Understanding Noether Current and Its Internal Symmetry in Lagrangian Dynamics

[latentcorpse]

Given the Lagrangian

$L=i \psi^\star \dot{\psi} - \frac{1}{2m} \nabla \psi^\star \nabla \psi$
which has an internal symmetry $\psi \rightarrow e^{i \alpha} \psi$ so $\delta \psi = i \psi$ (am I correct in saying that we omit the infinitesimal paramater $\alpha$ here because we don't want it appearing in the Noether current?)

Anyway I want to show the Noether current is given by
$j^\mu = \left( \psi^\star \psi , \frac{i}{2m} ( \psi^star \nabla \psi - \psi \nabla \psi^star ) \right)$

so my formula for the noether current is

$j^\mu = \frac{\partial L}{\partial ( \partial_\mu \psi)} \delta \psi - F^\mu$ where [itex] F^\mu satisfies $\delta L = \partial_\mu F^\mu$

now if i set $\mu =0$ and work through i get $j^0=- \psi^star \psi$ does anyone know how to get rid of this minus sign?

and if i work on the $\mu=i$ components i get it to work out using the formula for $j=i$ \bove added to the formula for $j=i$ where I replace $\psi$ with $\psi^\star$

so my only real problems are
a) how to get rid of the minus sign in the 0th component
b)why do we get rid of the infinitesimal parameter when writing out $\delta \psi$ - is it so it doesn't appear in $j^\mu$ and if so, why do we not want it in j^\mu[/itex]

 

[latentcorpse]

anybody?