# Measure of Functional Integral in Path Integral Formulation

I have a question regarding the prefactor $$\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}$$ in $$\left = \int D[x] \exp(\frac{i}{\hbar}\int dt' L(x, \dot{x})),$$ where $$D[x]=\lim_{N\rightarrow\infty}\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}^N \Pi dx_n .$$

From http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-13-functint.pdf, page 929, I found that for $$\int \mathcal{D}[j]\exp{\frac{1}{2}\int dxdx'j(x)A(x,x')j(x')}$$ to be $$1/\sqrt{\det A}$$, $$\mathcal{D}[j]$$ should be $$\lim_{N\rightarrow\infty}\Pi_n\frac{1}{\sqrt{2\pi/\epsilon^D}}$$.

Applying the above result to the path integral formulation, it seems that $$\mathcal{D}[x]$$ should be such that $$\mathcal{D}[x]\propto\lim_{N\rightarrow\infty}\Pi_n\frac{1}{\sqrt{2\pi/(\Delta t)}}\propto\lim_{N\rightarrow\infty}\left(\frac{\Delta t}{2\pi}\right)^\frac{N}{2}.$$ To me, the mathematical flow seems okay, so I feel the formula is correct.

Therefore, it appears that $$\mathcal{D}[x]$$ does not yield the proper Gaussian functional integration value but needs some sort of normalization. Could somebody please let me know what I am thinking wrong?

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Edit) The problem I am now stuck by is the stationary phase approximation of path integral. The book I am now referring to is the Altland and Simons textbook (Condensed Matter Field Theory), which asserts that $$\left \approx \det(\frac{i}{2\pi\hbar}\frac{\partial^2 S[q]}{\partial q_i \partial q_f})^{1/2}e^{\frac{i}{\hbar}S},\tag{3.28}$$ and that this formula is obtainable using $$\int Dx e^{-F[x]}\approx e^{-F[x_{st}]}\det({\frac{A}{2\pi}})^{-1/2}\tag{3.25}$$ under the stationary phase approximation. The $$Dx$$ is $$Dx=\lim_{N\rightarrow\infty}\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}^N \Pi dx_n$$ and $$A$$ is the second functional derivative $$A(t, t')=\frac{\delta^2 F[x]}{\delta x(t) \delta x(t')}$$.

I am struggling in deriving (3.25) since I do not get how $$\Delta t$$ in $$\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}$$ gets canceled when following the same procedure (discretizing first, and then taking the continuum limit) as in http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-13-functint.pdf.

• It is unclear what the context in which you are asking this is. $\mathrm{D}x$ in what formula? What is $m$? what is $\Delta t$? At which point are you taking the limit $N\to\infty$? Sep 19, 2022 at 7:29
• @ɪdɪətstrəʊlə Sorry, I forgot to add taking the limit. m is a particle mass, and $\Delta t = t/N$ Sep 19, 2022 at 7:33
• Write the action, the path integral that you are trying to evaluate, and your attempt at discretising and where it fails. We can't help by guessing the missing clues in your question. Sep 19, 2022 at 7:36
• @ɪdɪətstrəʊlə Thank you for pointing this out. I have clarified what exactly I am asking. Sep 19, 2022 at 7:57

## 1 Answer

• $$\sqrt{\frac{m}{2\pi ih \Delta t}}$$ is the famous Feynman fudge factor. It can be derived from the Gaussian momentum integrations in the corresponding Hamiltonian phase space path integral, see e.g. Section V of my Phys.SE answer here.

• Altland and Simons eq. (3.25) is essentially the method of steepest descent, while eq. (3.28) is derived in my Phys.SE answer here.

• Thank you for your answer. But actually, I wanted to know how that famous Feynman fudge factor yields a nondivergent value ($\int Dx e^{-F[x]}\approx e^{-F[x_{st}]}\det{\frac{A}{2\pi}}$) under stationary phase approximation. This is because, following the logic of the link I referred to, it seems the $Delta T$ should be at the numerator, not the denominator to yield a nondivergent integral value. Since the Feynman fudge factor is famous as you said, I am definitely thinking wrong in some way, and I asked what I am thinking wrong. Sep 19, 2022 at 8:09