Timeline for Are the Gell-Mann matrices extremal when used as Kraus operators for a quantum channel?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2010 at 21:49 | vote | accept | sep332 | ||
| Jan 31, 2010 at 17:25 | comment | added | Ian Durham | Thanks Jon. I actually did use Flatten in one of my iterations and I indeed did get 9 but then I did something else that ended up checking linear independence and returning "True" but I think I implemented it wrong. Oh well! Something good at least came out of it - I found this website! | |
| Jan 31, 2010 at 6:56 | comment | added | Jon Yard | No problem Ian. Though, I think that the way you are checking linear independence might not be the best way (since it gives wrong results). To see if m matrices are linearly independent, you should just vectorize them (using Flatten) and make them the rows of a matrix A (as I think you've done). They are linearly independent iff Rank(A) = m. For this channel Mathematica tells me the rank is 9 (the max possible value) (without resorting to numerics), and this is obviously much smaller than the 64 you'd need for extremality. | |
| Jan 30, 2010 at 22:24 | comment | added | Ian Durham | It's just curious that Mathematica thinks the set \{A_{k}^{\dag}A_{l}\}_{k,l\ldots N} is linearly independent no matter how I check it. Well, thanks for the assistance. | |
| Jan 30, 2010 at 22:05 | comment | added | Jon Yard | Okay then. The theorem you are referring to is quoted in Landau & Streater but is really due to Choi: "Completely positive linear maps on complex matrices", Lin. Alg. Appl. 1975. Anyway, a simple corollary is that any CPTP map on d x d matrices is non-extremal as soon as it requires more than d Kraus operators. So there's no way this map is extremal because it requires 8 \geq 3 Kraus operators since the Gell-Mann matrices are lineraly independent. | |
| Jan 30, 2010 at 3:23 | comment | added | Ian Durham | OK, nevermind. I thought you could do it with complex vectors (i.e. I thought the theorem was only true for reals) but apparently not. Nevertheless, in dealing with really large data sets, I rely on Mathematica and, for it not to pick up on something as simple as this, is seriously worrying. | |
| Jan 30, 2010 at 0:50 | comment | added | j.c. | (1) I'd like to hear your "counterexample". I don't see how you can have more than n^2 linearly independent n by n matrices, unless our definitions of linear independence don't agree... (2) A good argument for never blindly trusting the results of Mathematica, I guess. | |
| Jan 29, 2010 at 21:45 | comment | added | Ian Durham | Just checked, and, indeed, I get lots of zeroes (in fact, that's what I did originally, several weeks ago). But I would argue it's not a silly question for two reasons: (1) I'm not convinced by your dimensionality argument since I can think of a simple counterexample (I seem to be running out of characters...?) and (2) even if I were wrong, it would still imply a serious limitation in Mathematica - it ought to pick up on something as ridiculously silly as that. | |
| Jan 29, 2010 at 21:14 | comment | added | j.c. | The problem is in the line reading linearIndependeceQ. It currently compares the last row of the row-reduced V matrix to {0,0,0}. However, the rows of V are 9-vectors (as they come from flattening 3 by 3 matrices). Hence you could change the Table[0,{3}] to Table[0,{9}]. This however would fail if there were numerical error (always a problem when comparing real numbers with equality in a computer). A better way to check is just to look at RowReduce[V] and see that you get a whole bunch of rows of zeros. But, like I said - the dimensionality argument makes this all kind of silly. | |
| Jan 29, 2010 at 21:04 | history | answered | Ian Durham | CC BY-SA 2.5 |