Fidelity is measure of distance between density operators. it is a measure of the "closeness" of two quantum states, the input state $\rho_{in}$ and the teleported state $\rho_{out}$. where it's generally defined as : $ F(\rho_{in} , \rho_{out})$= ${ [tr(\sqrt{\sqrt{\rho_{in}} \rho_{out} \sqrt{\rho_{in}}}]^2} $,

for $ F(\rho_{in} , \rho_{out}) = 1$ one can said that $\rho_{in} $ and $\rho_{out}$ are exactly the same, but when $ F(\rho_{in} , \rho_{out}) = 0$ we can conclude that: $\rho_{in} $ and $\rho_{out}$ are totally diferente.

However, quantum correlations means the expected change in physical characteristics as one quantum system passes through an interaction site, i'm asking what the relation betwenn the fidelity of teleportation and the quantum correlations ?


1 Answer 1


Here are my two cents:

(a) The (usual) fidelity of quantum states, $F(\rho,\sigma) \equiv \lVert\sqrt{\rho} \sqrt{\sigma} \rVert_{1}^{2}$, where $\lVert \cdot \rVert_{1}$ is the $1$-norm distance, has, per se, no notion of "quantum correlations", since there is no subsystem involved. As an example, consider the case of pure states where fidelity reduces to the simple relation: $F( | \psi \rangle \langle \psi | , | \phi \rangle \langle \phi | ) = \left| \langle \psi | \phi \rangle \right|^{2}$. Now consider, $| \psi \rangle = \frac{1}{\sqrt{2}} \left( | 00 \rangle + | 11 \rangle \right), | \phi \rangle = \frac{1}{\sqrt{2}} \left( | 00 \rangle - | 11 \rangle \right)$, then their fidelity is zero; and so is for the pair of states $| \psi \rangle = | 00 \rangle, | \phi \rangle = | 11 \rangle$. That is, the usual fidelity is zero if two states are orthogonal, without any reference to their entanglement content and, in general, cannot tell us anything about the correlations between the subsystems.

(b) One can think about the entanglement fidelity (see, for example, Sec. 9.5.2 of From Classical to Quantum Shannon Theory), defined as the following: given a state $\rho_{\mathrm{in}}$, its purification $| \psi \rangle_{AB}$, a channel $\mathcal{E}$, and, $\sigma = (\mathrm{id} \otimes \mathcal{E}) (| \psi \rangle_{AB})$, we have, the entanglement fidelity is $\langle \psi_{AB} | \sigma | \psi_{AB} \rangle$. That is, the entanglement fidelity measures how well a quantum channel $\mathcal{E}$ preserves entanglement with a reference state (a similar defintion is used in fidelity of quantum teleportation). However, note that by choosing $\mathcal{E}$ to be a simple local Pauli, one can transform one Bell state into another and make the entanglement fidelity zero -- even though, the entanglement between systems is still maximal.

Therefore, entanglement fidelity, in general seems to be a poor measure of quantum correlations. The relationship between entanglement fidelity and measures of entanglement was also studied here. If you have something more specific in mind then please add in the comments and I will update my answer.


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