  Quantum teleportation with an arbitrary two-qubit state can be appropriately
characterized in terms of maximal fidelity and fidelity deviation. The former
quantifies optimality of the process and is defined as the maximal average
fidelity achievable within the standard protocol and local unitary strategies,
whereas the latter, defined as the standard deviation of fidelity over all
input states, is a measure of fidelity fluctuations. The maximal fidelity for a
two-qubit state is known and is given by a simple formula that can be exactly
computed, but no such formula is known for the fidelity deviation. In this
paper, we derive an exact computable formula for the fidelity deviation in
optimal quantum teleportation with an arbitrary state of two qubits. From this
formula, we obtain the dispersion-free condition, also known as the
universality condition: the condition that all input states are teleported
equally well and provide a necessary and sufficient condition for a state to be
both useful (maximal fidelity larger than the classical bound) and universal
(zero fidelity deviation). We also show that for any given maximal fidelity,
larger than the classical bound, there always exist dispersion-free or
universal states and argue that such states are the most desirable ones within
the set of useful states. We illustrate these results with well-known families
of two-qubit states: pure entangled states, Bell-diagonal states, and subsets
of $X$ states.
