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Qmechanic ♦
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An integral in Peskin'sPeskin & Schroeder's Quantum Field Theory Pp. 27

In Peskin'sPeskin & Schroeder's QFT Pp. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

HeThey said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

An integral in Peskin's Quantum Field Theory P. 27

In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

An integral in Peskin & Schroeder's Quantum Field Theory p. 27

In Peskin & Schroeder's QFT p. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

They said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

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Qmechanic ♦
Qmechanic ♦
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In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$$$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$

1.I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.$$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

2.If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$

1.I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

2.If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$

In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

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ZHANG Juenjie
ZHANG Juenjie
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An integral in Peskin's Quantum Field Theory P. 27

In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$

1.I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

2.If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$