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?


1 Answer 1


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).

  • $\begingroup$ What a wonderful answer! I have never found such an elaborate discussion on this topic in the literature. Thank You. $\endgroup$
    – Archimedes
    Commented Dec 13, 2017 at 18:09

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