What are the conserved currents and charges in QFT - operator form of Noether currents?

Question

I can't figure out how to define/compute conserved charges and currents in Quantum Field Theory. I am following Peskin & Schroeder's Introduction to Quantum Field Theory, and in the second chapter problem set there is a question about the conserved charges $Q$ of the complex scalar field defined by the following Lagrangian: $$ \mathcal L = \partial_\mu \phi^* \partial^\mu \phi - m^2 \phi^* \phi \, .$$ It seems to be assumed in this problem set that I know how to compute Noether currents $j^\mu$ and conserved charges $Q = \int \mathrm d^3 x \, j^0$ of this Lagrangian, but the book does not explain the theory of conserved currents and charges in quantum field theory. It explains it for classical field theory, which is not helpful because the usual derivations for the Noether current and charge assume that everything commutes.

Classical field theory explanation

In classical field theory, we'd identify a symmetry as a transformation under which the Lagrangian varies by a divergence: $$ \delta \mathcal L = \partial_\mu K^\mu \, .$$ Then, we use the fact that, on-shell, any Lagrangian also varies by the divergence of its symplectic potential density: $$ \delta \mathcal L = \frac{\partial \mathcal L}{\partial (\partial_\mu \varphi_a)} \partial_\mu (\delta \varphi_a) + \frac{\partial \mathcal L}{\partial \varphi_a} \delta \varphi_a = \partial_\mu \left( \frac{\partial \mathcal L}{\partial (\partial_\mu \varphi_a)} \delta \varphi_a \right) + \left\lbrace \frac{\partial \mathcal L}{\partial \varphi_a} - \partial_\mu \frac{\partial \mathcal L}{\partial (\partial_\mu \varphi_a)} \right\rbrace \delta \varphi_a \, .$$ The right-hand term is zero on-shell, leaving the left-hand term which is the divergence of the symplectic potential density: $$ \delta \mathcal L \cong \partial_\mu \Theta^\mu \quad \text{where} \quad \Theta^\mu = \frac{\partial \mathcal L}{\partial (\partial_\mu \varphi_a)} \delta \varphi_a \, .$$ (I am using $\cong$ to denote equality on-shell.) Then, we can define the Noether current as $$ j^\mu = K^\mu - \Theta^\mu $$ and it follows that $$ \partial_\mu j^\mu \cong 0 \, .$$ The conserved charge is then $$ Q = \int \mathrm d^3 x \, j^0 $$ which, as it can be shown, must have a total time-derivative of zero on-shell: $$ \frac{\mathrm d Q}{\mathrm d x^0} \cong 0 \, .$$

Quantum field theory question

In quantum field theory, the above working fails at the first step. This equation is incorrect, $$ \delta \mathcal L = \frac{\partial \mathcal L}{\partial (\partial_\mu \varphi_a)} \partial_\mu (\delta \varphi_a) + \frac{\partial \mathcal L}{\partial \varphi_a} \delta \varphi_a \, ,$$ because the operators $\partial_\mu \varphi_a$ and $\varphi_b$ do not generally commute. Nonetheless, I know that I can take the variation using the product rule, $$ \delta \mathcal L = \partial_\mu ( \delta \phi^*) \partial^\mu \phi + \partial_\mu \phi^* \partial^\mu ( \delta \phi ) - m^2 ( \delta(\phi^*) \phi + \phi^* \delta(\phi) ) \, .$$ But how do I derive a 'conserved current operator' $\partial_\mu j^\mu = 0$ from this, and an associated 'conserved charge' $Q$? I am not even sure what the conserved charge would mean in this case, since quantum theory does not require anything to be on-shell. (I am vaguely aware of Ward identities, I don't want to get into that here.)

Fallacious argument

Too often I have seen this fallacious approach. The conserved charge for the classical theory is derived, then the objects are silently replaced with their operator variants. This is fallacious, because the classical derivation uses the fact that these objects can be commuted, and by commuting the objects different quantum variants of the conserved charge will be obtained.

For example, in the classical theory of the complex scalar field where we treat $\phi$ and $\phi^*$ as separate fields, the following transformation $$ \phi \mapsto e^{i \alpha} \phi, \quad \phi^* \mapsto e^{-i \alpha} \phi^*$$ leads to the conserved/Noether current: $$ j^\mu = i ( \phi^* \partial^\mu \phi - \phi \partial^\mu \phi^*) \, .$$ The corresponding charge is $$ Q = i \int \mathrm d^3 x \, (\phi^* \pi^* - \phi \pi) \, .$$ However, according to Schroeder & Peskin, the conserved charge for the quantum field theory of the complex scalar field is $$ Q = i \int \mathrm d^3 x \, (\phi \pi - \pi^* \phi^*) \, .$$ Well, never mind the factor of $-1$. The problem is that $\pi^*$ and $\phi^*$ have been commuted compared to the way I've written the classical charge above, and there's no way to know which way the variables should be commuted before replacing the classical objects (position & momenta) with their operator variants. I do not know how they arrived at this charge. I also don't know the significance of this quantity we call 'charge', in the quantum theory.

Summary

I would appreciate it if someone could explain how the Noether currents and charges are defined for quantum field theories, and how to derive them. I have not been able to find a derivation which wasn't fallacious, by deriving a classical variant of the quantity first which just so happens to have the objects permuted in the right way before replacing them with their operator variants.

I also would appreciate a brief explanation of the significance of these quantities in quantum field theory. (I think that in the absence of a quantum anomaly, the charges which satisfy $\partial_\mu j^\mu \cong 0$ also satisfy $\langle \partial_\mu j^\mu \rangle = 0$ through the Ward identity; I do not know anything about the charge $Q$ however.)

Answer 1

Noether's theorem is the correspondence between conserved quantities in a physical system and symmetries. The original theorem concerns classical Lagrangian mechanics and is described by OP, but there are versions of this in Hamiltonian mechanics and in quantum theories.

In the path integral formalism, the theory is described by its Lagrangian. You can then adapt the classical calculations to derive Ward identities. Importantly, you are then working with some path integrals over fields which commutes (at least if there are no fermions). From a symmetry of the Lagrangian, you can the derive (if there are no anomalies) conserved currents and quantities in the correlation functions of the theory.

In canonical quantization, on the other hand, you are working with fields which are operators acting on a Hilbert space. A symmetry is realized by unitary transformation on our Hilbert space which commutes with the Hamiltonian. So you shouldn't really expect to easily derive conserved quantities from the Lagrangian expressed in terms of operators.

For the complex scalar field, the global $U(1)$ symmetry means that there are unitary operators $U(\theta)$ such that : \begin{align} U(\theta)\phi(x) U(\theta)^\dagger &= e^{i\theta}\phi(x) \\ U(\theta)\pi(x) U(\theta)^\dagger &= e^{i\theta}\pi(x) \end{align}

We then know that there exists a hermitian operator (which we will write $Q$ for reasons that will become clearer below), such that : $$U(\theta) = e^{i\theta Q}$$

The transformation laws for $\phi(x)$ and $\phi^\dagger(x)$ now become : \begin{align} [Q,\phi(x)] & = i\phi(x)\\ [Q,\pi(x)] &= i\pi(x) \end{align}

The fact that this is actually a symmetry is the statement that $U(\theta) H U(\theta)^\dagger = H$ or equivalently $[Q,H]= 0$. This then implies that $Q$ is a conserved quantity.

"But why can't I derive the form of this operator from the Lagrangian ?" Because when you use the Lagrangian, you have no information on the ordering of operators. If you try writing $Q$ with different ordering of the operators, not all of them will be hermitian or conserved. Deriving it directly from the operators allows you to find an operator which is really a symmetry at the quantum level.

Another way of saying this is that to quantize the theory, you need a coherent way to assign non-commuting operators to commuting functions. One of the coherence conditions will be that the symmetry generator is Hermitian and commutes with the Hamiltonian.