How to calculate the average fidelity of an amplitude damping channel

Question

An answer to this question shows how to calculate the average fidelity of a depolarizing channel. How would one go about calculating this for an amplitude dampening channel? I tried working out the math myself but had no luck. The tricks used in the previous answer can't be applied in this new scenario it seems...

Answer 1

An elementary method is to simply carry out the integration

$$ \begin{align} \overline{F} &= \int\langle\psi|\mathcal{N_\gamma}(|\psi\rangle\langle\psi|)|\psi\rangle d\psi\\ &=\int\langle\psi|K_0|\psi\rangle\langle\psi|K_0^\dagger|\psi\rangle + \langle\psi|K_1|\psi\rangle\langle\psi|K_1^\dagger|\psi\rangle d\psi\\ & =\frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\left|\begin{pmatrix}\cos\frac{\theta}{2}&e^{-i\phi}\sin\frac{\theta}{2}\end{pmatrix}\begin{pmatrix}1 & 0 \\0 & \sqrt{1 - \gamma}\end{pmatrix}\begin{pmatrix}\cos\frac{\theta}{2}\\e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}\right|^2\sin\theta \\ & + \left|\begin{pmatrix}\cos\frac{\theta}{2}&e^{-i\phi}\sin\frac{\theta}{2}\end{pmatrix}\begin{pmatrix}0 & \sqrt{\gamma} \\0 & 0\end{pmatrix}\begin{pmatrix}\cos\frac{\theta}{2}\\e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}\right|^2\sin\theta d\phi d\theta \\ &=\frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\left|\cos^2\frac{\theta}{2}+\sqrt{1-\gamma}\sin^2\frac{\theta}{2}\right|^2\sin\theta + \left|\sqrt{\gamma}e^{i\phi}\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right|^2\sin\theta d\phi d\theta \\ &=\frac{1}{2}\int_0^\pi\left(\cos^4\frac{\theta}{2}+(1-\gamma)\sin^4\frac{\theta}{2}+\frac{\sqrt{1-\gamma}}{2}\sin^2\theta + \frac{\gamma}{4}\sin^2\theta\right)\sin\theta d\theta \\ &=\frac{1}{2}\int_0^\pi\sin\theta\cos^4\frac{\theta}{2}+(1-\gamma)\sin\theta\sin^4\frac{\theta}{2}+\frac{\gamma+2\sqrt{1-\gamma}}{4}\sin^3\theta d\theta \\ &=\frac{1}{2}\left(\frac{2}{3} + (1-\gamma)\frac{2}{3} + \frac{\gamma+2\sqrt{1-\gamma}}{4}\frac{4}{3}\right) \\ &=\frac{1}{2}\left(\frac{4}{3} - \frac{\gamma}{3} + \frac{2\sqrt{1-\gamma}}{3}\right) \\ &=\frac{2}{3}-\frac{\gamma}{6} + \frac{\sqrt{1-\gamma}}{3}. \end{align} $$

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A computationally easier, but conceptually more sophisticated approach is based on the fact that the eigenstates of the Pauli operators, i.e. $S=\{|0\rangle, |1\rangle, |+\rangle, |-\rangle, |{+i}\rangle, |{-i}\rangle\}$ form a spherical $2$-design and thus averaging any expression of the form $\langle\psi|A|\psi\rangle\langle\psi|B|\psi\rangle$ over the six states gives the same result as averaging it over the Haar measure (see e.g. this paper). Therefore,

$$ \begin{align} \overline{F} &= \int\langle\psi|\mathcal{N_\gamma}(|\psi\rangle\langle\psi|)|\psi\rangle d\psi \\ &=\frac{1}{|S|}\sum_{\psi\in S}\langle\psi|\mathcal{N_\gamma}(|\psi\rangle\langle\psi|)|\psi\rangle \\ &=\frac{1}{6}\left[1 + 1 - \gamma + 4 \cdot \left(\frac{1}{2} + \frac{\sqrt{1-\gamma}}{2}\right)\right] \\ &= \frac{2}{3} - \frac{\gamma}{6} + \frac{\sqrt{1-\gamma}}{3} \end{align} $$

where individual fidelities

$$ \begin{align} \langle 0|\mathcal{N_\gamma}(|0\rangle\langle 0|)|0\rangle &= 1 \\ \langle 1|\mathcal{N_\gamma}(|1\rangle\langle 1|)|1\rangle &= 1 - \gamma \\ \langle +|\mathcal{N_\gamma}(|+\rangle\langle +|)|+\rangle &= \frac{1}{2} + \frac{\sqrt{1-\gamma}}{2} \\ \langle -|\mathcal{N_\gamma}(|-\rangle\langle -|)|-\rangle &= \frac{1}{2} + \frac{\sqrt{1-\gamma}}{2} \\ \langle {+i}|\mathcal{N_\gamma}(|{+i}\rangle\langle {+i}|)|{+i}\rangle &= \frac{1}{2} + \frac{\sqrt{1-\gamma}}{2} \\ \langle {-i}|\mathcal{N_\gamma}(|{-i}\rangle\langle {-i}|)|{-i}\rangle &= \frac{1}{2} + \frac{\sqrt{1-\gamma}}{2} \\ \end{align} $$

are easily computed using

$$ \mathcal{N_\gamma}\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right) = \begin{pmatrix} a+d\gamma & b\sqrt{1-\gamma} \\ c\sqrt{1-\gamma} & d(1-\gamma) \end{pmatrix}. $$