# Noether's Current in QFT with position dependent variations?

Setup

Consider a mapping $F$ that takes every point $x$ on the manifold $M$ to the point $x'$ on the same manifold. Under this mapping the field $\phi(x)$ evaluated at the point $x$ changes to $\phi'(x)$ when evaluated at the same point $x$ on the manifold or $\phi'(x')$ when evaluated at the mapped point $x'$. The action before mapping is given by: $S=\int d^Dx \mathcal{L}(\phi(x), \partial_\mu \phi(x),x)$ whilst that after mapping is: $$S'=\int d^D x' \mathcal{L}(\phi'(x'), \partial'_\mu \phi'(x'),x')$$ I am focusing here on the case of QFT, meaning intergrals are over the whole of Minkowski space.

Noether's Theorem

According to Noether's theorem a continuous symmetry which leaves the action invariant: $$\Delta S=S'-S=0$$ corresponds to a conserved quantity.

The two forms of Noether's Current

I have come across two forms of Noether's current (Peskin & Schroeder, $\S$2.2): $$j^\mu(x)=\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)} \delta \phi - \mathcal{J}^\mu\tag{1}$$ where $\mathcal{J}^\mu$ is defined by the mapping of $\mathcal{L}$: $$\mathcal{L}(x) \mapsto \mathcal{L}(x)+\alpha \partial_\mu \mathcal{J}^\mu(x)$$ and (Goldstein, 3rd ed, $\S$13.7): $$j^\nu =\left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)}\partial_\sigma \phi-\mathcal{L} \delta^\nu_\sigma\right) X^\sigma-\frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \Psi \tag{2}$$ Where $\delta x^\nu=\epsilon X^\nu$ and $\delta \phi=\epsilon \Psi$.

Problem with form (1)

Consider the case of dilation $x^\mu \mapsto (1+\delta\lambda )x^\mu$ then: $$\mathcal{L}(x) \mapsto \mathcal{L}(x)+\delta \lambda x^\mu \partial_\mu \mathcal{L}$$ here the change in $\mathcal{L}$ can not be written as an exact divergence (also the metric on integration will change). This does not therefore seem compatible with (1).

Problem with form (2)

In the derivation of (2) we get the following expression: $$\int \epsilon \frac{d}{d x^\nu} \left(\left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)}\partial_\sigma \phi-\mathcal{L} \delta^\nu_\sigma\right) X^\sigma-\frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \Psi\right)d^4x =0\tag{13.147}$$ from this Goldstein seems to infer that $$\frac{d}{d x^\nu} \left(\left( \frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)}\partial_\sigma \phi-\mathcal{L} \delta^\nu_\sigma\right) X^\sigma-\frac{\partial \mathcal{L}}{\partial (\partial_\nu \phi)} \Psi\right)=0\tag{13.148}$$ which given that we have a fixed range of integration (the whole of space) I cannot see any reason why this should hold.

Question

My question is what is therefore the most general form of Noether's current which can deal with things like scaling? And are my two concerns above justified?

• In (1) you are forgetting the transformation of $d^dx$ ($d$ is dimension of spacetime). Under $x^\mu \to ( 1 + \delta \lambda ) x^\mu$, we have $d^dx {\cal L} \to d^d x ( 1 + d \delta \lambda) ( {\cal L} + \delta \lambda x^\mu \partial_\mu {\cal L} = d^d x [ {\cal L} + \delta \lambda \partial_\mu ( x^\mu {\cal L} ) ]$. – Prahar Nov 19 '17 at 15:25
• @Prahar Ok I thought it was something to do with this. I guess this therefore means that the contribution $(1+d\delta \lambda)$ is included within the new $\mathcal{L}$? Should we therefore be defining a knew Lagrangian $\mathcal{L}'$ defined as: $$\mathcal{L}'=(1+d \partial_\mu(\delta x^\mu))\mathcal{L}(\phi',\partial'_\mu \phi',x')$$ and will this deal with all such cases? – Quantum spaghettification Nov 19 '17 at 15:33
• The reason for the second current is because "$\epsilon$ is arbitrary". It's the usual thing where we just set the thing proportional to a variation inside the integral to be zero. – childofsaturn Nov 19 '17 at 15:47
• @childofsaturn - that only works if "$\epsilon$ is an arbitrary function". Here, it seems to be an arbitrary constant, which is certainly not good enough. – Prahar Nov 19 '17 at 15:48
• @childofsaturn I agree with Prahar. From what I can tell $\epsilon$ does have to be constant. – Quantum spaghettification Nov 19 '17 at 15:49

2. Goldstein's formula (2) for the bare Noether current $j^{\mu}$ holds for combined horizontal and vertical transformations. The full Noether current $J^{\mu}=j^{\mu}- k^{\mu}$ has a possible improvement term $k^{\mu}$ in case of a quasi-symmetry.