# How to calculate the average fidelity of an amplitude damping channel

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...

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}

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

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

• I really admire the details and rigorousness you put in all your answers! Feb 17, 2021 at 17:28
• Thank you, @KAJ226! I do enjoy doing such calculations :-) Feb 17, 2021 at 18:32
• awesome, thanks again :) why did you choose the 6 eigenvectors of the pauli operators as a spherical 2-design? The paper you referenced doesn't mention the pauli eigenvectors as a 2-design. it says you only need $d^2 = 4$ vectors for a 2-design. did you find the 2-design you mentioned from another paper? Feb 17, 2021 at 19:50
• also another question: in the first solution you provided why did you add a sin theta to each term in the first integration step? is this because you are integrating using states evenly distributed along the surface of the bloch sphere representation? is this a Haar measure distribution? I have thought about this before actually, but i've never been able to prove it. again, thanks for all the help. you seem to really know this topic well! :P Feb 17, 2021 at 19:55
• This is a good question! After all, polynomials are not usually defined as acting on kets! I did try to clarify this matter somewhat in my answer by suggesting a specific form of the expression to be averaged $\langle\psi|A|\psi\rangle\langle\psi|B|\psi\rangle$, though I did not go into an explanation on how this relates to polynomials and their degree. See text around equation $(9)$ on p.3 of the paper I cited. Feb 17, 2021 at 22:52