An m-qubit state, inferring some inequality in QIT

Thread starter billtodd
Start date Jun 14, 2024
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Quantum information

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The discussion revolves around demonstrating an inequality related to m-qubit states in quantum information theory (QIT). The main focus is on the relationship between the trace of the squared density matrix, Trace(σ²), and the trace of the density matrix, Trace(σ). A proposed inequality is presented, linking the distance D(σ, I/2^m) to the fidelity F(σ, I/2^m). Clarification is sought on how to effectively proceed with proving this inequality. The conversation highlights the complexities of quantum state analysis and the importance of these relationships in QIT.

Jun 14, 2024

billtodd

Homework Statement: Let ##\sigma## be an ##m##-qubit such that ##Trace(\sigma^2)\le \frac{1+\epsilon^2}{2^m}##, then ##D(\sigma , I/2^m)\le \epsilon##

Relevant Equations: ##D(\sigma,\rho)=1/2 Trace(|\sigma-\rho|)## where ##\rho,\sigma## are quantum states; and there's an inequality between this and the fidelity, i.e. ##D(\sigma,\rho)\le \sqrt{1-F(\sigma,\rho)^2}##, where ##F(\sigma ,\rho)=Trace(\sqrt{\sigma^{1/2}\rho \sigma^{1/2}})##.

I am not sure how to show this inequality, I guess what is the relation between ##Trace(\sigma^2)## and ##Trace(\sigma)##?
Here's my latest attempt, let me know how to proceed?
$$D(\sigma, I/2^m)\le \sqrt{1-F(\sigma,I/2^m)^2}=\sqrt{1-(Tr(\sqrt{\sigma }))^2/2^m}$$

Last edited: Jun 14, 2024

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