# Noether Charges as symmetry generators: the boundary term

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?

1. OP is correct that the full Noether charge $$Q$$ is the bare Noether charge (28.5) plus boundary contributions.
2. At the classical level, in order for the Noether charge $$Q$$ to generate symmetry of the action, the action should be the Hamiltonian action.