Path Integral Quantization in Peskin and Schroeder

Question

I'm trying to learn the Feynman path integral quantization of scalar fields using Peskin and Schroeder, above eq. (9.14) on p. 282 in section 9.2 the book says:

$$\langle \phi_b({\bf x})|e^{-iHT}|\phi_a({\bf x})\rangle$$ $$=\int\,\mathcal{D}\phi\,\mathcal{D}\pi\,\text{exp}\left[i\int_0^T d^4x\,\left(\pi\dot{\phi}-\frac{1}{2}\pi^2-\frac{1}{2}(\nabla\phi)^2-V(\phi)\right)\right]\tag{p.282}$$ where the function $\phi(x)$ over which we integrate are constrained to be specific configurations $\phi_a({\bf x})$ at $x^0=0$ and $\phi_b({\bf x})$ at $x^0=T$.

While I now understand $\mathcal{D}\pi$ and $\mathcal{D}\phi$, I'm still bewildered by this sentence.

1. What are $\phi_a({\bf x})$ and $\phi_b({\bf x})$? How can we impose boundary conditions like this? In the second quantization approach, $\phi(x)$ is completely fixed, given by some integral expression.

2. How can we deduce the canonical quantization expression from this? Can we do this like in nonrelativistic quantum mechanics where we derived path integral and Schrödinger's equation from each other?

Answer 1

1. Perhaps P&S's notation would have been clearer if they had written $$\langle \phi_b(\cdot)|e^{-iHT}|\phi_a(\cdot)\rangle$$ $$~=~\int_{\phi(\cdot,0)=\phi_a(\cdot)}^{\phi(\cdot,T)=\phi_b(\cdot)}\,\mathcal{D}\phi\,\mathcal{D}\pi\,\text{exp}\Big[i\int_0^T d^4x\,\big(\pi\dot{\phi}-{\cal H} \big)\Big].$$

2. The above formula is the natural field theoretic generalization of the point mechanical formula $$\langle q_b|e^{-iHT}|q_a\rangle ~=~U(q_b,q_a,T)\tag{9.8}$$ $$~=~\int_{q(0)=q_a}^{q(T)=q_b}\,\mathcal{D}q\,\mathcal{D}p\,\text{exp}\Big[i\int_0^T dt\,\big(p_k\dot{q}^k- H \big)\Big].\tag{9.12} $$ P&S derives the Schrödinger equation in eq. (9.7).

   See also e.g. this & this related Phys.SE posts.