Short answer, yes, but it doesn't directly give you the thermal conductivity.
You can write down a lattice dynamical Hamiltonian as a sum of potential and kinetic energies of the atoms in the system:
$$H =\sum_i\frac{p_i^2}{2m_i} + U$$
We can use Newton's second law to write the equations of motion:
$$m_i\ddot{\mathbf{u}}_i(t) = -\sum_{i'}\phi_{ii'}\mathbf{u}_i(t),$$
where $\mathbf{u}$ is the displacement of atom $i$ from equilibrium, and $\phi$ is the second order matrix of interatomic force constants. These have plane wave solutions of the form,
$$\mathbf{u}_i(t) = \sum_{\mathbf{k},\nu}U(i,\mathbf{k},\nu)\exp{i[\mathbf{k}\cdot\mathbf{r}_i-\omega(\mathbf{k},\nu)t]}$$
Where $U$ is an amplitude of collective motions. This can be recast as a set of linear equations:
$$m\omega^2(\mathbf{k},\nu)\mathbf{e}(i,\mathbf{k},\nu) = \mathbf{D}(\mathbf{k})\mathbf{e}(\mathbf{k},\nu),$$
where $\mathbf{D}$ is the dynamical matrix, the Fourier transform of $\phi$. You can get the derivation from any lattice dynamics textbook.
The problem is, the solution is only analytic in the harmonic approximation, hence the second order force constant matrix. Essentially, it is assumed that the potential energy of the equilibrium configuration of atoms is locally harmonic, which is generally valid at low temperatures (or for small atomic displacements). In the harmonic approximation, phonon lifetimes are infinite, so there is no heat dissipation in a solid, and you have to expand the potential energy to higher orders, at which point the solution is no longer nice and analytic. Essentially, you need to solve the phonon-Boltzmann equation, which will allow you to recover phonon lifetimes, and this requires at the very minimum, the third order interatomic force constants.
