In A Modern Introduction to Quantum Field Theory by Maggiore (as well as in my quantum field theory course), the Noether current for an internal symmetry is found to be

$$j^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu} \phi)}\delta\phi\tag{3.32}$$

where the fields transform as $\phi\rightarrow\phi+\alpha\delta\phi$ for infinitesimal $\alpha$ (and the coordinates are unchanged). However, in An Introduction to Quantum Field Theory by Peskin and Schroeder, if the Lagrangian transforms as $$\mathcal{L}\rightarrow\mathcal{L}+\alpha\partial_{\mu}\mathcal{J}^{\mu},\tag{2.10}$$ it is given in general by

$$j^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu} \phi)}\delta\phi-\mathcal{J}^{\mu},\tag{2.12}$$

which is clearly not the same. Initially I thought that $\mathcal{J}^{\mu}=0$ for an internal symmetry so that they would match. However, answers on my last question have shown that this is not true. Both books assume the action is invariant, not necessarily the Lagrangian. So, what is going on here?

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    $\begingroup$ Maggiore is just starting with a special case. And of course, there are version of Noether's theorem even more general than Peskin's. $\endgroup$
    – knzhou
    Commented Mar 25, 2022 at 19:49

1 Answer 1

  1. No, Peskin & Schroeder between eqs. (2.9) & (2.10) assume that the infinitesimal transformation is a quasi-symmetry of the action, i.e. the action can change by a boundary term; while Maggiore below eq. (3.20) assumes a strict symmetry of the action.

  2. Correspondingly, the Noether current gets modified with an improvement term in Peskin & Schroeder's eq. (2.12) as compared to Maggiore's eq. (3.32).

  3. Btw, both Noether currents (2.12) & (3.32) assume so-called vertical transformations, cf. e.g. this related Phys.SE post.

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    $\begingroup$ But Maggiore also assumes a quasi-symmetry, i.e. that the action is invariant but not necessarily the Lagrangian. $\endgroup$ Commented Mar 25, 2022 at 19:49
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    $\begingroup$ That's not the definition of a quasi-symmetry of the action. $\endgroup$
    – Qmechanic
    Commented Mar 26, 2022 at 12:24
  • $\begingroup$ Thanks, I was thinking the boundary term vanishes for some reason $\endgroup$ Commented Mar 26, 2022 at 12:48

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