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In Zee's solution for (I.3.1) on p. 483 at the back of the book (2003 edition) he writes:

$$D(x)=-\frac{1}{2(2\pi)^2 r}\int_0^\infty \frac{dk~k}{\sqrt{k^2 +m^2}}(e^{ikr}-e^{-ikr})=-\frac{1}{8\pi^2 r}\int_{0}^\infty \frac{dk~k}{\sqrt{k^2 +m^2}}e^{ikr}.$$

I have seen from someone else's question on SE "QFT Propagator across spacelike separation" that the sign of $k$ can be switched. But that does not explain it. By switching the sign of one of the $k$'s how would this not just equal zero?

I looked at another posting on SE about this same problem "Exponential decay of Feynman propagator outside the lightcone", but it did not answer my question.

Reference: Quantum field theory in a nutshell, Anthony Zee, 2003

user202623
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1 Answers1

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The 2003 edition of the book has a typo. In rightmost expression, the correct limits are $\int_{-\infty}^{\infty}$, not $\int_{0}^{\infty}$.

The second part of \begin{equation} \int_0^\infty \frac{dk \ k}{\sqrt{k^2+m^2}} (e^{ikr}-e^{-ikr}) \quad\quad -(1) \end{equation}

is

\begin{equation} -\int_0^\infty \frac{dk \ k}{\sqrt{k^2+m^2}} e^{-ikr} \end{equation}

which after a change of variables $k \rightarrow -k$ becomes:

\begin{equation} -\int_0^{-\infty} \frac{dk \ k}{\sqrt{k^2+m^2}} e^{ikr} = \int_{-\infty}^0 \frac{dk \ k}{\sqrt{k^2+m^2}} e^{ikr} \end{equation}

Putting this back in (1), you get:

\begin{equation} \int_{-\infty}^{\infty} \frac{dk \ k}{\sqrt{k^2+m^2}} e^{ikr} \end{equation}

This is the correct expression, which is in the second edition of the book on page 545.

Errata for Zee's book can be verified here: https://www.kitp.ucsb.edu/zee/books/quantum-field-theory-nutshell/errata-and-addenda

Avantgarde
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