Friday, August 3, 2012

Keith Kenemer on LinkedIn physics forum wrote:•" Jack,

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

|Alice, Bob> = a0 |AB> + a1 |A'B'>

where |AB> = |A>(x)|B> and |A'B'> = |A'> (x) |B'>

[ and (x) = tensor product ]

density matrix rho = p = |Alice,Bob><Alice,Bob|
p = |a0|^2 |AB><AB| +a0a1*|AB><A'B'| + a1a0*|A'B'><AB|+|a1|^2|A'B'><A'B'|

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

P(|AB>)=<AB| p |AB> =

|a0|^2<AB|AB><AB|AB> + a0a1*<AB|AB><A'B'|AB>+a1a0*<AB|A'B'><AB|AB>+
|a1|^2 <AB|A'B'><A'B'|AB>

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

P(|AB>) = |a0|^2

Next, P(|A'B>) = <A'B| p | A'B>

= |a0|^2 <A'B|AB><AB|A'B> + a0a1*<A'B|AB><A'B'|A'B>+a0*a1<A'B|A'B'><AB|A'B>

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

P(|A'B>)=|a0|^2 |<A'|A>|^2


P(|B>) = P(|AB>)+P(|A'B>) = |a0|^2 + |a0|^2 <A'|A>|^2

which for the special case of a0 = 1/sqrt(2) is:

P(|B>) = 1/2 [ 1+ |<A|A'>|^2 ]

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

So at least there is another COMPETENT guy out there who has checked my calculation and gets the same answer.

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