# General Bell state expression: What condition for mixture of Bell states to be entangled?

Convention: $$|qubit_{A}, qubit_{B}\rangle$$

The general Bell state equation: $$|\beta(a,b)\rangle = \frac{1}{\sqrt{2}}\sum_{k=0}^{1}(-1)^{ka}|k, k\oplus b\rangle = \frac{1}{\sqrt{2}}[|0,0 \oplus b\rangle + (-1)^{a}|1,1 \oplus b\rangle]$$.

In mixing the states and expressing them as a matrix:

$$\rho = \sum_{a,b = 0}^{1}p_{a,b}|\beta(a,b)\rangle \langle\beta(a,b)|$$

where $$p_{a,b} \geq 0$$ and $$\sum_{a,b=0}^{1}p_{a,b} = 1$$.

How do I determine when $$\rho$$ is entangled?

I do know that entanglement is checked using either the Schmidt number (for state vector) or the partial transposition (density matrix). Since $$\rho$$ is a matrix, we can use take the partial transposition and apply it to the qubit B so that we have $$\rho \rightarrow \rho^{T_{B}}$$ More explicitly: $$\rho = \sum_{a,b,a,b'}p_{a,b,a',b'}|A,B\rangle \langle A',B'| \rightarrow \rho^{T_{B}} = \sum_{a,b,a,b'}p_{a,b,a',b'}|A,B'\rangle\langle A,B|$$

Any help is greatly appreciated.

• Didn't we have pretty much the same question last week? Oct 31, 2023 at 8:45
• Yes but this involves entanglement condition. I left it out my mind last week to think about it but I can't seem to get past it. Oct 31, 2023 at 8:46