# Replacing pole with delta function in Peskin and Schroeder

Hi on page 234 of their book P&S replaced a pole with a delta function to evaluate a contour integral (see scan of pages below), but I don't quite get how you can reproduce (7.54) this way? I assume with the delta function you can replace $$m^2$$ with $$(k/2+q)^2$$ in the first fraction, but that doesn't seem to work. Since they do not simply evaluate the integral using good old contour integration I'm assuming this delta function trick is easier, but how exactly do you use it? Any help is appreciated.

$$\def\d{\delta} \def\e{\varepsilon} \def\p{\pi} \def\l{\lambda} \def\vq{\mathbf{q}}$$The pole of interest

We solve $$(k/2\mp q)^2-m^2+i\e=0$$ with the result $$q^0 = \begin{cases} +k^0/2 + E_\vq-i\e \\ +k^0/2 - E_\vq+i\e \\ -k^0/2 + E_\vq-i\e \\ -k^0/2 - E_\vq+i\e. \end{cases}$$ (Here we use that $$q^2 = (q^0)^2-\vq^2 +E_\vq^2-E_\vq^2 = (q^0)^2-E_\vq^2+m^2$$.) The text claims that the only contribution to the discontinuity comes from $$q^0 = q^0_{-+} = -k^0/2 + E_\vq$$ which should be straightforward to verify using the techniques described below. We have \begin{align*} i\d M \rightarrow \frac{\l^2}{2} \int \frac{d^3q}{(2\p)^4}\int dq^0 \frac{1}{(k/2-q)^2-m^2}(-2\p i)\d((k/2+q)^2-m^2) [q^0=q^0_{-+}], \end{align*} where $$[P]$$, the Iverson bracket, is 1 if statement $$P$$ is true and $$0$$ if $$P$$ is false. Note that if $$g$$ has simple zeros, $$x_i$$, then \begin{align*} \d(g(x)) &= \sum_i \frac{\d(x-x_i)}{|g'(x_i)|}. \end{align*} That is, \begin{align*} \d(g(x))[x=x_j] &= \frac{\d(x-x_j)}{|g'(x_j)|}. \end{align*} Thus, the factor $$[q^0=q^0_{-+}]$$ is necessary to pick out only the contribution from $$q^0_{-+}$$ (and to ignore that from $$q^0_{--}$$). Continuing,
\begin{align*} \frac{d}{dq^0}\left[(k/2+q)^2-m^2\right]|_{q^0 = q^0_{-+}} &= 2(k/2+q)\cdot(1,\mathbf{0})|_{q^0 = q^0_{-+}} \\ &= 2(k^0/2+q^0)|_{q^0 = q^0_{-+}} \\ &= 2E_\vq. \end{align*} Also, \begin{align*} (k/2-q)^2-m^2|_{q^0=q^0_{-+}} &= (k/2+q)^2-2k\cdot q-m^2|_{q^0=q^0_{-+}} \\ &= m^2-2k^0(-k^0/2+E_\vq)-m^2 \\ &= k^0(k^0-2E_\vq). \end{align*} Thus, $$i\d M \rightarrow -2\p i \frac{\l^2}{2} \int \frac{d^3q}{(2\p)^4} \frac{1}{2E_\vq}\frac{1}{k^0(k^0-2E_\vq)}$$ as claimed.

Appearance of the delta function

Consider the integral $$I = \int_{-\infty}^\infty \frac{f(x)}{g(x)}dx,$$ where $$f$$ is holomorphic on the upper half-plane, $$g$$ has one simple zero in the upper half-plane at $$z=x_0+i\e$$ ($$x_0\in\mathbb{R}$$, $$0<\e\ll1$$), and where the integral on the upper half-circle vanishes. Then \begin{align*} I &= \int_\gamma\frac{f(z)}{g(z)}dz \\ &= 2\pi i\frac{f(x_0)}{g'(x_0)}\quad (\e\rightarrow 0) \end{align*} where $$\gamma$$ is the upper semicircular contour, which is equivalent to \begin{align*} I &= 2\pi i \int_{-\infty}^\infty f(x) \frac{\d(x-x_0)}{g'(x_0)}dx \\ &= 2\pi i \int_{-\infty}^\infty f(x) \frac{\d(x-x_0)}{|g'(x_0)|e^{i\arg g'(x_0)}}dx \\ &= 2\pi i \int_{-\infty}^\infty f(x)\d(g(x)) e^{-i \arg g'(x_0)} dx. \end{align*} Thus, up to a phase one can replace $$1/g(x)$$ with $$2\pi i \d(g(x))$$ in the integral. If instead $$f$$ is holomorphic on the lower half-plane, $$g(x)$$ has a simple zero in the lower half-plane, and the integral on the lower half-circle vanishes a minus sign will appear due to the change in sense of the contour.

• I am still confused, but I am probably just not getting something... My confusion comes from the fact that we have the formula $\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$, where the $x_i$ are the zeroes of $g$. We are integrating $\int_{-\infty}^{+\infty} dq^0 \delta((k^0/2 + q^0)^2 - E_q^2) ...$ and the function within the deltafunction has two zeroes, i.e. $q^0 = -k^0/2 \pm E_q$. Thus we would also have another term, which corresponds to picking up the pole at $q^0 = -k^0/2-E_q$, what we do not want. So wouldn't we need a heavyside function like it is explained in my post? Commented May 19, 2019 at 19:37
• @lomby: Since the only contribution to the discontinuity comes from $q^0_{-+}$ I ignore the other pole. I use $\rightarrow$ instead of $=$ to indicate this, though Peskin has no qualms about using $=$ in this scenario! Commented May 19, 2019 at 20:21
• Ah ok. I wonder why they did that... I don't think it is too obvious :D Commented May 19, 2019 at 20:28
• Peskin and Schroeder's cutting rule is SIMPLY INCORRECT. There needs to be a heaviside function in it, see jkb1603's answer. Commented Nov 5, 2022 at 6:29
• @Simplyorange: Thanks for the close reading. I have addressed this in my latest edit. Commented Aug 19, 2023 at 19:12

I don't know if understand your question right. Do you mean that you can not reproduce (7.54) by making the change (7.53) in (7.52)? I am actually not quite sure if (7.53) is entirely correct. Normally this cutting rule should have a Heaviside function in it, i.e.

$$\frac{1}{(k/2+q)^2-m^2+i\varepsilon} \rightarrow -2\pi i \delta((k/2+q)^2-m^2) \theta(k^0/2 + q^0)$$

Otherwise you would be also picking up the pole at $$-k^0/2-E_q$$, right? This would not be correct since we are closing the contour in the LHP. You can check this by doing the contour integration explicitly, which should not be that hard. Other than that it is straightforward:

$$\int \frac{d^4q}{(2\pi)^4} \frac{1}{(k/2-q)^2-m^2} (-2\pi i) \delta((k/2+q)^2-m^2) \theta(k^0/2 + q^0)$$

Now make a change of variable $$q \rightarrow q - k/2$$:

$$\int \frac{d^4q}{(2\pi)^4} \frac{1}{(k-q)^2-m^2} (-2\pi i) \delta(q_0^2 - E_q^2) \theta(q^0) = (-2\pi i) \int \frac{d^3q}{(2\pi)^4} \frac{1}{2E_q} \frac{1}{(k^0-E_q)^2 - E_q^2}$$

where I used that $$\vec{k} = 0$$ in the center of mass frame.

• (+1) for noticing the issue with the other zero. Commented Aug 19, 2023 at 19:13

I think there is an easier way to evaluate the integral when you replace the second factor with a delta function as prescribed by (7.53) in Peskin & Schroeder. You should use the transformation property of the delta function in a 4-momentum integral as shown by (2.40) in Chapter 2 :

This leads you straight to the desired result :