You have to interchange the integration over paths and over time to obtain

\(G(\mathbf{x}_1,\mathbf{x}_0;t_1,t_0)=
\int_{t_0}^{t_1} dt\int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_{t_0}^{t_1} ds ~ \frac{1}{2} m \mathbf{\dot x^2} ] (- \frac{i}{\hbar} ~V(\mathbf{x}(t)))\)

The inner integral can be interpreted as a propagator again, except that the particle is now scattered at the potential *V* at the place **x**(t).

The inner path integral corresponds to the propagator of a free particle, and we can write

\(\int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_{t_0}^{t_1} ds ~ \frac{1}{2} m \mathbf{\dot x^2} ]
= G_0(\mathbf{x_1},\mathbf{x_0};t_1,t_0)
\)\(= \int dxdt' G_0(\mathbf{x_1},\mathbf{x};t_1,t')
\cdot G_0(\mathbf{x},\mathbf{x_0};t',t_0)
\delta(\mathbf{x}-\mathbf{x}(t))\delta(t-t')\)

This is the amplitude for a particle traveling from the point **x0** to some point **x** and then to **x1**.

Another way of writing this is

\(\langle x_1 |U| x_0\rangle
= \int dx \langle x_1 |U_0| x\rangle
\langle x |U_0| x_0\rangle
(-\frac{i}{\hbar}V(x))\)

Now, the Born scattering amplitude is just the Fourier transform of this. The free propagator \(U_0\) will give the terms involving energy, whereas the addition multiplication with \(V(x)\)corresponds to the additional matrix element.