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Trying to understand the Chiral anomaly, I decided to explore the simplest example of a holomorphic fermion in 2D in a background electromagnetic field Adz+A¯dz¯

. The Euclidean action looks like
S(ψ¯,ψ)=∫d2zψ¯D¯ψ,
with D¯=∂¯+A¯.

The classical symmetry ψ↦eiϵψ
and ψ¯↦e−iϵψ¯
should be anomalous. In order to check this, I wanted to explore the renormalization of the conserved current j=ψ¯ψ
. In the literature I've seen the difficulties associated to using Pauli-Villars or dimensional regularization, so that I wanted to try something different. I decided to use heat kernel regularization, so that I replaced the A¯=0
propagator
⟨ψ¯(z)ψ(0)⟩=1z=∫[infinity]0dtt2e−|z|2/tz¯↦∫[infinity]ϵdtt2e−|z|2/tz¯.

Of course, since we are treating the electromagnetic field as a background field, the theory is free. Thus, renormalizing j
simply corresponds to normal ordering it, i.e. removing the divergent part of the loop diagram obtained by contracting the two fermions in j
. However, since A¯
spoils translation invariance, we can instead treat the ψ¯A¯ψ
term as an interaction, transforming the single loop diagram of the full theory to a sum of single loop diagrams with the A¯=0
propagator and photons coming out of the loop. The number of photons in the loop is the power of A¯
in the final result.

Since [j]=[A]=1
, only the diagram with one photon coming out is potentially divergent (the diagram with zero photons is equal to zero since in this regularization ⟨ψ¯(0)ψ(0)⟩=0
) and the divergence should be logarithmic ln(ϵ/μ)
, with μ
some subtraction scale. However, when I compute the diagram I obtain
−∫[infinity]ϵdt1t21∫[infinity]ϵdt2t22∫d2ze−|z|2(1t1+1t2)z¯2A¯(z).
As we explained below, the divergence can only be proportional to A¯
, not its derivatives. Thus, expanding A¯
in a Taylor series, we need only keep its constant part A¯(z)↦A¯(0).
The remaining integral is however a Gaussian integral that vanishes. Indeed,
z¯2=x2−y2−2ixy.
The xy
term won't contribute since there is no propagator for it. And then the x2
and the y2
contributions will vanish due to rotational invariance.

It seems like j
receives no quantum corrections and thus I don't see how it can become anomalous. Furthermore, even if I made a mistake in here, even if j∝A¯
, the conservation law is going to become ∂¯j∝∂¯A¯
, Once the antiholomorphic moving fermion is added, the axial current would be obtained by subtracting the two currents together. This would leave us with an anomalous term proportional to ∂¯A¯−∂A
, instead of proportional to F=∂A¯−∂¯A
. Furthermore, I don't see how the anomaly would cancel for the sum of the two currents, so that the vector symmetry can be gauged. I feel like there is a conceptual point that I am missing about how these anomalies work.
A) The current j becomes anomalous due to quantum corrections, aligning with the electromagnetic field F = ∂A¯ − ∂¯A.
B) The regularization method employed fails to capture the quantum corrections affecting the conserved current j in this scenario.
C) An anomaly proportional to ∂¯A¯ − ∂A emerges, affecting the conservation law for the axial current after incorporating the antiholomorphic fermion.
D) The observations indicate that the renormalization scheme successfully nullifies any impact of the electromagnetic field A¯ on the conserved current j, leading to no quantum corrections.

1 Answer

4 votes

Final answer:

The observations indicate that the renormalization scheme successfully nullifies any impact of the electromagnetic field A¯ on the conserved current j, leading to no quantum corrections.

Step-by-step explanation:

The observations indicate that the renormalization scheme successfully nullifies any impact of the electromagnetic field A¯ on the conserved current j, leading to no quantum corrections. Therefore, option D) The observations indicate that the renormalization scheme successfully nullifies any impact of the electromagnetic field A¯ on the conserved current j, leading to no quantum corrections. is correct.

User Matt Bryson
by
7.4k points
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