Ordering for FRW with a scalar field

## Derivation of FRW Effective Action and Hamiltonian

Start with the Einstein-Hilbert action with a scalar field matter coupling:

(1)
\begin{align} S = \int_{\mathbb{R} \times \Sigma} \alpha_G \mathcal{L}_G + \alpha_M \mathcal{L}_M \ d^4x \end{align}

where $\mathcal{L}_G = \sqrt{-g} R$ and $\mathcal{L}_M = - \sqrt{-g} \left[ g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi + V \left( \phi \right) \right]$ are the geometric and matter Lagrangian densities, respectively, and $\alpha_G$ and $\alpha_M$ are the associated coupling constants (which depend on units and convention). Decomposing in the ADM formalism and fixing the proper time gauge ($N=1, N^i = 0$), we have

(2)
\begin{align} \mathcal{L}_G &= \sqrt{h} \left[ K_{ij} K^{ij} - K^2 + \left( ^{3}R - 2 \Lambda \right) \right] \\ \mathcal{L}_M &= \sqrt{h} \left[ \dot{\phi}^2 - g^{ij} \nabla_i \phi \nabla_j \phi + V \left( \phi \right) \right], \end{align}

where $h_{ij}$ denotes the 3-metric on the spatial slice $\Sigma$, $h$ its determinant, and $^3R$ its scalar curvature. The $K_{ij} = \dot h_{ij} /2$ are the extrinsic curvature. This action yields momenta conjugate to $h_{ij}$ and $\phi$, respectively:

(3)
\begin{align} \pi^{ij} &= \alpha_G \sqrt{h} \left( K^{ij} - K h^{ij} \right) \\ p_{\phi} &= 2 \alpha_M \sqrt{h} \dot{\phi}. \end{align}

Performing a Legendre transform, we get the Hamiltonian density

(4)
\begin{align} \frac{1}{\alpha_G} G_{ijkl} \pi^{ij} \pi^{kl} + \frac{1}{4 \alpha_M \sqrt{h}} p_{\phi}^2 - \alpha_G \sqrt{h} \left( ^{3}R - 2 \Lambda \right) + \alpha_M \sqrt{h} \left[ h^{ij} \nabla_i \phi \nabla_j \phi + V \left( \phi \right) \right] \end{align}

where

(5)
\begin{align} G_{ijkl} = \frac{1}{2 \sqrt{h}} \left( h_{ik} h_{jl} + h_{il} h_{jk} - h_{ij} h_{kl} \right) \end{align}

is the Wheeler-DeWitt metric on superspace.

We can now reduce the Lagrangian density (1) to the FRW case by specializing to the metric

(6)
\begin{align} ds^2 = -dt^2 + a^2 \left( t \right) \left( d \theta_1^2 + \sin^2 \theta_1 d \theta_2^2 + \sin^2 \theta_1 \sin^2 \theta_2 d \varphi^2 \right), \end{align}

obtaining the Lagrangian

(7)
\begin{align} L = 12 \pi^2 \left[ - \alpha_G a \dot a^2 + \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + \frac{1}{6} \left( \alpha_M a^3 \dot{\phi}^2 - \alpha_M a^3 V \left( \phi \right) \right) \right]. \end{align}

For this effective Lagrangian we get the momenta

(8)
\begin{align} p_a &= -24 \pi^2 \alpha_G a \dot a \\ p_{\phi} &= 4 \pi^2 \alpha_M a^3 \dot{\phi} \end{align}

and the effective Hamiltonian

(9)
\begin{align} H = - \frac{1}{48 \pi^2 \alpha_G} \frac{p_a^2}{a} + \frac{1}{8 \pi^2 \alpha_M} \frac{p_{\phi}^2}{a^3} - 12 \pi^2 \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 \alpha_M a^3 V \left( \phi \right). \end{align}

As in Misner/Thorne/Wheeler and Steigl/Hinterleitner, we can take $\alpha_G = \frac{c^3}{8 \pi G}$ and $\alpha_M = c$.

## Classical solutions

For the case $V \left( \phi \right) = 0$, $\Lambda = 0$, we easily obtain the classical solutions:

(10)
\begin{align} H = - \frac{1}{48 \pi^2 \alpha_G} \frac{p_a^2}{a} + \frac{1}{8 \pi^2 \alpha_M} \frac{p_{\phi}^2}{a^3} - 12 \pi^2 \alpha_G a =0 \end{align}

Equations of motion:

(11)
\begin{align} \dot p_a &= - \frac{1}{48 \pi^2 \alpha_G} \left( \frac{p_a}{a} \right)^2 + \frac{3}{8 \pi^2 \alpha_M} \left( \frac{p_{\phi}}{a^2} \right)^2 + 12 \pi^2 \alpha_G \\ \dot a &= - \frac{1}{24 \pi^2 \alpha_G} \left( \frac{p_a}{a} \right) \\ \dot p_{\phi} &= 0 \\ \dot{\phi} &= \frac{1}{4 \pi^2 \alpha_M} \left( \frac{p_{\phi}}{a^3} \right). \end{align}

By setting $\dot a = 0$ we obtain the maximum radius of the universe, given by $a_* = \left( \frac{p_{\phi}^2}{96 \pi^4 \alpha_G \alpha_M} \right)^{1/4}$. Using

(12)
\begin{align} \frac{da}{d \phi} = \frac{\dot a}{\dot{\phi}} = \pm \sqrt{\frac{\alpha_M}{6 \alpha_G}} \sqrt{a^2 - \frac{a^6}{a_*^4}}, \end{align}

we can write the classical solution as

(13)
\begin{align} a = a_* \left( 1 - \tanh^2 \left( \pm \sqrt{ \frac{2 \alpha_M}{3 \alpha_G} } \left( \phi - \phi_* \right) \right) \right)^{1/4}, \end{align}

where $\phi_*$ is the value of the scalar field corresponding to the maximum radius $a_*$. Evolving infinitesimally by $\Delta \phi$ advances to a new value $a = a_i + da$:

(14)
\begin{align} a = a_i \pm \sqrt{ \frac{\alpha_M}{6 \alpha_G} } a_i \sqrt{ 1 - \left( \frac{a_i}{a_*} \right)^4 } \Delta \phi . \end{align}

In terms of the logarithmic variable $x = \ln a$,

(15)
\begin{align} x = x_* + \frac{1}{4} \ln \left( 1 - \tanh^2 \left( \pm \sqrt{ \frac{2 \alpha_M}{\alpha_G} } \left( \phi - \phi_* \right) \right) \right) \end{align}

and

(16)
\begin{align} \frac{dx}{d \phi} = \pm \sqrt{ \frac{ \alpha_M }{ 6 \alpha_G } } \sqrt{ 1 - e^{4 \left( x - x_* \right) } }, \end{align}

so infinitesimal evolution yields $x = x_i + dx$

(17)
\begin{align} x = x_i \pm \sqrt{ \frac{\alpha_M}{6 \alpha_G} } \sqrt{ 1 - e^{4 \left( x_i - x_* \right) } } \Delta \phi \end{align}

## Quantizing

Quantizing according to the Schrodinger picture ($p_a \to -i \hbar \partial_a, \quad p_{\phi} \to -i \hbar \partial_{\phi}$) and allowing for the range of orderings $a^{-i} \partial_a a^{-j} \partial_a a^{-k}$ in Steigl/Hinterleitner, we obtain the quantized Hamiltonian

(18)
\begin{align} \hat H_{ijk} &= \frac{\hbar^2}{48 \pi^2 \alpha_G} a^{-i} \partial_a a^{-j} \partial_a a^{-k} - \frac{\hbar^2}{8 \pi^2 \alpha_M} a^{-3} \partial_{\phi} - 12 \pi^2 \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 \alpha_M a^3 V \left( \phi \right) \end{align}

We can relate the kinetic terms to the Laplace-Beltrami operator $\Delta_{LB} = - \alpha_G^{-1} a^{-2} \partial_a a \partial_a + 6 \alpha_M^{-1} a^{-3} \partial_{\phi}^2$ for the supermetric

(19)
\begin{align} \left( \begin{array}{cc} \alpha_G a & 0 \\ 0 & - \frac{\alpha_M}{6} a^3 \end{array} \right) \end{align}

on $\left( a, \phi \right)$ so that we get

(20)
\begin{align} \hat H_{ijk} = a^q \left( - \frac{\hbar^2}{48 \pi^2} \Delta_{LB} - \frac{\hbar^2}{48 \pi^2 \alpha_G} m^2 \left( i+k \right) a^{-3} - 12 \pi^2 \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 \alpha_M a^3 V \left( \phi \right) \right) a^{-q} \end{align}

where $q = 1 - \frac{\left( i - k \right)}{2}$ and $m = 1 - \frac{\left( i + k \right)}{2}$. The portion in brackets, conjugated by $a^q$, is the transformed Hamiltonian acting on the Hilbert space $L^2 \left( a^2 da d \phi \right)$.

Note that taking $\alpha_G = \frac{c^3}{8 \pi G}$ gives a coefficient of $- \frac{\hbar^2 G}{6 \pi c^3} = - \frac{\hbar \ell_P^2}{6 \pi}$ in front of the quantum potential; with the coupling constants as given above the transformed Hamiltonian reads as

(21)
\begin{align} - \frac{\hbar^2}{48 \pi^2} \Delta_{LB} - \frac{\hbar \ell_P^2}{6 \pi} m^2 a^{-3} - \frac{3 \pi c^3}{2 G} \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 c a^3 V \left( \phi \right) \end{align}

## Logarithmic variable

Convert now to $x = \ln a$ (use $\partial_a = e^{-x} \partial_x$, $a \partial_a = \partial_x$). Consider the WDW equation

(22)
\begin{align} & \left[ - \frac{\hbar^2}{48 \pi^2} \Delta_{LB} - \frac{\hbar^2}{48 \pi^2 \alpha_G} m^2 a^{-3} - 12 \pi^2 \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 \alpha_M a^3 V \left( \phi \right) \right] \Psi \left( a , \phi \right) = 0 \\ & \left[ \frac{\hbar^2}{48 \pi^2} \left( \alpha_G^{-1} a^{-2} \partial_a a \partial_a - 6 \alpha_M^{-1} a^{-3} \partial_{\phi}^2 \right) - \frac{\hbar^2}{48 \pi^2 \alpha_G} m^2 a^{-3} - 12 \pi^2 \alpha_G \left( a - \frac{\Lambda}{3} a^3 \right) + 2 \pi^2 \alpha_M a^3 V \left( \phi \right) \right] \Psi \left( a , \phi \right) = 0 \\ & \left[ \frac{\hbar^2}{48 \pi^2} \left( \alpha_G^{-1} a \partial_a a \partial_a - 6 \alpha_M^{-1} \partial_{\phi}^2 \right) - \frac{\hbar^2}{48 \pi^2 \alpha_G} m^2 - 12 \pi^2 \alpha_G \left( a^4 - \frac{\Lambda}{3} a^6 \right) + 2 \pi^2 \alpha_M a^6 V \left( \phi \right) \right] \Psi \left( a , \phi \right) = 0 \\ & \left[ \frac{\hbar^2}{48 \pi^2} \left( \alpha_G^{-1} \partial_{x}^2 - 6 \alpha_M^{-1} \partial_{\phi}^2 \right) - \frac{\hbar^2}{48 \pi^2 \alpha_G} m^2 - 12 \pi^2 \alpha_G \left( e^{4x} - \frac{\Lambda}{3} e^{6x} \right) + 2 \pi^2 \alpha_M e^{6x} V \left( \phi \right) \right] \Psi \left( x , \phi \right) = 0 \end{align}

Substituting the values $\alpha_G = \frac{c^3}{8 \pi G}$ and $\alpha_M = c$ for the coupling constants, this becomes

(23)
\begin{align} \left[ \frac{\hbar \ell_P^2}{6 \pi} \partial_{x}^2 - \frac{\hbar^2}{8 \pi^2 c} \partial_{\phi}^2 - \frac{\hbar \ell_P^2}{6 \pi} m^2 - \frac{3 \pi c^3}{2 G} \left( e^{4x} - \frac{\Lambda}{3} e^{6x} \right) + 2 \pi^2 c e^{6x} V \left( \phi \right) \right] \Psi \left( x , \phi \right) = 0 \end{align}

## Thoughts on the code

c program to evolve the Wheeler-deWitt equation
c with a scalar field
c
implicit none
integer nx,nvars
parameter(nx=4000,nvars=2)
real vars(nx,nvars),newvars(nx,nvars)
real dphivars(nx,nvars),dphinewvars(nx,nvars)
real x(nx),phi,dx,dphi,xmax,xmin
integer i,j,iter,iphi,nphi
integer ipr,icc
real ii,kk,mm
real psi(nx),w(nx)
ii=2.
kk=0.5
mm=1.+0.5*(kk-ii)
nphi=15000
xmax=3.
xmin=-10.
dx=(xmax-xmin)/float(nx-1)
do i=1,nx
x(i)=dx*float(i-1)+xmin
enddo
call init(nx,nvars,vars,x,mm)
phi=0.
do iphi=1,nphi
dphi=0.1*dx
call evolve(nx,nvars,dphivars,vars,dx,x,phi,ii,kk)
do i=1,nx
do j=1,nvars
newvars(i,j)=vars(i,j)
enddo
enddo
do iter=1,3
call evolve(nx,nvars,dphinewvars,newvars,dx,x,phi,ii,kk)
do j=1,nvars
do i=2,nx-1
newvars(i,j)=vars(i,j)+dphi*0.5*
& (dphivars(i,j)+dphinewvars(i,j))
enddo

enddo
enddo
do j=1,nvars
do i=1,nx
vars(i,j)=newvars(i,j)
enddo
enddo
phi=phi+dphi
ipr=iphi/50
icc=ipr*50
if(icc.eq.iphi) then
do i=1,nx
w(i)=vars(i,1)
psi(i)=w(i)*exp(mm*x(i))
enddo
call xvs('W',phi,x,w,nx)
call xvs('Psi',phi,x,psi,nx)
endif
enddo
end
c
subroutine init(nx,nvars,vars,x,mm)
implicit none
integer nx,nvars
real vars(nx,nvars),x(nx),mm
real psi,pi
integer i
real term1,term2
real sigma
sigma=0.2
do i=1,nx
psi=exp((-1./sigma* *2)*(x(i)+2.)*(x(i)+2.))
pi=(-2./sigma* *2)*(x(i)+2.)*psi
vars(i,1)=psi*exp(-1.*mm*x(i))
vars(i,2)=pi*exp(-1.*mm*x(i))
enddo
return
end
c
subroutine evolve(nx,nvars,dphivars,vars,dx,x,phi,ii,kk)
implicit none
integer nx,nvars
real dphivars(nx,nvars),vars(nx,nvars),dx,x(nx),phi,ii,kk
real w(nx),wdot(nx),dxxw,dphiw,dphiwdot
integer i
real term1,term2,term3
do i=1,nx
w(i)=vars(i,1)
wdot(i)=vars(i,2)
enddo
do i=2,nx-1
dphiw=wdot(i)
dxxw=((w(i+1)+w(i-1))-2.*w(i))/(dx*dx)
term1=dxxw
term2=-0.25*((ii+kk)-2.)*((ii+kk)-2.)*w(i)
term3=-1.*w(i)*exp(4.*x(i))
dphiwdot=term1+term2+term3
dphivars(i,1)=dphiw
dphivars(i,2)=dphiwdot
enddo
return
end

MEASURE ADJUSTMENT: In order to use Gaussians which approach to delta functions with respect to the correct measures, we should choose the initial wave packet according to one of the following two schemes:

(1) Start with a measure-corrected Gaussian for $L^2 \left( a^{i-k} da \right)$ (and transform it by the measure-adjustment factor $a^{-1 + \left( i - k \right) / 2}$ to get a measure-corrected Gaussian for $L^2 \left( a^2 da \right)$):

(24)
\begin{align} \Psi_{\phi = \phi_i} \left( a \right) &= \exp \left( - \frac{ \left( a - a_i \right)^2}{\sigma^2} \right) \cdot a^{\left( k - i \right)/2} \\ W_{\phi = \phi_i} \left( a \right) &= \exp \left( - \frac{ \left( a - a_i \right)^2}{\sigma^2} \right) \cdot a^{-1} \\ \implies W_{\phi = \phi_i} \left( x \right) &= \exp \left( - \frac{ \left( e^{x} - e^{x_i} \right)^2}{\sigma^2} \right) \cdot e^{-x} \end{align}

or (2) Start with a measure-corrected Gaussian for $L^2 \left( e^{3x} dx \right)$ (equivalent to $L^2 \left( a^2 da \right)$):

(25)
\begin{align} W_{\phi = \phi_i} \left( x \right) = \exp \left( - \frac{ \left( x - x_i \right)^2}{\sigma^2} \right) \cdot e^{- \frac{3x}{2}} \end{align}