The Inverse Eigenvalue Problem for Partial Transposes of Two-Qubit States

Abstract:For a bipartite state $\rho$, information about the spectrum of its partial transpose $\rho^{\Gamma_B}$ can be inferred from measurements on multiple copies of $\rho$, without full state tomography. This raises a natural question: which eigenvalue lists can arise as $\operatorname{spec}(\rho^{\Gamma_B})$ for a density operator $\rho$? We completely solve this inverse eigenvalue problem for two qubits. Every nonnegative trace-one spectrum is realized as $\operatorname{spec}(\rho^{\Gamma_B})$ by some PPT state $\rho$, whereas an ordered candidate eigenvalue list $(x,y,z,-q)$, with $x\ge y\ge z\ge0$, $q>0$, and $x+y+z-q=1$, is realized by an NPT state iff $q\le y$ and $qy\le xz$. Sufficiency in the latter case is established by an explicit $X$ state whose quantum steering ellipsoid has center $c=(y-q)/(1-z)$ and normalized volume $V/V_{\max}(c)=qy/(xz)$, providing a geometric interpretation of the inequalities $q\le y$ and $qy\le xz$ as the allowed ellipsoid-center region and the fixed-center volume bound. Beyond this geometric picture, the two-qubit inverse theorem also yields exact negativity bounds from the two lowest nontrivial PT moments. Given fixed values of $p_2=Tr[(\rho^{\Gamma_B})^2]$ and $p_3=Tr[(\rho^{\Gamma_B})^3]$, we determine the exact minimum and maximum negativity over all two-qubit states subject to these moment constraints. When no PPT state is consistent with the pair $(p_2,p_3)$, the minimum is attained either at $x=y$ or $qy=xz$, while the maximum is attained either at $y=z$ or $q=y$. Finally, we show how the two-qubit inequalities persist as necessary constraints for the inverse eigenvalue problem in qubit--qudit systems.

arXiv:2608.29068 [quant-ph]

(or arXiv:2608.29068v1 [quant-ph] for this version)

https://doi.org/10.48550/arXiv.2608.29068 Focus to learn more arXiv-issued DOI via DataCite (pending registration)

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