Equivalence of Schwingers action principle and path integral formalism At the moment I'm working with Schwingers quantum action principle. For the harmonic oscillator I understood why this action principle is equivalent to the path integral formalism for quantum mechanics. 
But why is it equivalent to the path integral formalism for relativistic fields? I can't find anything really usefull to answer this question. 
 A: Consider a path integral of some functional $\Omega[\varphi]$ called the expectation of $\Omega$:
$$ \int D\varphi\, e^{i \hbar^{-1} S[\varphi]} \, \Omega[\varphi] = \left<\Omega\right>. $$
Now consider a small variation of the dummy integration variable:
$$ \varphi(x) \rightarrow \varphi(x) + \delta\varphi(x). $$
How does the expectation value $\left<\Omega\right>$ change? Obviously, it does not, since it does not depend on $\varphi$. But we could calculate the variation of the path integral:
$$ 0 = \delta \left<\Omega\right> = \delta \int D\varphi\, e^{i \hbar^{-1} S[\varphi]} \, \Omega[\varphi] = \int D\varphi\, \left( i \, \hbar^{-1} \delta S \, \Omega \, e^{i \hbar^{-1} S[\varphi]} + \delta \Omega \, e^{i \hbar^{-1} S[\varphi]} \right) = \left< \delta \Omega + i \hbar^{-1} \, \Omega \, \delta S \right>. $$
This is precisely the so called quantum action principle:
$$ \left<\delta\Omega\right> = -i \hbar^{-1}  \, \left<\Omega\,\delta S\right>. $$
Note that this equation holds even if we let $\delta \varphi$ to depend on $\varphi$, the only requirement being that the path integral measure $D\varphi$ is invariant under $\varphi \rightarrow \varphi + \delta\varphi$. In case it is not, an additional term will appear, which is called the quantum anomaly.
