I'll give as example the case of a scalar field. We assume the Einstein-Hilbert action:
\begin{equation}
S_{\text{grav}}=\int d^4x\ \sqrt{|g|}R
\end{equation}
Now, we would like to consider a quantum field in the spacetime:
$$S=S_{\text{grav}}+S_{\text{matter}}+S_{\text{coupling}}$$
To first order in the curvature, the only scalar that couples gravity and the quantum field, which we can build out of a scalar field $\phi$ and curvature-related tensor objects, is $R \phi$. In general, there will be higher order terms if one considers higher energies. We will take the standard Lagrangian for a scalar field. The total action is now
\begin{equation}
S=\int d^4x \sqrt{|g|}\bigl(R+\frac{1}{2}g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi+m^2\phi^2+\xi R\phi\bigr)
\end{equation}
Where $\xi$ is the coupling constant. Minimal coupling amounts to setting $\xi=0$. As you can imagine, this is the simplest (and perhaps most natural?) case. Another reasonably popular choice seems to be $\xi=\frac{1}{6}$. In this case, we say that the field is *conformally* coupled to gravity, because the action is now invariant under conformal transformations of the metric:
$$ g_{\mu\nu}\rightarrow \Omega^2(x)g_{\mu\nu}$$
Any possible $\xi\neq 0$ is a case of non-minimal coupling. Basically, minimal coupling means avoiding introducing any extra terms in the action.

This post imported from StackExchange Physics at 2014-04-18 05:47 (UCT), posted by SE-user Danu