Is the Noether current physically observable in experiments?

[kau]

I have very simple questions. Although these are simple but I am confused. So if I start with a lagrangian of the following form $$ \mathcal{L} = \partial^{\mu}\phi \partial_{\mu} \phi^{*} -m^{2} \phi^{*} \phi $$ then I get a current for the global invariance of the lagrangian and that is of the form $$ J^{\mu}=i(\partial^{\mu} \phi \phi^{*} -\partial^{\mu} \phi^{*} \phi) ... (1 )$$Then if I demand local gauge invariance then definitely we get gauge field terms in the lagrangian. Then if we find an expression of current for the global invariance of $\phi$ field we get expression $$ J^{\mu}=i(D^{\mu}\phi \phi^{*} -D^{\mu}\phi^{*} \phi) ...(2) $$ And there is other expression for current which is conserved on local gauge invariance but they are not physical since they are not gauge invariant. But global ones are gauge invariant and therefore it's ok to have their physical presence in the theory. But I am confused that which one is physical 1 or 2 or both? One more thing is these currents or the global charge defined from it is due to the symmetry of the theory ,not external ones. So in principle we don't see them. Is this right?
Thanks.

 

[samalkhaiat]

Noether currents are symmetry currents. They depend on the fields (in the Lagrangian) and their transformations laws under the symmetry group of the Lagrangian. Your first current follows from the Free fields Lagrangian $$\mathcal{L}_{0} = \mathcal{L}_{0}(\varphi_{a} , \partial_{\mu}\varphi_{a}) ,$$ using $$J_{(1)}^{\mu} = \frac{\partial \mathcal{L}_{0}}{\partial (\partial_{\mu}\varphi_{a})} \delta \varphi_{a} .$$
Your second current follows from a different Lagrangian $$\mathcal{L} = \mathcal{L} = \mathcal{L}(\varphi_{a} , \partial_{\mu}\varphi_{a} - i A_{\mu}\varphi_{a}) .$$ This Lagrangian describes an interacting field theory, its symmetry current is $$J_{(2)}^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{a})} \delta \varphi_{a} = \frac{\partial{L}}{\partial A_{\mu}} .$$ Comparing the two currents is like comparing apples with oranges. They are the global symmetry currents of two different theories. Therefore, each current is the physical current of the corresponding theory.

Sam

 

[kau]

Noether currents are symmetry currents. They depend on the fields (in the Lagrangian) and their transformations laws under the symmetry group of the Lagrangian. Your first current follows from the Free fields Lagrangian $$\mathcal{L}_{0} = \mathcal{L}_{0}(\varphi_{a} , \partial_{\mu}\varphi_{a}) ,$$ using $$J_{(1)}^{\mu} = \frac{\partial \mathcal{L}_{0}}{\partial (\partial_{\mu}\varphi_{a})} \delta \varphi_{a} .$$
Your second current follows from a different Lagrangian $$\mathcal{L} = \mathcal{L} = \mathcal{L}(\varphi_{a} , \partial_{\mu}\varphi_{a} - i A_{\mu}\varphi_{a}) .$$ This Lagrangian describes an interacting field theory, its symmetry current is $$J_{(2)}^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{a})} \delta \varphi_{a} = \frac{\partial{L}}{\partial A_{\mu}} .$$ Comparing the two currents is like comparing apples with oranges. They are the global symmetry currents of two different theories. Therefore, each current is the physical current of the corresponding theory.

Sam

Ok. thanks for the response. I have another concern about this. Does this current have anything to do with experiments? can you see them in the experiment or something like that. Or this current just signify the symmetry of the theory and to some extent we can infer if there is some conserved thing like that then there is some matter present that is like having finite chemical potential. That's it.
Another thing is in Schwartz's qft book it is mentioned that in fixed spacetime dimension this thing is true although if that is not the case ( e.g. AdS/CFT) then we can have the global current conserved in gauge theory. Can you elucidate this part?