# Quick 2 time Homework

###### tags: `2t`

## Questions
1. Write down the most general Lagrangian for the conformally invariant Standard Model coupled to gravity that includes the Higgs doublet, the ghost singlet phi and the dark matter candidate chi. Discuss the properties of the functionals that depend on the three scalar fields and make sure you include all possible terms permitted by Weyl symmetry.
1. In part 1 specialize to only conformally coupled scalars and only simplest form of kinetic energy and the most general form that is symmetric under reflections of chi replaced by -chi.
2. Now choose gauges: first the Higgs doublet reduced to a single scalar that we call s, while s will later be replaced by the vacuum value plus fluctuation s(x)=v+h(x). Second choose the c-gauge for the Weyl symmetry, which means you will replace $\phi(x)$ by the constant $\phi_0$ everywhere in spacetime.
3. In part 3, Consider the E-gauge as an alternative.
4. In the c-gauge write out the details of the scalar part of the full Lagrangian (including gravity) and discuss the vacuum configuration (minimum of the potential).

## Answer 1
The most general form of a Weyl lifted standard model with additional scalars is
\begin{equation}\mathcal{L}\ =\ \sqrt{-g}\left( \frac{1}{12}U(\phi^{i})R(g) - \frac{1}{2}G_{ij}g^{\mu\nu} \partial_\mu \phi^i \partial_\nu \phi^j -V(\phi^i)
\right).\end{equation}
However, this form does not tell us much. Thankfully, eqn(1) is constrained by: \begin{equation}\begin{split}
\partial_i U = -2 G_{ij}\phi^j \\
\phi^i \partial_i U = 2U\\
G_{ij}\phi^{i}\phi^{j} = -U
\end{split}\end{equation}
and the other homogeneity constraints
\begin{equation}
G_{ij}(t\phi^i)=t^0G_{ij}(\phi^i)\\
V(t\phi^i) = t^4 V(\phi^i).
\end{equation}
Plugging these constraints into eqn(1), we can rewrite our general lagrangian as:
\begin{equation}
\mathcal{L}\ =\ \sqrt{-g}\begin{bmatrix} \frac{1}{12}U(\phi^{i})R(g) -V(\phi^i)\\+\frac{1}{2}\left(U-s^I\frac{\partial U}{\partial s^I} -s^Is^JG_{IJ}\right)\left(\partial^\mu \mathrm{ln}\phi\ \partial_\mu \mathrm{ln}\phi \right)\\ -\frac{1}{2}G_{IJ}\left(D^\mu s^I D_\mu s^J\right) + \frac{1}{2}\left( 2G_{IJ}s^J +\frac{\partial U}{\partial s^I}\right)\partial^\mu s^I \partial_\mu \mathrm{ln}\phi\end{bmatrix}.
\end{equation}
The homothety conditions provide n+1 constraints, where n is the number of scalars beyond the $\phi$. In eqn(3), these are used to eliminate the $G_{00}$ and $G_{0I}$ components. Thus we are left with $\frac{d(d+1)}{2}$ independent components of $G$ and additional background fields from $U$ and $V$
Before we get into the 3 scalar model we care about, I'll write the 2 scalar model with a general higgs. Copying form Itzhak's paper we have:
\begin{equation}
\mathcal{L}\ = \sqrt{-g}\begin{bmatrix}
\frac{1}{12}\phi^2u(\frac{\sqrt{2H^\dagger H}}{\phi})R(g) -\phi^4 f(\frac{\sqrt{H^\dagger H}}{\phi})\\ + \frac{1}{2} \left(u-\frac{\sqrt{2H^\dagger H}}{\phi} u' -\left(\frac{2H^\dagger H}{\phi^2} \right) \tilde{G}\right)\left(\partial^\mu \phi\ \partial_\mu\phi \right)\\
-\frac{1}{2}\tilde{G}\left(D^\mu H^\dagger D_\mu H \right) + \left( 2\tilde{G}\frac{\sqrt{2H^\dagger H}}{\phi} + u'\right)\partial^\mu (\sqrt{2H^\dagger H}) \partial_\mu \phi
\end{bmatrix}.
\end{equation}

There are more constraints on the 2 scalar model, but instead we're going to add in the $\chi$. I'm also going to define $s\equiv \sqrt{2H^\dagger H}$ which could either be thought of as gauge fixing, or just notational shorthand. We know the higgs will only appear as an invariant of $SU(2)\times U(1)$, so we might as well simplify our model now. Explicitly written out, we have:
\begin{equation}
\mathcal{L}= \sqrt{-g}\begin{pmatrix}\frac{1}{12} \phi ^ 2 u(\frac{s}{\phi}, \frac{\chi}{\phi}) R(g) - \phi^4 f(\frac{s}{\phi}, \frac{\chi}{\phi})\\+ \frac{1}{2} \left(u - s \frac{\partial u}{\partial s} -\chi\frac{\partial u}{\partial \chi} -\frac{s^2}{\phi^2} G_{ss} - 2\frac{s}{\phi}\frac{\chi}{\phi} G_{\chi s} - \frac{\chi^2}{\phi^2}G_{\chi\chi}\right)\left( \partial^\mu \phi\ \partial_\mu\phi \right)\\-\frac{1}{2}\left(D^\mu sD_\mu s G_{ss} + 2 D^\mu s \partial_\mu \chi G_{s\chi} + \partial^\mu \chi \partial_\mu \chi G_{\chi\chi}\right)\\
+\frac{1}{\phi}\left(2 G_{ss} s+2 G_{s\chi} \chi +\phi^2\frac{\partial u}{\partial s}\right)D^\mu s \partial_\mu \phi\\
+\frac{1}{\phi}\left(2 G_{\chi s} s+2 G_{\chi\chi} \chi +\phi^2\frac{\partial u}{\partial\chi}\right)\partial^\mu \chi \partial_\mu \phi
\end{pmatrix}.
\end{equation}

By going from $\{s,\chi\}\xrightarrow{} \{z^1, z^2\} \equiv \{\frac{s}{\phi}, \frac{\chi}{\phi}\}$, this equation simplifies markedly. Furthermore $z^1$ and $z^2$ are both Weyl invariant quantities. Rewriting we have:
\begin{equation}
\mathcal{L}= \sqrt{-g}\begin{bmatrix}\frac{1}{12} \phi ^ 2 u(z^I) R(g) +\frac{1}{2} \partial^\mu\phi\left[ \partial_\mu\left(\phi u\right)\right]
\\-\frac{1}{2}G_{IJ}\left(\partial^\mu z^I\partial_\mu z^J\right) - \phi^4 v(z^I)
\end{bmatrix}.
\end{equation}
The first line only depends on $\phi$ and $u(z^I)$, while the second line depends on $z^1\ \mathrm{and }\ z^2$ explicitly. Now, we have 5 independent background fields(3 in $G_{IJ}$, 1 in u, and 1 in v). However, we can eliminate 3 of these through field reparameterization.

Using the field in your section A here is what we find for \begin{equation}
U = \phi^2 -\chi ^2 -s^2 +\frac{\phi^3}{\chi}
\end{equation}
This function is homogeneous, so it fits our requirements. Plugging in, we find\begin{equation} G_{ij} =
\begin{pmatrix}
-\frac{\phi^2 -\frac{3\phi^3}{\chi} +\frac{9\phi^4}{4 \chi^2} -\chi^2}{\phi^2 -\chi^2 +\frac{\phi^2}{\chi}} & \frac{\frac{3\phi^5}{4\chi^3} +\frac{\phi^4}{2\chi^2} +\frac{3\phi^2}{2} +2\phi\chi}{\phi^2 -\chi^2 +\frac{\phi^2}{\chi}} & 0 \\
\frac{\frac{3\phi^5}{4\chi^3} +\frac{\phi^4}{2\chi^2} +\frac{3\phi^2}{2} +2\phi\chi}{\phi^2 -\chi^2 +\frac{\phi^2}{\chi}} & -\frac{\chi^2 +\frac{\phi^6}{4\chi^4} +\frac{\phi^3}{\chi} -\phi^2}{\phi^2 -\chi^2 +\frac{\phi^2}{\chi}} & 0\\ 0 & 0 & 1
\end{pmatrix}
\end{equation}

Here's another choice that is homogeneous, but might be too simple. If we take
\begin{equation} U = \phi s +\chi s\end{equation}
we find
\begin{equation} G_{ij} =
\begin{pmatrix}
-\frac{\phi s + \chi s + 8 \chi ^2}{2(\phi + \chi)^2} & -\frac{\phi s + \chi s - 8 \phi \chi }{2(\phi + \chi)^2} & 0 \\
-\frac{\phi s + \chi s - 8 \phi \chi }{2(\phi + \chi)^2} & -\frac{\phi s + \chi s + 8 \phi^2 }{2(\phi + \chi)^2} & 0 \\
0 & 0 & -\frac{2s}{\chi + \phi}
\end{pmatrix}
\end{equation}

## Answer 2

The most general lagrangian consistent with the above requirements is
\begin{equation}
\mathcal{L}_{\mathrm{EW}}= \sqrt{-g}\begin{bmatrix}\frac{1}{12}\left(\phi^2 -\chi^2 -s^2\right) R(g) -\frac{1}{2}\left( \partial^\mu \phi \partial_\mu \phi -\partial^\mu \chi \partial_\mu \chi -\partial^\mu s \partial_\mu s\right)\\ -\left[\frac{\lambda}{4}(s^2 - \alpha^2 \phi^2 )^2 + \frac{\lambda'}{4}\phi ^4 +\frac{\beta'}{4}\chi^4 + \frac{\gamma}{2}\phi^2\chi^2 +\frac{\beta}{2}s^2 \chi ^2 \right] \\
-\frac{1}{4}W^{\mu\nu}_a W_{\mu\nu}^a -\frac{1}{4}B_{\mu\nu}B^{\mu\nu}
\end{bmatrix}.
\end{equation}

## Answer 3
Specifying to the c-gauge and expanding $s(\textbf{x}) = v + h(\textbf{x})$, we have
\begin{equation}
\mathcal{L}= \sqrt{-g_c}\begin{bmatrix}\frac{1}{12}\left(\phi_0^2- v^2 -\chi^2 -h^2-2vh\right) R(g_c)\\ +\frac{1}{2} \partial^\mu \chi \partial_\mu \chi +\frac{1}{2}\partial^\mu h \partial_\mu h\\ -\left( \frac{\lambda}{4} h^4 + \lambda v h^3 + \frac{\lambda}{2} (3v^2 -\alpha^2 \phi_0^2)h^2 +\lambda v(v^2 - \alpha ^2 \phi_0^2)h \\ \frac{\gamma}{2}\chi^2 + \beta v h\chi^2 + \frac{\beta}{2}h^2 \chi^2 + \frac{\beta'}{4} \chi^4 +\frac{ \lambda'}{4}\phi_c^4\right)
\end{bmatrix}.
\end{equation}
Where all fields are implicitly in the c gauge ($\chi = \chi_c$, $h = h_c$), and all of the cosmological constant is absorbed into the term with the free parameter, $\lambda'$, and the mass of the $\chi$ is absorbed into $\gamma$.
## Answer 4
Going to the E gauge for 2 scalars(as a sanity check) we set $\phi = \phi_0 \cosh(\frac{\sigma}{\phi_0})$, $s = \phi_0 \sinh(\frac{\sigma}{\phi_0})$ and find
\begin{equation}
\mathcal{L} = \sqrt{-g_E}\left( \frac{\phi_0^2}{12}R(g_E)+ \frac{1}{2} \partial^\mu \sigma \partial_\mu \sigma - V(\sigma) \right),
\end{equation} which I believe matches your notes.
If we parameterize our 3 scalars as
\begin{equation}
\phi = \phi_0 \cosh(\frac{\sigma}{\phi_0}),\\
\chi = \phi_0 \sinh(\frac{\sigma}{\phi_0})\sin(\theta)\\
s = \phi_0 \sinh(\frac{\sigma}{\phi_0})\cos(\theta)
\end{equation}
(with dimensionless $\theta$), we find \begin{equation} U = \phi_0^2\end{equation}
and

\begin{equation}
\mathcal{L} = \sqrt{-g_E}\left( \frac{\phi_0^2}{12}R(g_E)+ \frac{1}{2} \partial^\mu \sigma \partial_\mu \sigma + \frac{1}{2}\phi_0^2 \sinh^2(\frac{\sigma}{\phi_0})\partial_\mu \theta \partial^\mu \theta\right) \\-\phi_0^4\sqrt{-g_E}\left[\frac{\lambda}{4}\bigg((\alpha^2-1)\cosh^2(\frac{\alpha}{\phi_0})+1\bigg)^2 + \frac{\lambda'}{4} \cosh^2(\frac{\alpha}{\phi_0}) + \frac{\beta'}{4}\sinh^4(\frac{\alpha}{\phi_0})\sin^4(\theta)\\+ \frac{\gamma}{2}\sinh^2(\frac{\sigma}{\phi_0})\cosh^2(\frac{\sigma}{\phi_0}) + \frac{\beta-\gamma}{2}\sinh^2(\frac{\alpha}{\phi_0})\cosh^2(\frac{\alpha}{\phi_0}) \cos^2(\theta)\right].
\end{equation}
## Answer 5
The general lagrangian in the c gauge is \begin{equation}
\mathcal{L} = \sqrt{-g_c}\left( \frac{\phi_0^2}{12} u(\frac{s}{\phi_0}, \frac{\chi}{\phi_0})R(g_c)+\frac{1}{2}G_{IJ}(D_\mu\phi^I D^\mu\phi^J)\\
-\left[\frac{\lambda}{4}(s^2 - \alpha^2 \phi_0^2 )^2 +\frac{\beta'}{4}\chi^4 + \frac{\gamma}{2}\phi_0^2\chi^2 +\frac{\beta}{2}s^2\chi^2+ \frac{\lambda'}{4}\phi_0^4\right]\right),\end{equation}
where the covariant derivative is only nontrivial for the higgs.
Varying to find the minimum of the potential we find
\begin{equation}
\frac{\delta V}{\delta \chi} = \chi\left(\gamma\phi_0^2+\beta s^2 +\beta' \chi^2 \right)= -\frac{R(g)}{6}\chi\approx 0.
\end{equation}
Assuming $\beta$, $\beta'$, and $\gamma$ are all positive, this means \begin{equation}
\langle\chi\rangle = 0.
\end{equation}
Similarly for $s$ we find
\begin{equation} \frac{\delta V}{\delta s} = s\left( \frac{\lambda}{2}(s^2-\alpha^2\phi_c^2) +\beta \chi^2 \right) = 0 \\\Rightarrow s= \frac{R(g)}{6}s \approx 0\ \mathrm{or }\ s^2 = \alpha ^2 \phi_0^2
\end{equation}given that $\langle \chi \rangle = 0$.
$V(s = 0) =\frac{\lambda\alpha^4\phi_0^4}{4} + \frac{\lambda'}{4}\phi_0^4,$ while $V(s=\pm \alpha \phi_0) =\frac{\lambda'}{4}\phi_0^4$ so $s = \pm \alpha \phi_0$ is the true minimum.

Finally being careful about $\phi$,\begin{equation}
\frac{1}{\sqrt{-g_c}}\frac{\delta \mathcal{L}}{\delta \phi}\bigg\vert_{\phi = \phi_0} =\left[ \frac{R(g_c)}{12}\frac{\partial U }{\partial \phi} - \phi(\frac{\alpha^2\lambda}{2}(s^2-\alpha^2\phi^2) +\gamma \chi^2 +\lambda' \phi^2) \right]\bigg\vert_{\phi = \phi_0}=0.
\end{equation}

Plugging in for conformally coupled scalars and above results, \begin{equation}
\frac{R(g_c)}{6} = \lambda' \phi_0^2
\end{equation}
meaning $\lambda'$ is incredibly tiny.