TL;DR: The main reason why the naive path integral derivation of SD eqs. works well beyond what could be expected is that both concepts employ the same underlying notion of time ordering, namely the covariant time ordering $T_{\rm cov}$.

I) In the main part of this answer we would like to investigate in more detail some formal aspects of the correspondence between

$$ \text{Path integral formalism}\qquad \longleftrightarrow \qquad
\text{Operator formalism},\tag{1} $$

in particular the cohabitation of, on one hand the Schwinger-Dyson (SD) equations, and on another hand, the Heisenberg's equations of motion (eom).
The correspondence (1) is notoriously subtle, see e.g. Ref. 1. and this Phys.SE post. For a general interacting field theory, both sides of the path integral/operator correspondence (1) are typically not rigorously defined, cf. e.g. Ref. 1. Both sides of the correspondence may in principle receive quantum corrections due to operator ordering problems. So it is difficult to come with reliable statements at this formal level.

To simplify the discussion and gain intuition, we are going to make some assumptions.

We are going to consider *free* (quadratic) theories only. OP's example is covered by this. Free theories have the advantage that we can present explicit formulas.

It is most economically to argue via the *Hamiltonian* (as opposed to the Lagrangian) formulation, since we then are only going to have one (as opposed to two) time derivatives. So that's what we are going to do. (Also for simplicity we ignore cases with singular Legendre transformations. Then it is always possible to (Gaussian) integrate out the momenta to get to the corresponding Lagrangian formulation, if that's what one wants.)

Also we treat field theory formally like *point mechanics.* All spatial coordinates are suppressed via DeWitt condensed notation. Only the time-variable $t$ is manifestly kept. Thus our variables are
$$ z^I, \qquad I~\in~ \{1, \ldots, 2N\},\tag{2} $$
and they depend on time $t$, where $N$ could be infinite.

Also to avoid annoying sign factors, we restrict attention to theories with only *Grassmann-even* variables.

Furthermore, we assume for simplicity (or via Darboux's Theorem) that the equal-time Poisson bracket is constant and $z$-independent
$$\omega^{IK}
~=~ \{ z^I(t),z^K(t) \}_{PB}, \qquad
I,K~\in~\{1, \ldots, 2N\}. \tag{3} $$

Much of the above assumptions can be relaxed, but we will not discuss that here.

II) Classically, with the above assumptions, the Hamiltonian action reads

$$ S_H[z;J]~=~ \int dt\ L_H(z(t);\dot{z}(t);J(t)),\tag{4} $$

where the Hamiltonian Lagrangian is

$$ L_H(z;\dot{z};J)~=~\frac{1}{2} z^I\omega_{IK} \dot{z}^K - H(z,J),\tag{5} $$

and the Hamiltonian is quadratic

$$ H(z,J)~=~H_0(z) -J_I z^I, \qquad
H_0(z)~:=~ \frac{1}{2} z^I h_{IK} z^K. \tag{6} $$

The Euler-Lagrange derivatives (with an index raised by the symplectic metric $\omega^{IL}$) correspond to Hamilton's eom

$$ 0~\approx~\omega^{IK}\frac{\delta S_H[z;J]}{\delta z^K(t)}
~=~\dot{z}^I(t) - \{z^I(t),H(z(t),J(t))\}_{PB} $$
$$ ~=~\left(\delta^I_K\frac{d}{dt}- h^I{}_K\right)z^K(t) + \omega^{IK} J_K(t). \tag{7} $$

(Here the $\approx$ sign means equality modulo classical equations of motion.) Also we have defined the matrix

$$ h^I{}_L~:=~\omega^{IK} h_{KL}. \tag{8} $$

In the corresponding operator formalism, the Hamilton's eom (7) turns into Heisenberg's eom, which is an operator identity.

III) The Hessian reads

$$ {\cal H}_{IK}(t,t^{\prime})~:=~\frac{\delta^2 S_H[z;J]}{\delta z^I(t)\delta z^K(t^{\prime})} ~=~
\omega_{IK}\delta^{\prime}(t-t^{\prime})
-h_{IK}\delta(t-t^{\prime})$$
$$~=~\omega_{IL}\left(\delta^L_K\frac{d}{dt}- h^L{}_K\right)\delta(t-t^{\prime}).\tag{9} $$

The corresponding Green's function $G={\cal H}^{-1}$ is the inverse of the Hessian (9):

$$ G^{IK}(t,t^{\prime})~=~\frac{1}{2} {\rm sgn}(t-t^{\prime})\left(e^{(t-t^{\prime})h}\right)^I{}_L ~\omega^{LK}, \tag{10} $$

$$ \left(\delta^I_L\frac{d}{dt}- h^I{}_L\right)G^{LK}(t,t^{\prime})~=~\omega^{IK}\delta(t-t^{\prime}). \tag{11} $$

IV) We define the partition function

$$ Z[J]~:=~ \int\![dz] e^{\frac{i}{\hbar}S_H[z;J]}
~=~ e^{\frac{i}{\hbar}W_c[J]} $$
$$~=~ \int\![dz] \exp\left[\frac{i}{2\hbar}\iint \! dt~dt^{\prime}~z^I(t) ~{\cal H}_{IK}(t,t^{\prime})~ z^K(t^{\prime}) + \frac{i}{\hbar}\int \! dt~J_I(t) z^I(t)\right]$$
$$~=~Z[0]\exp\left[-\frac{i}{2\hbar}
\iint \! dt~dt^{\prime}~J_I(t) ~G^{IK}(t,t^{\prime})~ J_K(t^{\prime})\right] ,\tag{12} $$

and the generator of connected Feynman diagrams

$$ W_c[J]~:=~\frac{\hbar}{i}\ln Z[J]
~=~ W_c[0]- \frac{1}{2} \iint \! dt~dt^{\prime}~J_I(t) ~G^{IK}(t,t^{\prime})~ J_K(t^{\prime}) ,\tag{13} $$

whose explicit form follows from Gaussian integration. The quantum-average/expectation-value in the Heisenberg picture is defined as

$$\left< T_{\rm cov}\{F[\widehat{z}]\} \right>_J
~=~ \frac{1}{Z[J]}\int\![dz]~F[z]~e^{\frac{i}{\hbar}S_H[z;J]}
~=~\frac{1}{Z[J]} F\left[ \frac{\hbar}{i} \frac{\delta}{\delta J} \right] Z[J] .\tag{14} $$

Here $T_{\rm cov}$ is covariant time-ordering, i.e. time-differentiations inside its argument should be taken *after/outside* the usual time ordering $T$. Here time-ordering $T$ is defined as

$$T\left\{\widehat{A}(t)\widehat{B}(t^{\prime})\right\}~:=~ \theta(t-t^{\prime})\widehat{A}(t)\widehat{B}(t^{\prime})+\theta(t^{\prime}-t)\widehat{B}(t^{\prime})\widehat{A}(t). \tag{15} $$

It is crucial that time-differentiation and time-ordering do *not* commute:

$$T_{\rm cov}\left\{\frac{d\widehat{A}(t)}{dt}\widehat{B}(t^{\prime})\right\} ~:=~\frac{d}{dt}T\left\{\widehat{A}(t)\widehat{B}(t^{\prime})\right\}$$
$$~=~ T\left\{\frac{d\widehat{A}(t)}{dt}\widehat{B}(t^{\prime})\right\}+\delta(t-t^{\prime})[\widehat{A}(t),\widehat{B}(t^{\prime})], \tag{16}$$

but instead give rise to equal-time commutator (contact) terms.

It is important to realize that the time derivatives inside the Boltzmann factor $e^{\frac{i}{\hbar}S_H[z;J]}$ in the path integral should respect the underlying time slicing procedure. See e.g. this and this Phys.SE answer. This induces the covariant time-ordering $T_{\rm cov}$ in eq. (14).

The 1-point function reads

$$ \left< \widehat{z}^I(t)\right>_J
~=~ \frac{\delta W_c[J]}{\delta J_I(t)}
~=~ -\int \!dt^{\prime} ~G^{IK}(t,t^{\prime})~J_K(t^{\prime}), \tag{17} $$

while the time-ordered 2-point function reads

$$ \left< T_{\rm cov}\{ \widehat{z}^I(t)\widehat{z}^K(t^{\prime})
\}\right>_J
~=~ -\frac{\hbar^2}{Z[J]}
\frac{\delta^2 Z[J]}{\delta J_I(t)\delta J_K(t^{\prime})}$$
$$~=~\frac{\delta W_c[J]}{\delta J_I(t)}\frac{\delta W_c[J]}{\delta J_K(t^{\prime})}+\frac{\hbar}{i}\frac{\delta^2 W_c[J]}{\delta J_I(t)\delta J_K(t^{\prime})}$$
$$ ~=~\left< \widehat{z}^I(t)\right>_J
\left< \widehat{z}^K(t^{\prime})\right>_J
+i\hbar G^{IK}(t,t^{\prime}). \tag{18}$$

The corresponding 2-point function *without time-ordering* reads:

$$\left< \widehat{z}^I(t)\widehat{z}^K(t^{\prime}) \right>_J
~=~\left< \widehat{z}^I(t)\right>_J
\left< \widehat{z}^K(t^{\prime})\right>_J
+\frac{i\hbar}{2} \left(e^{(t-t^{\prime})h}\right)^I{}_L ~\omega^{LK}. \tag{19} $$

V) The Schwinger-Dyson (SD) equations read

$$ i\hbar\left< T_{\rm cov}\left\{\frac{\delta F[\widehat{z}]}{\delta z^I(t)} \right\}\right>_J
~=~\left< T_{\rm cov}\left\{ F[\widehat{z}]\frac{\delta S_H[\widehat{z};J]}{\delta z^I(t)}\right\}\right>_J.\tag{20} $$

The SD eqs. (20) are here formally written in the operator language, but their justification is most easily argued via the path integral formalism, cf. e.g. this Phys.SE post. The SD eqs. (20) simply reflect the fact that a path integral of a total derivative vanishes if the boundary contributions are zero

$$ 0~=~\int [dz]\frac{\delta}{\delta z^I(t)}\left\{F[z] e^{\frac{i}{\hbar}S_H[z;J]}\right\}, \tag{21} $$

cf. this Phys.SE post.

VI) Naively the rhs. of the SD-eqs. (20) is proportional to the Heisenberg eom (7), which is an operator expression that vanishes identically, so why then there is a non-zero quantum correction on the lhs. of the SD-eqs. (20)? The resolution to this apparent paradox is hidden in the fact that time-differentiation and time-ordering do not commute, cf. eq. (16). To see how this works, pick for simplicity the functional $F[z]=z^{K}(t^{\prime})$ of a single variable. (Then it is enough to use the Poisson bracket rather than the Moyal-Groenewold $\star$-product.) After raising an index by the symplectic metric $\omega^{IL}$, the SD-eqs. (20) become

$$i\hbar\omega^{IK} \delta(t-t^{\prime})
~\stackrel{(20)}{=}~\left< T_{\rm cov}\left\{\omega^{IL}\frac{\delta S_H[\widehat{z};J]}{\delta z^L(t)}\widehat{z}^K(t^{\prime})\right\} \right>_J$$
$$ ~\stackrel{(7)}{=}~\left< \frac{d}{dt}T\left\{\widehat{z}^I(t)
\widehat{z}^K(t^{\prime})\right\}\right>_J
- \left< T\left\{\{\widehat{z}^I(t),H(\widehat{z}(t),J(t))\}_{PB} \widehat{z}^K(t^{\prime})\right\}\right>_J$$
$$ ~=~\left(\delta^I_L\frac{d}{dt}- h^I{}_L\right)\left< T\left\{\widehat{z}^L(t)
\widehat{z}^K(t^{\prime})\right\}\right>_J +\omega^{IL} J_L(t)\left< \widehat{z}^K(t^{\prime})\right>_J $$
$$ ~\stackrel{(16)}{=}~i\hbar\omega^{IK} \delta(t-t^{\prime}) +\left< T\left\{ \underbrace{\left(\frac{d\widehat{z}^I(t)}{dt} - \{\widehat{z}^I(t),H(\widehat{z}(t),J(t))\}_{PB} \right)}_{=0}\widehat{z}^K(t^{\prime})\right\}\right>_J$$
$$~\stackrel{(7)}{=}~i\hbar\omega^{IK} \delta(t-t^{\prime}).\tag{22} $$

Eq. (22) shows how SD-eqs. (20) and Heisenberg's eom (7) can co-exist.

References:

- F. Bastianelli and P. van Nieuwenhuizen,
*Path Integrals and Anomalies in Curved Space,* 2006.

This post imported from StackExchange Physics at 2017-11-26 12:33 (UTC), posted by SE-user Qmechanic