper. Jonathan A Jones (talk) 19:34, 26 September 2023 (UTC)[reply]
I have been adding stuff at the stackexchange page as you suggested, as has Adrian Muller. Let me know if there's anything more I can help with on possible edits to the Wikipedia page. Jonathan A Jones (talk) 18:09, 13 November 2023 (UTC)[reply]

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