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

Question

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?

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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 1

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.

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

Answer 2

To an extent, the ambiguity is just a matter of notation. Some people would write a generic one-parameter transformation as $$ \phi(x) \to \phi(x) + \alpha\delta\phi(x) + \mathcal O(\alpha^2) $$ so that $\delta\phi(x)$ is the coefficient of the small parameter $\alpha$ as with expressions in ordinary calculus like $$ f(x+\alpha) = f(x) + \alpha f'(x) + \mathcal O(\alpha^2) $$ With this notation, the transformation $$ \phi(x) \to \phi(x) + \alpha $$ would have $\delta\phi(x) = 1$, and in this case you would recover the Peskin result. The change in notation is really just a matter of taste when you are working with one-parameter families of transformations. As you point out, for any real number $\alpha$, both $\alpha\partial_\mu\phi$ and $\partial_\mu\phi$ are conserved, and their conservation equations are the same.

Addendum. When it comes down to it, in a physical context all that really matters is the equivalence class corresponding to all expressions for a conserved quantity that differ by a constant, nonzero multiplicative factor (since their corresponding conservation equations are the same), so the issue here is really just a matter of semantics.

Cheers!