One of the cornerstones of point splitting technique of calculating chiral anomaly (Peskin and Schroeder 19.1, p.655) is a symmetric limit $\epsilon \rightarrow 0$. And this is the point that I don't get. Is it really possible to take such a limit? For example, consider the expression $$ \text{symm}\,\text{lim}_{\epsilon \rightarrow 0} \Bigl\{\frac{\epsilon^{\mu}\epsilon^{\nu}}{\epsilon^2}\Bigr\} = \frac{g^{\mu \nu}}{d} \tag{19.23} $$ in $d=2$ spacetime dimensions. Let $\mu = \nu = 0$. Then $$ \frac{\epsilon^{0}\epsilon^{0}}{\epsilon^2} = \frac{1}{1-(\frac{\epsilon^1}{\epsilon^0})^2}. $$ But the latter expression either greater than 1 or less than 0 and so can't be equal to $$\frac{g^{00}}{2} = \frac{1}{2}.$$
3 Answers
The symmetric limit (19.23)
$$S^{\mu\nu}~:=~ \text{symm}\,\lim_{\epsilon \rightarrow 0} \left\{\frac{\epsilon^{\mu}\epsilon^{\nu}}{\epsilon^2}\right\}, \qquad \epsilon^2~:=~\epsilon^{\mu}g_{\mu\nu}\epsilon^{\nu},$$ should be thought of as a regularization prescription. It is part of a symmetric point splitting regularization scheme, cf. Ref. 1. The traditional notion of limit $$\lim_{\epsilon \rightarrow 0} \left\{\frac{\epsilon^{\mu}\epsilon^{\nu}}{\epsilon^2}\right\}$$ does not exist. The prescription $$S^{\mu\nu}~:=~\frac{g^{\mu\nu}}{d}$$ is motivated by three things:
The prescription $S^{\mu\nu}$ can only depend on the metric.
$S^{\mu\nu}$ should be a (2,0) tensor wrt. Lorentz transformations.
If we contract $S^{\mu\nu}$ with $g_{\mu\nu}$, the result should be $1$.
References:
- M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 19.1, p. 655.

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1But since such a limit does not exist, this regularization scheme is not mathematically correct, is it? How can we trust the results obtained with the help of it? – user43283 May 01 '15 at 16:03
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- Regularization assigns a value to a mathematically ill-defined expression. Often this is done with the help of a regulator, but not always, cf. e.g. zeta function regularization. 2. In a physically meaningful theory, the result should not depend on regularization scheme.
– Qmechanic May 01 '15 at 16:43 -
Ok, then, it turns out that the only criterion that this particular regularization scheme is valid is agreement of theory with experiment? – user43283 May 01 '15 at 18:28
Same logic used in (7.87) in Peskin itself: $l^{\mu}l^{\nu} \rightarrow \frac{1}{d}l^2 g^{\mu\nu}$.
Ref. Peskin & Schroeder, Section 7.5, p. 251.

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Sorry, but this is wrong. Eq. 7.87 uses dimensional regularisation. Here OP is asking about point-splitting regularisation. Those are different methods. – AccidentalFourierTransform Jul 21 '18 at 17:57
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I mentioned the logic, not the method. The symmetry property of Lorentz transformation is used to reach (19.23) and (7.87). I don't see the difference caused by the two "different methods". – Luxtau Jul 21 '18 at 18:18
Taking the 'symmetric limit' means averaging the expression over the $S^{d-1}$ sphere surrounding $\epsilon^\mu=0$ and then taking the size of the sphere to zero. That should give you the expression you're looking for.

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