4

This question is related to this and this.

I am working on Qiskit to design QEC schemes.

My model works with Pauli errors.

I would like to give to my Pauli error channel probabilities $p_x,p_y,p_z$ corresponding to a fidelity $F$.

I could find here a way to map a depolarizing channel $\mathcal{D}_{\lambda}$, to its equivalent Pauli channel $\mathcal{P}_{p_x,p_y,p_z}$ with $p_x=p_y=p_z$. Hence, for this case, the first question would be enough.

However, I may need to customise the probabilities.

Stemming from the related questions, I need to solve the following equation:

$$\bar{F}(\mathcal{P}_{p_x,p_y,p_z}) = \int \langle\varphi|\mathcal{P}_{p_x,p_y,p_z}(|\varphi\rangle\langle\varphi|)|\varphi\rangle\ d\varphi$$

luciano
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Daniele Cuomo
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1 Answers1

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TL;DR: Two steps. First, we find the entanglement fidelity $F_e(\mathcal{P}_{p_x,p_y,p_z})=1-(p_x+p_y+p_z)$. Next, we use Horodecki's formula that connects the average and entanglement fidelities

$$ \overline{F}(\mathcal{E})=\frac{NF_e(\mathcal{E})+1}{N+1}\tag1 $$

to get

$$ \overline{F}(\mathcal{E})=1-\frac23(p_x+p_y+p_z).\tag2 $$

There is also an alternative approach that exploits the fact that Pauli eigenstates form a spherical 3-design.

Entanglement fidelity

There are many ways to quickly find the entanglement fidelity $F_e(\mathcal{E})$ of a channel $\mathcal{E}$. For example, it can be shown directly from the definition that $F_e(\mathcal{E})$ can be read off of the channel's Pauli transfer matrix (aka process matrix) $\chi(\mathcal{E})$. Namely, $F_e(\mathcal{E})$ is the diagonal element corresponding to the identity. In the present case, we have $\chi(\mathcal{P}_{p_x,p_y,p_z})=\mathrm{diag}(1-(p_x+p_y+p_z), p_x, p_y, p_z)$, so $F_e(\mathcal{P}_{p_x,p_y,p_z})=1-(p_x+p_y+p_z)$.

Alternatively, we can use the formula for $F_e(\mathcal{E})$ in terms of the channel's Kraus operators

$$ F_e(\mathcal{E})=\frac{1}{N^2}\sum_i|\mathrm{tr}K_i|^2\tag4 $$

c.f. equation $(9.135)$ on page $421$ in Nielsen & Chuang. For a Pauli channel all but one operators in $(4)$ have zero trace and once again we obtain $F_e(\mathcal{P}_{p_x,p_y,p_z})=1-(p_x+p_y+p_z)$.

Horodecki's formula

You can find a beautiful proof of formula $(1)$ in this paper. The key fact established in the course of the proof is that twirling$^1$ any channel produces a depolarizing channel with the same entanglement fidelity as the original channel.

You can also find a more technical but shorter proof of $(1)$ in this paper.

Spherical 2-designs

Alternatively, we can use a spherical 2-design to simplify the evaluation of the integral

$$ \overline{F}(\mathcal{P}_{p_x,p_y,p_z})=\int\langle\psi|\mathcal{P}_{p_x,p_y,p_z}(|\psi\rangle\langle\psi|)|\psi\rangle d\psi.\tag5 $$

Tetrahedron is the smallest spherical 2-design, but it is also not the most convenient to work with in this context. Instead, we can use an octahedron which is the smallest spherical 3-design. See this paper. This is convenient because the eigenstates of the Pauli operators form an octahedron. We get

$$ \overline{F}(\mathcal{P}_{p_x,p_y,p_z})=\frac{1}{|S|}\sum_{|\psi\rangle\in S}\langle\psi|\mathcal{P}_{p_x,p_y,p_z}(|\psi\rangle\langle\psi|)|\psi\rangle\tag6 $$

where $S=\{|0\rangle, |1\rangle, |+\rangle, |-\rangle, |{+i}\rangle, |{-i}\rangle\}$. We then calculate

$$ \begin{align} \overline{F}(\mathcal{P}_{p_x,p_y,p_z})&=\frac{1}{6}\left(6\cdot(1-(p_x+p_y+p_z)+2p_x+2p_y+2p_z\right)\\ &=\frac{1}{6}\left(6-4(p_x+p_y+p_z)\right)\\ &=1-\frac23(p_x+p_y+p_z)\tag7 \end{align} $$

in agreement with $(2)$.

$^1$ Twirl takes a channel and produces another channel by averaging the effect of the conjugation of the input channel by a Haar-random unitary.

Adam Zalcman
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