In the Book "Quantum Field Theory I" by Manoukian, in section 4.3, from what I understood, he derived the quantum-action-principle of Schwinger only by using unitary time-evolution of the field Operators. At least I assume it to be like this, because for the proof he looked at the fields operators $\hat{\Phi}(x)$ (which obey some unitary time evolution generated by $\hat{H}$, and defined a field $\Pi$ which is supposed to generate variations of $\hat{\Phi}(x)$ and which time evolution is supposed to be the same. With the variations simply being c-numbers, which are proportional to unity, he then derives Schwinger's variational principle.

In short: He states that any infinitesimal variation of fields $\Phi(\vec{x}) \rightarrow \Phi(\vec{x}) + \delta \Phi(\vec{x})$ (where $\delta \Phi(\vec{x})$ is a c-number) can be written with a generator
$$ G(t) = \int d^3\vec{x} \delta \Phi(x) \Pi(\vec{x}) $$ With $\Pi$ being the canonical momentum of the field. It should be explicitly noted that the proof only uses c-number variations (which I translate to "$\delta \Phi$ is an ordinary complex number). The author then shows that you can write $$ \frac{d}{dt} G(t) = \int d^3\vec{x} \delta ( \dot{\hat{\Phi}} \hat{\Pi} - \mathscr{H}(\hat{\Phi}, \hat{\Pi})) $$ And following is the variational principle: $$ G(t_2) - G(t_1) = \delta \int d^3\vec{x} dt ( \dot{\hat{\Phi}} \hat{\Pi} - \mathscr{H}(\hat{\Phi}, \hat{\Pi})) $$ To be more specific (Manoukian doesn't write the steps down like that, I just assume them to be like this, with the "generator of the full variation " being "$\delta \Phi \hat{\Pi} - \delta \Pi \hat{\Phi}$. I denoted operators with a hat to separate them from numerical variations: $$ \int d^3\vec{x} \delta ( \dot{\hat{\Phi}} \hat{\Pi} - \mathscr{H}(\hat{\Phi}, \hat{\Pi})) = \int d^3\vec{x} \dot{(\delta \Phi )} \hat{\Pi} + \hat{\dot{\Phi}} \delta \Pi - \frac{i}{\hbar}[\delta \Phi \hat{\Pi} - \delta \Pi \hat{\Phi}, \mathscr{H}] = \int d^3\vec{x} \dot{(\delta \Phi )} \hat{\Pi} + \hat{\dot{\Phi}} \delta \Pi + \delta \Phi \hat{\dot{\Pi}} - \delta \Pi \hat{\dot{\Phi}} = \int d^3\vec{x} \dot{(\delta \Phi )} \hat{\Pi} + \delta \Phi \hat{\dot{\Pi}} = \frac{d}{dt} \int d^3\vec{x} \delta \Phi \hat{\Pi} $$ Where $\delta$ is a simultaneous variation of the fields AND the canonical momenta. This proof uses that $\delta \Phi$ is just a c-number, since the 2nd equality uses that $\delta \Phi$ and $\delta \Pi$ can simply be pulled out of the commutator. However, later on, the author talks about also using grassmann variables as Field variations, which leads to anticommutators instead of commutators for the field. My Question here would be: Why can we also use grassmann variable variations, without breaking the derivation here?

To make it more explicit: Why would $$\mathscr{H}(\hat{\Phi}+ \delta \Phi, \hat{\Pi} + \delta \Pi) - \mathscr{H}(\hat{\Phi}, \hat{\Pi}) = \hat{\dot{\Phi}} \delta \Pi + \delta \Phi \hat{\dot{\Pi}}$$ hold, even if the variation is not commuting, but anticommuting?


Even in the case of grassman variables, you will still be able to pull the variations to the left of the commutator. This is because the Hamiltonian - while being an operator - is necessarily a c-number valued operator.

This implies that you can pull the grassman valued variations through the Hamiltonian in the second term of the commutator. Remember, they are themselves not operator-valued. The important point is that c-numbers and grassman numbers commute. Therefore, a c-number valued operator also commutes with the unit operator multiplied with a grassman number.

  • $\begingroup$ Where do I know that the Hamiltonian is a C-number valued operator? I mean it contains terms that consists of $\Phi$ and $\Pi$, which aren't so necessarilly. $\endgroup$ – Quantumwhisp Nov 7 '17 at 22:49

Grassmann-odd/fermionic variables constitute no problem for the Schwinger action principle (SAP) per se. A much bigger issue is operator ordering ambiguities, which are already present in the Grassmann-even/bosonic sector. The prescription of the SAP is incomplete in the sense that it is not fully explained how a general action $S$ (which is a function) is promoted to an operator in a manner that is consistent with, e.g., unitarity. Schwinger and his school just assume it is possible, and perhaps give some examples where it works, cf. OP's title question (v5).

  • $\begingroup$ So you say that I need to give additional information in the Sense of fixing an operator ordering? $\endgroup$ – Quantumwhisp Nov 13 '17 at 10:19
  • $\begingroup$ In a nutshell, yes. $\endgroup$ – Qmechanic Nov 13 '17 at 10:26

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