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  Why is it OK to keep the quadratic term in the small $\hbar$ approximation?

+ 3 like - 0 dislike

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

asked Aug 6, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

One wants to have a perturbation of a Gaussian integral, since the latter can be evaluated using Wick's theorem.

yeah, but is the justification really based on re-scaling the integration variable so that we have a power expansion on $\hbar$? Isn't there another way of motivating this?

1 Answer

+ 1 like - 0 dislike

The proper justification follows from the quasi-classical approximation to the Schrodinger equation solution. The explicit quasi-classical solution shows what it is about and how $\hbar$ enters it.

P.S. The correct dimensional rescaling is $\hbar^{1/2}\tilde{y}$. Written abstractly it looks silly. The right way is to use dimensionless variables where the small parameter is a ratio of the particle (de Broglie) wavelength to the characteristic distance $L$ where the potential changes essentially.

answered Aug 6, 2016 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Aug 7, 2016 by Vladimir Kalitvianski

Could you please give some more detail?

@ArnoldNeumaier: I mean that it is only clear "a posteriori" how to prepare the path integral expressions for expanding in powers of $\hbar$. It is not clear from an abstract expression above and this causes the OP question.

Another way is to read the passage by Dirac in his textbook where more details are given and the path integral expressions are better explained.

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