lagrangian formalism - Does a constant factor matter in the definition of the Noether current?

This is a very basic Lagrangian Field Theory question, it is about a definition convention. It takes much more time to typeset it than answering, but here it is:

Consider a field Lagrangian with only a kinetic term,

$$L = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi $$

Consider the very simple transformation $\phi \rightarrow \phi + \alpha$ ($\alpha$ constant), and so I understand here that $\alpha$ plays the role of $\delta\phi$. I determine the Noether current as $$\frac{\partial L}{\partial[\partial_{\mu}\phi]}\delta\phi$$

and the result is $$\alpha\partial_\mu\phi$$

But in Peskin & Schroeder (just above eq 2.14), the result they give is:

$$\partial_\mu\phi$$

And it doesn't seem to be an erratum. I don't care that "localized" Lagrangian very much (hey, wait before closing, please), but a very general question arises:

Is $\alpha$ dropped simply because $\partial_\mu\phi$ is too a conserved quantity (and so under "conserved current" one understands the general concept, momentum, energy or whatever, regardless of its value), or am I missing some other very basic detail that is assumed to be known by the reader?

---

Later edit: I have eventually understood this question and more, by reading the beginning of chapter 22 of Srednicki. I am finding that book (well, the free preprint for the moment) crystal clear, it seems excellent.

Answer

I) Let us for simplicity address OP's question in the context of point mechanics where $q^i$ are generalized position coordinates on some manifold $M$ [instead of considering field theory with fields $\phi^{\alpha}(x)$]. OP's question is rooted in the difference between

1. 
   

   on one hand, an infinitesimal variation $$\tag{1} q^i~\rightarrow \widetilde{q}^i~=~ q^i+\delta q^i $$ of the generalized position coordinates, or equivalently, $$ \tag{2} \delta q^ ~:=~ \widetilde{q}^i-q^i; $$

   
   


2. 
   

   and on the other hand, that of a generator/Lie algebra element/vector field $$ \tag{3} Y~=~Y^i\frac{\partial}{\partial q^i},\qquad Y^i~=~Y^i(q),$$ which is not infinitesimal (although $Y$ is sometimes confusingly referred to as an 'infinitesimal generator' in the literature).

   
   

Both concepts $\delta$ and $Y$ are linear derivations that satisfy Leibniz rule, and the interrelation between the two is given by

$$ \tag{4} \delta q^i~=~\epsilon Y^i, $$

where $\epsilon$ in eq. (4) is an infinitesimal parameter. The mathematical concept of a vector field $Y$ is tied in a bijective manner to the concept of a flow$^1$

$$ \tag{5} \sigma:~]\!-\!c,c[ ~\times~ M~\to~ M, \qquad ]\!-\!c,c[ ~\subseteq~ \mathbb{R},$$

where

$$ \tag{6} \frac{d}{d\epsilon}\sigma^i(\epsilon,q)~=~Y^i(\sigma(\epsilon,q)), \qquad\sigma^i(\epsilon=0,q)~=~q^i. $$ A flow $\sigma$ satisfies $$ \tag{7} \sigma^i(\epsilon,\sigma(\epsilon^{\prime},q))~=~\sigma^i(\epsilon+\epsilon^{\prime},q).$$ Note that in eq. (7), it is understood that $\epsilon$ and $\epsilon^{\prime}$ are real numbers in the interval $ ]\!-\!\frac{c}{2},\frac{c}{2}[ \subseteq \mathbb{R}$, and not infinitesimal.

II) The (bare) Noether charge

$$ \tag{8} Q ~=~p_i Y^i$$

is (in this case) momentum

$$ \tag{9} p_i ~:= ~\frac{\partial L}{\partial \dot{q}^i} $$

times generator $Y^i$. In particular, the definition (5) of the Noether charge $Q$ does not depend on the $\epsilon$ parameter.

--

$^1$ We ignore the possibility that the domain $]\!-\!c,c[$ could depend on the initial position $q\in M$.