The specific work you cite deals with Ramsey interferometry. In that case, there is a two-level atom with some states $|g\rangle$ and $|e\rangle$, which have different energies. Since it never matters where we set the $0$ of energy, people typically say that the ground state $|g\rangle$ 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
$$H_a=\hbar\omega_a|e\rangle\langle e|.$$ This Hamiltonian has two energy eigenstates: $|g\rangle$ is an eigenstate with eigenvalue $E_g=0$, and $|e\rangle$ is an eigenstate with eigenvalue $E_e=\hbar\omega_a$, as might be anticipated. Thus, if the initial state of the system is
$$|\psi(0)\rangle=c_g(0)|g\rangle+c_e(0)|e\rangle,$$ it evolves under the standard rules of quantum mechanics (the Schrödinger equation) to
$$|\psi(t)\rangle=c_g(t)|g\rangle+c_e(t)|e\rangle,$$ where, as usual,
$$c_g(t)=e^{-i E_g t/\hbar}c_g(0)=c_g(0)\quad\mathrm{and}\quad c_e(t)=e^{-i E_e t/\hbar}c_e(0)=e^{-i\omega_a t}c_e(0).$$ 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 $\{|g\rangle,|e\rangle\}$-basis, as $$|\psi(t)\rangle= \begin{pmatrix}c_g(t)\\c_e(t)\end{pmatrix}=\begin{pmatrix}c_g(0)\\e^{-i\omega_a t}c_e(0)\end{pmatrix}=\begin{pmatrix}1&0\\0&e^{-i\omega_a t}\end{pmatrix}\begin{pmatrix}c_g(0)\\c_e(0)\end{pmatrix}=\begin{pmatrix}1&0\\0&e^{-i\omega_a t}\end{pmatrix}|\psi(0)\rangle,$$ the matrix you described is equivalent to the evolution of the state under the Hamiltonian $H_a$.