# 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?

• 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. Aug 6, 2016 at 12:17
• @garyp that's exactly my point. Does it all really boil down to how we define $\tilde{y}$. Isn't there a better justification? Aug 6, 2016 at 12:48
• The link at the top of this question appears to be broken. Would another source be available?
– Nat
May 25, 2018 at 22:17

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.

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.