Multiple-Qubit Systems

[Genady]

(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.

[Genady]

1/2*(|00>+|11>) + 1/2*(|00>-|11>) = |00>

Edited October 10, 2023 by Genady

[Genady]

 

Let's assume it is not entangled with respect to such decomposition. Then it can be factorized (I skip the normalization factors for simplicity):

|W_n> = (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,

|W_n> = 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

[Genady]

|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

[Genady]

I will ask, with respect to what decomposition?

Edited October 11, 2023 by Genady

[Genady]

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⁺>)