Noether current: commute or not commute?

Question

I have two related questions.

1) Before promoting the fields in a theory (for example complex scalar $\mathcal{L}=\partial_{\mu}\phi^{\dagger}\partial^{\mu}\phi$) to operators one can commute the fields freely, for instance in the Noether current \begin{equation} j^{\mu}=i\left(\phi\partial^{\mu}\phi^{\dagger}-(\partial^{\mu}\phi)\phi^{\dagger}\right)=i\left((\partial^{\mu}\phi^{\dagger})\phi-\phi^{\dagger}\partial^{\mu}\phi\right) \end{equation} Why doesn't this affect the operator $j^{\mu}$ after quantization?

If I wanted to determine the current for a Lagrangian already in terms of operators (e.g. when taking a condensed matter Hamiltonian (operator) and Legendre transforming it), there the order of the operators is suddenly important and I get conflicting results for the current.

2) In the explicit scenario of a non-relativistic free particle $H=\frac{p^2}{2m}$ i.e. in second quantized form \begin{equation} L=\int \textrm{d}x\mathcal{L} =\int \textrm{d}x\ \Psi^{\dagger}(x,t)(i\partial_t+\frac{1}{2m}\partial_x^2)\Psi(x,t)=\frac{-1}{2m}\int \textrm{d}x\ \partial_x\Psi^{\dagger}(x)\partial_x\Psi(x) \end{equation} where I dropped the time dependence because it is not relevant to this question, can I use \begin{equation} j(x)=\frac{\partial\mathcal{L}}{\partial(\partial_x\Psi(x))}\Delta\Psi(x)+\frac{\partial\mathcal{L}}{\partial(\partial_x\Psi^{\dagger}(x))}\Delta\Psi^{\dagger}(x)=\frac{-i}{2m}\left[(\partial_x\Psi^{\dagger}(x))\Psi(x)-(\partial_x\Psi(x))\Psi^{\dagger}\right] \end{equation} to define the spatial component of the Noether current? This seems to be conflicting with the literature in terms of order of the operators and signs (see e.g. Mahan page 24).

Answer 1

Yes, the quantisation process has ordering ambiguities, not only in QFT but in standard QM too. You have to postulate some ordering, and the resulting theory usually depends (though rather trivially) on the order of factors.

Due to the trivial nature of the canonical commutation relations, different orderings for Noether charges are usually the same modulo a constant shift: $$ Q_\mathrm{ordering 1}=Q_\mathrm{ordering\ 2}+\text{constant} $$

so the standard procedure is to determine this arbitrary constant by declaring that the vacuum has zero charge: $$ Q|0\rangle\equiv 0 $$ which, in effect, just means to normal order the Noether charge.

I don't own a copy of Mahan, but I don't see anything wrong about your current, except perhaps for the global coefficient (which is not really fixed by Noether's theorem, inasmuch if $j$ is conserved so is $cj$ for any $c\in\mathbb C$). You can check if your normalisation of $j$ is the standard one by checking whether $$ [Q,\Psi]=\Psi $$ is satisfied as is, or if it comes with some funny coefficient in front of $\Psi$.

Answer 2

Noether's theorem is a statement about a classical theory with a classical action $S$. To quantize the theory, we should replace the classical commutative expressions, such as, e.g. the Noether current $J^{\mu}$ with non-commutative operators $\hat{J}^{\mu}$. Be aware that quantization is not a unique procedure (cf. e.g. this and this Phys.SE posts), and that quantum anomalies may appear in the quantized conservation laws.