Equivalence of Schrödinger operator formalism and Path Integral Formulations for Scalar Field Theory

Question

I'm exploring the deep connections between different formulations of quantum field theories and have a specific question about the equivalence between the Schrödinger representation and the path integral representation for scalar field theory.

Since the path integral is derived from the Schrödinger equation, I recon that the relation should be bi-directional:

$$ \langle \phi_f |e^{-iHt/\hbar} | \phi_i \rangle = \int D[\phi] e^{iS[\phi]/\hbar}. $$

Consider a free scalar field theory with the Lagrangian density:

$$ {\cal L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2 . $$

From this, we can derive the corresponding Hamiltonian density:

$$ {\cal H} = \frac{1}{2} \pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2 $$

where $ \pi $ is the canonical momentum associated with $ \phi $.

Now, in the Schrödinger representation, we would describe the quantum field by a wavefunctional, $\Psi[\phi(x), t]$, evolving according to:

$$ i\hbar \frac{\partial}{\partial t} \Psi[\phi(x), t] = H \Psi[\phi(x), t]. $$

Given that the wavefunctionals in the Schrödinger picture are elements of a Fock space, is the Schrödinger representation using the derived Hamiltonian equivalent to the path integral representation?

Answer 1

TL;DR: This is mainly an exercise in transcribing the standard QM formulas into QFT.

1. The standard convention to write the wavefunctional is$^1$ $$ \Psi_S[\phi,t]~\equiv~{}_H\langle\phi,t|\Psi\rangle_H~\equiv~{}_H\langle\phi|\Psi(t)\rangle_S. \tag{1} $$

2. Therefore OP's TDSE $$i\hbar \frac{d}{dt}\Psi_S[\phi, t] ~=~ \hat{H}_S \Psi_S[\phi, t]\tag{2}$$ in ket-form becomes $$i\hbar \frac{d}{dt}| \Psi(t)\rangle_S ~=~ \hat{H}_S | \Psi(t)\rangle_S. \tag{3} $$

3. Here the Hamiltonian $$\hat{H}_S ~=~\int d^3{\bf x}~ \hat{\cal H}_S ({\bf x})\tag{4}$$ is given in terms of the Hamiltonian density $$\hat{\cal H}_S({\bf x}) ~=~ \frac{1}{2}\left\{ \hat{\pi}_S({\bf x})^2 + (\nabla\hat{\phi}_S({\bf x}))^2 + m^2 \hat{\phi}_S({\bf x})^2 \right\} +{\cal V}(\hat{\phi}_S({\bf x})).\tag{5}$$

4. The CCR $$[\hat{\phi}_S({\bf x}), \hat{\pi}_S({\bf y})] ~=~i\hbar\hat{\bf 1}\delta^3({\bf x}\!-\!{\bf y})\tag{6}$$ is the first principle of canonical quantization.

5. The Schrödinger representation reads $$\hat{\phi}_S({\bf x})~=~\phi({\bf x}), \qquad \hat{\pi}_S({\bf x})~=~\frac{\hbar}{i}\frac{\delta}{\delta\phi({\bf x})}. \tag{7}$$

6. The time evolution operator is $$\hat{U}(t)~:=~\exp\left(-\frac{i}{\hbar}t\hat{H}_S\right). \tag{8}$$

7. The phase space path integral is derived from the operator formalism in the standard way by inserting infinitely many completeness relations $$\begin{align}{}_H\langle \phi_f|&\exp\Big[-\frac{i}{\hbar}(t_f\!-\!t_i)\hat{H}_S\Big]|\phi_i\rangle_H \cr ~=~&\int_{\phi(\cdot,t_i)=\phi_i(\cdot)}^{\phi(\cdot,t_f)=\phi_f(\cdot)}\, \mathcal{D}\frac{\phi}{\sqrt{\hbar}}\,\mathcal{D}\frac{\pi}{\sqrt{\hbar}}\,\exp\Big[\frac{i}{\hbar}\int_{t_i}^{t_f} d^4x\,\big(\pi(x)\dot{\phi}(x)-{\cal H}_S(x) \big)\Big], \end{align}\tag{9}$$ cf. e.g. this related Phys.SE post.

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$^1$ The subscripts $S$ and $H$ refer to the Schrödinger and Heisenberg picture, respectively. The notation $|\phi,t\rangle_H$ denotes a Heisenberg instantaneous eigenstate: $$ \hat{\phi}_H({\bf x},t)|\phi,t\rangle_H~=~\phi({\bf x})|\phi,t\rangle_H .\tag{10}$$ Here $$ |\phi\rangle_H~\equiv~|\phi,t\!=\!0\rangle_H.\tag{11}$$