# The interpretation of noether currents in scalar QED

1. Dec 16, 2013

### karlzr

In scalar QED, there are two noether currents $J_{global}$ and $J_{local}$corresponding to the global and local gauge transformations respectively.
In QED, the two currents are exactly the same. But in scalar QED, they are totally different.
$$J_{global}^\mu=i e (\phi^\dagger \partial^\mu\phi-\phi\partial^\mu\phi^\dagger)$$ and
$$J_{local}^\mu=i e (\phi^\dagger D^\mu\phi-\phi D^\mu\phi^\dagger)$$
where $D^\mu$ is covariant derivative. So my question is how to interpret these two quantities? which one represents the charge of the scalar field?

2. Dec 17, 2013

### fzero

It is the global symmetries that lead to the conserved quantities in a theory. The local symmetries don't define new conserved quantities. Heuristically, this is a consequence of the notion that gauge symmetries indicate a redundancy in the degrees of freedom being used to describe the physics.

In more detail, we can note that, in order to derive conserved quantities from the global symmetries, we typically must use the equations of motion. So we can say that they lead to quantities which are conserved on-shell. This is the content of Noether's first theorem. The local symmetries lead to quantities that are conserved without imposing the equations of motion, so we can say that these are off-shell conserved quantities. These results are contained in what's called Noether's second theorem. For a brief and clear review, you might look at http://arxiv.org/abs/hep-th/0009058.

However, the physical quantities in a quantum theory are precisely the on-shell observables. So the physical conserved charges are the ones that we compute from Noether's first theorem on global symmetries. The charges computed from the local symmetries will in any case correspond to these when placed on-shell.

In the present case, we can see the effect of placing the currents on-shell by computing the expectation value in some physical state describing the system. In a quantum mechanical system, we have to do this anyway in order to derive an observable quantity. In order for a quantity like $\langle \alpha | \mathcal{O} | \alpha \rangle$ to be nonvanishing, we must have equal numbers of creation and annihilation operators, so $\mathcal{O}$ must contain an even number of free fields. In particular, a quantity like $\langle \alpha |\phi^\dagger A^\mu \phi | \alpha\rangle =0$, so we can see that $\langle \alpha |J_{local}^\mu - J_{global}^\mu | \alpha\rangle =0$. Therefore $J_{local}^\mu$ leads to the same conserved charge as $J_{global}^\mu$.

3. Dec 17, 2013

### karlzr

Thanks a lot. Your reply is really informative.
Another question is : in the free field case, one can find the solution to KG equation and then $Q=\int d x^3 J^0=N^{+}-N^{-}$. So for the interaction case, can we do the same thing? this is not evident since the solution is quite complicated.

4. Dec 17, 2013

### fzero

In the interacting case, the global symmetry implies a Ward identity that prevents the multiplicative renormalization of the current. It turns out that there is actually an additive renormalization of the current if the photon is massless. The Ward identity is generally discussed in most QFT textbooks, but both it and the additive renormalization are discussed in the paper http://arxiv.org/abs/hep-th/0512187. I think the discussion there is fairly brief and clear, but still a bit too detailed to succinctly summarize here.