Does the commutator between the total derivative term of a symmetry generator and a quantum field always vanish?
I am trying to understand the following derivation in <PERSON> 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).
However this derivation neglects the possibility of a total derivative term in the current. For full generality we need to modify the charge expression to $$ Q=\int{d^3x}\frac{\delta L}{\delta\dot{\phi}}\frac{\delta\phi}{\delta\alpha}+\Lambda^0. 818
$$
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 question is: for every example I have seen in QFT so far, $[\Lambda^0,\phi]=0$ and so the formula $[Q,\phi]=-i\frac{\delta\phi}{\delta\alpha}$ seems to hold. Is this true in general, and if so, is there a proof of this? Without this missing element I feel that the derivation in <PERSON> is incomplete, as many currents have boundary terms.
To give an example, consider the super-current in the free Weiss Zumino Model, which contains the following total derivative term $$ \Lambda^\nu=-\theta\sigma^\mu\bar{\sigma}^{\nu}\psi\partial_{\mu}\bar{\phi} + \theta\psi\partial^{\nu}\bar{\phi}+\bar{\phi}\bar{\theta}\partial^{\nu}\phi $$ We see that $$ [\Lambda^0,\phi]=[-\theta\sigma^\mu\bar{\sigma}^{0}\psi\partial_{\mu}\bar{\phi} + \theta\psi\partial^{0}\bar{\phi},\phi]=[-\theta\sigma^0\bar{\sigma}^{0}\psi\partial_{0}\bar{\phi} + \theta\psi\partial^{0}\bar{\phi},\phi]=[-\theta\psi\partial_{0}\bar{\phi} + \theta\psi\partial_{0}\bar{\phi},\phi]=0 $$ where we used the fact that $\partial_0\bar\phi$ is the conjugate momentum to $\phi$, and $$ [\Lambda^0,\psi]=0 $$ as none of the terms in the current contain the conjugate momentum to $\psi$ ($\pi=-i\sigma^{0}\bar\psi$).
Is there a fundamental reason why this should always be true for any theory and any current? 818