Consider a quantum field theory with action $I[\phi]$, and suppose we 'turn on' a set of classical currents $J_r(x)$ coupled to the fields $\phi^r(x)$ of the theory. The complete vacuum vacuum amplitude in the presence of these currents is then

\(Z[J]\equiv\left< \text{VAC, out}|\text{VAC, in}\right>_J\\ =\int \Bigg[\prod_{s,y}d\phi^s(y)\Bigg]\text{exp}\bigg(iI[\phi]+i\int d^4x\phi^r(x)J_r(x)+\epsilon \text{ terms}\Bigg).\)

The Feynman rules for calculating $Z[J]$ are just the same as for calculating the vacuum-vacuum amplitude $Z[0]$ in the absence of the external current, except that the Feynman diagrams now contain vertices of a new kind, to which a $single$ $\phi^r$-line is attached. Such a vertex labelled with a coordinate x contributes a 'coupling' factor $iJ_r(x)$ to the integrand of the position-space Feynman amplitude. Equivalently, we could say that in the expansion of $Z[J]$ in powers of $J$, the coefficient of the term proportional to $iJ_r(x)iJ_s(y)\cdots$ is just the sum of diagrams with external lines (including propagators) corresponding to the fields $\phi^r(x), \phi^s(y,)$ etc. In particular, the first derivative gives the vacuum matrix element of the quantum mechanical operator $\Phi^r(x)$ corresponding to $\phi^r(x)$:

\(\Bigg[\frac{\delta}{\delta J_r(y)}Z[J]\Bigg]_{J=0}=\int \Bigg[\prod_{r,x}d\phi^r(x)\Bigg]\phi^r(y)\text{exp}\{iI[\phi]+\epsilon \text{ terms}\}\\ =i\left< \text{VAC, out}|\Phi^r(x)|\text{VAC, in}\right>_{J=0}.\)

What's the difference between quantum mechanical operator $\Phi^r(x)$ and its corresponding quantum fields $\phi^r(x)$ ?