Schwinger's quantum action principle states, that if you "vary the field Operators" $$\hat{\phi'}(x') = \hat{\phi}(x) + \delta \hat{\phi}(x)$$ (which changes the Action Operator), then "the variation of the transition Amplitude" is equal to the the variation of the Action. In the Heisenberg picture, given $$ x' = x + \delta x(x) $$ and $$ \hat{\phi'}(x') = \hat{\phi}(x) + \delta\hat{ \phi}(x) $$ you can calculate what the variation of the action $\delta S$ is supposed to be. Further more, you can give some kind of meaning to these mathematical operations: Consider the state of the field being $| \Psi \rangle$, then the transformation changes the mean value of the field accordingly to $$ \langle\phi'\rangle(x') = \langle\phi\rangle(x) + \langle\delta \phi\rangle(x) $$ In the same manner, the mean value of the action that "this path" has will change by $\delta S$.
So much for the right hand side. Now for the left hand side, where I always read something like:
$$ \delta \langle \phi_1, \tau_1| \phi_2, \tau_2 \rangle $$
Question: What is the meaning of $\delta$ in this expression?
To be more specific: Usualy a $\delta$ indicates a difference between two values, presumably between $ \langle \phi_1, \tau_1| \phi_2, \tau_2 \rangle$ and $ \langle \phi_1, \tau_1|' |\phi_2, \tau_2 \rangle'$. Which leads me to the question, how $|\phi_2, \tau_2 \rangle'$ is connected to $|\phi_2, \tau_2 \rangle$. Further more, since there is some transformation mapping $|\phi_2, \tau_2 \rangle$ to $|\phi_2, \tau_2 \rangle'$, how is this transformation connected to the transformation of the operator?
I'm asking this because I'm somehow lacking an intuitive understanding of what exactly I am varying here, and what exactly the principle says, and it seems that nobody ever dared to write it down in it's whole explicitly.