The time-dependent Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \psi(x,t)=\hat{H} \psi(x,t).$$
Let's assume for simplicity that
$$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x}),$$
i.e., that ##\hat{H}## is not explicitly time dependent. Then the formal solution of the equation above is
$$\psi(x,t)=\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \psi(x,0).$$
This you can write in the form
$$\psi(x,t)=\int_{\mathbb{R}} \mathrm{d} x' \left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle \psi(x',0)= \int_{\mathbb{R}} \mathrm{d} x' G(x,x',t) \psi(x',0),$$
i.e., the propator is
$$G(x,x',t)=\left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle.$$
Usually it's of course difficult to really calculate the propagator.
For a free particle, where ##\hat{H}=\hat{p}^2/(2m)## you can use the momentum eigenstates to evaluate it:
$$G(x,x',t)=\int_{\mathbb{R}} \mathrm{d} p \langle x |\exp[-\mathrm{i} \hat{p}^2 t/(2m \hbar)]|p \rangle \langle p|x' \rangle.$$
