Jack Sarfatti's Loophole Of No.communication Theorem

[Aethelwulf]

As calculated by Keith Kenemer

 

''I get the same result you get if I do the calculation as follows:

 

[math]|Alice, Bob> = a_0 |AB> + a_1 |A'B'>[/math]

 

where [math]|AB> = |A>\otimes|B>[/math] and [math]|A'B'> = |A'> \otimes |B'>[/math]

 

[ and [math]\otimes[/math] = tensor product ]

 

density matrix [math]\rho = |A,B><A,B|[/math]

 

[math]\rho = |a_0|^2 |AB><AB| +a_0a_1*|AB><A'B'| + a_1a_0*|A'B'><AB|+|a_1|^2|A'B'><A'B'|[/math]

 

and consider [math]P(|B>) = P(|AB>) + P(|A'B>)[/math] since non-zero overlap of [math]<A|A'>[/math] means that state [math]A'[/math] has a non-zero probability of actually being [math]A[/math] (which is entangled with B and would contribute to [math]P(|B>)[/math]...

 

 

P|AB>= <AB|\rho|AB> = |a_0|^2 <AB|AB> <AB|AB> + a_0a_1* <AB|AB> <A'B'|AB> + a_1a_0* <AB|A'B'> <AB|AB> + |a_1|^2 <AB|A'B'><A'B'|AB>

 

Using [math][X\otimes Y] [Z\otimes W] = XZ\otimes YW[/math], and assuming [math]<B|B'>=0[/math], all terms except the first will drop out, so

 

[math]P(|AB>) = |a_0|^2[/math]

 

Next,

 

P(|A'B>) = <A'B|\rho| A'B> = |a_0|^2 <A'B|AB> <AB|A'B> + a_0a_1* <A'B|AB> <A'B'|A'B> + a_0a_1* <A'B|A'B'> <AB|A'B> + |a_1|^2 <A'B|A'B'> <A'B'|A'B>

 

and similarly, due to orthogonality of [math]B[/math],[math]B'[/math], all terms except the first drop out, so I get:

 

[math]P(|A'B>)=|a_0|^2 |<A'|A>|^2[/math]

 

Combining:

 

[math]P(|B>) = P(|AB>)+P(|A'B>) = |a_0|^2 + |a_0|^2 |<A'|A>|^2[/math]

 

which for the special case of [math]a_0 = \frac{1}{\sqrt{2}}[/math] is:

 

[math]P(|B>) = \frac{1}{2} (1+ |<A|A'>|^2)[/math]

 

Previously, I had always reduced p first using tensor-product identities and then projected onto [math]|B>[/math], which I thought was equivalent--I guess the coherent states break some of the equivalencies as we've already discussed."

Edited August 4, 2012 by Aethelwulf

[JackSarfatti]

That's correct.

[Aethelwulf]

That's good. I'm glad I wrote it out correctly, shame the other two equations where not showing.

[Aethelwulf]

I'm just trying to follow the math the best I can at the moment, it's been a while since I have actually read up or forced myself to do entanglement equations.

 

So the first equation, would I be right in thinking this is the maximally entangled state, is the trace in your equation over [math](\alpha, \beta)[/math] should give you a density operator?