Lower Bound to Measure Entanglement in Mixed States

Question

Given a mixed state with a $n$ qubit density matrix of the following structure:
$$
rho=pmatrix{lambda_1&&&&nu\&lambda_2\ &&lambda_{dots}\ &&&lambda_2\ nu&&&&lambda_1},
$$
so a mixture of a GHZ state and a sum of diagonal contributions of the form: $prod Z_k^{x_k}$, where $x_k$ is the $k$th bit of the binary number $xin{0,1}^n$, which is restricted to contain an even number of $1$s. The latter restriction ensures that the matrix is symmetric w.r.t. the anti-diagonal.
If all $lambda_k$ are equal, there is a way to measure exponentially small amounts of the GHZ state, as proposed here. It looks to me, that my states don't respect the graph diagonal state requirement given in the comments.
What is the best lower bound for $nu$, that is efficiently measurable (involving polynomially many measurements)?

DaftWullie · Answer

As soon as you impose that $lambda_x=lambda_{bar x}$, your state is of the form that my previous results apply to. This is because you can write
$$
|xranglelangle x|+|bar xranglelanglebar x|=frac{1}{2}(|xrangle+|bar xrangle)(langle x|+langlebar x|)+frac{1}{2}(|xrangle-|bar xrangle)(langle x|-langlebar x|),
$$
and so this state is diagonal in the GHZ-state basis. This is easily rotated into the graph state basis of a star graph using Hadamards.
Let $yin{0,1}^n$ be the choice that gives the largest value of $lambda_y$ ($yneq 000ldots 0$ or $111ldots 1$). Choose $y$ to specify your bipartition of the qubits. This is entangled when $lambda_y<nu$.

Lower Bound to Measure Entanglement in Mixed States

One Answer

Add your own answers!

Ask a Question