Bell diagonal state

The Bell diagonal state is defined as the probabilistic mixture of Bell states:

${\displaystyle |\phi ^{+}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |0\rangle _{B}+|1\rangle _{A}\otimes |1\rangle _{B})}$

${\displaystyle |\phi ^{-}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |0\rangle _{B}-|1\rangle _{A}\otimes |1\rangle _{B})}$

${\displaystyle |\psi ^{+}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |1\rangle _{B}+|1\rangle _{A}\otimes |0\rangle _{B})}$

${\displaystyle |\psi ^{-}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |1\rangle _{B}-|1\rangle _{A}\otimes |0\rangle _{B})}$

In density operator form, a Bell diagonal state is defined as

${\displaystyle \varrho ^{Bell}=p_{1}|\phi ^{+}\rangle \langle \phi ^{+}|+p_{2}|\phi ^{-}\rangle \langle \phi ^{-}|+p_{3}|\psi ^{+}\rangle \langle \psi ^{+}|+p_{4}|\psi ^{-}\rangle \langle \psi ^{-}|}$

where ${\displaystyle p_{1},p_{2},p_{3},p_{4}}$is a probability distribution. Since ${\displaystyle p_{1}+p_{2}+p_{3}+p_{4}=1}$, a Bell diagonal state is determined by three real parameters. The maximum probability of a Bell diagonal state is defined as ${\displaystyle p_{max}=\max\{p_{1},p_{2},p_{3},p_{4}\}}$.

1. A Bell-diagonal state is separable if all the probabilities are less or equal to 1/2, i.e., ${\displaystyle p_{max}\leq 1/2}$.[2]

2. Many entanglement measures have a simple formulas for entangled Bell-diagonal states:[1]

Relative entropy of entanglement: ${\displaystyle S_{r}=1-h(p_{max})}$,[3] where ${\displaystyle h}$ is the binary entropy function.

Entanglement of formation: ${\displaystyle E_{f}=h({\frac {1}{2}}+{\sqrt {p_{max}(1-p_{max})}})}$,where ${\displaystyle h}$ is the binary entropy function.

Negativity: ${\displaystyle N=p_{max}-1/2}$

Log-negativity: ${\displaystyle E_{N}=\log(2p_{max})}$

3. Any 2-qubit state where the reduced density operators are maximally mixed, ${\displaystyle \rho _{A}=\rho _{B}=I/2}$, is bell-diagonal on some local basis. Viz., there exist local unitaries ${\displaystyle U=U_{1}\otimes U_{2}}$ such that ${\displaystyle U\rho U^{\dagger }}$is bell-diagonal.[2]