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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?

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  • $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Commented Oct 25, 2023 at 9:42
  • $\begingroup$ @DaftWullie I am getting 2 probability coefficients along the diagonal for each Bell state matrix. $\endgroup$
    – librarian_
    Commented Oct 25, 2023 at 10:08
  • $\begingroup$ Yes, and each will be $p_{ab}/2$, so they add to $p_{ab}$, and when you sum over all $a,b$, you get $\sum p_{ab}=1$. $\endgroup$
    – DaftWullie
    Commented Oct 25, 2023 at 10:11
  • $\begingroup$ @DaftWullie You are right. I was an idiot for missing the $\frac{1}{2}$ $\endgroup$
    – librarian_
    Commented Oct 25, 2023 at 10:12

1 Answer 1

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In answer to 2: Yes, this is correct. More carefully stated: it at least one eigenvalue is negative, the state is entangled. If all eigenvalues are non-negative, the state is separable.

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 $\max_{a,b}p_{a,b}>\frac12$.

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  • $\begingroup$ How does one arrive at this expression for the generalised Bell state matrix? Could you provide an example of reading off the eigenvalues here? $\endgroup$
    – librarian_
    Commented Oct 25, 2023 at 10:44
  • $\begingroup$ 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. $\endgroup$
    – Physkid
    Commented Nov 3, 2023 at 15:18
  • $\begingroup$ 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). $\endgroup$
    – DaftWullie
    Commented Nov 3, 2023 at 15:35

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