3

I'm working on the Conservation Laws in Peskin (page 309), but I was confused for it.

In last section, I know that

Classical: the action is stationary.i.e. $\delta S =0$ when $\phi(x)\rightarrow \phi(x)+\epsilon(x)$, so we obtain Euler-Lagrange equation

Quantum: the generating functional is invariant.i.e. $\delta Z[J]=0$ when $\phi(x)\rightarrow \phi(x)+\epsilon(x)$, so we obtain Dyson-Schwinger equations

However, when I started to derive Noether's theorem, I got confused.

When $\phi(x)\rightarrow \phi(x)+\epsilon(x)\Delta\phi_a(x)$ (Eq.9.93), in the classical case and in the quantum case, which is stationary?

What's the difference between $\phi(x)\rightarrow \phi(x)+\epsilon(x)$ and $\phi(x)\rightarrow \phi(x)+\epsilon(x)\Delta\phi(x)$, they're both infinitesimal transformations. Why would one give DS equations and one give Noether theorem? What are the differences and connections between them?

Qmechanic
  • 201,751
sky
  • 169

1 Answers1

1
  1. The core of OP's question seems to be the following question:

    What's the difference between (1) the infinitesimal variations to derive EL equations, and (2) the infinitesimal symmetry variations in Noether's theorem?

    This is explained in this related Phys.SE post.

  2. Concerning the Schwinger-Dyson (SD) equations, they are derived using a vertical translation symmetry (which is formally of type 2) in the target space of the fields.

    Other SD-like equations can be derived using other symmetry variations (2) if they exist. This is e.g. useful to derive Ward identities.

Qmechanic
  • 201,751