# Path integral in QM vs QFT

On page 282 of Peskin and Schroeder discussing functional quantization of scalar fields, the authors use expression 9.12, the path integral in ordinary quantum mechanics

$$U(q_a,q_b;T)=$$ $$\bigg(\prod_i\int\mathcal{D}q(t)\mathcal{D}p(t)\bigg)\exp\bigg[i\int_0^T\,dt\big(\sum_ip^i\dot{q}^i-H(q,p)\big)\bigg]\tag{9.12}$$

then since $$H=\int\,d^3x[\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+V(\phi)],$$ we have $$\langle\phi_b(\boldsymbol{x})|e^{-iHT}|\phi_a(\boldsymbol{x})\rangle=$$ $$\int\,\mathcal{D}\phi\mathcal{D}\pi\exp\bigg[i\int_0^T\,d^4x\Big(\pi\dot{\phi}-\frac{1}{2}\pi^2-\frac{1}{2}(\nabla\phi)^2-V(\phi)\Big)\bigg].\tag{9.14}$$

My question is: what does 9.14 even mean? Previously we derived 9.12 from ordinary quantum mechanics, and it has a well defined meaning, in the sense that I can in theory write down a precise equation for symbols like $$\mathcal{D}p(t)$$; in fact Peskin and Schroeder does it in 9.11. But how can one write down the precise meaning of $$\mathcal{D}\pi$$ and $$\mathcal{D}\phi$$?

Well, the path integral is a heuristic construction. Formally P&S identify the index $$i ~=~{\bf x}$$ with a point in 3-space, so that $$q^i(t)~=~\phi({\bf x},t) \qquad\text{and}\qquad p^i(t)~=~\pi({\bf x},t).$$ To complete the transition from point mechanics (9.12) to field theory (9.14), one often imagines that 3-space and time is discretized. Then spacetime derivatives are replaced by appropriate finite differences, and the path integral measures become $${\cal D}q ~=~\prod_{i,t} \mathrm{d}q^i(t) \qquad\text{and}\qquad {\cal D}\phi ~=~\prod_{{\bf x},t} \mathrm{d}\phi({\bf x},t).$$
Re "My question is: what does 9.14 even mean? " I will just add a footnote to the answer above which is fine. For the QM itself and, as expected, under some technical assumptions on the potential $$V$$, there are rigorous ways to give meaning to path integral. Things get trickier when labeling the spacetime points using $$\mathbb{R}^{d - 1} \times \mathbb{R} \cong \mathbb{R}^d$$ with $$d > 1$$ (as you'd expect in QFT) when one has to switch to distributions.