# Finding entanglement in matrix that is a sum of 4 Bell states

A general Bell state:

$$|\beta(a,b)\rangle = \frac{1}{\sqrt{2}}[|0,0 \oplus b\rangle + (-1)^{a}|1,1 \oplus b \rangle]$$

$$|\beta(0,0)\rangle = \frac{1}{2}[|00\rangle \langle 00| + |00\rangle \langle 11| + |11\rangle\langle00| + |11\rangle\langle11|]$$

$$|\beta(0,1)\rangle = \frac{1}{2}[|01\rangle\langle01| + |01\rangle\langle10| + |10\rangle\langle01| + |10\rangle\langle10|]$$

$$|\beta(1,0)\rangle = \frac{1}{2}[|00\rangle \langle00| - |00\rangle \langle11| - |11\rangle\langle00| + |11\rangle \langle11|]$$

$$|\beta(1,1)\rangle = \frac{1}{2}[|01\rangle \langle01| - |01\rangle\langle10| - |10\rangle\langle01| + |10\rangle\langle10|]$$

Using the above one can mix the four bell state to arrive at $$\rho = \sum_{a,b}p_{ab}|\beta(a,b)\rangle\langle\beta(a,b)|$$ where $$\sum_{ab}p_{ab} = 1$$, $$p_{ab} \geq 1$$

Few questions:

1. Each Bell state is a density matrix. If I take the expression for $$\rho$$: the trace of $$\rho$$ is no longer 1 - it fails the density operator test. What is going on?

2. I want to determine if $$\rho$$ is entangled: Since $$\rho$$ is mixed by hypothesis, I can take the positive partial transposition of $$\rho$$ with respect to $$b$$. This is equivalent to taking the positive partial transpose of each of the Bell states above and sum them. Then compute the eigenvalues. If the eigenvalues are negative, $$\rho$$ is an entangled state. Is this correct?

extension to 2. I am informed that $$|\beta((a,b)\rangle\ \langle\beta(a,b)| = \frac{1}{2}-|\beta(a,b)\rangle \langle\beta(a,b)||$$ together with

$$|\psi_{-}\rangle\langle\psi_{-}|^{T_{B}} = \frac{1}{2} - |\phi_{+}\rangle \langle \phi_{+}|$$, $$|\psi_{+}\rangle \langle \psi_{+}|^{T_{B}} = \frac{1}{2}-|\phi_{-}\rangle \langle \phi_{-}|$$, $$|\phi_{+}\rangle \langle \phi_{+}|^{T_{B}} = \frac{1}{2}-|\psi_{-}\rangle\langle\psi_{-}|$$, $$|\phi_{-}\rangle \langle \phi_{-}|^{T_{B}} = \frac{1}{2}-|\psi_{+}\rangle \langle \psi_{+}|$$.

How does this relates to positive partial transposition and entanglement?

• Why do you say the trace is not 1? Can you show us your calculation? So long as $\sum p_{ab}=1$, it should be 1. Commented Oct 25, 2023 at 9:42
• @DaftWullie I am getting 2 probability coefficients along the diagonal for each Bell state matrix. Commented Oct 25, 2023 at 10:08
• Yes, and each will be $p_{ab}/2$, so they add to $p_{ab}$, and when you sum over all $a,b$, you get $\sum p_{ab}=1$. Commented Oct 25, 2023 at 10:11
• @DaftWullie You are right. I was an idiot for missing the $\frac{1}{2}$ Commented Oct 25, 2023 at 10:12

In terms of doing the calculation, it looks like you've been told $$|\beta(x,y)\rangle\langle\beta(x,y)|^{T_B}=\frac12-|\beta(\bar x,\bar y)\rangle\langle\beta(\bar x,\bar y)|.$$ Thus, $$\rho^{T_B}=\frac12-\sum_{a,b}p_{\bar a,\bar b}|\beta(a,b)\rangle\langle\beta(a,b)|.$$ Your states $$|\beta(a,b)\rangle$$ are eigenvectors, so you can read off the eigenvalues. For example, $$|\beta(0,0)\rangle$$ is an eigenvector with eigenvalue $$\frac12-p_{11}$$. Thus, there is entanglement if $$p_{11}>\frac12$$. There's a lot of symmetry here, so the condition on entanglement becomes $$\max_{a,b}p_{a,b}>\frac12$$.
• Do you mean eigenvalue $\frac{1}{2} - p_{00}$? I couldn't even verify the claim made by the OP regarding the partial transpose is correct. Commented Nov 3, 2023 at 15:18
• No, I meant what I wrote ($x$ going to $\bar x$ is important). To verify the partial transpose claim, you can just write out the states one at a time. Personally, I like to use Pauli matrices: $|\beta(x,y)\rangle\langle\beta(x,y)|=\frac14(I+(-1)^yZZ+(-1)^xXX+(-1)^{x+y}YY)$. Since $X^T=X$, $Z^T=Z$ and $Y^T=-Y$, we can evaluate how these change very easily (without having to go through all $x$ and $y$ values separately). Commented Nov 3, 2023 at 15:35