# Vacuum energy density after spontaneous symmetry breaking

• A
• Geonaut
In summary, the conversation discusses the Lagrangian and scalar field in an interacting theory, and the calculation of vacuum energy density using Noether's theorem. The issue of divergent vacuum energy density and its relation to gravity is brought up, and the solution of using a constant counter term in the Lagrangian to renormalize the divergent term is mentioned. The potential problem of fine tuning in this approach is also mentioned.

#### 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.

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

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:

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

Using Noether's theorem I find

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

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

<\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}}.

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.

These are just the regular divergences that you encounter in Green's functions', they can be renormalized via a constant counter term in the Lagrangian.

You have for example a perturbation to your free Lagrangian,
$$\delta H = \int d^3 x \frac{\lambda}{4!}\phi^4 (x)$$
and the first-order correction to the ground state energy is then
$$\delta E^{(1)} = \int d^3 x \frac{\lambda}{4!} <0|\phi^4 (x)|0> = \frac{\lambda}{8} D_F (0)^2 \int d^3 x .$$
This is just a vacuum to vacuum diagram (aka. A bubble diagram).

Last edited:
Geonaut and vanhees71
Thank you for the response! That approach doesn't work in a theory that includes gravity though, does it? Also, how can a cow be homogenous if it's covered in spots?

... is it a shaved cow?... I wonder.