There many articles, in which non-abelian current commutators are computed. The general result is that quantum corrections lead to additional term in commutator

$$[J^a_\mu (x), J^b_\nu (y)] \delta (x^0 - y^0) = [J^a_\mu (x), J^b_\nu (y)]_{classical} \ \delta (x^0 - y^0) + A_{\mu \nu} (x, y),$$ where $A_{\mu \nu}$ is anomalous term and is proportional to $\partial_k \delta (x-y)$.

This result is obtained from calculation of diagrams. But is there any easy way, to show that quantum corrections lead to additional term proportional to $\partial_k \delta (x-y)$ with undefined coefficient (functional of gauge fields)?


Restrictions can be imposed on the anomalous terms from general considerations even without fully solving the Feynman diagrams. In fact, the restrictions on the Schwinger term in the commutator given in the question are explicitly described in detail by Roman Jackiw in his review article: Field theoretic investigations in current algebra (section 2.2.).

His arguments are summarized here:

Since the non-Abelian charges are the generators of symmetry, they must satify the following (equal time) commutation relations with the currents: $$[Q^a, J_i^b(0)] = f_{ab}^c J_i^c(0)$$ The local versions compatible with this relation must have the forms:

$$[J_0^a(x), J_i^b(y)] = f_{ab}^c J_i^c(x)\delta(x-y)+S_{ij}^{ab}(x)\partial^j\delta(x-y)+...$$

Where the dots indicate the possibility of higher order delta function derivatives. In addition, the function $S_{ij}^{ab}$, must satisfy conservation relations:

$$\partial^jS_{ij}^{ab} = 0$$

to satisfy the transformation equation after integration over $y$.

Now, as Jackiw mentions in equation (2.18), the commutator of the charge density with the $00$ component of the energy momentum tensor can be evaluated using the canonical commutation relations:

$$[\Theta^{00}(x), J_0^a(y)] = \partial_{\mu}J_{\mu}^a\delta(x-y)+J_i(x)\partial^i\delta(x-y)$$

Finally commuting the $00$ component of the energy momentum tensor with the density current commutator results that the only possible Schwinger term should be the one with a single delta function derivative.


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