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?

  • $\begingroup$ You are asking for something like this question on PI/canonical quantization equivalence but for the Klein Gordon field, right? $\endgroup$
    – Quillo
    Commented Oct 20, 2023 at 6:34
  • $\begingroup$ Possible duplicate: Path Integral Quantization in Peskin and Schroeder $\endgroup$
    – Qmechanic
    Commented Oct 20, 2023 at 6:41
  • $\begingroup$ That question and its (useful and closely related) answer do not say anything about the functional Schrödinger equation. I think the OP would like to know how to go from the functional Schrödinger equation to the PI and vice versa. Ofc, it is very likely to be a duplicate, but of another question. @Qmechanic $\endgroup$
    – Quillo
    Commented Oct 20, 2023 at 7:18
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Andrew
    Commented Oct 22, 2023 at 15:25

1 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 $$ \Psi[\phi,t] ~\equiv~ \langle \phi | \Psi(t)\rangle. $$

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

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

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

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

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

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


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