User:DavidBoden/sandbox

Separating the Bell State $|\Phi ^{+}\rangle$

When the CNOT gate acts upon two qubits that are perfectly correlated in the $|\Phi ^{+}\rangle$ state, the outputs are the unentangled states $|+\rangle _{A}$ and $|0\rangle _{B}$. The CNOT gate is its own inverse.

To demonstrate this, we show that in any chosen basis the perfect correlation and the operation of the CNOT gate combine to produce a constant output.

Selecting the computational basis $\{|0\rangle ,|1\rangle \}$ we have:

Qubit A's effect on qubit B

Based on qubit B correlating exactly with qubit A and then qubit B being subjected to the CNOT X-rotation depending on qubit A:

$|0\rangle _{A}$ correlates to $|0\rangle _{B}$ which results in $|0\rangle _{B}$

$|1\rangle _{A}$ correlates to $|1\rangle _{B}$ which results in $|0\rangle _{B}$

Qubit B's effect on qubit A

The basis vectors that we've chosen, represented by Hadamard basis vectors are:

$\{{\frac {1}{\sqrt {2}}}(|+\rangle +|-\rangle ),{\frac {1}{\sqrt {2}}}(|+\rangle -|-\rangle )\}$

${\frac {1}{\sqrt {2}}}(|+\rangle _{B}+|-\rangle _{B})$

Separates into:

${\frac {1}{2}}(|+\rangle _{A}+|-\rangle _{A})$ and ${\frac {1}{2}}(|-\rangle _{A}+|+\rangle _{A})$

The other basis vector:

${\frac {1}{\sqrt {2}}}(|+\rangle _{B}-|-\rangle _{B})$

Separates into:

${\frac {1}{2}}(|+\rangle _{A}-|-\rangle _{A})$ and ${\frac {-1}{2}}(|-\rangle _{A}-|+\rangle _{A})$

So the resulting state of summing the results of the basis transformations (and dividing by 2) is the constant:

$|+\rangle _{A}$

Further worked example

Using an arbitrarily-selected basis of: $\{{\frac {1}{5}}(3|0\rangle +4|1\rangle ),{\frac {1}{5}}(3|1\rangle -4|0\rangle )\}$

Qubit A's effect on qubit B

Based on qubit B correlating exactly with qubit A and then qubit B being subjected to the CNOT X-rotation depending on qubit A:

${\frac {1}{5}}(3|0\rangle _{A}+4|1\rangle _{A})$

Separates into:

${\frac {3}{25}}(3|0\rangle _{B}+4|1\rangle _{B})$ and ${\frac {4}{25}}(3|1\rangle _{B}+4|0\rangle _{B})$ which equals ${\frac {25}{25}}|0\rangle _{B}+{\frac {24}{25}}|1\rangle _{B}$

The other basis vector:

${\frac {1}{5}}(3|1\rangle _{A}-4|0\rangle _{A})$

Separates into:

${\frac {3}{25}}(3|0\rangle _{B}-4|1\rangle _{B})$ and ${\frac {-4}{25}}(3|1\rangle _{B}-4|0\rangle _{B})$ which equals ${\frac {25}{25}}|0\rangle _{B}-{\frac {24}{25}}|1\rangle _{B}$

So the resulting state of summing the results of the basis transformations (and dividing by 2) is the constant:

$|0\rangle _{B}$

Qubit B's effect on qubit A

The basis vectors that we've chosen, represented by Hadamard basis vectors are: $\{{\frac {1}{5{\sqrt {2}}}}(7|+\rangle -|-\rangle ),{\frac {-1}{5{\sqrt {2}}}}(|+\rangle +7|-\rangle )\}$

${\frac {1}{5{\sqrt {2}}}}(7|+\rangle _{B}-|-\rangle _{B})$

Separates into:

${\frac {7}{50}}(7|+\rangle _{A}-|-\rangle _{A})$ and ${\frac {-1}{50}}(7|-\rangle _{A}-|+\rangle _{A})$ which equals ${\frac {50}{50}}|+\rangle _{A}-{\frac {14}{50}}|-\rangle _{A}$

The other basis vector:

${\frac {-1}{5{\sqrt {2}}}}(|+\rangle _{B}+7|-\rangle _{B})$

Separates into:

${\frac {1}{50}}(|+\rangle _{A}+7|-\rangle _{A})$ and ${\frac {7}{50}}(|-\rangle _{A}+7|+\rangle _{A})$ which equals ${\frac {50}{50}}|+\rangle _{A}+{\frac {14}{50}}|-\rangle _{A}$

So the resulting state of summing the results of the basis transformations (and dividing by 2) is the constant:

$|+\rangle _{A}$

Bell basis

The four Bell states form a Bell basis. A two-qubit state that is described as a sum of Bell states describes a correlation between any two bases on the individual qubits. For example, ${\frac {1}{\sqrt {2}}}(|0+\rangle +|1-\rangle )$ is maximally entangled; it represents a correlation between the basis $b_{1}$ and $b_{2}=H.b_{1}$. It can be rewritten as ${\frac {1}{\sqrt {2}}}(|\Phi ^{-}\rangle +|\Psi ^{+}\rangle )$ using Bell basis states.