# Relation between Noether's charge and the generator of a $U(1)$ symmetry

Consider a $U(1)$ symmetry of a complex scalar field realized as $$\phi\to\phi^\prime=e^{iQ\theta}\phi.$$ where $Q$ is the generator of the symmetry. The conserved Noether's charge (in $D$-spatial dimensions) corresponding to this symmetry is given by $$Q_N=\int j^0(\textbf{x},t) d^D\textbf{x}=iQ\int d^D\textbf{x}[(\partial_0\phi)\phi^*-(\partial_0\phi^*)\phi].$$ So it turns out that $Q_N\propto Q$ but not equal to $Q$ itself.

But in A. Zee's book QFT in a Nutshell, page 198, the Noether charge corresponding to a symmetry $$U(\theta)=e^{iQ\theta}$$ is written as $$Q_N=\int j^0(\textbf{x},t)d^D\textbf{x}=Q$$ But as I've shown $Q_N\propto Q$. So is it a mistake there? Or am I missing something?

Your confusion resides in the notation. To make it clear I'll use a hat over the operators. The Noether symmetry is $$\hat\phi'\equiv\mathrm e^{i\theta\hat Q}\hat\phi\mathrm e^{-i\theta\hat Q}\equiv \mathrm e^{i\theta Q}\hat\phi$$ where $Q\in\mathbb R$ is a scalar. The generator of this symmetry is $\hat U(\theta)=\mathrm e^{i\theta\hat Q}$. The infinitesimal change in $\phi$ is $$\theta\delta\hat\phi=[\hat Q,\hat\phi]=iQ\phi$$

The conserved charge is $$\hat Q=\int \hat j^0(\textbf{x},t) d^D\textbf{x}=iQ\int d^D\textbf{x}[(\partial_0\hat\phi)\hat\phi^*-(\partial_0\hat\phi^*)\hat\phi].$$

Sometimes some authors drop the factor of $Q$ in the definition of $\hat Q$, so that the infinitesimal change in $\phi$ reads $$\theta\delta\hat \phi=i\phi$$

• @AFT Isn't the U(1) transformation on a classical field is implemented by $\phi^\prime=e^{iQ\theta}\phi$ where $\phi$ is not an operator? I don't understand why you added hats in this relation. My understanding was that $\hat{\phi}^\prime=e^{i\hat Q\theta}\hat{\phi}e^{-i\hat Q\theta}$ relation holds in quantum theory where $\hat{\phi}$ is an operator. Is that wrong?
– SRS
Commented Jan 17, 2017 at 13:08
• I though that your question was about quantum fields, not classical fields (because of the tag quantum-field-theory and because you mentioned the unitary operator $U(\theta)=\mathrm e^{i\theta Q}$). But now I am confused: in the OP is $\phi$ a function or an operator? Commented Jan 17, 2017 at 13:11
• You're right. My question was about quantum fields. But the transformation I wrote in the question, now I think, is valid only in classical field theory. But in your answer, you wrote $\hat{\phi}^\prime=e^{iQ\theta}\hat{\phi}$ even for quantum fields with $Q$ being a number. In short, you wrote two ways of transforming the quantum field $\hat{\phi}$ in your first equation. I think the latter implementation holds for classical $\phi$ only.
– SRS
Commented Jan 17, 2017 at 13:24
• I didnt really write two different ways of transforming $\hat\phi$. Recall that any symmetry transformation can always be written as $\hat\phi'=U(\alpha)\hat\phi U^\dagger(\alpha)$ for some unitary operator $U$ and some parameters $\alpha$. For example, the translation symmetry reads $\hat\phi'(x)= e^{iPa}\hat\phi(x)e^{-iPa}=\hat\phi(x-a)$. In this sense, $\hat\phi'=e^{i\hat Q\theta}\hat\phi e^{-i\hat Q\theta}$ is the general expression for the transformation; and $\hat\phi'=\mathrm e^{iQ\theta}\phi$ is the specific rule for the U(1) symmetry. They are not two different transformations. Commented Jan 17, 2017 at 14:54
• @SRS you can't derive a transformation law. The transformation $\phi'=e^{i\hat Q\theta}\hat\phi e^{-i\hat Q\theta}$ is just the general expression of a unitary transformation. All symmetries can be written as $\phi'=e^{i\hat Q\theta}\hat\phi e^{-i\hat Q\theta}$ for suitable operators $\hat Q$ and paramenters $\theta$. The particular case of a $U(1)$ symmetry is realised as $\phi'=\mathrm e^{iQ\theta}\phi$, and this defines the symmetry transformation. Commented Mar 23, 2017 at 21:24