Channel Fidelity

How faithfully does a channel transmit a state? The standard answer is the fidelity, a measure of overlap between the state you sent and the state that came out. It is the workhorse figure of merit for benchmarking gates, memories, and communication links.

Fidelity between two states

For states $\rho$ and $\sigma$ the (Uhlmann) fidelity is

F(\rho,\sigma) = \Big\lVert \sqrt{\rho}\,\sqrt{\sigma}\,\Big\rVert_1 = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\,\sqrt{\rho}},

ranging from $0$ (orthogonal, perfectly distinguishable) to $1$ (identical). It is symmetric, $F(\rho,\sigma)=F(\sigma,\rho)$ , and invariant under any unitary applied to both. (Some authors square this; we use the $F\in[0,1]$ convention where $F=1$ means equal.)

The formula simplifies dramatically when one argument is pure, $\rho=|\psi\rangle\langle\psi|$ :

F\big(|\psi\rangle\langle\psi|,\,\sigma\big) = \sqrt{\langle\psi|\,\sigma\,|\psi\rangle}.

The messy operator square roots collapse because $\sqrt{|\psi\rangle\langle\psi|} = |\psi\rangle\langle\psi|$ is its own square root. This is the case that matters for channels probed with pure inputs.

Fidelity through a channel

Send a pure state $|\psi\rangle$ through $\mathcal{E}$ and compare input to output. The input–output fidelity is

F\big(|\psi\rangle,\,\mathcal{E}(|\psi\rangle\langle\psi|)\big) = \sqrt{\langle\psi|\,\mathcal{E}(|\psi\rangle\langle\psi|)\,|\psi\rangle}.

The quantity under the root, $\langle\psi|\mathcal{E}(|\psi\rangle\langle\psi|)|\psi\rangle$ , is the probability that the output still "looks like" $|\psi\rangle$ . Two summary numbers are built from it:

the minimum fidelity $F_{\min} = \min_{|\psi\rangle} F$ , a worst-case guarantee;
the average fidelity $\bar F = \int F^2\,d\psi$ , the Haar-average over input states, which for a qubit channel relates to the entanglement fidelity $F_e$ by $\bar F = \tfrac{2F_e + 1}{3}$ .

Useful properties

Monotonicity (data processing). Channels can only blur states together, never separate them: $F\big(\mathcal{N}(\rho),\mathcal{N}(\sigma)\big)\ge F(\rho,\sigma)$ for any channel $\mathcal{N}$ .
Relation to trace distance. Fidelity and trace distance $D=\tfrac12\lVert\rho-\sigma\rVert_1$ bound each other: $1 - F \le D \le \sqrt{1-F^2}$ , so high fidelity forces small trace distance and vice versa.
Concavity / joint concavity, which makes averaging over input ensembles well behaved.

Try it

Send $|0\rangle$ through the qubit depolarizing channel $\mathcal{E}_p(\rho)=(1-p)\rho+p\,\mathbb{1}/2$ with $p=0.36$ . Compute the input–output fidelity $F = \sqrt{\langle0|\mathcal{E}_p(|0\rangle\langle0|) |0\rangle}$ . Since $\langle0|\mathcal{E}_p(|0\rangle\langle0|)|0\rangle = (1-p) + p/2 = 1 - p/2 = 0.82$ , the fidelity is $\sqrt{0.82}\approx0.9055$ — build $\mathcal{E}_p(\rho)$ and compute it.

Run your code to see the quantum state.