This is a slightly different version of the extended WKB-like approximation, useful for when the ground state energy is zero. Since in the finite-dimensional problem we want to allow for nonzero ground state energy, this is probably best thought of as a model for the case of $\phi^4$ theory.

Quantum Hamiltonian:

\begin{align} \hat H = - \frac{\hbar^2}{2m} \partial_x^2 + V \left( x \right) \end{align}

(where for AHO we want to take $V \left( x \right) = \frac{1}{2} \omega^2 x^2 + a x^4$).

Consider a state $\Omega \left( x \right)$ satisfying $\hat H \Omega = 0$ and represented by the expansion $\Omega \left( x \right) = N \exp \left\{ -S / \hbar \right\} \sum_{n=0}^{\infty} \hbar^n \phi_n \left( x \right)$. Then applying $\hat H$ to $\Omega \left( x \right)$ we obtain order by order in $\hbar$ the following set of differential recursion relations:

\begin{align} &\hbar^0 &&\implies -\frac{1}{2m} \left( \partial_x S \right)^2 + V \left( x \right) = 0 \\ &\hbar^1 &&\implies \left[ \left( \partial_x S \right) \partial_x + \frac{1}{2} \left( \partial_x^2 S \right) \right] \phi_0 \left( x \right) = 0 \\ &\vdots&& \\ &\hbar^n &&\implies \left[ \left( \partial_x S \right) \partial_x + \frac{1}{2} \left( \partial_x^2 S \right) \right] \phi_{n+1} \left( x \right) = \frac{1}{2} \partial_x^2 \phi_n \left( x \right) \end{align}

This means that at each stage past the $\phi_0$ stage we have a linear first-order ODE whose solution is

\begin{align} \phi_{n+1} \left( x \right) = C_{n+1} e^F + e^F \int e^{-F} \frac{\left(\partial_x \phi_n \right)}{2 \left( \partial_x S \right)} \end{align}

where $C_{n+1}$ is a constant to be determined and $F = -\frac{1}{2} \int \frac{d_x^2 S}{d_x S} = - \frac{1}{2} \ln S'$, so that $e^F = \left( \partial_x S \right)^{-1/2} = \left[ 2m V \left( x \right) \right]^{-1/4}$.


\begin{align} \phi_{n+1} \left( x \right) = C_{n+1} \left( S' \right)^{-1/2} + \frac{ \left( S' \right)^{-1/2} }{2} \int \left( S' \right)^{-1/2} \partial_x^2 \left[ \phi_n \right] dx \end{align}

Name the integro-differential operator in the second term

\begin{align} A f = \frac{ \left( S' \right)^{-1/2} }{2} \int \left( S' \right)^{-1/2} \partial_x^2 \left[ f \right] dx \end{align}

so we have

\begin{align} \phi_{n+1} \left( x \right) = C_{n+1} \left( S' \right)^{-1/2} + A \phi_n \left( x \right) \end{align}

and thus

\begin{align} \phi_{n} \left( x \right) = C_n \left( S' \right)^{-1/2} + C_{n-1} A \left( S' \right)^{-1/2} + \dots + C_1 A^{n-1} \left( S' \right)^{-1/2} + A^n \phi_0 \left( x \right) \end{align}

From here we now have to determine the values of the $C_n$ which make the ground state decay nicely.

How to vary the factor ordering

With the standard ordering as used above, the kinetic term is

\begin{align} - \frac{\hbar^2}{2m} \partial_x^2 \end{align}

We could instead have varied the ordering, taking the kinetic term to be anything of the form

\begin{align} - \frac{\hbar^2}{2m}& f \left( x \right) \partial_x g \left( x \right) \partial_x h \left( x \right) = \\ & - \frac{\hbar^2}{2m} \left[ \partial_x^2 + \left( f (x) g' (x) h (x) + 2 f (x) g (x) h' (x) \right) \partial_x + \left( f (x) g' (x) h' (x) + f (x) g (x) h '' (x) \right) \right] \end{align}

where we require $f (x) g (x) h (x) = 1$.

We want to rewrite this as something of the form

\begin{align} - \frac{\hbar^2}{2m} \left[ F (x) \partial_x^2 F^{-1} (x) + \mathrm{(Q.P.)} \right] \end{align}

where (Q.P.) is some function of $x$ which we regard as a quantum potential. Then by conjugating our original Hamiltonian by $F$, we need only study the Hamiltonian with a standard kinetic term and an added quantum potential (of the form we determine) in order to study all the orderings encompassed by 9.

By expanding 10 and matching terms with like orders of derivatives, we find that we need

\begin{align} \frac{d}{dx} \left[ \ln F \right] = - \frac{1}{2} \frac{d}{dx} \left[ \ln g + 2 \ln h \right] \\ \implies F (x) = \left( g (x) \right)^{-1/2} \left( h (x) \right)^{-1} \end{align}

and for the quantum potential we need

\begin{align} \mathrm{(Q.P.)} = - \frac{\rho^2 (x)}{4} - \frac{\rho' (x)}{2} \end{align}

where $\rho (x) = \frac{g' (x)}{g (x)}$.

Thus adding

\begin{align} \frac{\hbar^2}{2m} \left( \frac{\rho^2 (x)}{4} + \frac{\rho' (x)}{2} \right) \end{align}

to the potential accounts for the varying choice of ordering. Note that it is only $g (x)$ which has an actual physical effect (i.e. contributes to the quantum potential). The function $f (x)$ has no effect at all, and $h (x)$ only contributes to the combination used to conjugate the Hamiltonian (thus it does not alter the expectation values of observables).

We can now effectively vary the factor ordering merely by adding the appropriate Q.P.

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