I am following this http://itp.epfl.ch/webdav/site/itp/users/174685/private/RevisedLectureNotesV2.pdf set of notes on path integrals, page 21. I am having some issues to understand the small $\hbar$ expansion.

Consider the path integral in quantum mechanics giving the amplitude for a spinless particle to go from point $x_i$ to point $x_f$ in the time interval $T$

$$

\int D[x]e^{i\frac{S[x]}{\hbar}}=\ldots

$$

where

$$

S[x]=\int_{0}^{T}dt\,\mathcal{L}

$$

let's assume now that the action has one stationary point $x_0$. Let's change the variable of integration in the path integral from $x$ to fluctuations around the stationary point

$$

x=x_0+y

$$

$$

\ldots=\int D[y]e^{i\frac{S[x_0+y]}{\hbar}}=\ldots

$$

Let's Taylor expand the action around $x_0$

$$

S[x_0+y]=S[x_0]+\frac{1}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots

$$

which leaves us with

$$

\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2\hbar}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots}=\ldots

$$

this is where the author considers the rescaling

$$

y=\hbar\tilde{y}

$$

which leaves us with

$$

\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)+\mathcal{O}(\hbar^{1/2})}

$$

and we "obviously" have an expansion in $\hbar$, so when $hbar$ is small we may keep the first term

$$

e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)}

$$

I do not like this rationale at all. It's all based on the rescalig of $y$ we have itroduced, but had we done

$$

y=\frac{1}{\hbar^{500}}\tilde{y}

$$

we wouldn't have obtained an expansion on powers of $\hbar$ on the exponent. What is the proper justification for keeping the quadratic term?