# One-particle irreducibles in P&amp;S

• auditor
In summary, the authors have defined D \equiv \ell^2 - \Delta +i\epsilon, so in my case the numerator will be D^2. The integrand is odd under l_\mu \rightarrow - l_\mu, so it must vanish for the same reason that the integral $$\int dx~ \frac{x}{1+ x^3} from minus infinity to plus infinity vanishes.Thanks kdv! It's all clear now. auditor [SOLVED] One-particle irreducibles in P&amp;S I'm going through the derivation of the [tex]-i\Sigma_2 (p)$$ correction in Peskin & Shcroeder. On the top of page 218, eq. 7.17 they say
$$-i\Sigma_{2}\left(p\right) = -e^{2}\int_{0}^{1}dx\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{-2x pslash+4m_{0}}{\left[\ell^{2}-\Delta+i\epsilon\right]^{2}}$$
where the denominator is O.K. Have that $$\ell \equiv k-xp$$.

Before the Feynman parameter was introduced, the correction had the appearance (p. 217, eq. 7.16)
$$-i\Sigma_{2}\left(p\right)=\left(-ie\right)^{2}\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\gamma^{\mu}\frac{i\left(kslash+m_{0}\right)}{k^{2}-m_{0}^{2}+i\epsilon}\gamma_{\mu}\frac{-i}{\left(p-k\right)^{2}-\mu^{2}+i\epsilon}$$

To make a long story short: I get the numerator to be $$-2\gamma^{\mu}\ell_{\mu}-2x pslash+4m_{0}$$ as I suspect the authors did to. But they're dropping the linear terms in $$\ell$$, and I think the reason is to be found on p. 191, eq. 6.45.

"... This task is simplified by noting that since $$D$$ depends only on the magnitude of $$\ell$$,
$$\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0$$
The (...) identity follows from symmetry."
(They have defined $$D \equiv \ell^2 - \Delta +i\epsilon$$, so in my case the numerator will be $$D^2$$)

I don't really see how this follows from symmetry. Any suggestions?

Btw: sorry about the kslash/pslash notation.

Thanks!

auditor said:
I'm going through the derivation of the $$-i\Sigma_2 (p)$$ correction in Peskin & Shcroeder. On the top of page 218, eq. 7.17 they say
$$-i\Sigma_{2}\left(p\right) = -e^{2}\int_{0}^{1}dx\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{-2x pslash+4m_{0}}{\left[\ell^{2}-\Delta+i\epsilon\right]^{2}}$$
where the denominator is O.K. Have that $$\ell \equiv k-xp$$.

Before the Feynman parameter was introduced, the correction had the appearance (p. 217, eq. 7.16)
$$-i\Sigma_{2}\left(p\right)=\left(-ie\right)^{2}\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\gamma^{\mu}\frac{i\left(kslash+m_{0}\right)}{k^{2}-m_{0}^{2}+i\epsilon}\gamma_{\mu}\frac{-i}{\left(p-k\right)^{2}-\mu^{2}+i\epsilon}$$

To make a long story short: I get the numerator to be $$-2\gamma^{\mu}\ell_{\mu}-2x pslash+4m_{0}$$ as I suspect the authors did to. But they're dropping the linear terms in $$\ell$$, and I think the reason is to be found on p. 191, eq. 6.45.

"... This task is simplified by noting that since $$D$$ depends only on the magnitude of $$\ell$$,
$$\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0$$
The (...) identity follows from symmetry."
(They have defined $$D \equiv \ell^2 - \Delta +i\epsilon$$, so in my case the numerator will be $$D^2$$)

I don't really see how this follows from symmetry. Any suggestions?

Btw: sorry about the kslash/pslash notation.

Thanks!

You are integrating over all possible values of all the components of l. So if you have a term linear in $$l_\mu [/itex], all the positive contributions will cancel out all the negative contributions. In other words, the integrand is odd under [tex] l_\mu \rightarrow - l_\mu$$ so it must vanish for the same reason that the integral

$$\int dx~ \frac{x}{1+ x^3}$$ from minus infinity to plus infinity vanishes.

Thanks kdv! It's all clear now.

## 1. What are one-particle irreducibles (1PI) in P&S?

One-particle irreducibles (1PI) are a type of Feynman diagram in particle physics known as the "building blocks" of perturbative calculations. They represent the simplest possible interactions between particles and can be used to calculate physical quantities such as scattering amplitudes.

## 2. How do 1PI diagrams differ from other Feynman diagrams?

Unlike other Feynman diagrams, 1PI diagrams do not contain any internal loops. This means that they cannot be broken down into smaller diagrams, making them essential in perturbative calculations.

## 3. What is the significance of 1PI diagrams in particle physics?

1PI diagrams are important because they represent the most basic interactions between particles in a given theory. By calculating and summing these diagrams, physicists can make predictions about the behavior of particles and the outcomes of experiments.

## 4. Can 1PI diagrams be used in all theories of particle physics?

Yes, 1PI diagrams can be used in all theories of particle physics, as they represent the fundamental interactions between particles. However, the complexity and number of 1PI diagrams may vary depending on the specific theory being studied.

## 5. How are 1PI diagrams related to the renormalization process?

1PI diagrams play a crucial role in the renormalization process, which is a mathematical technique used to remove infinities from perturbative calculations in quantum field theory. By summing an infinite number of 1PI diagrams, physicists can eliminate divergences and obtain finite, meaningful results.

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