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 Answer 1


That's a valid question. General comments:

  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.

  3. Note that temporal boundary terms in the Lagrangian formulation may affect the Legendre transformation to the Hamiltonian formulation.

  4. As explained in statement 1 in my Phys.SE answer here, a symmetry in the Hamiltonian formulation is indeed still generated by its own Noether charge.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.