Vacuum energy density after spontaneous symmetry breaking

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  • Thread starter Geonaut
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  • #1
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.
 

Answers and Replies

  • #2
HomogenousCow
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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).
 
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  • #3
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.
 
  • #4
@HomogenousCow All stupid jokes aside... I made this post because I felt that it is important to know for sure what is the common solution that people use for this problem, and finding answers to really specific questions like this one can be very hard to find in dense books on QFT. I thought I remembered reading that the solution you mentioned doesn't work when we don't ignore gravity, but I asked myself why and I couldn't remember. So I went back and found where I read about this, and it turns out that I remembered things wrong. The book said that we can discard contributions to the vacuum energy density like this (they were talking about zero-point energy) because we measure energy differences, but in a side note they later write that this doesn't work if we don't ignore gravity. In a later chapter they mention the approach that you've pointed out (but instead applied to the zero-point energy problem), but it still seems that this is a problem. It looks like this should be considered to be a fine tuning problem.

Anyway, thanks again! I'm sorry that the question was a little trivial, but writing things down definitely helped me think things through.
 
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  • #5
HomogenousCow
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The problems with quantum gravity are so numerous and nebulous that I really wouldn't think too much about this while learning QFT, it's very likely that the correct (so to speak) theory of QG ends up being in some new language of physics that we haven't even seen.
 
  • #6
@HomogenousCow Well, although they don't clearly state it in the book I mentioned, I can't imagine how this could create a problem in a QG theory if we use the solution that you pointed out (I thought I read somewhere that it does, but I was mistaken). However, it does seem to create a fine tuning problem in the same way that the zero-point energy does. And you're right, QFT should be learned first, I feel like I know a lot about it at this point, but I didn't learn it in order or through a professor so my knowledge is probably weak in random places like this one.
 

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