for a single nuclear spin state, identified with qubit state $i=0$ or $i=1$, and discuss the case where we have multiple nuclear spin states below. We can first write the Hamiltonian for a two-state atom, with states ${|g,i \rangle}\equiv{|^1{\rm S}_0,m_I=i \rangle}$ and ${|e,i \rangle}\equiv{|^3{\rm P}_0,i \rangle}$ as ($\hbar\equiv 1$) $$\hat H=\hat H_M+\hat
H_0,$$ where $\hat{H}_M=\hat{\mathbf{p}}^2/2m$ is the kinetic energy, and $$\hat H_0=-\delta_i{|e,i \rangle}{\langlee,i |}+(\Omega_i(\mathbf{x})/2){|e,i \rangle}{\langleg,i |}+{\rm h.c.}$$ describes the near-resonant coupling field with $\Omega_i(\mathbf{x})$ and $\delta_i$ the Rabi frequency and detuning respectively.

Generation of adiabatic dressed potentials is then based on the validity of a Born-Oppenheimer-type assumption, where we assume that the kinetic energy of the atoms is small on a scale given by the separation of the resulting adiabatic potentials. The wavefunction $|\Phi(t)\rangle$ of a single atom satisfies the Schrödinger equation, $$i\hbar\frac{\partial}{\partial t}|\Phi(t)\rangle=(H_M+H_0)|\Phi(t)\rangle.$$ If we omit the kinetic energy term from the Hamiltonian, we obtain an equation for adiabatic eigenstates, ${|\Psi_{\pm}(t) \rangle}$, $$H_0(\mathbf{x})|\Psi_{\pm}\rangle=V^{\pm}(\mathbf{x})|\Psi_{\pm}\rangle,$$ Note here that as $H_0(\mathbf{x})$ is time-independent, there are only two such eigenstates ${|\Psi_{\pm} \rangle}$. If we consider the 1D case, and set $\Omega_i(x)=\Omega_i \sin(k_l x + \phi)$, representing the field of a standing wave (with $k_l$ the laser wavenumber and $\phi$ a phase), we find 