Equations of Motion and Noether's Theorem in the Functional Formalism

First, let us recall the derivation of the equations of motion and Noether's theorem in classical field theory. We have some action functional $S[\phi]$ defined by some local Lagrangian:
$$S[\phi] = \int L(\phi, \partial \phi) dx.$$
The classical equations of motion are just the Euler-Lagrange equations
$$\frac{\delta S}{\delta \phi(x)} = 0  \iff \partial_\mu \left( \frac{\partial L}{\partial(\partial_\mu\phi)} \right) = \frac{\partial L}{\partial \phi}$$

Now suppose that $S$ is invariant under some transformation $\phi(x) \mapsto \phi(x) + \epsilon(x) \eta(x)$, so that $S[\phi] = S[\phi+\epsilon \eta]$. Here we treat $\eta$ as a fixed function but $\epsilon$ may be an arbitrary infinitesimal function. The Lagrangian is not necessarily invariant, but rather can transform with a total derivative:
$$L(\phi+\epsilon \eta) = L(\phi) + \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu$$
For some unknown vector field $f^\mu$ (which we could compute given any particular Lagrangian). So let's compute
$$\begin{align}\delta_\epsilon S &= \int \delta_\epsilon L \\\ &= \int \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu \\\ &= \int \partial_\mu \left(f^\mu - \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \right) \epsilon \end{align}$$
Let us define the Noether current $J^\mu$ by
$$J^\mu = \frac{\partial L}{\partial (\partial_\mu \phi)} \eta - f^\mu.$$
Then the previous computation showed that
$$\frac{\delta S}{\delta \epsilon} = -\partial_\mu J^\mu.$$
If $\phi$ is a solution to the Euler-Lagrange equations, then the variation $dS$ vanishes, hence we obtain:

Theorem (Noether's theorem) The Noether current is divergence free, i.e.
$$\partial_\mu J^\mu = 0.$$

Functional Version

First, we derive the functional analogue of the classical equations of motion. Consider an expectation value
$$\langle \mathcal{O(\phi)} \rangle = \int \mathcal{O}(\phi) e^{\frac{i}{\hbar} S} \mathcal{D}\phi$$
We'll assume that $\phi$ takes values in a vector space (or bundle). Then we can perform a change of variables $\psi = \phi + \epsilon$, and since $\mathcal{D}\phi = \mathcal{D}\psi$ we find that
$$\int \mathcal{O}(\phi+\epsilon) \exp\left(\frac{i}{\hbar} S[\phi] \right) \mathcal{D}\phi$$
is independent of $\epsilon$. Expanding to first order in $\epsilon$, we have
$$0 = \int \left(\frac{\delta\mathcal{O}}{\delta \phi}  + \frac{i \mathcal{O}}{\hbar} \frac{\delta S}{\delta \phi} \right)  \exp \left( \frac{i}{\hbar} S \right) \mathcal{D}\phi$$
So we find the quantum analogue of the equations of motion:
$$\left\langle \frac{\delta \mathcal{O}}{\delta \phi} \right\rangle + \frac{i}{\hbar} \left\langle \mathcal{O} \frac{\delta S}{\delta \phi} \right\rangle = 0$$

Next, we move on to the quantum version of Noether's theorem. Suppose there is a transformation $Q$ of the fields leaving the action invariant. Assuming the path integral measure is invariant, we obtain
$$\left\langle QF \right \rangle + \frac{i}{\hbar} \left\langle F QS \right\rangle = 0$$
To compare with the classical result, consider $Q$ to be the (singular) operator
$$Q = \frac{\delta}{\delta \epsilon(x)}$$
Then by the previous calculations,
$$Q S = -\delta_\mu J^\mu,$$
so we obtain
$$\left\langle \frac{\delta \mathcal{O}}{\delta \epsilon(x)} \right\rangle = \frac{i}{\hbar} \left\langle \mathcal{O} \partial_\mu J^\mu \right\rangle.$$
This is the Ward-Takahashi identity, the quantum analogue of Noether's theorem.