The propagator for a scalar particle can be written as

$$ \frac{1}{x + i\epsilon} = {\rm PV}\left( \frac{1}{x} \right) - i\pi\delta(x), \quad x = p^2 - m^2, \tag{1} $$

where $p, m$ are the momentum and mass of the particle. My question is about how to compute ${\rm PV}(1/x)$. Gathering the post Principal value integral and Wikipedia's info (https://en.wikipedia.org/wiki/Cauchy_principal_value#Distribution_theory), it's written as

$$ {\rm PV}\left( \frac{1}{x} \right) [f] = \lim_{\epsilon \rightarrow 0^+}\int_{\mathbb{R} - [-\epsilon, \epsilon]} dx \frac{x}{x^2 + \epsilon^2}f(x)\tag{2} $$

But what is this $f(x)$: any function I would like to choose such as $f(x) \equiv 1$? How would you compute

$$ {\rm PV}\left( \frac{1}{x} \right) ?\tag{3} $$

If it's a distribution that it's applied to functions $f(x)$, how would you use PV in a Feynman diagram? In other words, let's say you compute a Feynman diagram and end up with something like

$$ \frac{1}{x + i\epsilon}f(x)\tag{4} $$

Would you say that

$$ \frac{1}{x + i\epsilon}f(x) = \lim_{\epsilon \rightarrow 0^+}\int_{\mathbb{R} - [-\epsilon, \epsilon]} dx \frac{x}{x^2 + \epsilon^2}f(x) - i\pi f(0)\ \ \ ?\tag{5} $$

This cannot be true because RHS is $f$ dependent and $x$ independent, while LHS is both $f$ and $x$ dependent. What am I losing?

  • $\begingroup$ Does this answer your question? Principal value of 1/x and few questions about complex analysis in Peskin's QFT textbook $\endgroup$ – Nihar Karve Nov 24 '20 at 4:04
  • $\begingroup$ Essentially, $\text{PV}(\frac{1}{x})$, like the scalar propagator, is a distribution, which means that it is designed to be integrated against another "test" function, $f(x)$ in this case. $\endgroup$ – Nihar Karve Nov 24 '20 at 4:06
  • $\begingroup$ @NiharKarve no, it does not solve my question. Actually I don't understand how you get a distribution out of a propagator which is simply a function. Are you telling me that it doesn't matter what function I shall pick when computing PV because (I guess) it is independent of $f(x)$? $\endgroup$ – Vicky Nov 24 '20 at 4:54
  • 2
    $\begingroup$ > The first equation, $$\frac{1}{x+i\epsilon}=P\frac{1}{x}-i\pi\delta(x)$$ is actually a shorthand notation for its correct full form, which is $$\underset{\epsilon\rightarrow0^+}{\lim}\int_{-\infty}^\infty\frac{f(x)}{x+i\epsilon}\,dx=P\int_{-\infty}^\infty\frac{f(x)}{x}\,dx-i\pi f(0)$$ and is valid for functions which are analytic in the upper half-plane and vanish fast enough that the integral can be constructed by an infinite semicircular contour. $\endgroup$ – Nihar Karve Nov 24 '20 at 5:17
  1. It seems that the main culprit behind OP's question is the mathematical notation $u[f]$ where $u$ is a distribution, and $f$ is a test function. It is not a product of $u$ and $f$.

  2. In particular, the LHS of OP's eq. (2) is not just a product of ${\rm PV}\left( \frac{1}{x} \right)$ and $f(x)$. For starters it also involves an implicitly written integration over $x$.

  3. A possible helpful example is the Dirac delta distribution $u=\delta$. Here $$\delta[f]~:=~f(0)~=:~\int_{\mathbb{R}}\! \mathrm{d}x~\delta(x)~f(x).$$

  4. Concerning the actual definition of the principal value, it seems to be already sufficiently clarified in the linked Phys.SE post.

  • $\begingroup$ Ok, but how do you use it in a Feynman diagram? Look at the difference in dependency form Eq. (5) where both the propagator and $f$ are part of a Feynman diagram $\endgroup$ – Vicky Nov 26 '20 at 10:59
  • $\begingroup$ Depends on the application. Normally one would leave the $i\epsilon$ on LHS of eq. (1) as it is (eventually performing a Wick rotation). $\endgroup$ – Qmechanic Nov 26 '20 at 11:06

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