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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 $$\frac{g^{00}}{2} = \frac{1}{2}.$$

Qmechanic
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user43283
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3 Answers3

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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:

  1. The prescription $S^{\mu\nu}$ can only depend on the metric.

  2. $S^{\mu\nu}$ should be a (2,0) tensor wrt. Lorentz transformations.

  3. If we contract $S^{\mu\nu}$ with $g_{\mu\nu}$, the result should be $1$.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 19.1, p. 655.
Qmechanic
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    But 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