(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.

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

Edited October 10, 20232 yr by 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, 20232 yr by 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, 20232 yr by Genady

I will ask, with respect to what decomposition?

Edited October 11, 20232 yr by 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⁺>)