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I am reading paper Channel Identification and its Impact on Quantum LDPC Code Performance (arXiv) where the authors discuss the scenario where the decoder of a Quantum LDPC code uses an estimation of the depolarization probability of the channel in order to correct the errors.

When doing so, they establish the quantum Cramer-Rao bound in terms of the Fisher information. The Fisher information is defined as

$J_f = \mathrm{Tr}(\rho_f L_f^2)$,

where $L_f$ is the symmetric logarithmic derivative $2\partial_f \rho_f=L_f\rho_f+\rho_f L _f$. They then state that when the probe for the estimation is an unnetangled pure quantum state, such Fisher information can be proved to be $J_f = [f(2-f)]^{-1}$, and if the probe is a maximally-entangled qubit pair, it equals $J_f=[f(\frac{4}{3}-f)]^{-1}$.

I have been trying to prove such equalities for those cases of the Fisher information, but I haven't been able to find something similar to it nor prove it.

Frederik vom Ende
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Josu Etxezarreta Martinez
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1 Answers1

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Let the input probe state be the pure state $\rho_s = |0\rangle\langle0|$ (unitary covariance of the depolarizing channel $\mathcal{E}_f(\rho_s) = (1-f)\rho_s + \frac{f}{2}I$ allows us this choice without loss of generality, see reference [15] in the paper you linked for a similar argument). The output state of the depolarizing channel is $\rho_f = \mathcal{E}_f(|0\rangle\langle0|) = (1-f)|0\rangle\langle0| + \frac{f}{2}I.$ The derivative with respect to $f$ is $\partial_f \rho_f = \begin{pmatrix} -1/2 & 0 \\ 0 & 1/2 \end{pmatrix}$. Solving for the elements of $L_f$ from $2\partial_f \rho_f=L_f\rho_f+\rho_f L _f$ gives $(L_f)_{00} = \frac{-1}{2-f}$ and $(L_f)_{11} = \frac{1}{f}$. Thus $$L_f = \begin{pmatrix} -1/(2-f) & 0 \\ 0 & 1/f \end{pmatrix}$$ and the QFI is \begin{align*} J(f) = \mathrm{Tr}(\rho_f L_f^2) &= \mathrm{Tr}\left(\begin{pmatrix} 1-f/2 & 0 \\ 0 & f/2 \end{pmatrix} \begin{pmatrix} 1/(2-f)^2 & 0 \\ 0 & 1/f^2 \end{pmatrix}\right) \\[1em] &= \frac{1}{f(2-f)}. \end{align*}

Let the input probe be a maximally entangled state $\rho_{AS} = |\Phi^+\rangle\langle\Phi^+|$, where $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. The output state $\rho_f = (I_A \otimes \mathcal{E}_f)(\rho_{AS})$ is diagonal in the Bell basis $\{|\Phi^+\rangle, |\Phi^-\rangle, |\Psi^+\rangle, |\Psi^-\rangle\}$ with eigenvalues: \begin{align*} \lambda_{\Phi^+}(f) = 1-3f/4, \quad \lambda_{\Phi^-}(f) = f/4, \quad \lambda_{\Psi^+}(f) = f/4, \quad \lambda_{\Psi^-}(f) = f/4. \end{align*}

The operator $2\partial_f \rho_f$ has diagonal elements $ -3/2, 1/2,1/2, 1/2.$ Simplifying $2\partial_f \rho_f=L_f\rho_f+\rho_f L _f$ we obtain diagonal elements of $L_f$: \begin{align*} (L_f)_{\Phi^+\Phi^+} &= \frac{-3}{4-3f}, \\ (L_f)_{\Phi^-\Phi^-} &= (L_f)_{\Psi^+\Psi^+} = (L_f)_{\Psi^-\Psi^-} = \frac{1}{f}. \end{align*} $L_f^2$ has diagonal entries $\frac{9}{(4-3f)^2}$ and $\frac{1}{f^2}$ (with multiplicity 3). The QFI is \begin{align*} J(f) &= \mathrm{Tr}(\rho_f L_f^2) = (1-3f/4) \frac{9}{(4-3f)^2} + 3 \cdot (f/4) \frac{1}{f^2} = \frac{1}{f(4/3-f)}. \end{align*}

Sujeet
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