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2

The calculation you are referring to is for the current induced by a fixed background electromagnetic field in two-dimensional massless QED. Having a background field means quantising the usual fermionic action $$\bar{\psi}(iD\!\!\!/)\psi$$ with $D_\mu=\partial_\mu + i e A_\mu(x)+i e B_\mu(x)$, where $A_{\mu}(x)$ is a fixed, classical gauge field ...

3

Basically the reason is that the classical conformal symmetry no longer holds at the quantum level due to the presence of the trace anomaly. More precisely, the tracelessness of the quantum stress-energy tensor is incompatible with the normal ordering needed to define it. By cohomological reasons, the trace of the stress-energy tensor, although ...

1

If the Fourier transform of $f(x)$ is $\tilde{f}(k)$, then the Fourier transform of $df/dx$ is $ik\tilde{f}(k)$. Proof: $$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$ This explains the momentum factors, so we've reduced the task to showing \Gamma(x, y, z) \sim ...

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A standard reference packed to the brim with other references to everything you ever wanted to know about anomalies is "Anomalies in Quantum Field Theory" by Bertlmann. This particular topic is what comprises part of chapter 11 there. I'll highlight the main points, but this is a technical topic for which you'll have to go to the references and follow all of ...

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