- #1

#### Geonaut

- TL;DR Summary
- My question is in regards to a calculation for the vacuum energy density for an interacting theory that's described in the post. I found what appears to be my first encounter with a divergent term that I haven't considered before.

If we start with the Lagrangian

\begin{equation} \begin{split} \mathcal{L} = & \frac{1}{2}(\partial_\mu \phi)^2 + \frac{1}{2}\mu^2 \phi^2 - \frac{1}{4}\lambda^2 \phi^4\\ \end{split} \end{equation}

and give the scalar field a VEV so that we can define the field ##\eta##, where

$$\eta = \phi \pm \frac{\mu}{\lambda}$$

then we have:

\begin{equation} \begin{split} \mathcal{L} = & \frac{1}{2}(\partial_\mu \eta)^2 \pm \frac{\mu^3}{\lambda}\eta \mp \frac{\mu^3}{\lambda}\eta + \frac{1}{2}\mu^2\eta^2 - \frac{3}{2}\mu^2\eta^2\\ & \mp \mu \lambda \eta^3 - \frac{1}{4} \lambda^2 \eta^4 + \frac{1}{2}\frac{\mu^4}{\lambda^2} - \frac{1}{4}\frac{\mu^4}{\lambda^2} \\ = & \frac{1}{2}(\partial_\mu \eta)^2 - \mu^2\eta^2 \mp \mu \lambda \eta^3 - \frac{1}{4} \lambda^2 \eta^4 +\frac{1}{4}\frac{\mu^4}{\lambda^2}. \end{split} \end{equation}

Using Noether's theorem I find

\begin{equation} \begin{split} H = & \frac{1}{2}(\partial_0 \eta)^2 + \frac{1}{2}(\bigtriangledown\eta)^2 + \mu^2\eta^2 \pm \mu \lambda \eta^3 + \frac{1}{4} \lambda^2 \eta^4 -\frac{1}{4}\frac{\mu^4}{\lambda^2}. \end{split} \end{equation}

So to find the vacuum energy density I calculate ## <\Omega| H |\Omega>##, where ##|\Omega>## is the ground state in this interacting theory, but I noticed something that I haven't considered before, namely, the term

\begin{equation}

<\Omega|\frac{1}{4} \lambda^2 \eta^4 |\Omega>= \frac{1}{4} \lambda^2 \frac{(4\pi)^2}{(2\pi)^6}\int_0^\infty dp \frac{p^2}{2 \sqrt{p^2 + m^2}} \int_0^\infty dq \frac{q^2}{2 \sqrt{q^2 + m^2}}.

\end{equation}

I figured this term might be thought of as the contribution due to vacuum bubbles, but it appears that this term gives us a divergent vacuum energy density, and so I fear that I made a mistake somewhere. Moreover, it appears that the constant term in this Hamiltonian gives us a negative contribution to the vacuum energy density, which is not what I was expecting, and so I'm even more suspicious.

\begin{equation} \begin{split} \mathcal{L} = & \frac{1}{2}(\partial_\mu \phi)^2 + \frac{1}{2}\mu^2 \phi^2 - \frac{1}{4}\lambda^2 \phi^4\\ \end{split} \end{equation}

and give the scalar field a VEV so that we can define the field ##\eta##, where

$$\eta = \phi \pm \frac{\mu}{\lambda}$$

then we have:

\begin{equation} \begin{split} \mathcal{L} = & \frac{1}{2}(\partial_\mu \eta)^2 \pm \frac{\mu^3}{\lambda}\eta \mp \frac{\mu^3}{\lambda}\eta + \frac{1}{2}\mu^2\eta^2 - \frac{3}{2}\mu^2\eta^2\\ & \mp \mu \lambda \eta^3 - \frac{1}{4} \lambda^2 \eta^4 + \frac{1}{2}\frac{\mu^4}{\lambda^2} - \frac{1}{4}\frac{\mu^4}{\lambda^2} \\ = & \frac{1}{2}(\partial_\mu \eta)^2 - \mu^2\eta^2 \mp \mu \lambda \eta^3 - \frac{1}{4} \lambda^2 \eta^4 +\frac{1}{4}\frac{\mu^4}{\lambda^2}. \end{split} \end{equation}

Using Noether's theorem I find

\begin{equation} \begin{split} H = & \frac{1}{2}(\partial_0 \eta)^2 + \frac{1}{2}(\bigtriangledown\eta)^2 + \mu^2\eta^2 \pm \mu \lambda \eta^3 + \frac{1}{4} \lambda^2 \eta^4 -\frac{1}{4}\frac{\mu^4}{\lambda^2}. \end{split} \end{equation}

So to find the vacuum energy density I calculate ## <\Omega| H |\Omega>##, where ##|\Omega>## is the ground state in this interacting theory, but I noticed something that I haven't considered before, namely, the term

\begin{equation}

<\Omega|\frac{1}{4} \lambda^2 \eta^4 |\Omega>= \frac{1}{4} \lambda^2 \frac{(4\pi)^2}{(2\pi)^6}\int_0^\infty dp \frac{p^2}{2 \sqrt{p^2 + m^2}} \int_0^\infty dq \frac{q^2}{2 \sqrt{q^2 + m^2}}.

\end{equation}

I figured this term might be thought of as the contribution due to vacuum bubbles, but it appears that this term gives us a divergent vacuum energy density, and so I fear that I made a mistake somewhere. Moreover, it appears that the constant term in this Hamiltonian gives us a negative contribution to the vacuum energy density, which is not what I was expecting, and so I'm even more suspicious.