Anomalous commutators in bosonization I am trying to learn about abelian bosonization, and I'm finding there to be very many subtle issues. One of these issues lies in the anomalous commutator, on which the constructive approach to abelian bosonization heavily relies. I'll take my notation from Bosonization for Beginners, Refermionization for Experts by von Delft and Schoeller. The anomalous commutator, proven in Eq. (18), is:
$$
[b_{q,\eta}, b^{\dagger}_{q' \eta'} ] = \delta_{\eta \eta'} \delta_{q q'} ,
$$
where $b^{\dagger}_{q \eta} = \frac{i}{n_q} \sum_k c^{\dagger}_{k+q,\eta} c_{k \eta}$ is essentially the Fourier transform of the density operator ($n_q$ here is $qL/2\pi$, not the number operator). Acted on the Fermi sea, it creates a linear superposition of particle-hole excitations with momentum $q > 0$. This commutator is called "anomalous", because it relies on normal-ordering with respect to the filled Fermi sea of the Dirac Hamiltonian. If we instead insisted on normal-ordering with respect to the vacuum, then this commutator would vanish.
Here's what I find so confusing: how is it that commutation relations for a set of operators depend on a choice of ground state, or even a choice of normal ordering? In the derivation of Eq. (18), the argument is that one must normal-order the Fermi operators before they can be subtracted, or else you are essentially subtracting infinite quantities. But how is it that the commutation relation of two operators, which are operators acting on the whole Hilbert space, "knows" about some chosen special state?
My best guess is that the Hilbert space in which operators are normal-ordered with respect to the Fermi sea, and the Hilbert space in which they are normal-ordered with respect to the vacuum, are actually somehow different Hilbert spaces. Otherwise, operators acting on these states differ by infinite c-numbers. But it's not at all clear whether this idea is correct, or how to make the statement more precise.
EDIT: Allow me to add a bit more detail, so that the post is self-contained and doesn't require leafing through the review article. The computation performed by von Delft and Schoeller works as follows:
$$
\begin{split}
[b_{q \eta}, b^{\dagger}_{q' \eta'}] &= \frac{1}{\sqrt{n_q n_{q'}}} \sum_{k k'} [c^{\dagger}_{k-q,\eta} c_{k\eta}, c^{\dagger}_{k'+q', \eta'} c_{k' \eta'} ] \\
&= \frac{\delta_{\eta \eta'}}{\sqrt{n_q n_{q'}}} \sum_{k k'} \Big( \delta_{k, k'+q' } c^{\dagger}_{k-q,\eta} c_{k' \eta} - \delta_{k-q,k'}c^{\dagger}_{k'+q',\eta}c_{k \eta} \Big) \\
&= \frac{\delta_{\eta \eta'}}{\sqrt{n_q n_{q'}}} \sum_{k} \Big(  c^{\dagger}_{k-q,\eta} c_{k-q' \eta} - c^{\dagger}_{k-q+q',\eta}c_{k \eta} \Big)
\end{split}
$$
Now comes the crucial step: according to von Delft and Schoeller, if $q \neq q'$ then the operators above are normal-ordered, and there is no reason we can't reindex the sum on the second term by $k \rightarrow k-q'$, so that the whole expression vanishes. But if $q = q'$, we cannot so naively reindex: each term in the sum alone would have an infinite expectation value, so we cannot separate the sum and reindex only one term. Instead, the authors propose to normal order each of the terms, at which point there is no issue with reindexing:
$$
\begin{split}
[b_{q,\eta}, b^{\dagger}_{q'\eta'}] &= \frac{\delta_{\eta \eta'} \delta_{q q'}}{n_q} \sum_k \Big( :c^{\dagger}_{k-q,\eta} c_{k-q,\eta}: + \langle FS | c^{\dagger}_{k-q,\eta} c_{k-q,\eta} | FS \rangle - :c^{\dagger}_{k,\eta} c_{k,\eta}: - \langle FS | c^{\dagger}_{k,\eta} c_{k,\eta} | FS \rangle \Big) \\
&= \frac{\delta_{\eta \eta'} \delta_{q q'}}{n_q} \sum_k \Big( \langle FS | c^{\dagger}_{k-q,\eta} c_{k-q,\eta} | FS \rangle - \langle FS | c^{\dagger}_{k,\eta} c_{k,\eta} | FS \rangle \Big) \\
&= \delta_{\eta \eta'} \delta_{q q'}
\end{split}
$$
where the last equality comes from noting that there are $n_q$ terms in the sum where the $k-q$ mode is occupied while the $k$ mode is empty.
Now, herein lies the problem: by normal-ordering with respect to the Fermi sea, we obtain the above anomalous correlator. But the sum instead vanishes if I chose to normal order with respect to the vacuum! How is it that the commutator of two operators can depend on an arbitrary state of my choosing? The operators $b_{q,\eta}$ and $b^{\dagger}_{q,\eta}$ aren't even themselves normal ordered!
 A: You can regard the Schwinger term as an approximation that arises because we can only excite  states relatively close to the Fermi-surface, and so only need those states in sums over intermediate states. The actual commutator of the current $j_n= \sum_k a^\dagger_{n+k}a_{-k}$ with $j_{n'}$ is a  operator containing bilinear  products of $a^\dagger, a$'s that involves $a_n$'s  only far from the Fermi surface where all states are either empty (high energy) or occupied (deep in the sea). In these regions  the operator can be replaced by its $c$-number expectation and that gives the "anomalous" $c$-number Schwinger  term in the current algebra
$$
[j_n,j_{n'}]= n \delta_{n+n'}.
$$
Incidently this formula  was found by Pascual Jordan long before Bosonization became a thing.
Comment added:
There are many ways to derive the commutator starting  different physical assumptions.  The fact that  all lead to the same answer indicates that the result is physically robust -- an essential property of you want to apply it to real-world situations.  In particlular it applies to condensed mater systems where the is a cutoff.  The point is that the Stone-Von Neumann theorem does not apply to Fock spaces based on an infinite set of of $a_n$'s.
The "obvious" Fock space (Van Hove's Universal Receptacle?) is   not irreducible and contains  inequivalent representations some of which have central extensions like the $n\delta_{n+n'}$.  The space with the empty vacuum, and the space with  a filled Fermi sea, are examples of such inequivalent representations.
