Path Integral from the Effective Action

Main

Since the BRST approach is somewhat problematic in this case, we just quantize from the (Lorentzian) effective action for FRW with proper time:

(1)
\begin{align} S \left[ a(t) \right] = 12 \pi^{2}\int dt \left( - a \dot a ^{2} + a - \frac{\Lambda}{3} a^{3} \right), \end{align}

which means we are completely in the situation of quantum mechanics on a curved background, and also that we can compare with the Steigl-Hinterleitner analysis.

First, we Euclideanize the effective action by making a Wick+conformal rotation via the substitution $\tau = -it,$ yielding

(2)
\begin{align} S \left[ a \left( \tau \right) \right] = 12 \pi^{2} i \int d \tau \left( a \dot a ^{2} + a - \frac{\Lambda}{3} a^{3} \right) \equiv i S_{E} \left[ a \left( \tau \right) \right]. \end{align}

Thus we have

(3)
\begin{align} \exp \left\{ \frac{i}{\hbar} S \left[ a \left( \tau \right) \right] \right\} = \exp \left\{- \frac{1}{\hbar} S_{E} \left[ a \left( \tau \right) \right] \right\} \end{align}

and can compute correlators in the measure $\mathcal D \left[ a(t) \right]$ corresponding to the Steigl/Hinterleitner operator ordering $\left( i, \frac{1}{2}, k \right)$:

(4)
\begin{align} \langle a \left( \tau_{1} \right) a \left( \tau_{2} \right) \rangle =& \int \mathcal D \left[ a \left( \tau \right) \right] \ a \left( \tau_{1} \right) a \left( \tau_{2} \right) \exp \left\{- \frac{1}{\hbar} S_{E} \left[ a \left( \tau \right) \right] \right\} \\ =& \int \mathcal D \left[ a \left( \tau \right) \right] \ a \left( \tau_{1} \right) a \left( \tau_{2} \right) \exp \left[ - \frac{12 \pi^{2}}{\hbar} \int_{\tau'}^{\tau''} d \tau \left( a \dot a + a - \frac{\Lambda}{3} a^{3} \right) \right] \end{align}

Correlators from propagators using composition law

For the ordering whose kinetic term is $a^{-i} \partial_{a} a^{-j} \partial_{a} a^{-k}$, the Hilbert space on which the Hamiltonian is self-adjoint is $L^{2} \left( \mathbb R^{+}, a^{i-k} da \right)$. On this Hilbert space we have the resolution of identity

(5)
\begin{align} I = \int_{ \mathbb R^{+}} a^{i-k} da \ \left| a \right> \left< a \right|. \end{align}

Using this we can expand the correlator as follows, assuming $t_{2} > t_{1}$:

(6)
\begin{align} \langle a'', t'' \rvert \hat a \left( t_{2} \right) \hat a \left( t_{1} \right) \lvert a', t' \rangle =& \int_{0}^{ \infty } a_{2}^{i-k} da_{2} \ \langle a'', t'' \vert a_{2}, t_{2} \rangle a_{2} \langle a_{2}, t_{2} \vert \hat a \left( t_{1} \right) \vert a', t' \rangle \\ \\ =& \int_{0}^{ \infty} a_{2}^{i-k} da_{2} \int_{0}^{ \infty} a_{1}^{i-k} da_{1} \ a_{1} \cdot a_{2} \langle a'', t'' \vert a_{2}, t_{2} \rangle \langle a_{2}, t_{2} \vert a_{1}, t_{1} \rangle \langle a_{1}, t_{1} \vert a', t' \rangle \\ \\ =& \int_{0}^{\infty} \int_{0}^{\infty} d a_{2} \ d a_{1} \ a_{1}^{1+i-k} a_{2}^{1+i-k} \ \langle a'', t'' \vert a_{2}, t_{2} \rangle \langle a_{2}, t_{2} \vert a_{1}, t_{1} \rangle \langle a_{1}, t_{1} \vert a', t' \rangle \end{align}

so that in principle we only need a means of computing propagators in order to obtain correlators as well.

page revision: 17, last edited: 11 Mar 2015 16:58