What's free evolution for a period T?

The specific work you cite deals with Ramsey interferometry. In that case, there is a two-level atom with some states $|grangle$ and $|erangle$, which have different energies. Since it never matters where we set the $0$ of energy, people typically say that the ground state $|grangle$ has energy $0$ and the excited state has energy $hbar omega_a$, where $hbar$ is the reduced Planck's constant and $omega_a$ is a frequency ($a$ stands for atomic).

In this case, we can write the atomic Hamiltonian as $$H_a=hbaromega_a|eranglelangle e|.$$ This Hamiltonian has two energy eigenstates: $|grangle$ is an eigenstate with eigenvalue $E_g=0$, and $|erangle$ is an eigenstate with eigenvalue $E_e=hbaromega_a$, as might be anticipated. Thus, if the initial state of the system is $$|psi(0)rangle=c_g(0)|grangle+c_e(0)|erangle,$$ it evolves under the standard rules of quantum mechanics (the Schrödinger equation) to $$|psi(t)rangle=c_g(t)|grangle+c_e(t)|erangle,$$ where, as usual, $$c_g(t)=e^{-i E_g t/hbar}c_g(0)=c_g(0)quadmathrm{and}quad c_e(t)=e^{-i E_e t/hbar}c_e(0)=e^{-iomega_a t}c_e(0).$$ If this evolution happens for a period $t=T$, the coefficient $c_g(0)$ does not change while the coefficient $c_e(0)$ aquires a phase of $omega_a T$. If we represent our state in the ${|grangle,|erangle}$-basis, as $$|psi(t)rangle= begin{pmatrix}c_g(t)c_e(t)end{pmatrix}=begin{pmatrix}c_g(0)e^{-iomega_a t}c_e(0)end{pmatrix}=begin{pmatrix}1&0&e^{-iomega_a t}end{pmatrix}begin{pmatrix}c_g(0)c_e(0)end{pmatrix}=begin{pmatrix}1&0&e^{-iomega_a t}end{pmatrix}|psi(0)rangle,$$ the matrix you described is equivalent to the evolution of the state under the Hamiltonian $H_a$.