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 $x\in\{0,1\}^n$, 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)?


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 $$ |x\rangle\langle x|+|\bar x\rangle\langle\bar x|=\frac{1}{2}(|x\rangle+|\bar x\rangle)(\langle x|+\langle\bar x|)+\frac{1}{2}(|x\rangle-|\bar x\rangle)(\langle x|-\langle\bar x|), $$ 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 $y\in\{0,1\}^n$ be the choice that gives the largest value of $\lambda_y$ ($y\neq 000\ldots 0$ or $111\ldots 1$). Choose $y$ to specify your bipartition of the qubits. This is entangled when $\lambda_y<\nu$.

  • $\begingroup$ thanks. When I apply $H\otimes H$ to $ZZ$, I get $XX$. To be diagonal in the star graph state bases, this should commute with all $K_y$. But for the latter I get $11,XZ,ZX,YY$. Looks like I got something wrong, right? $\endgroup$ – draks ... Jul 22 '20 at 6:49
  • $\begingroup$ The stabilizers for the GHZ states are $\{Z_iZ_{i+1}\}_{i=1}^{n-1}$ and $X^{\otimes n}$. If you apply Hadamards to all but the first qubit, you get $Z_1X_2$, $\{X_iX_{i+1}\}_{i=2}^{n-1}$ and $X_1Z_2Z_3\ldots Z_n$. These do commute with the stabilizer generators, which are $\{Z_1X_i\}_{i=2}^n$ and $X_1Z_2Z_3\ldots Z_n$ $\endgroup$ – DaftWullie Jul 22 '20 at 8:15
  • $\begingroup$ The way that I see this is that I use the graph state stabilizers to construct the stabilizers that I want. For example, $X_iX_{i+1}$ can be written as $(Z_1X_i)(Z_1X_{i+1})$. $\endgroup$ – DaftWullie Jul 22 '20 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.