I am trying to understand the following derivation in Schwartz section 28.2 as to how Noether Charges can be thought of as symmetry generators.
We start with the definition of $Q$ (for simplicity let's consider a single scalar field):
$$ Q=\int{d^3x}\frac{\delta L}{\delta\dot{\phi}}\frac{\delta\phi}{\delta\alpha}.\tag{28.5} $$ Using the fact that $\frac{\delta L}{\delta\dot{\phi}}(x)=\pi(x)$ and $[\pi(x),\phi(y)]$=$-i\delta^{(3)}(\overrightarrow{x}-\overrightarrow{y})$ we find $$ [Q,\phi]=-i\frac{\delta\phi}{\delta\alpha} $$ which is the desired result for $Q$ to be a symmetry generator (from my understanding).
My question is, what happens to this derivation when you have a total derivative term in your current? I am confused because we need to modify the charge expression to $$ Q=\int{d^3x}\frac{\delta L}{\delta\dot{\phi}}\frac{\delta\phi}{\delta\alpha}+\Lambda^0. $$ and we have no guarantee that $[\Lambda^0,\phi]=0$.
Thus I would expect the relation to be modified to $$ [Q,\phi]=-i\frac{\delta\phi}{\delta\alpha}+\int{d^3x[\Lambda^0,\phi]}. $$
My problem is that I know of situations where $[\Lambda^0,\phi]≠0$, but the formula $[Q,\phi]=-i\frac{\delta\phi}{\delta\alpha}$ still seems to hold. In such a case, I cannot see how the derivation above could still be correct, as it seems to be missing a piece.
Specifically I am thinking of the supercharges for the Wess-Zumino model where the current contains a total derivative term $$ \Lambda^\nu=-\theta\sigma^\mu\bar{\sigma}^{\nu}\psi\partial_{\mu}\bar{\phi}\; +\; ... $$ which does not commute with $\phi$, but is essential to recover the correct relation $$ [Q,\phi]=-i(\theta\psi). $$
Am I missing something obvious?
