On page 224 of P&S, they have the following expression (7.36),

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The integral over $d^3q$ gives us all the $q \to p$, then the integral over $dx^0$ is computed. The RHS given matches when only the $x^0 = T_+$ term of the integral is kept, what happens to the $x^0=\infty $ term?


1 Answer 1


I think I can answer your question. Some complex analysis is used in order to derive the above-mentioned result. Starting with the RHS expression,

$$\sum_{\lambda}\int_{T_+}^{\infty}dx^0\int\frac{d^3q}{(2\pi)^3} \frac{1}{2E_q(\lambda)} e^{i(p^0-q^0+i\epsilon)x^0}\langle\Omega|\phi(0)|\lambda_0\rangle (2\pi)^3\delta^3(\textbf{p}-\textbf{q}) \\\times\langle\lambda_{\textbf{q}}|T\{\phi(z_1)...\}|\Omega\rangle=\\ \sum_{\lambda}\int_{T_+}^{\infty}dx^0\int\frac{d^3q}{(2\pi)^3} \frac{1}{2E_q(\lambda)} \bigg[\frac{e^{i(p^0-q^0+i\epsilon)x^0}}{i(p^0-q^0+i\epsilon)}\bigg]_{T_+}^{\infty} \\\langle\Omega|\phi(0)|\lambda_0\rangle (2\pi)^3\delta^3(\textbf{p}-\textbf{q}) \times\langle\lambda_{\textbf{q}}|T\{\phi(z_1)...\}|\Omega\rangle$$ from which one only needs to consider the term in the brackets, namely $$\frac{e^{i(p^0-q^0+i\epsilon)x^0}}{i(p^0-q^0+i\epsilon)} \Bigg|_{T_+}^{\infty}= -i\lim_{x^0\to\infty} \frac{e^{i(p^0-q^0+i\epsilon)x^0}}{(p^0-q^0+i\epsilon)} +i\frac{e^{i(p^0-q^0+i\epsilon)T_+}}{(p^0-q^0+i\epsilon)}$$ The last term is the term you are looking for, see LHS of Eq. (7.36) in P&S. The first term vanishes, since $$-i\lim_{x^0\to\infty} \frac{e^{i(p^0-q^0+i\epsilon)x^0}}{(p^0-q^0+i\epsilon)}= -\frac{i}{(p^0-q^0+i\epsilon)} \lim_{x^0\to\infty}e^{i(p^0-q^0)x^0} \lim_{x^0\to\infty}e^{-\epsilon x^0}$$ This result is obviously zero, since the second limit vanishes, whereas the first one rapidly oscillates.

I hope this helps.


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