# Noether's theorem with infinite parameters

I'm trying to understand something regarding Noether's theorem - and with the given situation, my question isn't that much of a question, I'm rather just seeking confirmation whether I'm thinking right or not.

The situation:

Let $\mathcal L$ be a Lagrangian density, depending on some field $\phi$, and its first derivative. Noether's theorem (naively) says that if $\phi(x)\mapsto\phi(x)+\epsilon\delta\phi(x)$ is a specific infinitesimal deformation of the field (a more precise thing would be to say that this is a smooth 1-parameter family of finite deformations - and we're interested in behaviours under $d/d\epsilon|_{\epsilon=0}$), such that $\mathcal L$ changes by a divergence ($\delta\mathcal L=\partial_\mu K^\mu$), then the current $$j^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi-K^\mu$$is conserved on-shell.

It is known that this can be recast in a different form, by making $\epsilon$ be a function instead of a parameter. Then the action won't be invariant in general, but the deformation of the action gives the same current $j^\mu$, and its conservation can be shown.

The problem is that this doesn't make sense, imo. To show this, consider the following, let $\epsilon(x)$ be the "infinitesimal" functional parameter, the variation is $$\phi(x)\mapsto\phi(x)+\epsilon(x)\delta\phi(x).$$ Let us define $\epsilon'(x)$ and $\epsilon$ as $\epsilon(x)=\epsilon\epsilon'(x)$, where here only $\epsilon$ is "infinitesimal". Now the variation has the form $$\phi(x)\mapsto\phi(x)+\epsilon\epsilon'(x)\delta\phi(x).$$ Now we redefine $\delta\phi(x)$ to $\delta\phi'(x)=\epsilon(x)\delta\phi(x)$, then the variation has the form $$\phi(x)\mapsto\phi(x)+\epsilon\delta\phi'(x).$$

This is literally the same form we had before we assumed $\epsilon$ is a function.

So this begs the question - what do we mean on an "infinite-parameter" variation? The problem is clearly caused by the fact, that if $\delta\phi(x)$ is specific, but reasonably arbitrary, then this still contains as many "free parameters" as the different possible values for $x$. Essentially, $\delta\phi(x)$ already contains infinite parameters.

The resolution:

Looking at specific examples, such as an $U(1)$ transformation of the free, massive, complex Klein-Gordon field, the finite transformation is $$\phi(x)\mapsto e^{i\epsilon}\phi(x).$$ Infinitesimally, this is $$\phi(x)\mapsto \phi(x)+i\epsilon\phi(x),$$ so $$\delta\phi(x)=i\phi(x).$$

Here we see, that $\delta\phi(x)$ depends on $x$ only through the unperturbed field $\phi(x)$ itself, so here the variation is truly 1-parameter.

If we do this for another archetypical example - spacetime translations - we get the same results.

The question:

Am I right in saying that the usual form of the Noether's theorem should be stated that we consider variations of the form $$\phi(x)\mapsto\phi(x)+\epsilon\delta\phi[\phi(x),\partial\phi(x)],$$ where $\delta\phi$ is a specific function of the field $\phi$, and possibly, its derivatives, but not the coordinates $x$?

Because only then does it makes any sense to me to discuss whether the variation has finite or infinite amount of parameters.

In Noether's first theorem, can the infinitesimal variation depend explicitly on the spacetime point $x^{\mu}$?
Example: Lagrangian formulation. Consider the Lagrangian $$L~:=~T-V ,\qquad T~:=~\frac{m}{2}\dot{q}^2,\qquad V~:=~\frac{\alpha}{q^2}. \tag{L1}$$ where $\alpha$ is a constant. It is easy to check that $$\delta q ~=~ \varepsilon (q-2t\dot{q}), \qquad \delta t ~=~0, \tag{L2}$$ is an infinitesimal quasi-symmetry $$\delta L ~=~\ldots ~=~\varepsilon\frac{dk^{0}}{dt}, \qquad k^0~:=~-2t L . \tag{L3}$$ Here $\varepsilon$ is a constant infinitesimal parameter. The bare Noether charge is $$Q^0~=~m\dot{q}( q-2t\dot{q}). \tag{L4}$$ The full Noether charge $$Q~:=~Q^0-k^0~=~mq\dot{q} -2t(T+V) \tag{L5}$$ is a conserved quantity.
Example: Hamiltonian formulation. Consider the Hamiltonian Lagrangian $$L_H~=~p\dot{q}-H, \qquad H~:=~\frac{p^2}{2m}+\frac{\alpha}{q^2}, \tag{H1}$$ where $\alpha$ is a constant. It is easy to check that $$\delta q ~=~ \varepsilon \left( q-\frac{2t}{m}p\right), \qquad \delta p ~=~- \varepsilon \left( p+\frac{4\alpha t}{q^3}\right), \qquad \delta t ~=~0, \tag{H2}$$ is an infinitesimal quasi-symmetry $$\delta L_H ~=~\ldots ~=~\varepsilon\frac{dk^{0}}{dt}, \qquad k^0~:=~2t\left(\frac{\alpha}{q^2} -\frac{p^2}{2m}\right) . \tag{H3}$$ Here $\varepsilon$ is a constant infinitesimal parameter. The bare Noether charge is $$Q^0~=~p\left( q-\frac{2t}{m}p\right). \tag{H4}$$ The full Noether charge $$Q~:=~Q^0-k^0~=~qp -2tH \tag{H5}$$ is a conserved quantity. This example is further discussed in this Phys.SE post.