# Fidelity of unitary operators: is $||U-\tilde U|| < \delta$ a *necessary* and *sufficient* condition?

There's a notion of fidelity of quantum states. However, is there a standard notion of the fidelity of unitary operators? Say, I wish to approximate a unitary operator $$U$$ acting on $$n$$ qubits with a unitary $$\tilde U$$, to within some "infidelity" range (using some appropriate formula for fidelity); this is a common problem in the context of Hamiltonian simulation. A necessary and sufficient condition (?) might be that $$\tilde U$$ should be $$\delta$$-close to $$U$$, such that $$||U - \tilde U|| < \delta$$, in the supremum norm, for some small positive $$\delta$$ .

Questions:

1. Does $$||U - \tilde U|| < \delta$$ necessarily ensure that $$\tilde U| \Psi\rangle$$ is "close to" $$U|\Psi\rangle$$ in all fidelity measures of quantum states? ($$|\Psi\rangle$$ is any arbitrary $$n$$-qubit quantum state.)

2. Are there other better standard conditions for measuring fidelity of unitary operators? I'm looking for sufficiency conditions as well as necessary conditions (it'd be helpful if you could explicitly distinguish between the two types of conditions).

In case you're wondering why, say, just $$U|0\ldots 0\rangle$$ being close to $$\tilde U|0\ldots 0\rangle$$ isn't a good enough condition: the point here is that unitary operators form continuous groups (Lie groups: $$U(2^n)$$ or $$PU(2^n)$$), whose geometry is richer than the geometry of merely the space of unit. vectors. By the way, I'm not just concerned about finite-dimensional cases of qubits but also infinite-dimensional cases.

• Isn't that precisely the point of the operator norm? – Norbert Schuch Jan 6 at 21:53
• @NorbertSchuch Yes, I realized after asking. :) – S.D. Jan 6 at 21:54
• Then why don't you self-answer your question for the benefit of future visitors? – Norbert Schuch Jan 6 at 22:49
• @NorbertSchuch I will, when I get the time. – S.D. Jan 7 at 2:37