Let $\gamma$ be a completely trace preserving operator such that $\gamma(\rho) \to (1-\epsilon)\rho+\epsilon(|\phi\rangle \langle\phi|)$. Here $\rho$ is density matrix of two dimensional hilbert space and $|\phi\rangle$ a fixed pure state of 2 dimensional hilbert space. The average fidelity of the channel described by some trace preserving operator $\eta$ is given as $F(\eta)=\int \langle\psi| {\eta(|\psi\rangle)} | \psi\rangle d\psi$.

My doubts are

  1. Why is average fidelity defined such that we are only taking into account action of $\eta$ on density operator of pure states ( the part $\eta(|\psi\rangle)$ in the formula ). Is it because it would be enough to take average only how pure states are mapped as mixed states are just convex combination of density operator of pure states ?
  2. If I am not missing anything and understanding the formula correct then for $\gamma$ $F(\gamma) = (1-\epsilon)+ \epsilon\int\langle \psi|\phi\rangle \langle\phi|\psi\rangle d\psi$, but how do I calculate the second term ?
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    $\begingroup$ I don't want to revamp this from the dead but for future generations yes the formula in point 2 is correct and the "second term" equals $1/d$ where $d$ is the Hilbert's space dimension. This latter term is also called (for self-explanatory reasons) "fidelity of a random guess". $\endgroup$
    – lcv
    Commented Jun 5, 2017 at 22:49

1 Answer 1

  1. The action of η on the pure states is enough because the density operator is a convex combination of pure state projection operators and because η is linear, so all the contributions mixed states add to the average of <η> end up evening out.

  2. I think the arrow → was not meant to express an equality, but rather something about the infinitesimal behavior of a trace-preserving operator.

Edit: Originally, I wrote the following, which contained a novice mistake: "Note that when it is an equality, which you are considering here, for pure states ψ and Φ, <ψ∣Φ> = <Φ∣ψ> = δ(ψ,Φ) where δ is the Dirac delta, and ∫ δ² dψ = 1, so your average fidelity would also = 1."

Peter rightly pointed out that <ψ∣Φ> != <Φ∣ψ> != δ(ψ,Φ) in general. Heck, (1, 0) * (1/sqrt(2), 1/sqrt(2)) != 0, so I am clearly wrong here.

My initial impression now is that a formula may be derived with a triple integral, as the varying quantum state can be parameterized by the relative phases of the two coefficients of psi with respect to phi and the magnitude of one coefficient of psi (as the other coefficient is constrained by a normalization condition).

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    $\begingroup$ the arrow is not an equality it gives the action of the trace preserving map $\endgroup$ Commented Jun 15, 2015 at 9:37
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    $\begingroup$ Your equation <ψ∣Φ> = <Φ∣ψ> = δ(ψ,Φ) is incorrect. $\endgroup$ Commented Sep 18, 2016 at 12:32

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