In the path integral formulation of QFT, we should in principle be able to calculate the transition amplitude from a classical field configuration $\phi_{in}(x,t=0)$ to $\phi_{out}(x,t=T)$ using the path integral:
$$\int_{\phi_{in}(x,0)}^{\phi_{out}(x,T)} \mathcal D \phi e^{ iS[\phi]}.$$
For a free scalar field theory, is it possible to actually calculate this path integral and arrive at some closed form formula for the transition amplitude, as a functional of the input and output field configurations, namely $K[ \phi_{in}, \phi_{out};T]$?
In the free scalar field theory the integrals are Gaussian, so on the face of it seems that this should be possible (maybe by discretizing time, in analogy to the calculation in QM), but I don't think that I ever saw this actually being done. If this is impossible (or wrong to begin with), is there a particular reason why?
(this seems to be a similar question, but the answer is not so relevant)
UPDATE
Can we get it by treating the field as a infinite collection of harmonic oscillators, as hinted by @Qmechanic's comment ? applying a Fourier transform to the spatial coordinates, we can write
$$S = \int d^4x \frac{1}{2}\left( (\partial_{\mu} \phi)^2 - m^2 \phi^2 \right)=\int dt \int d^3 p \frac{1}{2}\left( \dot \phi(p)^2 - (\mathbf p ^2+m^2)\phi(p)^2\right)$$
(Using $\phi(p,t)$ to denote the 3-dimentionsal Fourier transform of $\phi(x,t)$, hopefully without confusion)
Then by somewhat naively taking an infinite product of harmonic oscillator propagators, we would get something like
$$K[ \phi_{in}, \phi_{out};T] = A(T) \exp \left( \frac{i}{2} \int d^3p \left( \omega_p (\phi_{in}(p)^2+\phi_{out}(p)^2)\cot \omega_p T - \frac{2\phi_{in}(p)\phi_{out}(p)}{\sin \omega_p T}\right) \right) $$
where $\omega_p^2=\mathbf p ^2+m^2$ and $A(T)$ is a normalization factor.
Does this make sense ?
