Genady Posted October 10, 2023 Share Posted October 10, 2023 (From Rieffel, Eleanor G.; Polak, Wolfgang H.. Quantum Computing: A Gentle Introduction.) Let's call vectors a=(1,0,0), b=(0,1,0), and c=(0,0,1) One answer is the basis {aa,ab,ac,ba,bb,bc,ca,cb,cc}. Let's take two different vectors in V: d=(0,sqrt(1/2),sqrt(1/2)) and e=(0,sqrt(1/2),-sqrt(1/2)). a, d, and e are orthonormal. The second answer is the basis {aa,ad,ae,da,dd,de,ea,ed,ee}. Of course, there are many other bases. Link to comment Share on other sites More sharing options...
Genady Posted October 10, 2023 Author Share Posted October 10, 2023 (edited) 1/2*(|00>+|11>) + 1/2*(|00>-|11>) = |00> Edited October 10, 2023 by Genady Link to comment Share on other sites More sharing options...
Genady Posted October 11, 2023 Author Share Posted October 11, 2023 (edited) Let's assume it is not entangled with respect to such decomposition. Then it can be factorized (I skip the normalization factors for simplicity): |Wn> = (a|0>+b|1>)...(y|0>+z|1>) Then, this state would have a term, a...y|0...0>. But there is no such term. It means that a...y=0. Let's say, a=0. Then, |Wn> = b|1>(...)...(y|0>+z|1>). Then, every term would have |1> for the first qubit, which is not so. Contradiction. Thus, such factorization is impossible, and the state is entangled with respect to such decomposition. Edited October 11, 2023 by Genady Link to comment Share on other sites More sharing options...
Genady Posted October 11, 2023 Author Share Posted October 11, 2023 (edited) |0>|+> + |1>|- > = |0>(|0>+|1>) + |1>(|0>-|1>) = |00>+|01>+|10>-|11> =(assume) (|0>+x|1>)(|0>+y|1>) = |00>+y|01>+x|10>+xy|11> This requires x=1, y=1, xy=-1, which is impossible. Thus, the state is entangled. Edited October 11, 2023 by Genady Link to comment Share on other sites More sharing options...
Genady Posted October 11, 2023 Author Share Posted October 11, 2023 (edited) I will ask, with respect to what decomposition? Edited October 11, 2023 by Genady Link to comment Share on other sites More sharing options...
Genady Posted October 11, 2023 Author Share Posted October 11, 2023 For the reference, the Bell basis: a. = 1/√2(|Φ+>+|Φ->) b. = 1/2(|0>+|1>)(|0>-|1>) = 1/2(|00>-|11>+|10>-|01>) = 1/√2(|Φ->-|Ψ->) c. = 1/√3(1/√2(|Φ+>+|Φ->)+√2|Ψ+>) Link to comment Share on other sites More sharing options...
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