# Why is it justified to discard off-shell momenta contributions in the exponent of the expression for a path integral amplitude?

Let us consider a free field theory with one field $$\phi$$. The Lagrangian density is $$L(\phi, \partial_{\mu} \phi)$$ and the corresponding Hamiltonian density is $$H(\phi,\pi,\partial_{\mu \neq 0}\phi)$$. If the system is in state $$|\phi_{i} \rangle$$ at time $$0$$, then the probability amplitude for the system to be in state $$| \phi_{f} \rangle$$ at a future time $$t$$ can be derived to be ($$\mathcal{H}= \int d^{3}x \, H(x)$$)

$$\begin{equation} Z=\langle \phi_{f}| \exp(-i \mathcal{H}t)|\phi_{i} \rangle = \int D \phi D \pi \, \exp \left(i \int d^{4}x \, \left[\partial_{0} \phi(x) \pi(x)-H(x) \right] \right) \end{equation}$$

At this stage one makes the stationary state approximation to claim that the integral over $$\partial_{0} \phi \pi-H$$ is overwhelmingly dominated by those values of $$\pi$$ that render it equal to the Lagrangian density $$L$$, and so we end up having

$$\begin{equation} Z \approx \int D \phi \, \exp \left(i \int d^{4}x \, L \right) \end{equation}$$

The derivation itself has some subtleties with operator ordering etc., but besides that, how can one convince oneself that this stationary state approximation is fully consistent with calculations, i.e. that this won't cause any calculations in this theory to go awry?