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Landau and Streater proved that a set of Kraus operators, Ai, is extremal if and only if the set

$\{A_{k}^{\dagger}A_{l}\}_{k,l \ldots N}$

are linearly independent. I have seen very convincing arguments both for and against. You can even see two PDFs of Mathematica notebooks "proving" both answers here:

What is missing from these proofs?

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I think you mean Streater, and not Streeter. Also, try using \dagger and not \dag - it may like the LATEX better. Anyway, I think the theorem says that the CPTP map $\rho \mapsto \sum_k A_k \rho A_k^\dagger$ is extremal in the convex set of all CPTP maps. The way you've written this is not really right, as it makes no sense to say the set of Kraus operators themselves are extremal and nobody outside quantum information will have any idea what you are asking. – Jon Yard Jan 29 '10 at 19:05
@Jon Thanks for the LaTeX hint and the Streater correction. I will try to rephrase the question so that it makes more sense. – sep332 Jan 29 '10 at 19:15
I just looked at your Mathematica code and your check for linear independence is wrong. You compare the last row of "V" with a row of three zeros, not nine. Actually the whole thing is moot since you are checking to see if 64 3 by 3 matrices are linearly independent - well the space of 3 by 3 matrices is only 9 dimensional, so there's no way that they could be linearly independent. – j.c. Jan 29 '10 at 19:48
up vote 1 down vote accepted

Hi Jon. Actually, the requirement that the set $A_{i}^{\dagger}A_{j} \oplus A_{j}A_{i}^{\dagger}$ be linearly independent is specifically for extremal unital channels. If the requirement on unitality is relaxed, Landau and Streater showed that only the set $A_{i}^{\dagger}A_{j}$ need be linearly independent.

jc: I see your point, but then can you tell me what is wrong with my Mathematica code? In other words, there must be something wrong with it if it returns a "True" for linear independence. Maybe someone has a better way to check these in Mathematica.

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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. – j.c. Jan 29 '10 at 21:14
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. – Ian Durham Jan 29 '10 at 21:45
(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. – j.c. Jan 30 '10 at 0:50
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. – Ian Durham Jan 30 '10 at 3:23
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. – Jon Yard Jan 31 '10 at 6:56

Actually, I think the channel is not extremal because I suspect you are misquoting the Landau-Streater result. So I will state it here.

To be precise, for anyone unfamiliar with the field, a quantum channel is a trace-preserving, completely-positive linear map on density matrices (positive semidefinite matrices with unit trace), of potentially different sizes. A basic theorem in quantum information says that every quantum channel from $m\times m$-dimensional to $n\times n$-dimensional density matrices can be written in Kraus form: $$ \rho \mapsto \sum_{i=1}^N A_i \rho A_i^\dagger, \text{ for linear operators } A_k \colon \mathbb{C}^m \to \mathbb{C}^n \text{ satisfying } \sum_k A_k^\dagger A_k = I_m. $$

It is easy to show that the set of quantum channels between systems of fixed dimension is convex. It also easy to show that the set of channels that map $\frac{1}{m} I_m$ to a fixed density matrix $\sigma$ is convex. Now the theorem of Landau-Streater says that if $m = n$, a channel with Kraus form as above is extremal in this latter set if and only if the $N^2$ linear operators $A_i^\dagger A_j \oplus A_j A^\dagger_i$ (of size $2m \times 2m$) are linearly independent. It seems you have instead been working with $m\times m$ matrices. But I think that even if you were to continue and apply the theorem correctly, you would only prove or disprove extremality in the convex subset of unital channels, i.e. those for which $\frac 1m I$ is a fixed point. So potentially you could strengthen Ben-Or's conclusion by showing non-extremality in this subset, or otherwise you might conclude extremality there, which would tell you nothing about extremality in the entire set of channels.

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This thread seems more or less dead, but there is one minor correction I must make to Jon's statement of the theorem. Landau and Streater's result assumes that the set of Kraus operators $\lbrace A_i\rbrace$ is linearly independent to start with; otherwise the theorem is not true. A trivial case of this would be to pad with 0 operators. Otherwise Jon's write up is excellent.

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