# Can you compute the average fidelity of two qubits by averaging over a finite number of states?

For a single qubit and a quantum operation, the fidelity $$F_{|\psi\rangle\!\langle \psi|} = \mathrm{Tr}\bigg(U | \psi \rangle\!\langle \psi | U^\dagger \mathcal M\big(| \psi \rangle\!\langle \psi |\big)\bigg),$$

averaged over all possible initial states $$|\psi\rangle$$ (i.e., over the Bloch sphere) can be calculated by the fidelity averaged over the six states $$|\pm\rangle,|\pm\rangle_x,|\pm \rangle_y$$ (see this paper).

Is there a similar result for two qubits or more?

• have you come across the concept of a quantum $t$-design? I'm not overly familiar with the details, but the broad concept is that a $t$-design gives a set of states on up to $t$ qubits that, for calculating averages, is as good as averaging over all possible states. IIRC the Clifford group (i.e. stabilizer states) serves your purpose. Commented Oct 27, 2023 at 7:48
To see it, start from expanding the given expression and observing that it is linear in $$\mathbb{P}_\psi\otimes\mathbb{P}_\psi$$, where $$\mathbb{P}_\psi\equiv|\psi\rangle\!\langle \psi|$$. More explicitly, this means that it can be reframed as an expression of the form $$F_{\psi}= \langle A, \mathbb{P}_\psi\otimes\mathbb{P}_\psi\rangle \equiv \operatorname{tr}(A^\dagger(\mathbb{P}_\psi\otimes\mathbb{P}_\psi)),$$ for some operator $$A$$ that you can find explicitly expanding all terms in your definition.
Therefore when averaging the fidelity over all states, you are effectively averaging over $$\mathbb{P}_\psi\otimes\mathbb{P}_\psi$$ over all (uniformly random) states $$\psi$$. A complex-projective 2-design is defined as a finite set of vectors such that averaging this quantity over these vectors equals the average over all states, hence the result. See e.g. https://en.wikipedia.org/wiki/Quantum_t-design for more info.