What constitutes a symmetry for Noether's Theorem?

Question

I have some confusion over what exactly constitutes a symmetry when trying to apply Noether's theorem. I have heard both that a symmetry in the action gives a conserved quantity, and that a symmetry in the Lagrangian gives a conserved quantity.

Both of these statements are confusing to me. From my understanding, a transformation is a symmetry of the action when $\delta S = 0$. However, on the classical path, I would expect this to be trivially true for any transformation, not just a few special ones, since the classical path by definition minimizes the action. Should I actually take $\delta S = 0$ to indicate a symmetry only if it is true for all paths, and not just the classical path? If so, keeping in mind the conserved quantity which this corresponds to is actually only conserved along the classical path, it seems weird that I would need to know something about the action on all paths in order to make a statement about a conserved quantity on the classical path. It seems like you should only need to know information about the action on the paths near the classical path in order to make a statement about conserved quantities on the classical path. Is this incorrect?

The statement that a symmetry of the Lagrangian gives a conserved quantity seems strange because there are transformations which give conserved quantities that do not leave the Lagrangian unchanged. For example this is the case for any transformation which changes the Lagrangian by a total derivative. It seems weird to call this a symmetry of the Lagrangian if the Lagrangian actually changes. Am I misunderstanding this terminology?

Answer 1

A symmetry of a classical theory described by an action $S[\varphi]$ where $\varphi$ is the set of all fields in the theory is a field redefinition $\varphi(x) \to \varphi'(x)$ such that $S[\varphi] = S[\varphi']$. Note that the field redefinition is such that it does not act on the coordinates $x$.$^\dagger$ Note also that this definition of a symmetry is off-shell.

The statement of Noether's theorem is that for every a connected (to the identity) continuous off-shell symmetry of an action, there exists a current $j_\mu(x)$ that is conserved on-shell. Thus, even though the symmetry exists at an off-shell level, the current is only conserved on-shell.

A symmetry of a quantum theory is a field redefinition $\varphi(x) \to \varphi'(x)$ such that the path integral measure is invariant $$ [d\varphi]e^{-S[\varphi]} = [d\varphi']e^{-S[\varphi']} $$ In this case, the statement of Noether's theorem (or in this case, known as Ward Identity) is $$ \partial^\mu \langle j_\mu(x) {\cal O}_1(x_1) \cdots {\cal O}_n(x_n) \rangle = 0 \quad \text{if}~x\neq x_1,x_2,\cdots,x_n~. $$ Strictly speaking, there is no notion of on-shell or off-shell in a quantum theory, but one could loosely speaking, say that all correlators are "on-shell" in the sense that correlators of fields at distinct points satisfy the equations of motion. One could also derive a more general Ward identity which tells us what happens when $x=x_i$ for some $i\in\{1,\cdots,n\}$, but I will not do that here.

$^\dagger$_{This is crucial and often a something that a lot of people confuse with. In field theories, all symmetry transformations act only on the fields, not on the coordinates. One often like to talk about spacetime symmetries which are described as acting on coordinates in some way $x \to x'$. However, it is crucial to remember that that is simply a tool to package information about how fields transform. For instance, you might like to talk about translations. This is described by the field redefinition $\phi(x) \to \phi'(x)$ where $\phi'(x+a) = \phi(x)$. Note that the equation $\phi'(x+a) = \phi(x)$ is to be understood as a way to deduce what is $\phi'(x)$ in terms of $\phi(x)$ and not as translations acting on the coordinates in some way.}

Answer 2

1. By a symmetry is meant an off-shell symmetry. An on-shell symmetry is a vacuous notion, as OP has already noted. See also this related Phys.SE post.

2. Noether's theorem may be generalized to quasi-symmetries, cf. my Phys.SE answers here & here.

Answer 3

The source of your confusion comes from the definition of symmetry of action. When Noether's theorem says that A symmetry operation is the one which remains the action invariant it does not mean $\delta S = 0$ in the sense of finding classical trajectories. Remember the proof of Noether's theorem uses the classical Euler-Lagrange equation. That means the calculation is done on shell. Here the invariance of action is examined for the particular symmetry operation. That includes External symmetry like $x -> x' = \lambda x$ or Internal Symmetry such as $\phi -> \phi'$. So you see the variation of action with respect to these variations. If the action is invariant then you call it a symmetry. On the other hand $\delta S = 0$ means variation of action w.r.t its parameters and setting them to zero.

Answer 4

Symmetry properties of the action functional are independent of the equations of motion (they are off-shell). It just means that under transformations such as $t\rightarrow t'=t+\epsilon f(q,t)$, $q\rightarrow q'=q+\epsilon g(q,t)$ the action is invariant, i.e., $S'=S$, which is normally denoted by $\delta S\equiv S'-S=0$

Both, (continuous) symmetries of the action and of the lagrangian, give conserved quantities. However, transformations which do not leave the lagrangian invariant may as well give conserved quantities. Namely, if the lagrangian is quasi-invariant, $L'=L+\epsilon\frac{d\sigma(q,t)}{dt}$, then the action (as well as the equations of motion) remains invariant and thre is a conserved quantity which depends on the surface term $\sigma$.

On the other hand, in order to prove Noether's Theorem one makes use of the equations of motions. Therefore a continuous symmetry of the action implies on a conserved quantity on-shell.