# Equality for two expressions of vacuum energy density

I have sometimes seen two expressions for the vacuum energy density of a scalar field. \begin{align} \frac{E}{V} &= \int \frac{d^3 k}{(2 \pi)^3} \frac{1}{2} \sqrt{\vec{k}^2 + m^2} \\ &= \frac{1}{2} \int \frac{d^4 k_E}{(2 \pi)^4} \log(k_E^2 + m^2) \end{align} where the second expression is written as an integral over Euclidean space.

Just at the basic level of the integrals... How are these two expressions equal? When I perform the $$k_E^0$$ integral for the second expression I don't get the first expression. I then wondered if the expressions asymptotically approach each other as the UV cut off $$\Lambda \to \infty$$ but I don't think they do that either? In what sense is this equation correct?

• Source for the second equation? Commented May 5, 2021 at 5:00
• One place I've seen it is Kiritsis, String Theory in a Nutshell, Equation 5.3.1 Commented May 5, 2021 at 5:38

They differ by an infinite constant. Write the logarithm as $$\log\left( \omega^2 + \vec{k}^2 + m^2 \right) = \log \omega^2 + \log\left( 1 + \frac{\vec{k}^2 + m^2}{\omega^2} \right).$$ Then your second integral is $$\frac{E}{V} = \frac{1}{2} \int \frac{d\omega d^3 k}{(2 \pi)^4} \log \omega^2 + \frac{1}{2} \int \frac{d\omega d^3 k}{(2 \pi)^4}\log\left( 1 + \frac{\vec{k}^2 + m^2}{\omega^2} \right).$$ You can perform the integral over $$\omega$$ in the second integral, and you get $$\frac{E}{V} = \frac{1}{2} \int \frac{d\omega d^3 k}{(2 \pi)^4} \log \omega^2 + \frac{1}{2} \int \frac{d^3 k}{(2 \pi)^3}\sqrt{\vec{k}^2 + m^2}.$$ Of course, you will still need to regulate the second integral in some way. But if you ask about physical quantities like Casimir energies you'll get the same answer using both equations.
• @user1379857 I believe the extra divergent factor comes from constructing the path integral expression for the ground-state energy. Your first expression is the usual ground state energy of a free Gaussian field computing by diagonalizing $\hat{H}$ exactly. One should get the same expression by computing $E_0 = -\lim_{\beta \rightarrow \infty} \log \mathrm{Tr} \, e^{- \beta \hat{H}}$. It is the process of converting the partition function into a continuum path integral which introduces extra factors, and if one does this carefully one presumably can see where the extra constant came in. Commented May 11, 2021 at 21:55
• Yes, I did it carefully for the 1D oscillator where the ground state energy does not diverge. Indeed, you basically have to divide $det(-\frac{d}{d \tau} + \omega^2)$ by $det( - \frac{d}{d\tau})$, which 'regulates' the answer in exactly the same way you did in your answer. This divergence has nothing to do with renormalization per se, it's just about getting the normalization constants right for the path integral Commented May 12, 2021 at 2:08