3
$\begingroup$

This question pertains to Cosmic Inflation, but I guess the answer lies in quantum field theory. And I am not being able to figure out the proper reasoning from the basic theory.

Let's consider the inflaton(the scalar field) Lagrangian

$$\mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi) - V(\phi) - g^2 \phi^2 \chi^2$$

At the end of inflation, when the scalar field oscillates around the minima, and we apply perturbative reheating i.e., consider the perturbative production of $\phi\phi \to \chi \chi$. Now, when $V(\phi) = \frac{1}{ 2} m^2 \phi^2$, the scalar field equation has a simple solution of the form(See, for example:http://www.damtp.cam.ac.uk/user/db275/TEACHING/INFLATION/Lectures.pdf , page.65) $$\phi(t) \approx \Phi(t) sin(mt)$$ where $\Phi(t) \sim \frac{M_p}{m t}$. The decay rate for the above scattering process can be computed as $$\Gamma_{\phi \phi \to \chi\chi} = \frac{g^4 \Phi^2}{8 \pi m}$$

What would occur if $V(\phi) = \frac{1}{4}\lambda\phi^4$. Isn't the perturbatibe reheating possible in this case? if yes, how to compute the $\Gamma_{\phi \phi \to \chi\chi}$ in terms of the inflaton solution?

$\endgroup$
4
$\begingroup$

It is. The basic picture is that the inflaton behaves as an oscillating classical field which can be thought of, more or less, as a collection of particles with zero momentum and mass $m\sim \omega$ where $\omega$ is the frequency of oscillation. In the case $V=\frac{1}{2}m^2\phi^2$ the frequency is given by the inflaton mass, $\omega = m$. The only difference in the case $V=\frac{\lambda}{4}\phi^4$ is that the oscillation is a little bit different. Now the frequency of the oscillation is $\omega \sim \sqrt{\lambda}\Phi$. So replacing $m$ with $\sqrt{\lambda}\Phi$ in the formula for the decay rate we would expect something like

\begin{equation} \Gamma(\phi\phi\rightarrow\chi\chi) \sim \frac{g^4\Phi}{\sqrt{\lambda}} \end{equation}

If we want a more accurate result we need to actually do the perturbative calculation. I have outlined it below.

Particle production by an oscillating field

Let us ignore the expansion of the universe and suppose that we have a classical field $\phi$ which is oscillating with a period $T$ and that it couples to a quantum field $\chi$ via interaction $V_I = g^2\phi(t)\chi^2$. We can expand the field in a harmonic series

\begin{equation} \phi(t) = \sum_{n=-\infty}^\infty \phi_n e^{-i n\omega t} \end{equation} where $\omega = 2\pi/T$ is the leading frequency. We are interested in producing a pair of $\chi$-particles out of the vacuum so we want to find the transition amplitude

\begin{equation} \mathcal A \equiv \langle \mathbf{k_1,k_2}|e^{-i\int\mathrm{d} t H_I}|0\rangle \simeq -ig^2 \int \mathrm{d}t\:\phi(t)\int \mathrm{d}^3\mathbf{x}\langle 0| \hat a_\mathbf{k_1}\hat a_\mathbf{k_2}\hat \chi^2|0\rangle. \end{equation} As usual we plug in the interaction picture expression for $\chi$

\begin{equation} \hat\chi = \int\frac{\mathrm{d}^3\mathbf{k}}{(2\pi)^{3/2}\sqrt{2\omega_k}}\left(\hat a_\mathbf{k}e^{-i\omega_k t + i\mathbf{k\cdot x}} + \mathrm{h.c.}\right), \end{equation} where $\omega_k^2 \equiv k^2+m_\chi^2$, and we get for the transition amplitude

\begin{equation} \mathcal{A} = -\frac{2i\pi g^2\delta(\mathbf{k_1+k_2})}{\omega_{k}}\int \frac{\mathrm{d}t}{2\pi}\phi(t)e^{2i\omega_kt} \end{equation} Plugging in our harmonic expansion for $\phi$ we get

\begin{equation} \mathcal{A} = -\frac{i\pi g^2}{\omega_{k}}\delta(\mathbf{k_1+k_2})\sum_n\phi_n \delta\left(\omega_k-\frac{n\omega}{2}\right) \end{equation} The transition probability is then

\begin{equation} |\mathcal A|^2 = (2\pi)^4\delta^4(0)\frac{g^4}{16\pi^2\omega_k^2}\delta(\mathbf{k_1+k_2})\sum_n |\phi_n|^2\delta\left(\omega_k-\frac{n\omega}{2}\right) \end{equation} As usual we interpret $(2\pi)^4\delta^4(0) = VT$ as the integral over all of spacetime and the transition probability per volume per unit time is

\begin{equation} \mathrm d P(\mathbf{k_1,k_2}) = \frac{g^4}{16\pi^2\omega_k^2}\delta(\mathbf{k_1+k_2})\sum_n |\phi_n|^2\delta\left(\omega_k-\frac{n\omega}{2}\right) \end{equation}

Decay of the oscillating field

We may calculate the decay rate of the oscillating field by energy conservation. In time $\mathrm{d}t$ and volume $V$ we expect to create $2\mathrm d P(\mathbf{k_1,k_2})V\mathrm{d}t$ particles with energy $\omega_k$. Thefore the change in energy is $\mathrm{d}E = 2\omega_k\mathrm d P(\mathbf{k_1,k_2})V\mathrm{d}t$. The oscillating field must therefore lose the same amount of energy so that

\begin{equation} \frac{\mathrm{d \rho_\phi(\mathbf{k_1,k_2})}}{\mathrm{dt}} = - \frac{g^4}{8\pi^2\omega_k}\delta(\mathbf{k_1+k_2})\sum_n |\phi_n|^2\delta\left(\omega_k-\frac{n\omega}{2}\right) \end{equation} This is the loss of energy due to production of pairs of two $\chi$ -particles with momenta $\mathbf{k_1,k_2}$. We get the total energy loss by integrating over the momenta

\begin{equation} \dot\rho_\phi = - \frac{g^4}{2\pi}\sum_n |\phi_n|^2\int_{m_\chi}^\infty \mathrm{d}\omega_k \omega_k\sqrt{1-\frac{m_\chi^2}{\omega_k^2}}\delta\left(\omega_k-\frac{n\omega}{2}\right) = - \frac{g^4\omega}{4\pi}\sum_{n=1}^{\infty} n|\phi_n|^2\sqrt{1-\frac{4m_\chi^2}{n^2\omega^2}} \end{equation} We define the decay width for the field through $\dot\rho_\phi = -\Gamma \rho_\phi$. In the limit $m_\chi \ll \omega$ the decay rate is

\begin{equation} \Gamma = \frac{g^4\omega}{4\pi\rho_\phi}\sum_{n=1}^{\infty}n|\phi_n|^2 \end{equation}

Potential $V=\frac{1}{2}m^2\phi^2$

If the inflaton oscillates in a harmonic potential then $\phi = \Phi\sin mt$ for a trilinear interaction and $\phi = \Phi^2\sin^2mt$ for a quartic interaction. Therefore it is trivial to check that for interactions $V_I = g^2\sigma \phi\chi^2$ and $V_I = g^2\phi^2\chi^2$ we get, respectively,

\begin{equation} \Gamma(\phi\rightarrow\chi\chi) = \frac{g^4\sigma^2}{8\pi m}, \qquad \Gamma(\phi\phi \rightarrow\chi\chi) = \frac{g^4\Phi^2}{16\pi m} \end{equation}

Potential $V = \frac{\lambda}{4}\phi^4$

If the inflaton oscillates in the quartic potential (ignoring expansion of the universe) the equation of motion for the inflaton is

\begin{equation} \ddot \phi + \lambda \phi^3 = 0 \end{equation} which has the solution

\begin{equation} \phi = \Phi \mathrm{cn}\left(\sqrt{\lambda}\Phi t, \frac{1}{\sqrt{2}}\right) \end{equation} where $\mathrm{cn}(x,k)$ is the Jacobi elliptic cosine function which has a period $T=4K/\sqrt{\lambda \Phi^2}$, $K\equiv K(k)$ being a complete elliptic integral of the first kind. We can look up the harmonic expansion of the elliptic cosine on, say, Wikipedia. The coefficients are

\begin{equation} \phi_n = 2\sqrt{2}\lambda^{-1/2}\Phi\omega \frac{e^{-\pi |n|/2}}{1+e^{-\pi |n|}} \end{equation} for odd $n$ and $\phi_n = 0$ for even $n$. The energy density of the inflaton is $\rho_\phi = \frac{1}{4}\lambda\Phi^4$. For the interaction $V_I = g^2\sigma\phi $ the decay rate is then

\begin{equation} \Gamma(\phi\rightarrow\chi\chi) = \frac{8g^4\sigma^2\omega^3}{\pi\lambda^2 \Phi^4}\sum_{n=1}^{\infty}(2n-1)\frac{e^{-(2n-1)\pi}}{\left(1+e^{-(2n-1)\pi}\right)^2} \simeq \frac{g^4\sigma^2\omega^3}{\pi^2\lambda^2 \Phi^4} = \frac{\sqrt{3}\pi^2}{K^3}\frac{g^4\sigma^2}{8\pi\sqrt{3\lambda \Phi^2}} \simeq 2.7 \frac{g^4\sigma^2}{8\pi m_\mathrm{eff}} \end{equation} (the sum is very well approximated by $1/(8\pi)$). The frequency of the inflaton oscillation is $\omega = 2\pi/T = \sqrt{\lambda}\Phi\pi/2K$ and in the last expression we defined an effective mass of the inflaton quanta $m_\mathrm{eff}^2 \equiv 3\lambda \Phi^2$. Thus, the decay rate is somewhat enhanced compared to that of a condensate of particles. For the quartic interaction $V_I = g^2\phi^2\chi^2$ we have to expand $\Phi^2\mathrm{cn}^2\left(\sqrt{\lambda \Phi^2}t,1/\sqrt{2}\right)$ in a harmonic series. I'm not aware of an exact expansion but we can do this numerically:

\begin{equation} \phi^2(t) = \sum_{n=-\infty}^{\infty}\alpha_n e^{-in\omega t} \qquad \Rightarrow \qquad \alpha_n = \frac{1}{T}\int_0^{T}\mathrm{d}t\:\phi^2e^ {in\omega t} \end{equation} Taking just a couple of leading terms we get

\begin{equation} \Gamma(\phi \rightarrow \chi\chi) \simeq 0.5 \times \frac{g^4m_\mathrm{eff}}{8\pi\lambda} \sim \frac{g^4\Phi}{\sqrt{\lambda}} \end{equation} just as we had initially suspected.

Expansion of the universe

So far we have neglected the expansion of the universe. In the case of the quadratic inflaton potential $V=\frac{1}{2}m^2\phi^2$ the behaviour of the inflaton is well enough approximated by $\phi_0 a^{-3/2}\sin mt$ and the time scale of oscillation is much smaller than the Hubble time so we can just replace $\Phi\sim \Phi a^{-3/2}$. This leaves $\Gamma(\phi\rightarrow \chi\chi)$ unaffected while $\Gamma(\phi\phi\rightarrow \chi\chi)$ now scales as $a^{-3}$, decaying faster than $H$, which means that this process never becomes effective.

In the case of the quartic potential we can rescale the fields $\phi\rightarrow \phi/a$, $\chi\rightarrow \chi/a$ and change our time coordinate to conformal time $\mathrm d \tau \equiv a^{-1}\mathrm{d}t$ in which case the solution that we had before $\phi = \Phi \mathrm{cn}(\sqrt{\lambda\Phi^2}\tau,1/\sqrt{2})$ is still valid and $\Phi$ is the amplitude at the onset of oscillation. Now any term which is quartic in the fields or quadratic in the derivatives scales in the same way, $\propto a^{-4}$, and so the expansion of the universe factors out (the situation is conformal to the Minkowski case) and $\Gamma(\phi\phi\rightarrow \chi\chi)$ is unchanged, only now it is understood as the decay rate in conformal time (in cosmic time $\Gamma(t) = \Gamma(\tau)/a$).

The trilinear interaction $V_I = g^2\sigma \phi \chi^2$ is more tricky. Because it only has three fields it scales as $a^{-3}$ rather than $a^{-4}$ and thus brakes conformal invariance. Then, in the rescaled variables the interaction becomes $V_I = g^2\sigma \phi(\tau)a(\tau) \chi^2$ and so the classical field responsible for particle production has a growing amplitude and is no longer periodic. However, assuming that $a$ changes very slowly compared to the oscillations we can just replace $\sigma\rightarrow a\sigma$ and so the decay rate goes $\Gamma\rightarrow a^2\Gamma$ in conformal time ($a\Gamma$ in cosmic time).

$\endgroup$
  • $\begingroup$ What a wonderful answer! I have never found such an elaborate discussion on this topic in the literature. Thank You. $\endgroup$ – Archimedes Dec 13 '17 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.