Algebraic detail
We can rewrite the expression
N(a|+>μ|+>eA|->eB
- a|+>μ|->eA|+>eB
+ b|->μ|+>eA|->eB
- b|->μ|->eA|+>eB)
in terms of the Bell states.
The Bell states are:
|
|ψ+> = N(|+>μ|->e + |->μ|+>e) |
|
|ψ-> = N(|+>μ|->e - |->μ|+>e) |
|
|ϕ+> = N(|+>μ|+>e + |->μ|->e) |
|
|ϕ-> = N(|+>μ|+>e - |->μ|->e) |
We write
N(a|+>μ|+>eA|->eB
- a|+>μ|->eA|+>eB
+ b|->μ|+>eA|->eB
- b|->μ|->eA|+>eB)
= (N/2)[(|+>μ|->eA
+ |->μ|+>eA)X1
+ (|+>μ|->eA
- |->μ|+>eA)X2
+ (|+>μ|+>eA
+ |->μ|->eA)X3
+ (|+>μ|+>eA
- |->μ|->eA)X4].
The Xi are different possible states of Bob’s electron. Let us solve for the Xi.
a|+>μ|+>eA|->eB
- a|+>μ|->eA|+>eB
+ b|->μ|+>eA|->eB
- b|->μ|->eA|+>eB
=
(1/2)[|+>μ|+>eA
(X3 + X4) - |+>μ|->eA
(X1 + X2) + |->μ|+>eA(
X1 - X2) - |->μ|->eA
(X3 - X4)].
The factors multiplying a particular combination of the spins of Alice's muon and electron on each side of the equation are equal to each other. We find
a|->eB = (X3 + X4)/2,
-a|+>eB = (X1 + X2)/2,
b|->eB = (X1 - X2)/2,
-b|+>eB = (X3 - X4)/2.
Therefore we have
X1 = -a|+>e + b|->e,
X2 = -a|+>e - b|->e,
X3 = a|->e - b|-+>e,
X4 = a|->e + b|+>e,
and
(a|+>μ + b|->μ) N(|+>e|->e - |->e |+>e)
= (1/2) [|ψ+>(-a|+>e + b|->e) + |ψ->(-a|+>e - b|->e) + |Φ+>(a|->e - b|-+>e) + |Φ->(a|->e + b|+>e)].