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.


The $\delta$ in this context is from the calculus of variations. It's used in analogy to ordinary calculus where $\operatorname{d}x$ is notation for an infinitesimal change in the quantity $x$, $\delta f(x)$ is an infinitesimal change to the function, $f$. In the notation of the Wikipedia article linked, $\delta f \equiv \epsilon \eta$. Think of it as being the function space analogue of $\mathrm{d}x_i$ for some vector $\vec{x}$.

When it is placed in a functional derivative, like $$\frac{\delta S[f]}{ \delta f(x)}$$ it is analogous to a gradient of the form $\frac{\partial h(\mathbf{x})}{\partial x_i}$.

Now, you may ask what is meant by an "infinitesimal function," and the answer to that depends on what which metric is chosen for function space. In physics, we usually just use the $L^p$ space with $p=2$, so the square distance between two functions $f$ and $g$ is given by: $$ D^2 = \int |f(x) - g(x)|^2 \operatorname{d}x.$$ So an infinitesimal function will be any function that is an infinitesimal distance from the $0$ function.

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    $\begingroup$ I'm sorry, but that doesn't answer my question at all, since here $\delta$ isn't used in front of a function, but instead in front of an inner product of states. $\endgroup$ – Quantumwhisp Jul 16 '17 at 6:50
  • $\begingroup$ @Quantumwhisp The states are represented by wave functionals (e.g. physics.stackexchange.com/questions/491136/… ), so the parameter is a function of space. $\endgroup$ – Sean E. Lake Nov 10 '20 at 2:52

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