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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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  • $\begingroup$ Don't you mean $\mathcal{O}(\hbar^2)$? Also, one would not choose $y=\hbar^{-500}\tilde{y}$ because in that case subsequent terms would get larger instead of smaller. $\endgroup$ – garyp Aug 6 '16 at 12:17
  • $\begingroup$ @garyp that's exactly my point. Does it all really boil down to how we define $\tilde{y}$. Isn't there a better justification? $\endgroup$ – Yossarian Aug 6 '16 at 12:48
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    $\begingroup$ The link at the top of this question appears to be broken. Would another source be available? $\endgroup$ – Nat May 25 '18 at 22:17
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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.

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  1. 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.

  2. 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.

  3. 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.

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

  5. 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.

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

  7. 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.

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$^1$ We use DeWitt condensed notation to not clutter the notation.

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