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Equation (7) in the 2012 paper, "Complementarity Reveals Bound Entanglement of Two Twisted Photons" of B. C. Hiesmayr and W. Löffler for a state $\rho_d$ in the "magic simplex" of Bell states \begin{equation} \rho_d= \frac{q_4 (1-\delta (d-3)) \sum _{z=2}^{d-2} \left(\sum _{i=0}^{d-1} P_{i,z}\right)}{d}+\frac{q_2 \sum _{i=1}^{d-1} P_{i,0}}{(d-1) (d+1)}+\frac{q_3 \sum _{i=0}^{d-1} P_{i,1}}{d}+\frac{\left(-\frac{q_1}{d^2-d-1}-\frac{q_2}{d+1}-(d-3) q_4-q_3+1\right) \text{IdentityMatrix}\left[d^2\right]}{d^2}+\frac{q_1 P_{0,0}}{d^2-d-1} \end{equation} yields "for $d=3$ the one-parameter Horodecki-state, the first found bound entangled state".

No explicit ranges — in which I am interested — for the four $q$ parameters are given, though. Any thoughts/insights?

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It looks like the only relation they say is $\text{IdentityMatrix}[3^2]=\sum P_{k,l}$.

You get a linear combination of $P_{k,l}$. Those are vertices of a $d^2-1$ simplex so the coefficents $c_{k,l}$ are baryocentric coordinates.

You can then match with the previous more general definition of $\rho_d$ term by term on each of the $c_{k,l}$.

The inequalities $0 \leq c_{k,l} \leq 1$ turn into $2d^2$ inequalities on the $q_i$.

It looks like that how the $q_i$ were constructed in the first place.

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  • $\begingroup$ For the $d=3$ case, the desired range conditions appear to be $0<q_1<5, 0<q_2<8, 0<q_3<3, q_1/5+q_2/8+q_3/3<1$. Utilizing this information, I have obtained for the Hilbert-Schmidt PPT-probability of the states $\rho_d$, as given in the formulation of the question, the result $-\frac{11}{35}+\frac{3 \pi }{8}-\frac{129 \sin ^{-1}\left(\frac{5 \sqrt{7}}{16}\right)}{49 \sqrt{7}}+\frac{8 \cos ^{-1}\left(\frac{11}{8 \sqrt{2}}\right)}{7 \sqrt{7}}-\frac{1}{2} \tan ^{-1}(7)+\frac{157 \tan ^{-1}\left(\frac{5 \sqrt{7}}{9}\right)}{49 \sqrt{7}} \approx 0.461554$. I will now try the $d=4$ case. $\endgroup$
    – Paul B. Slater
    Apr 20, 2019 at 14:35
  • $\begingroup$ Second thoughts about the above formula. $\endgroup$
    – Paul B. Slater
    Apr 20, 2019 at 18:37
  • $\begingroup$ Yes the formula does hold--using the range constraint $q_1>0, q_2>0, q_3>0, 0<1-q_1/5-q_2/4-q_3<1$. $\endgroup$
    – Paul B. Slater
    Apr 21, 2019 at 0:18
  • $\begingroup$ @PaulB.Slater I've edited the formatting in your comments a bit. $\endgroup$
    – Sanchayan Dutta
    Apr 21, 2019 at 7:19

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