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?
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