# Simplifying expression for scalar field vacuum energy

Let $$\varphi$$ be a scalar field whose particle has mass $$m$$. I have obtained the expression that gives the difference in energy relative to a particle with mass $$m'$$. The expression is

$$\frac{E}{V}=-\frac{i}{2}\int \frac{d^4k}{(2\pi)^4}\log\left( \frac{k^2-m^2+i\varepsilon}{k^2-m'^2+i\varepsilon} \right) .$$

We want to integrate out the $$k^0$$ dependence. Furthermore, we work in units where $$k^0=\omega$$. We introduce $$\omega_k^2=\vec k^2+m^2$$ to write

$$\frac{E}{V}=-\frac{i}{2}\int \frac{d^3k}{(2\pi)^3}\int \frac{d\omega}{2\pi}\log\left( \frac{\omega^2-\omega^2_k+i\varepsilon}{\omega^2-\omega'^2_k+i\varepsilon} \right) .$$

By an identity of the logarithm, $$\log(\frac{a}{b})=\log(a)-\log(b)$$, and since we are only concerned with the integral over $$dk^0=d\omega$$, it suffices to consider

$$\mathcal{I}= \frac{1}{2\pi}\int \!d\omega\,\log\left( \omega^2-\omega^2_k+i\varepsilon \right)$$

We integrate by parts with \begin{align} u=&\log\left( \omega^2-\omega^2_k+i\varepsilon \right)&\qquad\text{and}\qquad du&=\frac{2\omega\,d\omega}{\omega^2-\omega^2_k+i\varepsilon }\\ v=&\omega&\qquad\text{and}\qquad dv&= d\omega \end{align}

to obtain

$$\mathcal{I}= -\frac{1}{2\pi}\int\!d\omega\, \frac{2\omega^2}{\omega^2-\omega^2_k+i\varepsilon } .$$

In Zee's QFT 2nd Edition, page 125, he inserted a factor $$\frac{d\omega}{d\omega}$$ to do the integration by parts, as in

$$\mathcal{I}= \frac{1}{2\pi}\int \!d\omega\,\frac{d\omega}{d\omega}\,\log\left( \omega^2-\omega^2_k+i\varepsilon \right)$$

QUESTION: Why did he do that? Seem like it integrates just fine without inserting that. Continuing, we will try to apply the residue formula so we factor the denominator as

$$\mathcal{I}= -\frac{1}{2\pi}\int\!d\omega\, \frac{2\omega^2}{[\omega-(\omega_k-i\varepsilon)][\omega-(-\omega_k+i\varepsilon)]} .$$

We will choose the pole in the upper complex half plane. It does not appear to me that the integrand will trivially vanish along the semicircular path at infinity, call that path $$\gamma_\infty$$. Zee proceeds as if the integral along that path is zero, but I do not see it. QUESTION: How do we know that the path $$\gamma_\infty$$ contributes nothing? Perhaps this has something to do with the factor $$\frac{d\omega}{d\omega}$$ which I did not use? Why is the final term below equal to zero?

$$2\pi i\,\text{Res}(\omega_0^+)= \oint \!dz\,f(z) =\mathcal{I}+\underbrace{\int_{\gamma_\infty}\!dz\, \frac{2\omega^2}{\omega^2-\omega^2_k+i\varepsilon }}_{\text{???}}$$

If we set this equal to zero, then I get the correct answer for the energy.

• The $$d\omega/d\omega$$ is just equal to $$1$$. He is just using the chain rule to show you that it is an equivalence: $$\frac{\mathrm{d}\omega}{2\pi} = \frac{\color{red}{\mathrm{d}\omega}}{2\pi}\frac{\mathrm{d}\omega}{\color{red}{\mathrm{d}\omega}}.$$

• You forgot the $$-(\omega_k \rightarrow \omega_k')$$ bit that you have to add to your last integral.

The full integral contribution along the semi-circle is: $$\propto \int_{\gamma_\infty}\!dz\, \frac{z^2}{z^2-\omega^2_k+i\varepsilon } - \int_{\gamma_\infty}\!dz\, \frac{z^2}{z^2-\omega'^2_k+i\varepsilon }.$$

Take one of the integrands and do a partial fraction expansion: $$\frac{z^2}{z^2-\omega^2_k+i\varepsilon } = 1 - \frac{i\varepsilon-\omega_k^2}{i\varepsilon+ z^2-\omega_k^2}.$$

When integrating along a semi-circle of radius $$R$$, the first term gives you : $$\int_\gamma \mathrm{d}z\,1=\pi R$$ while the second term gives you: $$\int_\gamma \mathrm{d}z\, \frac{i\varepsilon-\omega_k^2}{i\varepsilon+ z^2-\omega_k^2} \leq \left |\int_\gamma \mathrm{d}z\, \frac{i\varepsilon-\omega_k^2}{i\varepsilon+ z^2-\omega_k^2} \right | \leq \left | \int_\gamma \mathrm{d}z \right | \cdot \left| \frac{i\varepsilon-\omega_k^2}{i\varepsilon+ z^2-\omega_k^2} \right | \sim \frac{\pi R}{R^2} \propto \frac{1}{R}.$$

So as $$R\rightarrow \infty$$, the second term goes to $$0$$. You are only left with the first term.

$$\require{cancel}$$ Going back to the full integral involving both $$\omega_k$$ and $$\omega'_k$$ then: $$\propto \int_{\gamma_\infty}\!dz\, \frac{z^2}{z^2-\omega^2_k+i\varepsilon } - \int_{\gamma_\infty}\!dz\, \frac{z^2}{z^2-\omega'^2_k+i\varepsilon },$$ $$\leq (\pi R) + \cancel{\mathcal{O}\left (\frac{1}{R} \right )}\color{red}{-} (\pi R) -\cancel{\mathcal{O}\left (\frac{1}{R} \right )} = 0.$$

• Thanks. I see it was not sufficient not consider $\mathcal{I}$ by itself and I needed both log terms for the residue formula. Very nice! That is pretty cool how you approximated the integral like that. What is a source where I can look at how that approximation scheme works? I never think to to do that, but now I've had at least a couple of questions I asked about integrals answered with concise bounding methods like what you did there. Seems highly useful. Commented Oct 18, 2020 at 23:09
• The standard way to do contour integration with the semi-circle contribution going to zero is by using Jordan's lemma. You can google that and find some examples. The inequality with the absolute values is just geometry: the absolute value (=length) of a sum is less than or equal to the absolute value of the individual terms. Commented Oct 18, 2020 at 23:12
• Yes, Jordan's lemma. That's what I was looking for. You are a cool guy. Commented Oct 18, 2020 at 23:17