Why are infinitesimal shifts sufficient to prove that a symmetry holds?

Question

Why are infinitesimal shifts in the Lagrangian sufficient to prove that a symmetry holds? Couldn't a lot of things happen at higher orders? Especially when I am introducing an infinitesimal shift of a non-commuting operator?

Answer 1

I don't know how these things appear in QFT, but if a "classical" answer is sufficient to you, then here are several things:

• First of all, you need to tell us what do you mean on "symmetry holds". For the purposes of Noether's theorem, what you need is a Lie algebra representation of a Lie algebra $\mathfrak g$ on the target space $V$ of a field $\psi$ (here I consider a field to be a map from spacetime to a fixed target space - for "global" symmetries to make sense, the field must be a section of a trivial vector bundle, so this is fine). You can basically forget about Lie groups, because Noether's theorem doesn't care about them. It's just usually what you have is a Lie group representation on the target space, and Noether's theorem applies to the tangent map of this representation (eg. the corresponding Lie algebra representation).
• The infinitesimal invariance actually implies finite invariance. To simplify the derivation, let us consider a "Lagrangian" that only depends on the field but not the field derivatives. This doesn't change our results, just makes it easier to calculate. $$ \\ $$ Let $\phi_\epsilon$ be a one-parameter subgroup of $G$ (our symmetry group) and let $\psi\mapsto\phi_\epsilon(\psi)$ be it's linear action on the field via the representation. Obviously we have due to the homomorphism property $$\phi_\epsilon(\phi_\tau(\psi))=\phi_{\epsilon+\tau}(\psi).$$ The variation of the Lagrangian is defined as $$\delta\mathcal L(\psi)=\frac{d}{d\epsilon}\mathcal L(\phi_\epsilon(\psi))|_{\epsilon=0}.$$ Assume that this variation vanishes (for simplicity, I am ignoring the case when the Lagrangian turns into a total divergence) for any $\psi$ and any one-parameter subgroup. Then the $\epsilon$-derivative also vanishes, when evaluated at anywhere, not just at 0. To see this, consider $$ \frac{d}{d\epsilon}\mathcal L(\phi_\epsilon(\psi))|_{\epsilon=\epsilon}=\frac{d}{d\tau}\mathcal L(\phi_\tau(\phi_\epsilon(\psi)))|_{\tau=0}=\delta\mathcal L(\phi_\epsilon(\psi)).$$ But we said that the variation vanishes at any field, including the field $\psi^\prime=\phi_\epsilon(\psi)$, so the above vanishes. $$\\$$ Hence, we obtain that if a Lagrangian is infinitesimally symmetric under $G$, it also satisfies $$\frac{d}{d\epsilon}\mathcal L(\phi_\epsilon(\psi))=0$$ for any one-parameter subgroup and field, with this derivative evaluated everywhere. But we know that a differentiable function whose derivative vanishes everywhere must be constant everywhere. Hence we also obtain that $$ \mathcal L(\psi)=\mathcal L(\phi_\epsilon(\psi)) $$ for any $\epsilon$ (and also one-parameter subgroup and field, etc.). $$\\$$ Now this result is strictly speaking only guaranteed for group elements that can be reached by a one-parameter subgroup. So given $G$, we can form $G^\prime=\exp(\mathfrak g)$, and what we have obtained is that if the Lagrangian is infinitesimally symmetric under $\mathfrak g$, then it is also finitely symmetric under $G^\prime=\exp(\mathfrak g)$. However most groups in physics that are considered for Noether symmetries, are such that all elements can be reached from the Lie algebra via exponentialization.