Musings

Maldacena’s result was already unobservably small, so these loop corrections are not of great observational interest. Still, it’s nice to see that there is a consistent formalism for calculating them, and that the late-time result, say for the loop correction to the 2-point function, depends only on the behaviour of the inflaton field at horizon-exit, and not on the whole intervening history¹.

The “In-In” formalism, as the name suggests, consists of computing the expectation values of operators in the $|in\rangle$ vacuum, in which the fluctuations behave as free fields in the far past. Take the Hamiltonian, expand it about the classical solution, and drop the terms linear in the fluctuations. This yields the fluctuation Hamiltonian, $\tilde{H}$. Working in the interaction picture for $\tilde{H}$, $$\langle in| Q(t) |in\rangle \equiv \left\langle \overline{T}\exp\left(i\int^t_{-\infty} H_I dt\right)\, Q^I(t)\, T \exp\left(-i\int^t_{-\infty} H_I dt\right) \right\rangle$$ where $T$ denotes time-ordering, $\overline{T}$ denotes anti-time-ordering and $Q^I(t)$ is a string of interaction picture fields.

Because there’s both time-ordering and anti-time-ordering, the Feynman rules that follow from this formalism are a bit baroque. Steve applies them to calculate the 1-loop correction to the 2-point function of scalar fluctuations. The well-known tree-level answer is $$\int d^3x e^{i\vec{q}\cdot(\vec{x}-\vec{x}')} \langle in|\zeta(\vec{x},t)\zeta(\vec{x}',t')|in\rangle_{\text{tree}} = \frac{8\pi G H^2(t_q)}{4(2\pi)^3 |\epsilon(t_q)|q^3}$$ The 1-loop correction is $$\int d^3x e^{i\vec{q}\cdot(\vec{x}-\vec{x}')} \langle in|\zeta(\vec{x},t)\zeta(\vec{x}',t')|in\rangle_{\text{1-loop}} = -\frac{\pi(8\pi G H^2(t_q))^2 N}{15(2\pi)^3 q^3}[\log q +C]$$ where $$\epsilon =-\frac{\dot{H}}{H^2},\qquad H= \frac{\dot{a}}{a}$$ $t_q$ is the time when the mode with wave number $q$ exits the horizon, and $N$ is a numerical constant.

Note that this is suppressed, relative to the tree-level answer, by a factor of the slow-roll parameter, $\epsilon$, and an additional factor of $H^2/M_p^4$. That makes it really tiny. But, interestingly, even in the very slow-roll limit, where $H(t_q),\epsilon(t_q)$ are nearly constant, it departs from the scale-invariant tree-level answer by a factor of $\log(q)$.

Anyway, lots of interesting discussion of the subtleties of field theory in (quasi-)de Sitter space. You know you meant to spend some time understanding that stuff.