Why is it OK to keep the quadratic term in the small $\hbar$ approximation? I am following this[link broken] set of notes: 

Riccardo Rattazzi, The Path Integral approach to Quantum Mechanics, Lecture Notes for Quantum Mechanics IV, 2009, 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
$$
\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2\hbar}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}y(t_1)y(t_2)+\ldots}=\ldots \tag{1.65}
$$
this is where the author considers the rescaling
$$
y=\sqrt{\hbar}\tilde{y}
$$
which leaves us with
$$
\ldots=e^{i\frac{S[x_0]}{\hbar}}\int D[y]e^{\frac{i}{2}\int_0^T dt_1dt_2\,\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}\bigg|_{x_0}\tilde{y}(t_1)\tilde{y}(t_2)+\mathcal{O}(\hbar^{1/2})} \tag{1.66}
$$
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 rescaling of $y$ we have introduced, 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?
 A: You are right to be concerned by this procedure, since a Taylor expansion in powers of $\hbar$ itself lacks a clear-cut physical meaning. This is because $\hbar$ is a dimensionful constant: its value depends on the unit system. For example, in SI units $\hbar$ is a very small number, while in many "natural" unit systems used in quantum physics $\hbar=1$. Clearly, if $\hbar=1$ a low-order Taylor expansion should be a very poor approximation.
In fact, what is happening here is not a Taylor expansion in terms of $\hbar$. You are Taylor expanding the dimensionless functional $S[x(t)]/\hbar$ about the "point" $x(t) = x_0(t)$ (really a function, not a point) up to quadratic order. This can be expected to converge quickly if successive terms in the expansion are small. What this means physically is that the action associated with quantum fluctuations $y(t) = x(t)-x_0(t)$ is assumed to be much less than $\hbar$. 
In the physics literature, it is often convenient to perform such expansions in terms of a dimensionful parameter. However, such a procedure only makes sense if an appropriate dimensionless expansion parameter can be identified and reasonably assumed to be small.
A: *

*First of all, recall that it is in general an open problem in mathematics to rigorously define a path integral. A heuristic motivation is given by a path integral generalization of the method of steepest descent around a non-degenerate stationary point, i.e. the Hessian $H_{jk}$ should be non-degenerate. 

*The action is a sum of a free/quadratic part and an interaction part
$$S[x] ~=~S_2[y] +S_{\rm int}[y],\qquad x~=~x_{\rm cl}+y,\tag{1} $$ 
where$^1$
$$ S_2[y]~=~S[x_{\rm cl}]+ \frac{1}{2}y^j H_{jk} y^k, \qquad  S_{\rm int}[y]~=~{\cal O}(y^3).\tag{2}$$
Let us for simplicity assume that the classical path $x_{\rm cl}$ is unique, i.e. no instantons.

*The free path integral $Z_2$ reads
$$\begin{align} Z_2
&~~:=~\int \!{\cal D}y~\exp\left\{\frac{i}{\hbar} S_2[y]  \right\}\cr
&~~\stackrel{(2)}{=}~ \exp\left\{\frac{i}{\hbar}S[x_{\rm cl}]\right\}
\int \!{\cal D}y~\exp\left\{\frac{i}{2\hbar} y^j H_{jk} y^k  \right\}\cr
&\stackrel{y=\sqrt{h}\tilde{y}}{=}~
{\cal N} I_2 \exp\left\{\frac{i}{\hbar}S[x_{\rm cl}]\right\},\end{align} \tag{3}$$
where $${\cal N}~=~\prod_x \sqrt{\hbar}~=~ {\sqrt{\hbar}^{\infty}}\tag{4}$$ is a formal normalization constant from the Jacobian factor, and where 
$$I_2~:=~ \int \!{\cal D}\tilde{y}~\exp\left\{ \frac{i}{2}\tilde{y}^j H_{jk} \tilde{y}^k\right\} ,\tag{5}$$
is a Gaussian path integral, which is independent of $\hbar$, and which is made convergent via Wick rotation/$i\epsilon$-prescription.

*The full path integral $Z$ is often defined perturbatively relatively to the free path integral
$$\begin{align}\frac{Z}{Z_2}&\quad:=\quad\frac{1}{Z_2}\int \!{\cal D}x~\exp\left\{\frac{i}{\hbar} S[x]  \right\}\cr 
&\stackrel{(1)+(2)+(3)}{=}~\frac{1}{{\cal N} I_2} \int\! {\cal D}y~\exp\left\{\frac{i}{\hbar}\left( \frac{1}{2}y^j H_{jk} y^k +{\cal O}(y^3)\right)\right\}\cr
&~~\stackrel{y=\sqrt{h}\tilde{y}}{=}~
\frac{1}{ I_2}\int\! {\cal D}\tilde{y}~\exp\left\{ \frac{i}{2}\tilde{y}^j H_{jk} \tilde{y}^k +{\cal O}(\sqrt{\hbar})\right\} \cr
&\quad\sim\quad 1+{\cal O}(\sqrt{\hbar})\quad\text{for}\quad \hbar~\to ~0.\end{align}\tag{6}$$

*Note in particular that it is crucial to use the substitution $y^j=\sqrt{h}\tilde{y}^j$ in order to make the quadratic part of the Boltzmann factor in eqs. (5) and (6) independent of $\hbar$. This choice makes manifest the dominant role of the quadratic part of the Boltzmann factor in the $\hbar$-expansion as compared to the subleading interaction part, cf. the method of steepest descent.

*The $\sim$ symbol in eq. (6) stands for an asymptotic series in $\sqrt{h}$. Often it is not convergent. 

*Finally let us mention that in practice in order not to have to deal explicitly with the normalization factor ${\cal N}$, the definitions of path integrals $Z_2$ and $Z$ are modified with a factor $1/\sqrt{h}$ inside the path integral measure, cf. e.g. this related Phys.SE post.
--
$^1$ We use DeWitt condensed notation to not clutter the notation.
